35894243 Energy Efficiency Assessment Book

1. ENERGY PERFORMANCE ASSESSMENTOF BOILERS

1Bureau of Energy Efficiency

1.1 Introduction

Performance of the boiler, like efficiency and evaporation ratio reduces with time, due topoor combustion, heat transfer fouling and poor operation and maintenance. Deteriorationof fuel quality and water quality also leads to poor performance of boiler. Efficiency test-ing helps us to find out how far the boiler efficiency drifts away from the best efficiency.Any observed abnormal deviations could therefore be investigated to pinpoint the problemarea for necessary corrective action. Hence it is necessary to find out the current level ofefficiency for performance evaluation, which is a pre requisite for energy conservationaction in industry.

1.2 Purpose of the Performance Test

� To find out the efficiency of the boiler � To find out the Evaporation ratio

The purpose of the performance test is to determine actual performance and efficien-cy of the boiler and compare it with design values or norms. It is an indicator for trackingday-to-day and season-to-season variations in boiler efficiency and energy efficiencyimprovements

1.3 Performance Terms and Definitions

1.4 Scope

The procedure describes routine test for both oil fired and solid fuel fired boilers using coal,agro residues etc. Only those observations and measurements need to be made which can bereadily applied and is necessary to attain the purpose of the test.

1.5 Reference Standards

British standards, BS845: 1987

The British Standard BS845: 1987 describes the methods and conditions under which a boil-er should be tested to determine its efficiency. For the testing to be done, the boiler shouldbe operated under steady load conditions (generally full load) for a period of one hour afterwhich readings would be taken during the next hour of steady operation to enable the effi-ciency to be calculated.

The efficiency of a boiler is quoted as the % of useful heat available, expressed as a per-centage of the total energy potentially available by burning the fuel. This is expressed on thebasis of gross calorific value (GCV).

This deals with the complete heat balance and it has two parts:

� Part One deals with standard boilers, where the indirect method is specified� Part Two deals with complex plant where there are many channels of heat flow. In this

case, both the direct and indirect methods are applicable, in whole or in part.

ASME Standard: PTC-4-1 Power Test Code for Steam Generating Units

This consists of

� Part One: Direct method (also called as Input -output method)� Part Two: Indirect method (also called as Heat loss method)

IS 8753: Indian Standard for Boiler Efficiency Testing

Most standards for computation of boiler efficiency, including IS 8753 and BS845 are designedfor spot measurement of boiler efficiency. Invariably, all these standards do not include blowdown as a loss in the efficiency determination process.

Basically Boiler efficiency can be tested by the following methods:

1) The Direct Method: Where the energy gain of the working fluid (water and steam) iscompared with the energy content of the boiler fuel.

2) The Indirect Method: Where the efficiency is the difference between the losses and theenergy input.

1.6 The Direct Method Testing

1.6.1 Description

This is also known as 'input-output method' due to the fact that it needs only the useful output(steam) and the heat input (i.e. fuel) for evaluating the efficiency. This efficiency can be eval-uated using the formula:

1. Energy Performance Assessment of Boilers


x 100

1.6.2 Measurements Required for Direct Method Testing

Heat input

Both heat input and heat output must be measured. The measurement of heat input requiresknowledge of the calorific value of the fuel and its flow rate in terms of mass or volume, accord-ing to the nature of the fuel.

For gaseous fuel:Agas meter of the approved type can be used and the measured volume shouldbe corrected for temperature and pressure. A sample of gas can be collected for calorific valuedetermination, but it is usually acceptable to use the calorific value declared by the gas suppliers.

For liquid fuel: Heavy fuel oil is very viscous, and this property varies sharply with tem-perature. The meter, which is usually installed on the combustion appliance, should beregarded as a rough indicator only and, for test purposes, a meter calibrated for the partic-ular oil is to be used and over a realistic range of temperature should be installed. Evenbetter is the use of an accurately calibrated day tank.

For solid fuel: The accurate measurement of the flow of coal or other solid fuel is verydifficult. The measurement must be based on mass, which means that bulky apparatusmust be set up on the boiler-house floor. Samples must be taken and bagged throughoutthe test, the bags sealed and sent to a laboratory for analysis and calorific value determi-nation. In some more recent boiler houses, the problem has been alleviated by mountingthe hoppers over the boilers on calibrated load cells, but these are yet uncommon.

Heat output

There are several methods, which can be used for measuring heat output. With steam boilers,an installed steam meter can be used to measure flow rate, but this must be corrected for tem-perature and pressure. In earlier years, this approach was not favoured due to the change in



accuracy of orifice or venturi meters with flow rate. It is now more viable with modern flowmeters of the variable-orifice or vortex-shedding types.

The alternative with small boilers is to measure feed water, and this can be done by previ-ously calibrating the feed tank and noting down the levels of water during the beginning andend of the trial. Care should be taken not to pump water during this period. Heat addition forconversion of feed water at inlet temperature to steam, is considered for heat output.

In case of boilers with intermittent blowdown, blowdown should be avoided during the trialperiod. In case of boilers with continuous blowdown, the heat loss due to blowdown should becalculated and added to the heat in steam.

1.6.3 Boiler Efficiency by Direct Method: Calculation and Example

Test Data and Calculation

Water consumption and coal consumption were measured in a coal-fired boiler at hourly inter-vals. Weighed quantities of coal were fed to the boiler during the trial period. Simultaneouslywater level difference was noted to calculate steam generation during the trial period. Blowdown was avoided during the test. The measured data is given below.



1.6.4 Merits and Demerits of Direct Method

Merits

� Plant people can evaluate quickly the efficiency of boilers� Requires few parameters for computation� Needs few instruments for monitoring

Demerits

� Does not give clues to the operator as to why efficiency of system is lower� Does not calculate various losses accountable for various efficiency levels� Evaporation ratio and efficiency may mislead, if the steam is highly wet due to water carryover

1.7 The Indirect Method Testing

1.7.1 Description

The efficiency can be measured easily by measuring all the losses occurring in the boilers usingthe principles to be described. The disadvantages of the direct method can be overcome by thismethod, which calculates the various heat losses associated with boiler. The efficiency can bearrived at, by subtracting the heat loss fractions from 100.An important advantage of thismethod is that the errors in measurement do not make significant change in efficiency.

Thus if boiler efficiency is 90% , an error of 1% in direct method will result in significantchange in efficiency. i.e. 90 ± 0.9 = 89.1 to 90.9. In indirect method, 1% error in measurementof losses will result in

Efficiency = 100 � (10 ± 0.1) = 90 ± 0.1 = 89.9 to 90.1

The various heat losses occurring in the boiler are:



The following losses are applicable to liquid, gas and solid fired boiler

L1� Loss due to dry flue gas (sensible heat)L2� Loss due to hydrogen in fuel (H2)L3� Loss due to moisture in fuel (H2O)L4� Loss due to moisture in air (H2O)L5� Loss due to carbon monoxide (CO)L6� Loss due to surface radiation, convection and other unaccounted*.

*Losses which are insignificant and are difficult to measure.The following losses are applicable to solid fuel fired boiler in addition to above

L7� Unburnt losses in fly ash (Carbon)L8� Unburnt losses in bottom ash (Carbon)

Boiler Efficiency by indirect method = 100 � (L1 + L2 + L3 + L4 + L5 + L6 + L7 + L8)

1.7.2 Measurements Required for Performance Assessment Testing

The following parameters need to be measured, as applicable for the computation of boiler effi-ciency and performance.

a) Flue gas analysis1. Percentage of CO2 or O2 in flue gas 2. Percentage of CO in flue gas3. Temperature of flue gas

b) Flow meter measurements for1. Fuel 2. Steam 3. Feed water 4. Condensate water 5. Combustion air

c) Temperature measurements for1. Flue gas 2. Steam 3. Makeup water 4. Condensate return 5. Combustion air 6. Fuel 7. Boiler feed water

d) Pressure measurements for1. Steam2. Fuel3. Combustion air, both primary and secondary4. Draft



1.7.3 Test Conditions and Precautions for Indirect Method Testing

A) The efficiency test does not account for:

� Standby losses. Efficiency test is to be carried out, when the boiler is operating under asteady load. Therefore, the combustion efficiency test does not reveal standby losses,which occur between firing intervals

� Blow down loss. The amount of energy wasted by blow down varies over a wide range.� Soot blower steam. The amount of steam used by soot blowers is variable that depends on

the type of fuel.� Auxiliary equipment energy consumption. The combustion efficiency test does not

account for the energy usage by auxiliary equipments, such as burners, fans, and pumps.

B) Preparations and pre conditions for testing

� Burn the specified fuel(s) at the required rate.� Do the tests while the boiler is under steady load. Avoid testing during warming up of boil-

ers from a cold condition� Obtain the charts /tables for the additional data.� Determination of general method of operation� Sampling and analysis of fuel and ash.� Ensure the accuracy of fuel and ash analysis in the laboratory.� Check the type of blow down and method of measurement� Ensure proper operation of all instruments.� Check for any air infiltration in the combustion zone.



TABLE 1.1 TYPICAL INSTRUMENTS USED FOR BOILER PERFORMANCE

ASSESSMENT.

Instrument Type Measurements

Flue gas analyzer Portable or fixed % CO2 , O2 and CO

Temperature indicator Thermocouple, liquid in Fuel temperature, flue gasglass temperature, combustion air

temperature, boiler surfacetemperature, steam temperature

Draft gauge Manometer, differential Amount of draft usedpressure or available

TDS meter Conductivity Boiler water TDS, feed water TDS,make-up water TDS.

Flow meter As applicable Steam flow, water flow, fuel flow,air flow

e) Water condition1. Total dissolved solids (TDS)2. pH3. Blow down rate and quantity

The various parameters that were discussed above can be measured with the instrumentsthat are given in Table 1.1.

C) Flue gas sampling location

It is suggested that the exit duct of the boiler be probed and traversed to find the location of thezone of maximum temperature. This is likely to coincide with the zone of maximum gas flowand is therefore a good sampling point for both temperature and gas analysis.

D) Options of flue gas analysis

Check the Oxygen Test with the Carbon Dioxide TestIf continuous-reading oxygen test equipment is installed in boiler plant, use oxygen reading.Occasionally use portable test equipment that checks for both oxygen and carbon dioxide. If the car-bon dioxide test does not give the same results as the oxygen test, something is wrong. One (or both)of the tests could be erroneous, perhaps because of stale chemicals or drifting instrument calibration.Another possibility is that outside air is being picked up along with the flue gas. This occurs if thecombustion gas area operates under negative pressure and there are leaks in the boiler casing.

Carbon Monoxide TestThe carbon monoxide content of flue gas is a good indicator of incomplete combustion with alltypes of fuels, as long as they contain carbon. Carbon monoxide in the flue gas is minimal withordinary amounts of excess air, but it rises abruptly as soon as fuel combustion starts to be incom-plete.

E) Planning for the testing

� The testing is to be conducted for a duration of 4 to 8 hours in a normal production day.� Advanced planning is essential for the resource arrangement of manpower, fuel, water and

instrument check etc and the same to be communicated to the boiler Supervisor andProduction Department.

� Sufficient quantity of fuel stock and water storage required for the test duration should bearranged so that a test is not disrupted due to non-availability of fuel and water.

� Necessary sampling point and instruments are to be made available with working condition.� Lab Analysis should be carried out for fuel, flue gas and water in coordination with lab per-

sonnel.� The steam table, psychometric chart, calculator are to be arranged for computation of boil-

er efficiency.

1.7.4 Boiler Efficiency by Indirect Method: Calculation Procedure and Formula

In order to calculate the boiler efficiency by indirect method, all the losses that occur in theboiler must be established. These losses are conveniently related to the amount of fuel burnt.In this way it is easy to compare the performance of various boilers with different ratings.



Conversion formula for proximate analysis to ultimate analysis%C = 0.97C + 0.7 (VM + 0.1A) � M(0.6 � 0.01M)%H2 = 0.036C + 0.086 (VM � 0.1xA) � 0.0035M2 (1 � 0.02M)%N2 = 2.10 � 0.020 VM

where C = % of fixed carbon A = % of ashVM = % of volatile matterM = % of moisture

The various losses associated with the operation of a boiler are discussed below withrequired formula.

1. Heat loss due to dry flue gas

This is the greatest boiler loss and can be calculated with the following formula:



However it is suggested to get a ultimate analysis of the fuel fired periodically from a reputed laboratory.

Theoretical (stoichiometric) air fuel ratio and excess air supplied are to be determined firstfor computing the boiler losses. The formula is given below for the same.

m x Cp x (Tf - Ta )L1 = x 100

GCV of fuelWhere,

L1 = % Heat loss due to dry flue gasm = Mass of dry flue gas in kg/kg of fuel

= Combustion products from fuel: CO2 + SO2 + Nitrogen in fuel +Nitrogen in the actual mass of air supplied + O2 in flue gas.(H2O/Water vapour in the flue gas should not be considered)

3. Heat loss due to moisture present in fuel

Moisture entering the boiler with the fuel leaves as a superheated vapour. This moisture loss ismade up of the sensible heat to bring the moisture to boiling point, the latent heat of evapora-tion of the moisture, and the superheat required to bring this steam to the temperature of theexhaust gas. This loss can be calculated with the following formula



Cp = Specific heat of flue gas in kCal/kg°CTf = Flue gas temperature in °CTa = Ambient temperature in °C

Note�1:

For Quick and simple calculation of boiler efficiency use the following.

A: Simple method can be used for determining the dry flue gas loss as given below.

m x Cp x (Tf � Ta) x 100a) Percentage heat loss due to dry flue gas =

GCV of fuel

Total mass of flue gas (m)/kg of fuel = mass of actual air supplied/kg of fuel + 1 kg of fuel

Note-2: Water vapour is produced from Hydrogen in fuel, moisture present in fuel and air dur-ing the combustion. The losses due to these components have not been included in the dry fluegas loss since they are separately calculated as a wet flue gas loss.

2. Heat loss due to evaporation of water formed due to H2 in fuel (%)

The combustion of hydrogen causes a heat loss because the product of combustion is water.This water is converted to steam and this carries away heat in the form of its latent heat.

9 x H2 x {584 + Cp (Tf � Ta )} L2 = x 100

GCV of fuel

WhereH2 = kg of hydrogen present in fuel on 1 kg basisCp = Specific heat of superheated steam in kCal/kg°CTf = Flue gas temperature in °CTa = Ambient temperature in °C584 = Latent heat corresponding to partial pressure of water vapour

M x {584 + Cp (Tf � Ta)} L3 = X 100

GCV of fuel

4. Heat loss due to moisture present in air

Vapour in the form of humidity in the incoming air, is superheated as it passes through the boil-er. Since this heat passes up the stack, it must be included as a boiler loss.

To relate this loss to the mass of coal burned, the moisture content of the combustion air andthe amount of air supplied per unit mass of coal burned must be known.

The mass of vapour that air contains can be obtained from psychrometric charts and typicalvalues are included below:



whereM = kg moisture in fuel on 1 kg basisCp = Specific heat of superheated steam in kCal/kg°CTf = Flue gas temperature in °CTa = Ambient temperature in °C584 = Latent heat corresponding to partial pressure of water vapour

Dry-Bulb Wet Bulb Relative Humidity Kilogram waterper Kilogram dry

Temp °C Temp °C (%) air (Humidity Factor)

20 20 100 0.016

20 14 50 0.008

30 22 50 0.014

40 30 50 0.024

AAS x humidity factor x Cp x (Tf � Ta ) x 100L4 =

GCV of fuel

whereAAS = Actual mass of air supplied per kg of fuelHumidity factor = kg of water/kg of dry airCp = Specific heat of superheated steam in kCal/kg°CTf = Flue gas temperature in °CTa = Ambient temperature in °C (dry bulb)

5. Heat loss due to incomplete combustion:

Products formed by incomplete combustion could be mixed with oxygen and burned again witha further release of energy. Such products include CO, H2, and various hydrocarbons and aregenerally found in the flue gas of the boilers. Carbon monoxide is the only gas whose concen-tration can be determined conveniently in a boiler plant test.

6. Heat loss due to radiation and convection:

The other heat losses from a boiler consist of the loss of heat by radiation and convection fromthe boiler casting into the surrounding boiler house.

Normally surface loss and other unaccounted losses is assumed based on the type and sizeof the boiler as given below

For industrial fire tube / packaged boiler = 1.5 to 2.5%

For industrial watertube boiler = 2 to 3%

For power station boiler = 0.4 to 1%

However it can be calculated if the surface area of boiler and its surface temperature areknown as given below :



%CO x C 5744L5 = x x 100

% CO + % CO2 GCV of fuel

L5 = % Heat loss due to partial conversion of C to COCO = Volume of CO in flue gas leaving economizer (%)CO2 = Actual Volume of CO2 in flue gas (%)C = Carbon content kg / kg of fuel

orWhen CO is obtained in ppm during the flue gas analysisCO formation (Mco) = CO (in ppm) x 10�6 x Mf x 28Mf = Fuel consumption in kg/hrL5 = Mco x 5744** Heat loss due to partial combustion of carbon.

L6 = 0.548 x [ (Ts / 55.55)4 � (Ta / 55.55)4] + 1.957 x (Ts � Ta)1.25 x sq.rt of[(196.85 Vm + 68.9) / 68.9]

whereL6 = Radiation loss in W/m2

Vm = Wind velocity in m/sTs = Surface temperature (K)Ta = Ambient temperature (K)

Heat loss due to unburned carbon in fly ash and bottom ash:

Small amounts of carbon will be left in the ash and this constitutes a loss of potential heat inthe fuel. To assess these heat losses, samples of ash must be analyzed for carbon content. Thequantity of ash produced per unit of fuel must also be known.

7. Heat loss due to unburnt in fly ash (%).

Total ash collected / kg of fuel burnt x G.C.V of fly ash x 100L7 =

GCV of fuel

8. Heat loss due to unburnt in bottom ash (%)

Total ash collected per kg of fuel burnt x G.C.V of bottom ash x 100L8 =

GCV of fuel

Heat Balance:

Having established the magnitude of all the losses mentioned above, a simple heat balancewould give the efficiency of the boiler. The efficiency is the difference between the energyinput to the boiler and the heat losses calculated.

Boiler Heat Balance:



Input/Output Parameter kCal / kg of fuel %

Heat Input in fuel = 100

Various Heat losses in boiler

1. Dry flue gas loss =

2. Loss due to hydrogen in fuel

3. Loss due to moisture in fuel =

4. Loss due to moisture in air =

5. Partial combustion of C to CO =

6. Surface heat losses =

7. Loss due to Unburnt in fly ash =

8. Loss due to Unburnt in bottom ash =

Total Losses =

Boiler efficiency = 100 � (1+2+3+4+5+6+7+8)

1.8 Example: Boiler Efficiency Calculation

1.8.1 For Coal fired Boiler

The following are the data collected for a boiler using coal as the fuel. Find out the boiler effi-ciency by indirect method.



Fuel firing rate = 5599.17 kg/hr

Steam generation rate = 21937.5 kg/hr

Steam pressure = 43 kg/cm2(g)

Steam temperature = 377 °C

Feed water temperature = 96 °C

%CO2 in Flue gas = 14

%CO in flue gas = 0.55

Average flue gas temperature = 190 °C

Ambient temperature = 31 °C

Humidity in ambient air = 0.0204 kg / kg dry air

Surface temperature of boiler = 70 °C

Wind velocity around the boiler = 3.5 m/s

Total surface area of boiler = 90 m2

GCV of Bottom ash = 800 kCal/kg

GCV of fly ash = 452.5 kCal/kg

Ratio of bottom ash to fly ash = 90:10

Fuel Analysis (in %)Ash content in fuel = 8.63

Moisture in coal = 31.6

Carbon content = 41.65

Hydrogen content = 2.0413

Nitrogen content = 1.6

Oxygen content = 14.48

GCV of Coal = 3501 kCal/kg

Boiler efficiency by indirect method

Step � 1 Find theoretical air requirement

Theoretical air required for = [(11.6 x C) + {34.8 x (H2 � O2/8)} + (4.35 x S)] /100complete combustion kg/kg of coal

= [(11.6 x 41.65) + {34.8 x (2.0413 � 14.48/8)} +(4.35 x 0)] / 100

= 4.91 kg / kg of coal



Step � 3 To find Excess air supplied

Actual CO2 measured in flue gas = 14.0%

7900 x [ ( CO2%)t � (CO2%)a]% Excess air supplied (EA) =

(CO2%)a x [100 � (CO2%)t ]

7900 x [20.37 � 14 ]=

14a x [100 � 20.37]

= 45.17 %

Step � 4 to find actual mass of air supplied

Actual mass of air supplied = {1 + EA/100} x theoretical air

= {1 + 45.17/100} x 4.91

= 7.13 kg/kg of coal



Step � 5 to find actual mass of dry flue gas

Mass of dry flue gas = Mass of CO2 + Mass of N2 content in the fuel +Mass of N2 in the combustion air supplied + Mass ofoxygen in flue gas

0.4165 x 44 7.13 x 77 (7.13�4.91) x 23Mass of dry flue gas = + 0.016 + +

12 100 100

= 7.54 kg / kg of coal

Step � 6 to find all losses

m x Cp x (Tf � Ta) 1. % Heat loss in dry flue gas (L1) = x 100

GCV of fuel

7.54 x 0.23 x (190 � 31)= x 100

3501

L1 = 7.88 %

2. % Heat loss due to formation = 9 x H2 x {584 + Cp (Tf � Ta)} of water from H2 in fuel (L2) x 100

GCV of fuel

9 x .02041 x {584 + 0.45(190 � 31)} = x 100

3501

L2 = 3.44 %



M x {584 + Cp ( Tf � Ta )}3. % Heat loss due to moisture in = X 100fuel (L3) GCV of fuel

0.316 x {584 + 0.45 ( 190 - 31) }= x 100

3501

L3 = 5.91 %

AAS x humidity x Cp x (Tf � Ta ) x 1004. % Heat loss due to moisture =in air (L4) GCV of fuel

7.13 x 0.0204 x 0.45 x (190 � 31) x 100=

3501

L4 = 0.29 %

%CO x C 57445. % Heat loss due to partial = x x 100conversion of C to CO (L5) % CO + (% CO2)a GCV of fuel

0.55 x 0.4165 5744= x x 100

0.55 + 14 3501

L5 = 2.58 %

6. Heat loss due to radiation and = 0.548 x [ (343/55.55)4 � (304/55.55)4] + 1.957 xconvection (L6) (343 � 304)1.25 x sq.rt of [(196.85 x 3.5 + 68.9) / 68.9]

= 633.3 w/m2

= 633.3 x 0.86= 544.64 kCal / m2



Total radiation and convection = 544.64 x 90loss per hour = 49017.6 kCal

49017.6 x 100% radiation and convection loss =

3501 x 5599.17

L6 = 0.25 %

7. % Heat loss due to unburnt in fly ash

% Ash in coal = 8.63

Ratio of bottom ash to fly ash = 90:10

GCV of fly ash = 452.5 kCal/kg

Amount of fly ash in 1 kg of coal = 0.1 x 0.0863

= 0.00863 kg

Heat loss in fly ash = 0.00863 x 452.5

= 3.905 kCal / kg of coal

% heat loss in fly ash = 3.905 x 100 / 3501

L7 = 0.11 %

8. % Heat loss due to unburnt in bottom ash

GCV of bottom ash = 800 kCal/kg

Amount of bottom ash in 1 kg of = 0.9 x 0.0863coal

= 0.077 kg

Heat loss in bottom ash = 0.077 x 800

= 62.136 kCal/kg of coal

% Heat loss in bottom ash = 62.136 x 100 / 3501

L8 = 1.77 %

Boiler efficiency by indirect = 100 � (L1 + L2 + L3 + L4 + L5 + L6 + L7 + L8)method

= 100 � (7.88 + 3.44 + 5.91 + 0.29 + 2.58 + 0.25+ 0.11 + 1.77)

= 100 � 22.23= 77.77 %

1.8.2 Efficiency for an oil fired boiler

The following are the data collected for a boiler using furnace oil as the fuel. Find out the boil-er efficiency by indirect method.



SUMMARY OF HEAT BALANCE FOR COAL FIRED BOILER

Input/Output Parameter kCal / kg of % losscoal

Heat Input = 3501 100

Losses in boiler

1. Dry flue gas, L1 = 275.88 7.88

2. Loss due to hydrogen in fuel, L2 = 120.43 3.44

3. Loss due to moisture in fuel, L3 = 206.91 5.91

4. Loss due to moisture in air, L4 = 10.15 0.29

5. Partial combustion of C to CO, L5 = 90.32 2.58

6. Surface heat losses, L6 = 8.75 0.25

7. Loss due to Unburnt in fly ash, L7 = 3.85 0.11

8. Loss due to Unburnt in bottom ash, L8 = 61.97 1.77

Boiler Efficiency = 100 � (L1 + L2 + L3 + L4 + L5 + L6 + L7 + L8) = 77.77 %

Ultimate analysis (%)

Carbon = 84Hydrogen = 12Nitrogen = 0.5Oxygen = 1.5Sulphur = 1.5Moisture = 0.5GCV of fuel = 10000 kCal/kgFuel firing rate = 2648.125 kg/hrSurface Temperature of boiler = 80 °CSurface area of boiler = 90 m2

Humidity = 0.025 kg/kg of dry airWind speed = 3.8 m/s

Flue gas analysis (%)

Flue gas temperature = 190°CAmbient temperature = 30°CCo2% in flue gas by volume = 10.8O2% in flue gas by volume = 7.4



a) Theoretical air required = [(11.6 x C) + [{34.8 x (H2 � O2/8)} + (4.35 x S)] /100kg/kg of fuel. [from fuel analysis]

= [(11.6 x 84) + [{34.8 x (12 � 1.5/8)}+ (4.35 x 1.5)] / 100

= 13.92 kg/kg of oil

b) Excess Air supplied (EA) = (O2 x 100) / (21 � O2)

= (7.4 x 100) / (21 � 7.4)

= 54.4 %

c) Actual mass of air supplied/ kg = {1 + EA/100} x theoretical airof fuel (AAS)

= {1 + 54.4/100} x 13.92

= 21.49 kg / kg of fuel

Mass of dry flue gas = Mass of (CO2 + SO2 + N2 + O2) in flue gas + N2in air we are supplying

0.84 x 44 0.015 x 64

12 32

= 21.36 kg / kg of oil

m x Cp x (Tf � Ta)% Heat loss in dry flue gas = x 100

GCV of fuel

21.36 x 0.23 x (190 � 30)= x 100

10000

L1 = 7.86 %

9 x H2 x{584 + Cp (Tf � Ta )}Heat loss due to evaporation of = x 100water due to H2 in fuel (%) GCV of fuel

9 x 0.12 x {584 + 0.45 (190 � 30)}= x 100

10000

L2 = 7.08 %

+ 0.005 + + 7.4x23

100

21.49 x 77

100+=



M x {584 + Cp ( Tf - Ta )}% Heat loss due to moisture = X 100in fuel GCV of fuel

0.005 x {584 + 0.45 (190 � 30)}= x 100

10000

L3 = 0.033%

AAS x humidity x Cp x (Tf � Ta ) x 100% Heat loss due to moisture in air =

GCV of fuel

21.36 x 0.025 x 0.45 x (190 � 30) x 100=

10000

L4 = 0.38 %

Radiation and convection loss = 0.548 x [ (Ts / 55.55)4 � (Ta / 55.55)4] + 1.957 (L6) x (Ts � Ta)1.25 x sq.rt of [(196.85 Vm + 68.9) / 68.9]

= 0.548 x [ (353 / 55.55)4 � (303 / 55.55)4] + 1.957x (353 � 303)1.25 x sq.rt of [(196.85 x 3.8 + 68.9)/ 68.9]

= 1303 W/m2

= 1303 x 0.86

= 1120.58 kCal / m2

Total radiation and convection = 1120 .58 x 90 m2

loss per hour = 100852.2 kCal

% Radiation and convection loss = 100852.2 x 100

10000 x 2648.125

L6 = 0.38 %Normally it is assumed as 0.5 to 1 % for simplicity

Boiler efficiency by indirect = 100 � (L1 + L2 + L3 + L4 + L6)method = 100 � (7.86 + 7.08 + 0.033 + 0.38 + 0.38)

= 100 � 15.73= 84.27 %

Summary of Heat Balance for the Boiler Using Furnace Oil



Input/Output Parameter kCal / kg of %Lossfurnace oil

Heat Input = 10000 100

Losses in boiler :

1. Dry flue gas, L1 = 786 7.86

2. Loss due to hydrogen in fuel, L2 = 708 7.08

3. Loss due to Moisture in fuel, L3 = 3.3 0.033

4. Loss due to Moisture in air, L4 = 38 0.38

5. Partial combustion of C to CO, L5 = 0 0

6. Surface heat losses, L6 = 38 0.38

Boiler Efficiency = 100 � (L1 + L2 + L3 + L4 + L6) = 84.27 %

Note:

For quick and simple calculation of boiler efficiency use the following.

A: Simple method can be used for determining the dry flue gas loss as given below.

m x Cp x (Tf � Ta ) x 100a) Percentage heat loss due to dry flue gas =

GCV of fuel

Total mass of flue gas (m) = mass of actual air supplied (ASS)+ mass of fuel supplied= 21.49 + 1=22.49

%Dry flue gas loss = 22.49 x 0.23 x (190-30) x 100 = 8.27%

10000

1.9 Factors Affecting Boiler Performance

The various factors affecting the boiler performance are listed below:

� Periodical cleaning of boilers � Periodical soot blowing � Proper water treatment programme and blow down control� Draft control� Excess air control� Percentage loading of boiler� Steam generation pressure and temperature � Boiler insulation� Quality of fuel

All these factors individually/combined, contribute to the performance of the boiler andreflected either in boiler efficiency or evaporation ratio. Based on the results obtained from thetesting further improvements have to be carried out for maximizing the performance. The testcan be repeated after modification or rectification of the problems and compared with standardnorms. Energy auditor should carry out this test as a routine manner once in six months andreport to the management for necessary action.

1.10 Data Collection Format for Boiler Performance Assessment



Sheet 1 � Technical specification of boiler

1 Boiler ID code and Make2 Year of Make3 Boiler capacity rating4 Type of Boiler5 Type of fuel used6 Maximum fuel flow rate7 Efficiency by GCV8 Steam generation pressure &superheat temperature9 Heat transfer area in m2

10 Is there any waste heat recovery device installed11 Type of draft12 Chimney height in metre

Sheet 2 � Fuel analysis details

Fuel FiredGCV of fuelSpecific gravity of fuel (Liquid)Bulk density of fuel (Solid)

Proximate Analysis Date of Test:

1 Fixed carbon %2 Volatile matter %3 Ash %4 Moisture %

Ultimate Analysis Date of Test:

1 Carbon %2 Hydrogen %3 Sulphur %

1.11 Boiler Terminology

MCR: Steam boilers rated output is also usually defined as MCR (Maximum ContinuousRating). This is the maximum evaporation rate that can be sustained for 24 hours and may beless than a shorter duration maximum rating

Boiler Rating

Conventionally, boilers are specified by their capacity to hold water and the steam generationrate. Often, the capacity to generate steam is specified in terms of equivalent evaporation (kgof steam / hour at 100°C). Equivalent evaporation- "from and at" 100°C. The equivalent of theevaporation of 1 kg of water at 100°C to steam at 100°C.

Efficiency : In the boiler industry there are four common definitions of efficiency:

a. Combustion efficiency

Combustion efficiency is the effectiveness of the burner only and relates to its ability to completely burnthe fuel. The boiler has little bearing on combustion efficiency. A well-designed burner will operatewith as little as 15 to 20% excess air, while converting all combustibles in the fuel to useful energy.

b. Thermal efficiency

Thermal efficiency is the effectiveness of the heat transfer in a boiler. It does not take into accountboiler radiation and convection losses - for example from the boiler shell water column piping etc.

c. Boiler efficiency

The term boiler efficiency is often substituted for combustion or thermal efficiency. True boil-er efficiency is the measure of fuel to steam efficiency.



4 Nitrogen %5 Ash %6 Moisture %7 Oxygen %

Water Analysis Date of Test:

1 Feed water TDS ppm2 Blow down TDS ppm3 PH of feed water4 PH of blow down

Flue gas Analysis Date of Test:

1 CO2 %2 O2 %3 CO %4 Flue gas temperature °C

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d. Fuel to steam efficiency

Fuel to steam efficiency is calculated using either of the two methods as prescribed by theASME (American Society for Mechanical Engineers) power test code, PTC 4.1. The firstmethod is input output method. The second method is heat loss method.

Boiler turndown

Boiler turndown is the ratio between full boiler output and the boiler output when operating atlow fire. Typical boiler turndown is 4:1. The ability of the boiler to turndown reduces frequenton and off cycling. Fully modulating burners are typically designed to operate down to 25% ofrated capacity. At a load that is 20% of the load capacity, the boiler will turn off and cycle fre-quently.

A boiler operating at low load conditions can cycle as frequently as 12 times per hour or 288times per day. With each cycle, pre and post purge airflow removes heat from the boiler andsends it out the stack. Keeping the boiler on at low firing rates can eliminate the energy loss.Every time the boiler cycles off, it must go through a specific start-up sequence for safety assur-ance. It requires about a minute or two to place the boiler back on line. And if there is a sud-den load demand the start up sequence cannot be accelerated. Keeping the boiler on line assuresthe quickest response to load changes. Frequent cycling also accelerates wear of boiler com-ponents. Maintenance increases and more importantly, the chance of component failureincreases.

Boiler(s) capacity requirement is determined by many different type of load variations inthe system. Boiler over sizing occurs when future expansion and safety factors are added toassure that the boiler is large enough for the application. If the boiler is oversized the ability ofthe boiler to handle minimum loads without cycling is reduced. Therefore capacity and turn-down should be considered together for proper boiler selection to meet overall system loadrequirements.

Primary air: That part of the air supply to a combustion system which the fuel first encoun-ters.

Secondary air: The second stage of admission of air to a combustion system, generally tocomplete combustion initiated by the primary air. It can be injected into the furnace of a boil-er under relatively high pressure when firing solid fuels in order to create turbulence above theburning fuel to ensure good mixing with the gases produced in the combustion process andthereby complete combustion

Tertiary air: A third stage of admission of air to a combustion system, the reactions of whichhave largely been completed by secondary air. Tertiary air is rarely needed.

Stoichiometric: In combustion technology, stoichiometric air is that quantity of air, and nomore, which is theoretically needed to burn completely a unit quantity of fuel. 'Sub-stoichio-metric' refers to the partial combustion of fuel in a deficiency of air

Balanced draught: The condition achieved when the pressure of the gas in a furnace is thesame as or slightly below that of the atmosphere in the enclosure or building housing it.



Gross calorific value (GCV): The amount of heat liberated by the complete combustion,under specified conditions, by a unit volume of a gas or of a unit mass of a solid or liquid fuel,in the determination of which the water produced by combustion of the fuel is assumed to becompletely condensed and its latent and sensible heat made available.

Net calorific value (NCV): The amount of heat generated by the complete combustion, underspecified conditions, by a unit volume of a gas or of a unit mass of a solid or liquid fuel, in thedetermination of which the water produced by the combustion of the fuel is assumed to remainas vapour.

Absolute pressure The sum of the gauge and the atmospheric pressure. For instance, if thesteam gauge on the boiler shows 9 kg/cm2g the absolute pressure of the steam is 10 kg/cm2(a).

Atmospheric pressure The pressure due to the weight of the atmosphere. It is expressed inpounds per sq. in. or inches of mercury column or kg/cm2. Atmospheric pressure at sea level is14.7 lbs./ sq. inch. or 30 inch mercury column or 760mm of mercury (mm Hg) or 101.325 kiloPascal (kPa).

Carbon monoxide (CO): Produced from any source that burns fuel with incomplete com-bustion, causes chest pain in heart patients, headaches and reduced mental alertness.

Blow down: The removal of some quantity of water from the boiler in order to achieve anacceptable concentration of dissolved and suspended solids in the boiler water.

Complete combustion: The complete oxidation of the fuel, regardless of whether it isaccomplished with an excess amount of oxygen or air, or just the theoretical amount requiredfor perfect combustion.

Perfect combustion: The complete oxidation of the fuel, with the exact theoretical (stoi-chiometric) amount of oxygen (air) required.

Saturated steam: It is the steam, whose temperature is equal to the boiling point corre-sponding to that pressure.

Wet Steam Saturated steam which contains moisture

Dry Steam Either saturated or superheated steam containing no moisture.

Superheated Steam Steam heated to a temperature above the boiling point or saturation tem-perature corresponding to its pressure

Oxygen trim sensor measures flue gas oxygen and a closed loop controller compares theactual oxygen level to the desired oxygen level. The air (or fuel) flow is trimmed bythe controller until the oxygen level is corrected. The desired oxygen level for each firingrate must be entered into a characterized set point curve generator. Oxygen Trim maintains



the lowest possible burner excess air level from low to high fire. Burners that don't haveOxygen Trim must run with Extra Excess Air to allow safe operation during variations inweather, fuel, and linkage.

Heat transfer mediums

There are different types of heat transfer medium e.g. steam, hot water and thermal oil. Steamand Hot water are most common and it will be valuable to briefly examine these common heattransfer mediums and associated properties.

Thermic Fluid

Thermic Fluid is used as a heat transfer mechanism in some industrial process and heatingapplications. Thermic Fluid may be a vegetable or mineral based oil and the oil may be raisedto a high temperature without the need for any pressurization. The relatively high flow andreturn temperatures may limit the potential for flue gas heat recovery unless some other systemcan absorb this heat usefully. Careful design and selection is required to achieve best energyefficiency.

Hot water

Water is a fluid with medium density, high specific heat capacity, low viscosity and relativelylow thermal conductivity. At relatively low temperature e.g. 70°C � 90°C, hot water is usefulfor smaller heating installations.

Steam

When water is heated its temperature will rise. The heat added is called sensible heat and theheat content of the water is termed its enthalpy. The usual datum point used to calculateenthalpy is 0°C.

When the water reaches its boiling point, any further heat input will result in some propor-tion of the water changing from the liquid to the vapour state, i.e. changing to steam. The heatrequired for this change of state is termed the 'latent heat of evaporation' and is expressed interms of a fixed mass of water. Where no change in temperature occurs during the change ofstate, the steam will exist in equilibrium with the water. This equilibrium state is termed 'satu-ration conditions'. Saturation conditions can occur at any pressure, although at each pressurethere is only one discrete temperature at which saturation can occur.

If further heat is applied to the saturated steam the temperature will rise and the steam willbecome 'superheated'. Any increase in temperature above saturated conditions will be accom-panied by a further rise in enthalpy.

Steam is useful heat transfer medium because, as a gas, it is compressible. At high pressureand consequently density, steam can carry large quantities of heat with relatively small volume.





QUESTIONS

1) Define boiler efficiency.

2) Why boiler efficiency by indirect method is more useful than direct method?

3) What instruments are required for indirect efficiency testing?

4) What is the difference between dry flue gas loss and wet flue gas loss?

5) Which is the best location for sampling flue gas analysis?

6) Find out the efficiency by direct method from the data given below.An oil fired package boiler was tested for 2 hours duration at steady state condition.The fuel and water consumption were 250 litres and 3500 litres respectively. Thespecific gravity of oil is 0.92. The saturated steam generation pressure is7 kg/cm2(g). The boiler feed water temperature is 30°C. Determine the boiler effi-ciency and evaporation ratio.

7) What is excess air? How to determine excess air if oxygen / carbon dioxide percent-age is measured in the flue gas?

8) As a means of performance evaluation, explain the difference between efficiency andevaporation ratio.

9) Testing coal-fired boiler is more difficult than oil-fired boiler. Give reasons.

10) What is controllable and uncontrollable losses in a boiler?

REFERENCES 1. Energy audit Reports of National Productivity Council2. Energy Hand book, Second edition, Von Nostrand Reinhold Company - Robert

L.Loftness 3. Industrial boilers, Longman Scientific Technical 1999

www.boiler.comwww.eng-tips.comwww.worldenergy.org

2. ENERGY PERFORMANCE ASSESSMENT OF FURNACES


2.1 Industrial Heating Furnaces

Furnace is by definition a device for heating materials and therefore a user of energy. Heatingfurnaces can be divided into batch-type (Job at stationary position) and continuous type (largevolume of work output at regular intervals). The types of batch furnace include box, bogie,cover, etc. For mass production, continuous furnaces are used in general. The types of continu-ous furnaces include pusher-type furnace (Figure 2.1), walking hearth-type furnace, rotaryhearth and walking beam-type furnace.(Figure 2.2)

The primary energy required for reheating / heat treatment (say annealing) furnaces are inthe form of Furnace oil, LSHS, LDO or electricity

Figure 2.1: Pusher-Type 3-Zone Reheating Furnace Figure 2.2: Walking Beam-Type Reheating Furnace


� To find out the efficiency of the furnace

� To find out the Specific energy consumption

The purpose of the performance test is to determine efficiency of the furnace and specificenergy consumption for comparing with design values or best practice norms. There are manyfactors affecting furnace performance such as capacity utilization of furnaces, excess air ratio,final heating temperature etc. It is the key for assessing current level of performances and find-ing the scope for improvements and productivity.

Heat Balance of a Furnace

Heat balance helps us to numerically understand the present heat loss and efficiency andimprove the furnace operation using these data. Thus, preparation of heat balance is a pre-requirement for assessing energy conservation potential.

2. Energy Performance Assessment of Furnaces



Heat outputHeat Input

Heat in stock (material) (kCals)Heat in Fuel /electricity (kCals)

Quantity of fuel or energy consumedQuantity of material processed.

1. Furnace Efficiency, η =

=

2. Specific Energy Consumption =


In addition to conventional methods, Japanese Industrial Standard (JIS) GO702 "Method ofheat balance for continuous furnaces for steel" is used for the purpose of establishing the heatlosses and efficiency of reheating furnaces.

2.5 Furnace Efficiency Testing Method

The energy required to increase the temperature of a material is the product of the mass, thechange in temperature and the specific heat. i.e. Energy = Mass x Specific Heat x rise intemperature. The specific heat of the material can be obtained from a reference manual anddescribes the amount of energy required by different materials to raise a unit of weight throughone degree of temperature.

If the process requires a change in state, from solid to liquid, or liquid to gas, then anadditional quantity of energy is required called the latent heat of fusion or latent heat ofevaporation and this quantity of energy needs to be added to the total energy requirement.However in this section melting furnaces are not considered.

The total heat input is provided in the form of fuel or power. The desired output is the heatsupplied for heating the material or process. Other heat outputs in the furnaces are undesirableheat losses.

The various losses that occur in the fuel fired furnace (Figure 2.3) are listed below.1. Heat lost through exhaust gases either as sensible heat, latent heat or as incomplete

combustion2. Heat loss through furnace walls and hearth3. Heat loss to the surroundings by radiation and convection from the outer surface of the

walls4. Heat loss through gases leaking through cracks, openings and doors.

x 100

x 100x 100



Furnace Efficiency

The efficiency of a furnace isthe ratio of useful output to heatinput. The furnace efficiency canbe determined by both direct andindirect method.

2.5.1 Direct Method Testing

The efficiency of the furnace canbe computed by measuring theamount of fuel consumed per unitweight of material produced fromthe furnace.

Thermal efficiency of the furnace =Heat in the stock

Heat in the fuel consumed

The quantity of heat to be imparted (Q) to the stock can be found from the formula

Q = m x Cp (t2 – t1)Where

Q = Quantity of heat in kCalm = Weight of the material in kgCp = Mean specific heat, kCal/kg°Ct2 = Final temperature desired, °Ct1 = Initial temperature of the charge before it enters the furnace, °C

2.5.2 Indirect Method Testing

Similar to the method of evaluating boiler efficiency by indirect method, furnace efficiency canalso be calculated by indirect method. Furnace efficiency is calculated after subtracting sensi-ble heat loss in flue gas, loss due to moisture in flue gas, heat loss due to openings in furnace,heat loss through furnace skin and other unaccounted losses from the input to the furnace.

In order to find out furnace efficiency using indirect method, various parameters that arerequired are hourly furnace oil consumption, material output, excess air quantity, temperatureof flue gas, temperature of furnace at various zones, skin temperature and hot combustion airtemperature. Efficiency is determined by subtracting all the heat losses from 100.

Measurement Parameters

The following measurements are to be made for doing the energy balance in oil fired reheatingfurnaces (e.g. Heating Furnace)

i) Weight of stock / Number of billets heatedii) Temperature of furnace walls, roof etciii) Flue gas temperatureiv) Flue gas analysisv) Fuel Oil consumption



Instruments like infrared thermometer, fuel consumption monitor, surface thermocoupleand other measuring devices are required to measure the above parameters. Reference manualshould be referred for data like specific heat, humidity etc.

Example: Energy Efficiency by Indirect Method

An oil-fired reheating furnace has an operating temperature of around 1340°C. Average fuel con-sumption is 400 litres/hour. The flue gas exit temperature after air preheater is 750°C. Air is pre-heated from ambient temperature of 40°C to 190°C through an air pre-heater. The furnace has 460mm thick wall (x) on the billet extraction outlet side, which is 1 m high (D) and 1 m wide. Theother data are as given below. Find out the efficiency of the furnace by both indirect and directmethod.

Flue gas temperature after air preheater = 750°CAmbient temperature = 40°CPreheated air temperature = 190°CSpecific gravity of oil = 0.92Average fuel oil consumption = 400 Litres / hr

= 400 x 0.92 =368 kg/hrCalorific value of oil = 10000 kCal/kgAverage O2 percentage in flue gas = 12%Weight of stock = 6000 kg/hrSpecific heat of Billet = 0.12 kCal/kg/°CSurface temperature of roof and side walls = 122 °CSurface temperature other than heating and soaking zone = 85 °C

Solution

1. Sensible Heat Loss in Flue Gas:

Excess air × 100

(Where O2 is the % of oxygen in flue gas = 12% )= 12 x 100 / (21 - 12)= 133% excess air

Theoretical air required to burn 1 kg of oil = 14 kg (Typical value for all fuel oil)Total air supplied = Theoretical air x (1 + excess air/100)Total air supplied = 14 x 2.33 kg / kg of oil

= 32.62 kg / kg of oilSensible heat loss = m x Cp x ∆T

m = Weight of flue gas= Actual mass of air supplied / kg of

fuel + mass of fuel (1kg)= 32.62 + 1.0 = 33.62 kg / kg of oil.

Cp = Specific heat of flue gas= 0.24 kCal/kg/°C

∆T = Temperature difference

O2%= ————

21–O2%



Heat loss = m x Cp x ∆T = 33.62 x 0.24 x (750- 40)= 5729 kCal / kg of oil

% Heat loss in flue gas =

2. Loss Due to Evaporation of Moisture Present in Fuel M {584 + 0.45 (Tfg–Tamb)}

% Loss = ——————————— × 100GCV of Fuel

Where,

M - kg of Moisture in 1 kg of fuel oil (0.15 kg/kg of fuel oil)Tfg - Flue Gas TemperatureTamb - Ambient temperatureGCV - Gross Calorific Value of Fuel

0.15 {584 +0.45 (750-40)}% Loss = -------------------------------- x 100

10000

= 1.36 %

3. Loss Due to Evaporation of Water Formed due to Hydrogen in Fuel

% Loss 9 x H2 {584 + 0.45 (Tfg-Tamb)}= --------------------------------------- x 100

GCV of Fuel

Where, H2 – kg of H2 in 1 kg of fuel oil (0.1123 kg/kg of fuel oil)

= 9 x 0.1123 {584 + 0.45 (750-40)}------------------------------------------ x 100

10000

= 9.13 %

4. Heat Loss due to Openings:

If a furnace body has an opening on it, the heat in the furnace escapes to the outside asradiant heat. Heat loss due to openings can be calculated by computing black body radiation atfurnace temperature, and multiplying these values with emissivity (usually 0.8 for furnace brickwork), and the factor of radiation through openings. Factor for radiation through openings canbe determined with the help of graph as shown in figure 2.4. The black body radiation lossescan be directly computed from the curves as given in the figure 2.5 below.

5729 x 100—————— = 57.29%

10000



Figure 2.5 Graph for Determining Black Body Radiation at a Particular Temperature

Figure 2.4 Factor for Determining the Equivalent of Heat Release from Openings to the Quality of HeatRelease from Perfect Black Body

The reheating furnace in example has 460mm thick wall (X) on the billet extraction outletside, which is 1m high (D) and 1m wide. With furnace temperature of 1340°C, the quantity (Q)of radiation heat loss from the opening is calculated as follows:

The shape of the opening is square and D/X = 1/0.46 = 2.17The factor of radiation (Refer Figure 2.4) = 0.71Black body radiation corresponding to 1340°C = 36.00 kCal/cm2/hr(Refer Figure 2.5 On black body radiation)

TO

TA

LB

LA

CK

BO

DY

RA

DIA

TIO

N (

kCal

/cm

2 /hr

)

Temperature (°C)



Area of opening = 100 cm x 100 cm= 10000 cm2

Emissivity = 0.8

Total heat loss = Black body radiation x area of opening x factor of radiation x emissivity

= 36 x 10000 x 0.71 x 0.8= 204480 kCal/hr

Equivalent Oil loss = 204480/10,000= 20.45 kg/hr

% of heat loss = 20.45 /368 x 100= 5.56 %

5. Heat Loss through Skin:

Method 1: Radiation Heat Loss from Surface of Furnace

The quantity of heat loss from surface of furnace body is the sum of natural convection andthermal radiation. This quantity can be calculated from surface temperatures of furnace. Thetemperatures on furnace surface should be measured at as many points as possible, and theiraverage should be used. If the number of measuring points is too small, the error becomeslarge.

The quantity (Q) of heat release from a reheating furnace is calculated with the followingformula:

where

Q : Quantity of heat release in kCal / W / m2

a : factor regarding direction of the surface of natural convection ceiling = 2.8, side walls = 2.2, hearth = 1.5

tl : temperature of external wall surface of the furnace (°C)t2 : temperature of air around the furnace (°C)E : emissivity of external wall surface of the furnace

The first term of the formula above represents the quantity of heat release by natural con-vection, and the second term represents the quantity of heat release by radiation.

Method 2 : Radiation Heat Loss from Surface of Furnace

The following Figure 2.6 shows the relation between the temperature of external wall surfaceand the quantity of heat release calculated with this formula.



Figure 2.6 Quantity of Heat Release at VariousTemperatures

From the Figure 2.6, the quantities of heat release from ceiling, sidewalls and hearth per unitarea can be found.

5a). Heat loss through roof and sidewalls:

Total average surface temperature = 122°CHeat loss at 122 °C = 1252 kCal / m2 / hrTotal area of heating + soaking zone = 70.18 m2

Heat loss = 1252 kCal / m2 / hr x 70.18 m2

= 87865 kCal/hrEquivalent oil loss (a) = 8.78 kg / hr

5b). Total average surface temperature of area other than heating and soaking zone = 85°CHeat loss at 85°C = 740 kCal / m2 / hrTotal area = 12.6 m2

Heat loss = 740 kCal / m2 / hr x 12.6 m2

= 9324 kCal/hrEquivalent oil loss (b) = 0.93 kg / hr Total loss of fuel oil = a + b = 9.71 kg/hrTotal percentage loss = 9.71 / 368

= 2.64%

6. Unaccounted Loss

These losses comprise of heat storage loss, loss of furnace gases around charging door andopening, heat loss by incomplete combustion, loss of heat by conduction through hearth, lossdue to formation of scales.



Furnace Efficiency (Direct Method)

Fuel input = 400 litres / hr= 368 kg/hr

Heat input = 368 x 10000 = 36,80,000 kCalHeat output = m x Cp x ∆T

= 6000 kg x 0.12 x (1340 – 40)= 936000 kCal

Efficiency = 936000 x 100 / (368 x 10000)= 25.43 %= 25% (app)

Total Losses = 75% (app)

Furnace Efficiency (Indirect Method)

1. Sensible heat loss in flue gas = 57.29%2. Loss due to evaporation of moisture in fuel = 1.36 % 3. Loss due to evaporation of water

formed from H2 in fuel = 9.13 %4. Heat loss due to openings = 5.56 %5. Heat loss through skin = 2.64%

Total losses = 75.98%

Furnace Efficiency = 100 - 75.98

= 24.02 %

Specific Energy Consumption = 400 litre /hour (fuel consumption)6 Tonnes/hour (Wt of stock)

= 66.6 Litre of fuel /tonne of Material (stock)

2.5.4 Factors Affecting Furnace Performance

The important factors, which affect the efficiency, are listed below for critical analysis.

� Under loading due to poor hearth loading and improper production scheduling � Improper Design� Use of inefficient burner� Insufficient draft/chimney� Absence of Waste heat recovery� Absence of Instruments/Controls� Improper operation/Maintenance� High stack loss � Improper insulation /Refractories



2.6 Data Collection Format for Furnace Performance Assessment

The field-testing format for data collection and parameter measurements are shown below

Stock

Charged amount in Charging Discharging Dischargefurnace temperature temperature material

Tons/hr °C °C kg/ton

Fuel Analysis

Fuel Consumption Components of heavy oil Gross Temperature

type C H2 O2 N2 S Water calorificcontent value

Kg/hr % % % % % % kCal/kg °C

Temperature Composition of dry exhaust gas

CO2 O2 CO

°C % % %

Amount of Water Inlet temperature Outlet temperature

kg/ton °C °C

Flue gas Analysis

Cooling water

Temperature of combustion air =Ambient air temperature =



The Table 2.1 can be used to construct a heat balance for a typical heat treatment furnace

TABLE 2.1 HEAT BALANCE TABLE

Heat Input Heat output

Item kCal/t % Item kCal/t %

Combustion heat of fuel Quantity of heat in steel

Sensible heat in flue gas

Moisture andhydrogen loss of fuel

Heat loss byIncomplete combustion(CO loss)

Heat loss incooling water

Sensible heatof scale

Heat Loss Due ToOpenings

Radiation and Otherunaccounted heat loss

Total = 100% Total = 100%

2.7 Useful Data

Radiation Heat Transfer

Heat transfer by radiation is proportional to the absolute temperature to the power 4.Consequently the radiation losses increase considerably as temperature increases.

°C1 °C2 K1 K2 (K1/K2)4 Relative(°C1 +273) (°C2 +273) Radiation

700 20 973 293 122 1.0

900 20 1173 293 255 2.1

1100 20 1373 293 482 3.96

1300 20 1573 293 830 6.83

1500 20 1773 293 1340 11.02

1700 20 1973 293 2056 16.91



In practical terms this means the radiation losses from an open furnace door at 1500°C are11 times greater than the same furnace at 700°C. A good incentive for the iron and steel meltersis to keep the furnace lid closed at all times and maintaining a continuous feed of cold chargeonto the molten bath.

Furnace Utilization Factor

Utilization has a critical effect on furnace efficiency and is a factor that is often ignored orunder-estimated. If the furnace is at temperature then standby losses of a furnace occur whetheror not a product is in the furnace.

Standby Losses

Energy is lost from the charge or its enclosure in the way of heat: (a) conduction, (b) convec-tion; or/and (c) radiation

Furnace Draft Control

Furnace pressure control has a major effect on fuel fired furnace efficiency. Running a furnaceat a slight positive pressure reduces air ingress and can increase the efficiency.

Theoretical Heat

Example of melting one tonne of steel from an ambient temperature of 20°C . Specific heat ofsteel = 0.186 Wh/kg/°C, latent heat for melting of steel = 40 Wh/kg/°C. Melting point of steel= 1600°C.

Theoretical Total heat = Sensible heat + Latent heat

Sensible Heat = 1000 kg x 0.186 Wh /kg °C x (1600-20)°C = 294 kWh/T

Latent heat = 40 Wh/ kg x 1000 kg = 40 kWh/T

Total Heat = 294 + 40 = 334 kWh/T

So the theoretical energy needed to melt one tonne of steel from 20°C = 334 kWh.Actual Energy used to melt to 1600°C is 700 kWh

Efficiency = 334 kWh x 100 = 48%700 kwh



Typical furnace efficiency for reheating and forging furnaces (As observed in few trialsundertaken by an Energy Auditing Agency on such furnaces)

Pusher Type Billet Reheating Furnace (for rolling mills)

Furnace Specific Fuel Thermal EfficiencyCapacity Consumption Achieved

Upto 6 T/hr 40-45 Ltrs/tonne 52%

7-8 T / hr 35-40 Ltrs/tonne 58.5%

10-12 T/hr 33-38 Ltrs/tonne 63%

15-20 T/hr 32-34 Ltrs/tonne 66.6%

20 T/hr & above 30-32 Ltrs/tonne 71%

Pusher type forging furnace

Furnace Specific Fuel Thermal EfficiencyCapacity Consumption Achieved

500-600 kg/hr 80-90 Ltrs/tonne 26%

1.0 T/hr 70-75 Ltrs/tonne 30%

1.5-2.0 T/hr 65-70 Ltrs/tonne 32.5%

2.5-3.0 T/hr 55-60 Ltrs/tonne 38%

The above fuel consumption figures were valid when the furnaces were found to be operat-ing continuously at their rated capacity.

Note: These are the trial figures and cannot be presumed as standards for the furnacesin question.



QUESTIONS

1) What is a heating Furnace and give two examples?

2) Define furnace efficiency.

3) How do you determine the furnace efficiency by direct method?

4) How do you determine the furnace efficiency by Indirect method?

5) Between efficiency and specific energy consumption, which is a better mean of com-paring furnaces?

6) List down the various heat losses taking place in oil-fired furnace.

7) What are the major factors affecting the furnace performance?

8) Apart from the furnace operating parameters, energy auditor needs certain data fromreference book/manual for assessing furnace. Name few of them

9) What will be the difference in approach for conducting efficiency testing of batchand continuous type furnace?

10) How will you measure the temperature of the stock inside the furnace?

REFERENCES1. Handbook of Energy Conservation for Industrial Furnaces, Japan Industrial Furnace

Association.2. Energy audit reports of National Productivity Council3. Industrial Furnace, Volume 1 and Volume 2, John Wiley & Sons - Trinks4. Improving furnace efficiency, Energy Management Journal

3. ENERGY PERFORMANCE ASSESSMENT OFCOGENERATION AND TURBINES (GAS, STEAM)


3.1 Introduction

Cogeneration systems can be broadly classified as those using steam turbines, Gas turbines andDG sets. Steam turbine cogeneration systems involve different types of configurations withrespect to mode of power generation such as extraction, back pressure or a combination of back-pressure, extraction and condensing.

Gas turbines with heat recovery steam generators is another mode of cogeneration.Depending on power and steam load variations in the plant the entire system is dynamic. A per-formance assessment would yield valuable insights into cogeneration system performance andneed for further optimisation.


The purpose of the cogeneration plant performance test is to determine the power output andplant heat rate. In certain cases, the efficiency of individual components like steam turbine isaddressed specifically where performance deterioration is suspected. In general, the plant per-formance will be compared with the base line values arrived at for the plant operating condi-tion rather than the design values. The other purpose of the performance test is to show themaintenance accomplishment after a major overhaul. In some cases the purpose of evaluationcould even be for a total plant revamp.


3. Energy Performance Assessment of Cogeneration and Turbine


3.4 Reference standards

Modern power station practices by British electricity International (Pergamon Press) ASMEPTC 22 - Gas turbine performance test.

3.5 Field Testing Procedure

The test procedure for each cogeneration plant will be developed individually taking into con-sideration the plant configuration, instrumentation and plant operating conditions. A method is

kCal/kgkCal/kg

outlined in the following section for the measurement of heat rate and efficiency of a co-generation plant. This part provides performance-testing procedure for a coal fired steambased co-generation plant, which is common in Indian industries.

3.5.1 Test Duration

The test duration is site specific and in a continuous process industry, 8-hour test data shouldgive reasonably reliable data. In case of an industry with fluctuating electrical/steam load pro-file a set 24-hour data sampling for a representative period.

3.5.2 Measurements and Data Collection

The suggested instrumentation (online/ field instruments) for the performance measurement isas under:

Steam flow measurement : Orifice flow metersFuel flow measurements : Volumetric measurements / Mass flow metersAir flow / Flue gas flow : Venturi / Orifice flow meter / Ion gun / Pitot tubesFlue gas Analysis : Zirconium Probe Oxygen analyserUnburnt Analysis : Gravimetric AnalysisTemperature : ThermocoupleCooling water flow : Orifice flow meter / weir /channel flow/

non-contact flow metersPressure : Bourdon Pressure GaugesPower : Trivector meter / Energy meterCondensate : Orifice flow meter

It is essential to ensure that the data is collected during steady state plant running conditions.Among others the following are essential details to be collected for cogeneration plant perfor-mance evaluation.



Step 1 :

Calculate the actual heat extraction in turbine at each stage,

Steam Enthalpy at turbine inlet : h1 kCal / kg Steam Enthalpy at 1st extraction : h2 kCal / kgSteam Enthalpy at 2nd extraction : h3 kCal / kg Steam Enthalpy at Condenser : h4* kCal / kg

* Due to wetness of steam in the condensing stage, the enthalpy of steam cannot be consideredas equivalent to saturated steam. Typical dryness value is 0.88 – 0.92. This dryness value can beused as first approximation to estimate heat drop in the last stage. However it is suggested to cal-culate the last stage efficiency from the overall turbine efficiency and other stage efficiencies.

II. Electrical Energy:

1. Total power generation for the trial period from individual turbines.2. Hourly average power generation3. Quantity of power import from utility ( Grid )*4. Quantity of power generation from DG sets.*5. Auxiliaries power consumption

* Necessary only when overall cogeneration plant adequacy and system optimization / upgra-dation are the objectives of the study.

3.5.3 Calculations for Steam Turbine Cogeneration System

The process flow diagram for cogeneration plant is shown in figure 3.1. The following calcu-lation procedures have been provided in this section.

• Turbine cylinder efficiency.• Overall plant heat rate



Figure 3.1 Process Flow Diagram for Cogeneration Plant



Heat extraction from inlet : h1 – h2 kCal / kgto stage –1 extraction (h5)

Heat extraction from : h2 – h3 kCal / kg1st –2nd extraction (h6)

Heat extraction from 2nd : h3 – h4 kCal / kgExtraction – condenser (h7)

Step 2:

From Mollier diagram (H-S Diagram) estimate the theoretical heat extraction for the conditionsmentioned in Step 1. Towards this:

a) Plot the turbine inlet condition point in the Mollier chart - corresponding to steampressure and temperature.

b) Since expansion in turbine is an adiabatic process, the entropy is constant. Hence drawa vertical line from inlet point (parallel to y-axis) upto the condensing conditions.

c) Read the enthalpy at points where the extraction and condensing pressure lines meetthe vertical line drawn.

d) Compute the theoretical heat drop for different stages of expansion.

Theoretical Enthalpy after 1st extraction : H1

Theoretical Enthalpy after 2nd extraction : H2

Theoretical Enthalpy at condenser conditions H3

Theoretical heat extraction from inlet to : h1 – H1

stage 1 extraction, h8

Theoretical heat extraction from : H1 – H2

1st – 2nd extraction, h9

Theoretical heat extraction from : H2 – H3

2nd extraction – condensation, h10

Step 3 :

Step 4 :

Calculate plant heat rate*

M x (h1 – h11)Heat rate, kCal / kWh =

P

M – Mass flow rate of steam in kg/hrh1 – Enthalpy of inlet steam in kCal/kgh11 – Enthalpy of feed water in kCal/kgP – Average Power generated in kW

*Alternatively the following guiding parameter can be utilised

Plant heat consumption = fuel consumed for power generation, kg/hrPower generated, kW

3.6 Example

3.6.1 Small Cogeneration Plant

A distillery plant having an average production of 40 kilolitres of ethanol is having a cogener-ation system with a backpressure turbine. The plant steam and electrical demand are 5.1Tons/hr and 100 kW. The process flow diagram is shown in figure 3.2.Gross calorific value ofIndian coal is 4000kCal/kg



Figure 3.2 Process Flow Diagram for Small Cogeneration Plant

Calculations :

Step 1 :

Total heat of steam at turbine inlet conditions at 15kg / cm2 and 250°C, h1 =698 kCal/kg

Step 2 :

Total heat of steam at turbine outlet conditions at 2 kg/cm2 and 130°C, h2 = 648 kCal/kg

Step 3 :

Heat energy input to turbine per kg of inlet steam (h1– h2) = (698-648) = 50 kCal/kg

Step 4 :

Total steam flow rate, Q1 = 5100 kg/hrPower generation = 100 kWEquivalent thermal energy = 100 x 860 = 86,000 kCal /hr

Step 5 :

Energy input to the turbine = 5100 x 50 = 2,55,000 kCal/hr.

Step 6 :

Energy outputPower generation efficiency of the turbo alternator = --------------------- x 100

Energy Input

86,000= ------------- x 100 = 34%

2,55,000

Step 7 :

Efficiency of the turbo alternator = 34% Efficiency of Alternator = 92 %Efficiency of gear transmission = 98 %

Step 8 :

Quantity of steam bypassing the turbine = Nil

Step 9 :

Coal consumption of the boiler = 1550 kg/hr.



Step 10:

Overall plant heat rate, kCal/kWh

= Mass flow rate of steam x ((Enthalpy of steam, kCal/kg – Enthalpy of feed water, kCal/kg)

Power output, kW= 5100 x (698 – 30)

100

= 34068 kCal/kWh*

*Note: The plant heat rate is in the order of 34000 kCal/kWh because of the use of backpres-sure turbine. This value will be around 3000 kcal/kWh while operating on fully condensingmode. However with backpressure turbine, the energy in the steam is not wasted, as it is utilisedin the process.

Overall plant fuel rate including boiler = 1550/100 = 15.5 kg coal / kW

Analysis of Results:

The efficiency of the turbine generator set is as per manufacturer design specification. There isno steam bypass indicating that the power generation potential of process steam is fully utilized.At present the power generation from the process steam completely meets the process electri-cal demand or in other words, the system is balanced.

Remarks: Similar steps can be followed for the evaluation of performance of gas turbinebased cogeneration system.





QUESTIONS

1. What is meant by plant heat rate? What is its significance?

2. What is meant by turbine cylinder efficiency? How is it different from turbo-genera-tor efficiency?

3. What parameters should be monitored for evaluating the efficiency of the turbine?

4. What is the need for performance assessment of a cogeneration plant?

5. The parameters for back pressure steam turbine cogeneration plant is given below

Inlet Steam: P =16 kg/cm2, T = 310°C, Q = 9000kg/hrOutlet Steam: P = 5.0 kg/cm2, T = 235°C, Q = 9000kg/hrFind out the turbine cylinder efficiency?

6. Explain why heat rate for back pressure turbine is greater than condensing turbine.

7. Explain the methodology of evaluating performance of a gas turbine with a heatrecovery steam generator.

REFERENCES 1. NPC report on 'Assessing cogeneration potential in Indian Industries'2. Energy Cogeneration Handbook, George Polimeros, Industrial Press Inc.

4. ENERGY PERFORMANCE ASSESSMENTOF HEAT EXCHANGERS


4.1 Introduction

Heat exchangers are equipment that transfer heat from one medium to another. The properdesign, operation and maintenance of heat exchangers will make the process energy efficientand minimize energy losses. Heat exchanger performance can deteriorate with time, offdesign operations and other interferences such as fouling, scaling etc. It is necessary toassess periodically the heat exchanger performance in order to maintain them at a high effi-ciency level. This section comprises certain proven techniques of monitoring the perfor-mance of heat exchangers, coolers and condensers from observed operating data of theequipment.


To determine the overall heat transfer coefficient for assessing the performance of the heatexchanger. Any deviation from the design heat transfer coefficient will indicate occurrence offouling.


Overall heat transfer coefficient, U

Heat exchanger performance is normally evaluated by the overall heat transfer coefficient Uthat is defined by the equation

When the hot and cold stream flows and inlet temperatures are constant, the heat transfercoefficient may be evaluated using the above formula. It may be observed that the heat pick upby the cold fluid starts reducing with time.

Nomenclature

A typical heat exchanger is shown in figure 4.1 with nomenclature.

4. Energy Performance Assessment Of Heat Exchangers


Heat duty of the exchanger can be calculated either on the hot side fluid or cold side fluidas given below.Heat Duty for Hot fluid, Qh = Wx Cph x (Ti–To) ………..Eqn–1, Heat Duty for Cold fluid, Qc = wx Cpc x ( to–ti) ………...Eqn–2 If the operating heat duty is less than design heat duty, it may be due to heat losses, fouling

in tubes, reduced flow rate (hot or cold) etc. Hence, for simple performance monitoring ofexchanger, efficiency may be considered as factor of performance irrespective of other para-meter. However, in industrial practice, fouling factor method is more predominantly used.

4.4 Methodology of Heat Exchanger Performance Assessment

4.4.1 Procedure for determination of Overall heat transfer Coefficient, U at field

This is a fairly rigorous method of monitoring the heat exchanger performance by calculatingthe overall heat transfer coefficient periodically. Technical records are to be maintained for allthe exchangers, so that problems associated with reduced efficiency and heat transfer can beidentified easily. The record should basically contain historical heat transfer coefficient dataversus time / date of observation. A plot of heat transfer coefficient versus time permits ratio-nal planning of an exchanger-cleaning program.

The heat transfer coefficient is calculated by the equation

U = Q / (A x LMTD)

Where Q is the heat duty, A is the heat transfer area of the exchanger and LMTD is tem-perature driving force.

The step by step procedure for determination of Overall heat transfer Coefficient aredescribed below

Density and viscosity can be determined by analysis of the samples taken from the flowstream at the recorded temperature in the plant laboratory. Thermal conductivity and specificheat capacity if not determined from the samples can be collected from handbooks.





4.4.2 Examples

a. Liquid - Liquid Exchanger

A shell and tube exchanger of following configuration is considered being used for oil cool-er with oil at the shell side and cooling water at the tube side.

Tube Side

• 460 Nos x 25.4mmOD x 2.11mm thick x 7211mm long• Pitch - 31.75mm 30° triangular• 2 Pass

Shell Side

• 787 mm ID • Baffle space - 787 mm• 1 Pass

Heat Duty: Actual duty differences will be practically negligible as these duty differencescould be because of the specific heat capacity deviation with the temperature. Also, there couldbe some heat loss due to radiation from the hot shell side.

Pressure drop: Also, the pressure drop in the shell side of the hot fluid is reported normal(only slightly less than the design figure). This is attributed with the increased average bulktemperature of the hot side due to decreased performance of the exchanger.

Temperature range: As seen from the data the deviation in the temperature ranges could bedue to the increased fouling in the tubes (cold stream), since a higher pressure drop is noticed.

Heat Transfer coefficient: The estimated value has decreased due to increased fouling that hasresulted in minimized active area of heat transfer.

Physical properties: If available from the data or Lab analysis can be used for verificationwith the design data sheet as a cross check towards design considerations.

Troubleshooting: Fouled exchanger needs cleaning.

b. Surface Condenser

A shell and tube exchanger of following configuration is considered being used for Condensingturbine exhaust steam with cooling water at the tube side.

Tube Side

20648 Nos x 25.4mmOD x 1.22mm thk x 18300mm longPitch - 31.75mm 60° triangular1 PassThe monitored parameters are as below:



Parameters Units Inlet Outlet

Hot fluid flow, W kg/h 939888 939888

Cold fluid flow, w kg/h 55584000 55584000

Hot fluid Temp, T °C No data 34.9

Cold fluid Temp, t °C 18 27

Hot fluid Pressure, P m Bar g 52.3 mbar 48.3

Cold fluid Pressure, p Bar g 4 3.6

Calculation of Thermal data:

Area = 27871 m2

1. Duty:Q = qS + qL

Hot fluid, Q = 576990 kWCold Fluid, Q = 581825.5 kW

2. Hot Fluid Pressure DropPressure Drop = Pi – Po = 52.3 – 48.3 = 4.0 mbar.

3. Cold Fluid Pressure DropPressure Drop = pi – po = 4 – 3.6 = 0.4 bar.

4. Temperature range hot fluidTemperature Range ∆T = Ti– To = No data

5. Temperature Range Cold FluidTemperature Range ∆t = ti – to = 27 – 18 = 9 °C.

6. Capacity RatioCapacity ratio, R = Not significant in evaluation here.

7. EffectivenessEffectiveness, S = (to – ti) / (Ti – ti) = Not significant in evaluation here.

8. LMTD

Calculated considering condensing part only

a). LMTD, Counter Flow = ((34.9 – 18)–(34.9–27))/ ln ((34.9–18)/(34.9–27)) = 11.8 deg C.

b). Correction Factor to account for Cross flow

F = 1.0.

9. Corrected LMTDMTD = F x LMTD = 1.0 x 11.8 = 11.8 deg C.

10. Heat Transfer Co-efficientOverall HTC, U = Q/ A ∆T = 576990/ (27871 x 11.8) = 1.75 kW/m2. K

Comparison of Calculated data with Design Data



Parameters Units Test Data Design Data

Duty, Q kW 576990 588430

Hot fluid side pressure drop, ∆Ph mBar 4 mbar 3.7 mbar

Cold fluid side pressure drop, ∆Pc Bar 0.4

Temperature Range hot fluid, ∆T °C

Temperature Range cold fluid, ∆t °C (27–18) = 9 (28–19) = 9

Capacity ratio, R -----

Effectiveness, S -----

Corrected LMTD, MTD °C 11.8 8.9

Heat Transfer Coefficient, U kW/(m2. K) 1.75 2.37

Heat Duty: Actual duty differences will be practically negligible as these duty differencescould be because of the specific heat capacity deviation with the temperature. Also, there couldbe some heat loss due to radiation from the hot shell side.

Pressure drop: The condensing side operating pressure raised due to the backpressurecaused by the non-condensable. This has resulted in increased pressure drop across the steamside

Temperature range: With reference to cooling waterside there is no difference in the rangehowever, the terminal temperature differences has increased indicating lack of proper heattransfer.

Heat Transfer coefficient: Heat transfer coefficient has decreased due to increased amount ofnon-condensable with the steam.

Trouble shooting: Operations may be checked for tightness of the circuit and ensure proper venting of the system. The vacuum source might be verified for proper functioning.

C. Vaporizer

A shell and tube exchanger of following configuration is considered being used for vaporizingchlorine with steam at the shell side.

Tube Side

200 Nos x 25.4mmOD x 1.22mm thick x 6000mm longPitch - 31.75mm 30° triangular2 PassArea = 95.7.m2



Calculation of Thermal data:

1. Duty:Q = qS + qL

Hot fluid, Q = 3130 kW

Cold Fluid, Q = qS + qL = 180.3 kW + 2948 kW = 3128.3 kW

2. Hot Fluid Pressure DropPressure Drop = Pi – Po = 0.4 – 0.3 = 0.1 bar

3. Cold Fluid Pressure DropPressure Drop = pi – po = 9 – 8.8 = 0.2 bar.

4. Temperature range hot fluidTemperature Range ∆T = Ti – To = 0 °C




8. LMTDCalculated considering condensing part only

a). LMTD, Counter Flow =((108 – 30)–(108–34))/ ln ((108–30)/(108–34)) = 76 °C.

b). Correction Factor to account for Cross flow

F = 1.0.

9. Corrected LMTDMTD = F x LMTD = 1.0 x 76 = 76 °C.

10. Heat Transfer Co-efficientOverall HTC, U = Q/ A ∆T = 3130/ (95.7 x 76) = 0.43 kW/m2. K






Hot fluid Temp, T °C 108 108


Hot fluid Pressure, P Bar g 0.4 0.3

Cold fluid Pressure, p Bar g 9 8.8

The monitored parameters are as below:




Duty, Q kW 3130 3130

Hot fluid side pressure drop, ∆Ph Bar 0.1 Neg

Cold fluid side pressure drop, ∆Pc Bar 0.2


Temperature Range cold fluid, ∆t °C 4 4



Corrected LMTD, MTD °C 76



Heat Duty: There is no difference inferred from the duty as the exchanger is performing as perthe requirement

Pressure drop: The steam side pressure drop has increased in spite of condensation at thesteam side. Indication of non-condensable presence in steam side

Temperature range: No deviations

Heat Transfer coefficient: Even at no deviation in the temperature profile at the chlorine side,heat transfer coefficient has decreased with an indication of overpressure at the shell side. Thisindicates disturbances to the condensation of steam at the shell side. Non-condensable suspect-ed at steam side.

Trouble shooting: Operations may be checked for presence of chlorine at the shell sidethrough tube leakages. Observing the steam side vent could do this. Alternately condensate pHcould be tested for presence of acidity.

d. Air heater

A finned tube exchanger of following configuration is considered being used for heating airwith steam in the tube side.

The monitored parameters are as below:






Hot fluid Temp, T °C 150 150


Hot fluid Pressure, P Bar g

Cold fluid Pressure, p mBar g 200 mbar 180 mbar

Calculation of Thermal data:Bare tube Area = 42.8 m2; Fined tube area = 856 m2

1.Duty:Hot fluid, Q = 1748 kW

Cold Fluid, Q = 1726 kW

2. Hot Fluid Pressure DropPressure Drop = Pi – Po = Neg

3. Cold Fluid Pressure DropPressure Drop = pi – po = 200–180 = 20 mbar.

4. Temperature range hot fluidTemperature Range ∆T = Ti – To = Not required.




8. LMTDCalculated considering condensing part only

a). LMTD, Counter Flow =((150 – 30)–(150–95)/ ln ((150–30)/(150–95)) = 83.3 °C.

b). Correction Factor to account for cross flow

F = 0.95

9. Corrected LMTD

MTD = F x LMTD = 0.95 x 83.3 = 79 °C.

10. Overall Heat Transfer Co-efficient (HTC)U = Q/ A ∆T = 1748/ (856 x 79) = 0.026 kW/m22. K





Duty, Q kW 1748 1800

Hot fluid side pressure drop, ∆Ph Bar Neg Neg

Cold fluid side pressure drop, ∆Pc Bar 20 15


Temperature Range cold fluid, ∆t °C 65 65



Corrected LMTD, MTD °C 79 79


Heat Duty: The difference inferred from the duty as the exchanger is under performing thanrequired

Pressure drop: The airside pressure drop has increased in spite of condensation at the steamside. Indication of choking and dirt blocking at the airside.

Temperature range: No deviations

Heat Transfer coefficient: Decreased because of decreased fin efficiency due to choking onair side.

Trouble shooting: Operations may be checked to perform pulsejet cleaning with steam / blowair jet on air side if the facility is available. Mechanical cleaning may have to be planned dur-ing any down time in the immediate future.

4.4.3 Instruments for monitoring:

The test and evaluation of the performance of the heat exchanger equipment is carried out bymeasurement of operating parameters upstream and downstream of the exchanger. Due careneeds to be taken to ensure the accuracy and correctness of the measured parameter. The instru-ments used for measurements require calibration and verification prior to measurement.



Parameters Units Instruments used

Fluid flow kg/h Flow can be measured with instruments like Orifice flow meter, Vortex flow meter, Venturi meters, Coriollis flow meters, Magnetic flowmeter as applicable to the fluid service and flow ranges

Temperature °C Thermo gauge for low ranges, RTD, etc.

Pressure Bar g Liquid manometers, Draft gauge, Pressure gauges Bourdon and diaphragm type, Absolute pressure transmitters, etc.

Density kg/m3 Measured in the Laboratory as per ASTM standards, hydrometer, etc

Viscosity MpaS Measured in the Laboratory as per ASTM standards, viscometer, etc.

Specific heat capacity J/(kg.K) Measured in the Laboratory as per ASTM standards

Thermal conductivity W/(m.K) Measured in the Laboratory as per ASTM standards

Composition+ %wt (or) % Vol Measured in the Laboratory as per ASTM standards using Chemical analysis, HPLC, GC, Spectrophotometer, etc.

Terminology Definition Unit

Capacity ratio Ratio of the products of mass flow rate and specific heat capacity of the cold fluid to that of the hot fluid. Also computed by the ratio of temperature range of the hot fluid to that of the cold fluid. Higher the ratio greater will be size of the exchanger

Co current flow An exchanger wherein the fluid flow direction of the exchanger cold and hot fluids are same

Counter flow Exchangers wherein the fluid flow direction of the cold and exchanger hot fluids are opposite. Normally preferred

Cross flow An exchanger wherein the fluid flow direction of the cold and hot fluids are in cross

4.4.4 Terminology used in Heat Exchangers

Density It is the mass per unit volume of a material kg/m3

Effectiveness Ratio of the cold fluid temperature range to that of the inlet temperature difference of the hot and cold fluid. Higher the ratio lesser will be requirement of heat transfer surface

Fouling The phenomenon of formation and development of scales and deposits over the heat transfer surface diminishing the heat flux. The process of fouling will get indicated by the increase in pressure drop

Fouling Factor The reciprocal of heat transfer coefficient of the dirt formed in the heat exchange process. Higher the factor lesser will be the overall heat transfer coefficient. (m2.K)/W

Heat Duty The capacity of the heat exchanger equipment expressed in terms of heat transfer rate, viz. magnitude of energy or heat transferred per time. It means the exchanger is capable of performing at this capacity in the given system W

Heat exchanger Refers to the nomenclature of equipment designed and constructed to transmit heat content (enthalpy or energy) of a comparatively high temperature hot fluid to a lower temperature cold fluid wherein the temperature of the hot fluid decreases (or remain constant in case of losing latent heat of condensation) and the temperature of the cold fluid increases (or remain constant in case of gaining latent heat of vaporisation). A heat exchanger will normally provide indirect contact heating. E.g. Acooling tower cannot be called a heat exchanger where water is cooled by direct contact with air

Heat Flux The rate of heat transfer per unit surface of a heat exchanger W/ m2

Heat transfer The process of transport of heat energy from a hot source to the comparatively cold surrounding

Heat transfer Refers to the surface area of the heat exchanger that surface or heat provides the indirect contact between the hot and cold Transfer area fluid in effecting the heat transfer. Thus the heat transfer

area is defined as the surface having both sides wetted with one side by the hot fluid and the other side by the cold fluid providing indirect contact for heat transfer m2

Individual The heat flux per unit temperature difference across Heat transfer boundary layer of the hot / cold fluid film formed Coefficient at the heat transfer surface. The magnitude of heat

transfer coefficient indicates the ability of heat conductivity of the given fluid. It increases with increase in density, velocity, specific heat, geometry of the film forming surface W/( m2.K)



LMTD Calculated considering the Capacity and effectiveness Correction of a heat exchanging process. When multiplied with factor LMTD gives the corrected LMTD thus accounting

for the temperature driving force for the cross flow pattern as applicable inside the exchanger

Logarithmic The logarithmic average of the terminal temperature Mean approaches across a heat exchangerTemperature difference, LMTD °C

Overall Heat The ratio of heat flux per unit difference in approach transfer across a heat exchange equipment considering the Coefficient individual coefficient and heat exchanger metal surface

conductivity. The magnitude indicates the ability of heat transfer for a given surface. Higher the coefficient lesser will be the heat transfer surface requirement W/(m2.K)

Pressure drop The difference in pressure between the inlet and outlet of a heat exchanger Bar

Specific The heat content per unit weight of any material per heat capacity degree raise/fall in temperature J/(kg.K)

Temperature The difference in the temperature between the hot and Approach cold fluids at the inlet / outlet of the heat exchanger.

The greater the difference greater will be heat transfer flux °C

Temperature The difference in the temperature between the inlet Range and outlet of a hot/cold fluid in a heat exchanger °C

Terminal The temperatures at the inlet / outlet of the hot / cold temperature fluid steams across a heat exchanger °C

Thermal The rate of heat transfer by conduction though any Conductivity substance across a distance per unit

temperature difference W/(m2.K)

Viscosity The force on unit volume of any material that will cause per velocity Pa





QUESTIONS

1. What is meant by LMTD ?

2. Distinguish between heat exchanger efficiency and effectiveness.

3. Explain the terms heat duty and capacity ratio.

4. What is meant by fouling?

5. List five heat exchangers used in industrial practice.

6. What are the parameters, which are to be monitored for the performance assessmentof heat exchangers?

7. In a heat exchanger the hot stream enters at 70°C and leaves at 55°C. On the otherside the cold stream enters at 30°C and leaves at 55°C. Find out the LMTD of theheat exchanger.

8. In a condenser what type of heats are considered in estimating the heat duty?a) Latent Heat b) Sensible heat c) Specific heat d) Latent heat and sensible heat

9. What is the need for performance assessment of a heat exchanger?

10. The unit of overall coefficient of heat transfer is a) kCal/hr/m2 °C b) kCal/kg °C c) kCal/m2 hr d) kCal/hg m2

REFERENCES1. "Process Heat Transfer" by D.Q.Kern, Edn. 1965.2. "Modern Power Station Practice" - British Electricity International- Volume - G;

Chapter - 7 - " Plant performance and performance monitoring.3. Coulsons & Richardson's CHEMICAL ENGINEERING Volume 3 third edition4. Scimod " Scientific Modeling Software", techno software International, India5. Ganapathy. V, "Fouling factor estimated quickly", O&G Journal, Aug 1992.6. Liberman, Norman P, Trouble shooting Process Operations, Penwell Books, Tulsa,

Oklahoma

5. ENERGY PERFORMANCE ASSESSMENT OF MOTORSAND VARIABLE SPEED DRIVES


5.1 Introduction

The two parameters of importance in a motor are effi-ciency and power factor. The efficiencies of inductionmotors remain almost constant between 50% to 100%loading (Refer figure 5.1). With motors designed toperform this function efficiently; the opportunity forsavings with motors rests primarily in their selectionand use. When a motor has a higher rating than thatrequired by the equipment, motor operates at part load.In this state, the efficiency of the motor is reduced.Replacement of under loaded motors with smallermotors will allow a fully loaded smaller motor to oper-ate at a higher efficiency. This arrangement is general-ly most economical for larger motors, and only whenthey are operating at less than one-third to one-halfcapacity, depending on their size.


Efficiency :

The efficiency of the motor is given by

Where Pout – Output power of the motorPin – Input power of the motorPLoss – Losses occurring in motor

Motor Loading :

Motor Loading % =

Pout Ploss

η = �� = 1 � ��Pin Pin

Actual operating load of the motorRated capacity of the motor

Figure 5.1 Efficiency vs. Loading

x 100

5. Energy Performance Assessment of Motors and Variable Speed Drives


5.3 Efficiency Testing

While input power measurements are fairly simple, measurement of output or losses need a labo-rious exercise with extensive testing facilities. The following are the testing standards widely used.

Europe: IEC 60034-2, and the new IEC 61972US: IEEE 112 - Method BJapan: JEC 37

Even between these standards the difference in efficiency value is up to 3%.For simplicity nameplate efficiency rating may be used for calculations if the motor load is inthe range of 50 -100 %.

Field Tests for Determining Efficiency

(Note: The following section is a repeat of material provided in the chapter-2 on ElectricalMotors in Book-3.)

No Load Test :

The motor is run at rated voltage and frequency without any shaft load. Input power, current,frequency and voltage are noted. The no load P.F. is quite low and hence low PF watt metersare required. From the input power, stator I2R losses under no load are subtracted to give thesum of Friction and Windage (F&W) and core losses. To separate core and F & W losses, testis repeated at variable voltages. It is worthwhile plotting no-load input kW versus Voltage; theintercept is F & W kW loss component.

F&W and core losses = No load power (watts) – (No load current)2 x Stator resistance

Stator and Rotor I2R Losses :

The stator winding resistance is directly measured by a bridge or volt amp method. The resis-tance must be corrected to the operating temperature. For modern motors, the operating tem-perature is likely to be in the range of 100°C to 120°C and necessary correction should be made.Correction to 75°C may be inaccurate. The correction factor is given as follows :

R2 235 + t2—– = ———– , where, t1 = ambient temperature, °C & t2 = operating temperature, °C.R1 235 + t1

The rotor resistance can be determined from locked rotor test at reduced frequency, but rotorI2R losses are measured from measurement of rotor slip.

Rotor I2R losses = Slip x (Stator Input - Stator I2R Losses - Core Loss)

Accurate measurement of slip is possible by stroboscope or non-contact type tachometer. Slipalso must be corrected to operating temperature.

Stray Load Losses :

These losses are difficult to measure with any accuracy. IEEE Standard 112 gives a complicatedmethod, which is rarely used on shop floor. IS and IEC standards take a fixed value as 0.5 % of



output. It must be remarked that actual value of stray losses is likely to be more. IEEE - 112 spec-ifies values from 0.9 % to 1.8 %.

Motor Rating Stray Losses

1 – 125 HP 1.8 %125 – 500 HP 1.5 %501 – 2499 HP 1.2 %2500 and above 0.9 %

Points for Users :

It must be clear that accurate determination of efficiency is very difficult. The same motor test-ed by different methods and by same methods by different manufacturers can give a differenceof 2 %.

Estimation of efficiency in the field can be summarized as follows:a) Measure stator resistance and correct to operating temperature. From rated current

value, I2R losses are calculated.b) From rated speed and output, rotor I2R losses are calculatedc) From no load test, core and F & W losses are determined for stray loss

The method is illustrated by the following example :

Example :

Motor Specifications

Rated power = 34 kW/45 HPVoltage = 415 VoltCurrent = 57 AmpsSpeed = 1475 rpmInsulation class = FFrame = LD 200 LConnection = Delta

No load test Data

Voltage, V = 415 VoltsCurrent, I = 16.1 AmpsFrequency, F = 50 HzStator phase resistance at 30°C = 0.264 OhmsNo load power, Pnl = 1063.74 Watts

a) Calculate iron plus friction and windage losses

b) Calculate stator resistance at 120°C

235 + t2

R2 = R1 x ————235 + t1



c) Calculate stator copper losses at operating temperature of resistance at 120°C

d) Calculate full load slip(s) and rotor input assuming rotor losses are slip times rotor input.

e) Determine the motor input assuming that stray losses are 0.5 % of the motor rated power

f) Calculate motor full load efficiency and full load power factor

Solution

a) Let Iron plus friction and windage loss, Pi + fwNo load power, Pnl = 1063.74 WattsStator Copper loss, P st-30°C (Pst.cu)= 3 x (16.1 / √3)2 x 0.264= 68.43 WattsPi + fw = Pnl - Pst.cu= 1063.74 – 68.43= 995.3 W

b) Stator Resistance at 120°C,

120 + 235R120°C = 0.264 x —————

30 + 235

= 0.354 ohms per phase

c) Stator copper losses at full load, Pst.cu 120°C= 3 x (57 / √3)2 x 0.354= 1150.1 Watts

d) Full load slipS = (1500 – 1475) / 1500

= 0.0167

Rotor input, Pr = Poutput/ (1-S)= 34000 / (1-0.0167)= 34577.4 Watts

e) Motor full load input power, P input = Pr + Pst.cu 120°C + (Pi + fw) + Pstray

= 34577.4 + 1150.1 + 995.3 + (0.005* x 34000)= 36892.8 Watts*where, stray losses = 0.5% of rated output (assumed)

f) Motor efficiency at full loadPoutput

Efficiency = ——– x 100Pinput

34000= ——–

36892.8

= 92.2%



= PinputFull Load PF = —————–

= √3 x V x Ifl

= 36892.8= ——————–= √3 x 415 x 57

= 0.90

Comments :

a) The measurement of stray load losses is very difficult and not practical even on test beds.b) The actual value of stray loss of motors up to 200 HP is likely to be 1 % to 3 % compared

to 0.5 % assumed by standards.c) The value of full load slip taken from the nameplate data is not accurate. Actual measure-

ment under full load conditions will give better results.d) The friction and windage losses really are part of the shaft output; however, in the above

calculation, it is not added to the rated shaft output, before calculating the rotor inputpower. The error however is minor.

e) When a motor is rewound, there is a fair chance that the resistance per phase wouldincrease due to winding material quality and the losses would be higher. It would be inter-esting to assess the effect of a nominal 10 % increase in resistance per phase.

5.4 Determining Motor Loading

1. By Input Power Measurements

• First measure input power Pi with a hand held or in-line power meterPi = Three-phase power in kW

• Note the rated kW and efficiency from the motor name plate

• The figures of kW mentioned in the name plate is for output conditions.So corresponding input power at full-rated load

Nameplate full rated kWPir = ————————————————

ηfl

ηfl = Efficiency at full-rated loadPir = Input power at full-rated load in kW

• The percentage loading can now be calculated as follows

PiLoad = — x 100%

Pir



Example

The nameplate details of a motor are given as power = 15 kW, efficiency η = 0.9. Using a powermeter the actual three phase power drawn is found to be 8 kW. Find out the loading of the motor.

Input power at full-rated power in kW, Pir = 15 /0.9= 16.7 kW

Percentage loading = 8/16.7= 48 %

2. By Line Current Measurements

The line current load estimation method is used when input power cannot be measured and onlyamperage measurements are possible. The amperage draw of a motor varies approximately lin-early with respect to load, down to about 75% of full load. Below the 75% load point, powerfactor degrades and the amperage curve becomes increasingly non-linear. In the low loadregion, current measurements are not a useful indicator of load. However, this method maybe used only as a preliminary method just for the purpose of identification of oversized motors.

Input load current% Load = ———————— *100 (Valid up to 75% loading)

Input rated current

3. Slip Method

In the absence of a power meter, the slip method can be used which requires a tachometer. Thismethod also does not give the exact loading on the motors.

SlipLoad = —— *100%

Ss–Sr

Where:Load = Output power as a % of rated powerSlip = Synchronous speed - Measured speed in rpmSs = Synchronous speed in rpm at the operating frequencySr = Nameplate full-load speed

Example: Slip Load Calculation

Given: Synchronous speed in rpm = 1500 at 50 HZ operating frequency.(Synchronous speed = 120f/P) f: frequency, P: Number of polesNameplate full load speed = 1450Measured speed in rpm = 1480Nameplate rated power = 7.5 kW

Determine actual output power.

1500 – 1480Load = ————— *100% = 40%

1500 – 1450

From the above equation, actual output power would be 40% x 7.5 kW = 3 kW



The speed/slip method of determining motor part-load is often favored due to its simplicityand safety advantages. Most motors are constructed such that the shaft is accessible to atachometer or a strobe light.

The accuracy of the slip method, however, is limited. The largest uncertainty relates tothe accuracy with which manufacturers report the nameplate full-load speed. Manufacturersgenerally round their reported full-load speed values to some multiple of 5 rpm. While 5 rpm isbut a small percent of the full-load speed and may be considered as insignificant, the slipmethod relies on the difference between full-load nameplate and synchronous speeds. Given a40 rpm "correct" slip, a seemingly minor 5 rpm disparity causes a 12% change in calculatedload.

Slip also varies inversely with respect to the motor terminal voltage squared. A voltage cor-rection factor can, also, be inserted into the slip load equation. The voltage compensated loadcan be calculated as shown

SlipLoad = ———————– x 100%

(Ss – Sr) x (Vr/V)2

Where:

Load = Output power as a % of rated power

Slip = Synchronous speed - Measured speed in rpm

Ss = Synchronous speed in rpm

Sr = Nameplate full-load speed

V = RMS voltage, mean line to line of 3 phases

Vr = Nameplate rated voltage

5.5 Performance Evaluation of Rewound Motors

Ideally, a comparison should be made of the efficiency before and after a rewinding. Arelatively simple procedure for evaluating rewind quality is to keep a log of no-load inputcurrent for each motor in the population. This figure increases with poor quality rewinds. Areview of the rewind shop's procedure should also provide some indication of the quality ofwork. When rewinding a motor, if smaller diameter wire is used, the resistance and the I2Rlosses will increase.



5.6 Format for Data Collection

The motor loading survey can be performed using the format given below:

Motor Field Measurement Format

Company_________________________ Location_______________________Date ________ Process________________________

Department_____________________

General Data

Driven Equipment__________________ Motor Operating Profile:

Motor Name Plate Data No of hours of operationManufacturer ______________________ I Shift _____________ Model ___________________________ II Shift _____________ Serial Number _____________________ III Shift _____________Type :Squirrel cage/Slp ring__________Size (hp/kW)______________________ Annual Operating Time ______ hours/yearSynchronous Speed (RPM) ___________Full-Load Speed (RPM) _____________ Type of load Voltage Rating _____________________ 1.Load is quite steady, motor "On" during shiftFull-Load Amperage ________________ 2.Load starts, stops, but is constant when "On"Full-Load Power Factor (%) __________ 3.Load starts, stops, and fluctuates when "On"Full-Load Efficiency (%) ____________Temperature Rise __________________Insulation Class ____________________

From Test Certificate

Load 100% 75% 25% No Load

Current

PF

Efficiency

Stator resistance per phase =

Measured DataSupply VoltageBy VoltmeterVRY ________VYB ________ V avg ______VBR ________Input AmpsBy AmmeterA a __________A b __________ A avg ______A c __________Power Factor (PF) _____________________Input Power (kW) ______________________

Motor Operating Speed ____________RPMAt frequency of __________Driven Equipment Operating Speed__________RPMType of Transmission (Direct/Gear/Fluid coupling)

Rewound � Yes ,if yes How manytimes rewound ?---

� NoMotor Loading %_________________



The monitoring format for rewound motor is given below:

5.7 Application of Variable Speed Drives (VSD)

Although there are many methods of varying the speeds of the driven equipment such ashydraulic coupling, gear box, variable pulley etc., the most possible method is one ofvarying the motor speed itself by varying the frequency and voltage by a variablefrequency drive.

5.7.1 Concept of Variable Frequency Drive

The speed of an induction motor is proportional to the frequency of the AC voltage applied toit, as well as the number of poles in the motor stator. This is expressed by the equation:

RPM = (f x 120) / p

Where f is the frequency in Hz, and p is the number of poles in any multiple of 2.

Therefore, if the frequency applied to the motor is changed, the motor speed changes indirect proportion to the frequency change. The control of frequency applied to the motor is thejob given to the VSD.

The VSD's basic principle of operation is to convert the electrical system frequency and volt-age to the frequency and voltage required to drive a motor at a speed other than its rated speed.The two most basic functions of a VSD are to provide power conversion from one frequency toanother, and to enable control of the output frequency.

VSD Power Conversion

As illustrated by Figure 5.1,there are two basic components,a rectifier and an inverter, toaccomplish power conversion.

The rectifier receives the50-Hz AC voltage and con-verts it to direct current (DC)voltage. A DC bus inside theVSD functions as a "parkinglot" for the DC voltage. The Figure 5.1 Components of a Variable Speed Drive



DC bus energizes the inverter, which converts it back to AC voltage again. The inverter canbe controlled to produce an output frequency of the proper value for the desired motor shaftspeed.

5.7.2 Factors for Successful Implementation of Variable Speed Drives

a) Load Type for Variable Frequency Drives

The main consideration is whether the variable frequency drive application require a variabletorque or constant torque drive. If the equipment being driven is centrifugal, such as a fan orpump, then a variable torque drive will be more appropriate. Energy savings are usually the pri-mary motivation for installing variable torque drives for centrifugal applications. For example,a fan needs less torque when running at 50% speed than it does when running at full speed.Variable torque operation allows the motor to apply only the torque needed, which results inreduced energy consumption.

Conveyors, positive displacement pumps, punch presses, extruders, and other similar typeapplications require constant level of torque at all speeds. In which case, constant torque vari-able frequency drives would be more appropriate for the job. A constant torque drive shouldhave an overload current capacity of 150% or more for one minute. Variable torque variablefrequency drives need only an overload current capacity of 120% for one minute since cen-trifugal applications rarely exceed the rated current.

If tight process control is needed, then you may need to utilize a sensor less vector, or fluxvector variable frequency drive, which allow a high level of accuracy in controlling speed,torque, and positioning.

b) Motor Information

The following motor information will be needed to select the proper variable frequency drive:

Full Load Amperage Rating. Using a motor's horsepower is an inaccurate way to size vari-able frequency drives.

Speed Range. Generally, a motor should not be run at any speed less than 20% of its specifiedmaximum speed allowed. If it is run at a speed less than this without auxiliary motor cooling,the motor will overheat. Auxiliary motor cooling should be used if the motor must be operatedat very slow speeds.

Multiple Motors. To size a variable frequency drive that will control more than one motor, addtogether the full-load amp ratings of each of the motors. All motors controlled by a single drivemust have an equal voltage rating.

c) Efficiency and Power Factor

The variable frequency drive should have an efficiency rating of 95% or better at full load.Variable frequency drives should also offer a true system power factor of 0.95 or better

across the operational speed range, to save on demand charges, and to protect the equipment(especially motors).

d) Protection and Power Quality

Motor overload Protection for instantaneous trip and motor over current.



Additional Protection: Over and under voltage, over temperature, ground fault, control ormicroprocessor fault. These protective circuits should provide an orderly shutdown of the VFD,provide indication of the fault condition, and require a manual reset (except under voltage)before restart. Under voltage from a power loss shall be set to automatically restart after returnto normal. The history of the previous three faults shall remain in memory for future review.

If a built-up system is required, there should also be externally-operated short circuit protec-tion, door-interlocked fused disconnect and circuit breaker or motor circuit protector (MCP)

To determine if the equipment under consideration is the right choice for a variable speeddrive:

The load patterns should be thoroughly studied before exercising the option of VSD. In effectthe load should be of a varying nature to demand a VSD ( refer figure 5.3 & 5.4).

Figure 5.3 Example of an excellent variablespeed drive candidate

Figure 5.4 Example of a poor variable speeddrive candidate

The first step is to identify the number of operating hours of the equipment at various loadconditions. This can be done by using a Power analyzer with continuous data storage or by asimple energy meter with periodic reading being taken.

5.7.3 Information needed to Evaluate Energy Savings for Variable Speed Application

1. Method of flow control to which adjustable speed is compared: o output throttling (pump) or dampers (fan) o recirculation (pump) or unrestrained flow (fan) o adjustable-speed coupling (eddy current coupling) o inlet guide vanes or inlet dampers (fan only) o two-speed motor.

2. Pump or fan data: o head v's flow curve for every different type of liquid (pump) or gas (fan) that is

handled o Pump efficiency curves.



3. Process information: o specific gravity (for pumps) or specific density of products (for fans) o system resistance head/flow curve o equipment duty cycle, i.e. flow levels and time duration.

4. Efficiency information on all relevant electrical system apparatus: o motors, constant and variable speed o variable speed drives o gears o transformers.

If we do not have precise information for all of the above, we can make reasonable assump-tions for points 2 and 4.



QUESTIONS

1) Define motor efficiency.

2) Why it is difficult to measure motor efficiency at site?

3) Describe the various methods by which you calculate motor loading.

4) If no instrument other than tachometer is available, what method you would suggestfor measuring the motor load?

5) A 20 kW rated motor is drawing actual measured power of 14 kW. If the rated effi-ciency is 92%, determine the motor loading?

6) What are the limitations of slip method in determining motor loading?

7) A 4 pole motor is operating at a frequency of 50 Hz. Find the RPM of the motor?

8) What are the two factors influencing the speed of induction motor?

9) A fan's operating hours and loading are given below:

15 hours at 100% load

8 hours at 95% load

1 hour at 40% load

Is the application suitable candidate for application of VSD?

11) The losses in a variable speed drive is a) 12% b) 8% c) <5% d) no losses at all

REFERENCES1. Motor challenge: Office of Industrial Technologies, Department of Energy, USA2. Energy audit Reports of National Productivity Council

6. ENERGY PERFORMANCE ASSESSMENT OFFANS AND BLOWERS


6.1 Introduction

This section describes the method of testing a fan installed on site in order to determine theperformance of the fan in conjunction with the system to which it is connected.


The purposes of such a test are to determine, under actual operating conditions, the volume flowrate, the power input and the total pressure rise across the fan.

These test results will provide actual value for the flow resistance of the air duct system,which can be compared with the value specified by supplier.


Static Pressure: The absolute pressure at a point minus the reference atmospheric pressure.

Dynamic Pressure: The rise in static pressure which occurs when air moving with specifiedvelocity at a point is bought to rest without loss of mechanical energy. It is also known as veloc-ity pressure.

Total Pressure: The sum of static pressures and dynamic pressures at a point.

Fan Shaft Power: The mechanical power supplied to the fan shaft

Motor Input Power: The electrical power supplied to the terminals of an electric motordrive.

6.4 Scope

The procedure describes field testing of centrifugal fans and blowers for assessing performanceand efficiency.


British Standard, BS 848 - Fans for general purposes Part 1, Methods of testing performance

6.6 Field Testing

6.6.1 Instruction for Site Testing

Before site tests are carried out, it should be ensured that:

• Fan and its associated equipment are functioning properly, and at the rated speed • Operations are at stable conditions, e.g. steady temperatures, densities, system resistance

etc.

6.6.2 Location of Measurement Planes

General: The flow measurement plane shall be located in any suitable straight length,(preferably on the inlet side of the fan) where the airflow conditions are substantially axial,symmetrical and free from turbulence. Leakage of air from or into the air duct shall be negligi-ble between the flow measuring plane and the fan. Bends and obstructions in an air duct candisturb the airflow for a considerable distance downstream, and should be avoided for the pur-poses of the test.

Test length: That part of the duct in which the flow measurement plane is located, is termedthe 'test length' and shall be straight, of uniform cross section and free from any obstructionswhich may modify the airflow. It shall have a length equal to not less than twice the equiva-lent diameter of the air duct (i.e. 2De). For rectangular duct, equivalent diameter, De is given by2 LW/(L + W) where L, W is the length and width of the duct. For circular ducts De is the sameas diameter of the duct.

Inlet side of the fan: Where the 'test length' is on the inlet side of the fan, its downstream endshall be at a distance from the fan inlet equal to atleast 0.75De. See figure 6.1. In the case of afan having an inlet box , the downstream end of the test length shall be at a distance from thenearest part of the inlet cone of the fan equal to at least 0.75De.

Outlet side of the fan: Where the 'test length' is on the outlet side of the fan, the upstreamend of the 'test length' shall be at a distance from the fan outlet of at least 3De. See figure 6.2.For this purpose, the fan outlet shall be considered as being the outlet of any expander on theoutlet side of the fan.

Location of the Flow Measurement Plane within the 'Test Length': The flow measure-ment plane shall be located within the 'test length' at a distance from the downstream end of the'test length' equal to at least 1.25De.

Location of Pressure Measurement Plane: For the purpose of determining the pressure riseproduced by the fan, the static pressure shall be measured at planes on the inlet and/or the out-let side of the fan sufficiently close to it to ensure that the pressure losses between the measur-ing planes and the fan are calculable in accordance with available friction factor data withoutadding excessively to the uncertainty of fan pressure determination.

If conveniently close to the fan, the 'test length' selected for air flow measurement shouldalso be used to pressure measurement. Other planes used for pressure measurement should be

6. Energy Performance Assessment of Fans and Blowers


no closer than 0.25De from the fan inlet and no closer than 4De from the fan outlet. The planeof pressure measurement should be selected at least 4De downstream of any bend, expander or



2

obstruction which are likely to cause separated flow or otherwise interfere with uniformity ofpressure distribution.



6.6.3 Measurement of Air Velocity on Site

Velocity shall be measured by either pitot tube or a rotating vane anemometer. When in use, thepitot tube shall be connected by means of airtight tubes to a pressure measuring instrument. Theanemometer shall be calibrated before the test.

Pitot Tube: In Figure 6.4, note that separate static connections (A) and total pressure con-nections (B) can be connected simultaneously across a manometer (C). Since the static pressureis applied to both sides of the manometer, its effect is canceled out and the manometer indicatesonly the velocity pressure.

In practice this type of measurement is usually made with a Pitot tube which incorporates bothstatic and total pressure sensors in a single unit. Essentially, a Pitot tube consists of an impact tube(which receives total pressure input) fastened concentrically inside a second tube of slightly largerdiameter which receives static pressure input from radial sensing holes around the tip. The air spacebetween inner and outer tubes permits transfer of pressure from the sensing holes to the static pres-sure connection at the opposite end of the Pitot and then, through connecting tubing, to the low ornegative pressure side of a manometer. When the total pressure tube is connected to the high pres-sure side of the manometer, velocity pressure is indicated directly. See Figure 6.5.

To ensure accurate velocity pressure readings, the Pitot tube tip must be pointed directly into(parallel with) the air stream. As the Pitot tube tip is parallel with the static pressure outlet tube,the latter can be used as a pointer to align the tip properly. When the Pitot tube is correctlyaligned, the pressure indication will be maximum.



Traverse readings: In practical situations, the velocity of the air stream is not uniform acrossthe cross section of a duct. Friction slows the air moving close to the walls, so the velocity isgreater in the center of the duct.

To obtain the average total velocity in ducts of 100 mm diameter or larger, a series ofvelocity pressure readings must be taken at points of equal area. A formal pattern of sensingpoints across the duct cross section is recommended. These are known as traverse readings.Figure 6.6 shows recommended Pitot tube locations for traversing round and rectangularducts.

Figure 6.4 Types of Pressure Measurement

Figure 6.5 Pitot tube senses total and static pressure. Manometer measuresvelocity pressure (Difference between total and static pressures)



In round ducts, velocity pressure readings should be taken at centers of equal concentric areas.At least 20 readings should be taken along two diameters. In rectangular ducts, a minimum of 16and a maximum of 64 readings are taken at centers of equal rectangular areas. Actual velocities foreach area are calculated from individual velocity pressure readings. This allows the readings andvelocities to be inspected for errors or inconsistencies. The velocities are then averaged.

By taking Pitot tube readings with extreme care, air velocity can be determined within anaccuracy of ± 2%. For maximum accuracy, the following precautions should be observed:

Example-Traverse point determination for round duct

Round duct: Let us calculate various traverse points for a duct of 1 m diameter. From Figure6.4, for round duct of 1 m diameter (D). The radius, R is 0.5 m. The various points from theport holes are given below:

Figure 6.6 Traverse on Round and Square Duct Areas

0.5 – 0.949 x 0.5 0.0255

0.5 – 0.837 x 0.5 0.0815

0.5 – 0.707 x 0.5 0.1465

0.5 – 0.548 x 0.5 0.226

0.5 – 0.316 x 0.5 0.342

0.5 + 0.316 x 0.5 0.658

0.5 + 0.548 x 0.5 0.774

0.5 + 0.707 x 0.5 0.8535

0.5 + 0.837 x 0.5 0.9185

0.5 + 0.949 x 0.5 0.9745

Example-Traverse point determination for rectangular duct

Rectangular duct: For 1.4 m x 0.8 m rectangular duct, let us calculate the traverse points. 16points are to be measured.

Dividing the area 1.4 x 0.8 = 1.12 m2 into 16 equal areas, each area is 0.07 m2. Takingdimensions of 0.35 m x 0.20 m per area, we can now mark the various points in the rectangu-lar duct as follows:



In small ducts or where traverse operations are otherwise impossible, an accuracy of ± 5%can frequently be achieved by placing Pitot in center of duct.

Calculation of Velocity: After taking velocity pressures readings, at various traverse points,the velocity corresponding to each point is calculated using the following expression.

Anemometer: The indicated velocity shall be measured at each traverse point in the crosssection by holding the anemometer stationary at each point for a period of time of not less than1 minute. Each reading shall be converted to velocity in m/s and individually corrected in accor-dance with the anemometer calibration. The arithmetic mean of the corrected point velocitiesgives the average velocity in the air duct and the volume flow rate is obtained by multiplyingthe area of the air duct by the average velocity.

6.6.4 Determination of Flow

Once the cross-sectional area of the duct is measured, the flow can be calculated as follows:

Flow, (m3/s) = Area (m2) x Velocity (m/s)



6.6.5 Determination of Fan Pressure

General: Precautions shall be taken so that the measurements of the static pressure on theinlet and outlet sides of the fan are taken relative to the atmosphere pressure.

Measurement of Static Pressure: This shall be done by using a manometer in conjunctionwith the static pressure connection of a pitot tube or a U tube manometer.

When using a pitot tube it is necessary to carry out a traverse in the pressure measurementplane taking individual point pressure readings in a manner similar to that for determining flowrate. In general, a smaller number of readings will be found adequate where individual readingsdo not vary by more than 2% from each other. The average of all the individual readings shallbe taken as the static pressure of that section.

6.6.6 Determination of Power Input

Power Measurement: The power measurements can be done using a suitable clamp- onpower meter. Alternatively by measuring the amps, voltage and assuming a power factor of 0.9the power can be calculated as below:

Transmission Systems: The interposition of a transmission system may be unavoidableintroducing additional uncertainties. The following values shall be used as a basis for transmis-sion efficiency in the case of drives rated at 20 kW and above unless other reliable informationis available:

Properly lubricated precision spur gears 98% for each stepFlat belt drive 97%V-belt drive 95%

Other Prime Movers: When the fan forms one unit with a non-electric prime mover it is rec-ommended that the fuel consumption (oil, steam, compressed air etc.) should be specified anddetermined in place of the overall power.

input to fan shaft in kW

6.7 Example: Performance Test Report on Cooling Air Fan

The following is a typical report on measurements taken and calculations made for adouble inlet fan in a palletizing plant.

A. Design Parameters:

Volume = 292 m3/sec.Static Pressure = 609.6 mmwc

B. Measurements:

Temperature = 32°CSpeed = 740 RPM



Inlet Suction Outlet Measured Volume Amps PowerDamper Pressure Pressure Velocity m3/Sec. (I) ConsumptionPosition (-) mmwc (+) mmwc Pressure (∆∆ p), (kW)% mmwc

80% ONE SIDE 455, 462, Average = 7025, 22, 20 480,478Average=22.33 Avg.=468.75

ANOTHER 166.6 220 2127 KWSIDE

15, 18, 23, 21 459, 464, 473Average=19.25 479, 480, 470

Avg.=470.83 Average = 70

Instruments used

a) Suction pressure, outlet pressure = 'U' tube manometerb) For differential pressure = Inclined tube manometerc) For temperature = Mercury in glass thermometerd) Fan speed = Tachometere) Line current = Tong tester

C. Performance calculations:

a) Gas Density = 273 x 1.293

(Corrected to NTP) 273 + T°C (at site condition)

= 273 x 1.293

273 + 32°C (at site condition)

= 1.15 kg/m3

d) P = Power input to the fan shaft

= Power input to the motor (kW) x Efficiency of motor (%) at the operating load x transmission efficiency

Motor efficiency = 0.94

P = 2263 x 0.94 x 1 (as motor was direct coupled)

= 2127 kW

Volume in m3 / Sec x total pressure in mmwce) Fan Efficiency % =

102 x Power input to the shaft in (kW)

Where 102 is a conversion constant

For double inlet fan,The total Volume ofair, m3 / Sec = 166.6 x 2 = 333.2



Total static pressure, = 468.75 – (–22.33) = 491mmwc (∆ pStatic,across the fan)

Fan Efficiency = 333.2 x 491 x 100102 x 2127

Static Fan Efficiency = 75%

6.8 Factors that Could Affect Performance

• Leakage, re-circulation or other defects in the system;• Inaccurate estimation of flow resistance; • Erroneous application of the standardized test data; • Excessive loss in a system component located too close to the fan outlet;• Disturbance of the fan performance due to a bend or other system component located too

close to the fan inlet; • Error in site measurement





QUESTIONS

1) What is the relationship between static pressure, dynamic pressure and total pres-sure?

2) For determining fan efficiency, why static pressure readings should be taken as closeto fan as possible?

3) What is the significance of having traverse points in velocity measurement?

4) What is fan efficiency?

5) Determine various traverse points for a round duct of 0.5 m diameter.

6) Why flow should not be measured very close to inlet and outlet of fan?

7) Calculate the flow rate for the following data:Diameter of duct: 0.5 m, differential pressure: 100mmWC, Density of air at 0°C: 1.293, Temperature of air in the duct: 100°C, pitot coefficient:0.85

8) How many traverse points you would propose for a rectangular duct of 1 m x 1 mdimensions?

9) What are the various ways of measuring the flow?

10) What are the various factors, which can affect fan performance?

REFERENCES 1. British Standard: BS 848 : Part 1 : 19802. Energy and Environmental Audit Reports of National Productivity Council

7. ENERGY PERFORMANCE ASSESSMENTOF WATER PUMPS


7.1 Introduction

Pumping is the process of addition of kinetic and potential energy to a liquid for the purpose of moving it from one point to another. This energy will cause the liquid to do worksuch as flow through a pipe or rise to a higher level. A centrifugal pump transforms mechan-ical energy from a rotating impeller into a kinetic and potential energy required by the system.

The most critical aspect of energy efficiency in a pumping system is matching of pumps toloads. Hence even if an efficient pump is selected, but if it is a mismatch to the system then thepump will operate at very poor efficiencies. In addition efficiency drop can also be expectedover time due to deposits in the impellers. Performance assessment of pumps would reveal theexisting operating efficiencies in order to take corrective action.


• Determination of the pump efficiency during the operating condition• Determination of system resistance and the operating duty point of the pump and compare

the same with design.


Pump Capacity, Q = Volume of liquid delivered by pump per unit time,m3/hr or m3/secQ is proportional to N, where N- rotational speed of the pump

Total developed head, H = The difference of discharge and suction pressure

The pump head represents the net work done on unit weights of a liquid in passing frominlet of the pump to the discharge of the pump.

There are three heads in common use in pumps namely

(i) Static head(ii) Velocity head (iii) Friction head.

The frictional head in a system of pipes, valves and fittings varies as a function (roughly asthe square) of the capacity flow through the system.

System resistance: The sum of frictional head in resistance & total static head.

Pump Efficiency: Fluid power and useful work done by the pump divided by the power inputin the pump shaft.

7. Energy Performance Assessment of Water Pumps


7.4 Field Testing for Determination of Pump Efficiency

To determine the pump efficiency, three key parameters are required: Flow, Head and Power.Of these, flow measurement is the most crucial parameter as normally online flow meters arehardly available, in a majority of pumping system. The following methods outlined below canbe adopted to measure the flow depending on the availability and site conditions.

7.4.1 Flow Measurement, Q

The following are the methods for flow measurements:

• Tracer method BS5857• Ultrasonic flow measurement• Tank filling method• Installation of an on-line flowmeter

Tracer Method

The Tracer method is particularly suitable for cooling water flow measurement because of theirsensitivity and accuracy.

This method is based on injecting a tracer into the cooling water for a few minutes at anaccurately measured constant rate. A series of samples is extracted from the system at a pointwhere the tracer has become completely mixed with the cooling water. The mass flow rate iscalculated from:

qcw = q1 x C1/C2

where qcw = cooling water mass flow rate, kg/sq1 = mass flow rate of injected tracer, kg/sC1 = concentration of injected tracer, kg/kg

C2 = concentration of tracer at downstream position during the 'plateau' period of constant concentration, kg/kg

The tracer normally used is sodium chloride.

Ultrasonic Flow meter

Operating under Doppler effect principle these meters are non-invasive, meaning measurementscan be taken without disturbing the system. Scales and rust in the pipes are likely to impact theaccuracy.

• Ensure measurements are taken in a sufficiently long length of pipe free from flow distur-bance due to bends, tees and other fittings.

• The pipe section where measurement is to be taken should be hammered gently to enablescales and rusts to fall out.

• For better accuracy, a section of the pipe can be replaced with new pipe for flow mea-surements.

Tank filing method

In open flow systems such as water getting pumped to an overhead tank or a sump, the flowcan be measured by noting the difference in tank levels for a specified period during whichthe outlet flow from the tank is stopped. The internal tank dimensions should be preferabletaken from the design drawings, in the absence of which direct measurements may beresorted to.

Installation of an on-line flowmeter

If the application to be measured is going to be critical and periodic then the best option wouldbe to install an on-line flowmeter which can get rid of the major problems encountered withother types.

7.4.3 Determination of total head, H

Suction head (hs)

This is taken from the pump inlet pressure gauge readings and the value to be converted in tometers (1kg/cm2 = 10. m). If not the level difference between sump water level to the center-line of the pump is to be measured. This gives the suction head in meters.

Discharge head (hd)

This is taken from the pump discharge side pressure gauge. Installation of the pressure gaugein the discharge side is a must, if not already available.

7.4.4 Determination of hydraulic power (Liquid horse power),

Hydraulic power Ph(kW) = Q x (hd – hs) x ρ x g / 1000Q = Volume flow rate (m3/s), ρ = density of the fluid (kg/m3), g = acceleration due to gravity(m/s2), (hd - hs) = Total head in metres



7.4.5 Measurement of motor input powerThe motor input power Pm can be measured by using a portable power analyser.

7.4.6 Pump shaft power

The pump shaft power Ps is calculated by multiplying the motor input power by motor effi-ciency at the existing loading.

Ps = Pm x ηMotor

7.4.7 Pump efficiency

This is arrived at by dividing the hydraulic power by pump shaft power

ηPump = Ph

Ps

Example of pump efficiency calculation

Illustration of calculation method outlinedA chemical plant operates a cooling water pump for process cooling and refrigeration

applications. During the performance testing the following operating parameters weremeasured;

Measured Data

Pump flow, Q 0.40 m3/ s Power absorbed, P 325 kWSuction head (Tower basin level), h1 +1 MDelivery head, h2 55 MHeight of cooling tower 5 M Motor efficiency 88 %Type of drive Direct coupledDensity of water 996 kg/ m3

Pump efficiency

Flow delivered by the pump 0.40 m3/s Total head, h2 –(+h1) 54 MHydraulic power 0.40 x 54 x 996 x 9.81/1000 = 211 kWActual power consumption 325 kWOverall system efficiency (211 x 100) / 325 = 65 %Pump efficiency 65/0.88 = 74 %

7.5 Determining the System resistance and Duty point

Determination of the system resistance curve and imposing the pump curve over it will give anidea of the operating efficiency of the pump and also the drop in efficiencies when the system



curve changes from normal / design. The example following from the earlier example outlinesthe method of constructing a system curve.

Example:

Location of equipments

The Refrigeration plant is located at +0.00 level and the Process plant condensers are located at+15 M level. One cooler having a design pressure drop of 1.9 kg/cm2 is located at the 0.00 level(ground level). Other relevant data can be inferred from the earlier section. See schematic inFigure 7.1.



The step-by-step approach for determining system resistance curve is given below.

Step-1 Divide system resistance into Static and dynamic head

Find static head;Static head (Condenser floor height) ; 15M

Find dynamic head;Dynamic Head = Total Head – Static HeadDynamic head = (54–15) = 39 M

Step-2 Check the maximum resistance circuit

Resistance in the different circuits is as under



S.no System Condenser loop Reactor loop Cooler loopresistance, M resistance, M resistance, M

1. Supply line from 15 10 15pump

2. Static head 15 5 Nil (cooler at ground level)

3. Equipment 5 5 19

4. Return line from 15 10 15equipment to CT

5. Tower head - - 5

6. Total 50 30 54

S.No. Flow (%) Dynamic head Static head M Total head M= 39x (%flow)2

1. 100 39 15 54

2. 75 21.9 15 36.9

3. 50 9.75 15 24.75

4. 25 2.44 15 17.44

It can be noted that at full load the condenser and cooler circuits offer the maximum resis-tance to flow.

Step 3; Draw system resistance curve

Choose the condenser loop as it offers maximum resistance and is also having a static headcomponent Static head: 15 MDynamic head at full load; 39 MCompute system resistance at different flow rates

Step 4 - Plot the system resistance against flow in the pump efficiency curves (see Figure 7.2)provided by the vendor and compare actual operating duty point and see whether it operates atmaximum efficiency. In the example provided it is found that the pump system efficiency islower by 4 % due to change in operating conditions.



QUESTIONS

1) How would you measure the flow by using tracer method?

2) What are the various ways of measuring flow?

3) A pump motor draws 75 A current. The voltage is 415 V. Assuming a power factor of0.9. Calculate the power drawn?

4) The suction head is 1m below the pump centerline. The discharge pressure shows 3kg/cm2. The flow is calculated to be 100 m3/hr. Find out the pump efficiency.

5) The pump efficiency is 70%. The hydraulic power is calculated to be 22 kW. Findout the motor power required to drive the pump.

REFERENCES 1. Pump handbook by Karassik2. Energy Audit Reports of National Productivity Council

8. ENERGY PERFORMANCE ASSESSMENTOF COMPRESSORS


8.1 Introduction

The compressed air system is not only an energy intensive utility but also one of the least ener-gy efficient. Over a period of time, both performance of compressors and compressed air sys-tem reduces drastically. The causes are many such as poor maintenance, wear and tear etc. Allthese lead to additional compressors installations leading to more inefficiencies. A periodic per-formance assessment is essential to minimize the cost of compressed air.


To find out:

• Actual Free Air Delivery (FAD) of the compressor• Isothermal power required • Volumetric efficiency • Specific power requirement

The actual performance of the plant is to be compared with design / standard values forassessing the plant energy efficiency.


8.4 Field Testing

8.4.1 Measurement of Free Air Delivery (FAD) by Nozzle method

Principle: If specially shaped nozzle discharge air to the atmosphere from a receiver gettingits supply from a compressor, sonic flow conditions sets in at the nozzle throat for a particular

ratio of upstream pressure (receiver) to the downstream pressure (atmospheric) i.e. Mach num-ber equals one.

When the pressure in the receiver is kept constant for a reasonable intervals of time, the air-flow output of the compressor is equal to that of the nozzle and can be calculated from theknown characteristic of the nozzle.

8.4.2 Arrangement of test equipment

The arrangement of test equipment and measuring device shall confirm to Figure 8.1.

8.4.3 Nozzle Sizes

The following sizes of nozzles are recommended for the range of capacities indicated below:Flow Nozzle: Flow nozzle with profile as desired in IS 10431:1994 and dimensions

8. Energy Performance Assessment of Compressors


Nozzle size (mm) Capacity (m3/hr)

6 3 – 910 9 – 3016 27 – 9022 60 – 17033 130 – 37550 300 – 45080 750 – 2000125 1800 – 5500165 3500 – 10000

8.4.4 Measurements and duration of the test.

The compressor is started with the air from the receiver discharging to the atmosphere throughthe flow nozzle. It should be ensured that the pressure drop through the throttle valve should beequal to or twice the pressure beyond the throttle. After the system is stabilized the followingmeasurements are carried out:

• Receiver pressure• Pressure and temperature before the nozzle• Pressure drop across the nozzle• Speed of the compressor• kW, kWh and amps drawn by the compressor

The above readings are taken for the 40%, 60%, 100% and 110% of discharge pressure values.

Measuring instruments required for test

• Thermometers or Thermocouple• Pressure gauges or Manometers• Differential pressure gauges or Manometers• Standard Nozzle

• Psychrometer• Tachometer/stroboscope• Electrical demand analyser



8.5 Calculation Procedure for Nozzle Method

k : Flow coefficient – as per ISd : Nozzle diameter MT1 : Absolute inlet temperature °KP1 : Absolute inlet pressure kg/cm2

P3 : Absolute Pressure before nozzle kg/cm2

T3 : Absolute temperature before nozzle °KRa : Gas constant for air 287.1 J/kg kP3–P4 : Differential pressure across the nozzle kg/cm2

II. Isothermal Efficiency = Isothermal power/Input power

Isothermal power(kW) = P1 x Qf x loger

36.7

P1 = Absolute intake pressure kg/ cm2

Qf = Free air delivered m3/hr.r = Pressure ratio P2/P1

III. Specific power consumption = Power consumption ,kW

at rated discharge pressure Free Air Delivered, m3/hr

IV. Volumetric efficiency = Free air delivered m3/min x 100

Compressor displacement, m3/min

Compressor Displacement = π

4

D = Cylinder bore, metreL = Cylinder stroke, metreS = Compressor speed rpmχ = 1 for single acting and

2 for double acting cylindersn = No. of cylinders

8.6 Example

Calculation of Isothermal Efficiency for a Reciprocating Air Compressor.

Step – 1 : Calculate Volumetric Flow Rate

k : Flow coefficient (Assumed as 1)d : Nozzle diameter : 0.08 metreP2 : Receiver Pressure – 3.5 kg / cm2 (a)P1 : Inlet Pressure – 1.04 kg / cm2(a)T1 : Inlet air temperature 30°C or 303°KP3 : Pressure before nozzle – 1.08 kg / cm2

T3 : Temperature before the nozzle 40°C or 313°KP3 – P4 : Pressure drop across the nozzle = 0.036 kg / cm2

Ra : Gas constant : 287 Joules / kg K



x D2 x L x S x χ x n

Step – 2 : Calculate Isothermal Power Requirement

Isothermal Power (kW) = P1 x Qf x loger36.7

P1 - Absolute intake pressure = 1.04 kg / cm2 (a)Qf - Free Air Delivered = 1407.6 m3 / h.Compression ratio r = 3.51 = 3.36

1.04Isothermal Power = 1.04 x 1407.6 x loge3.36 = 48.34 kW

36.7

Step – 3 : Calculate Isothermal Efficiency

Motor input power = 100 kWMotor and drive efficiency = 86 %Compressor input power = 86 kW

Isothermal efficiency = Isothermal Power x 100Compressor input Power

= 48.34 x 100 = 56%86.0

8.7 Assessment of Specific Power requirement

Specific power consumption = Actual power consumed by the compressorMeasured Free Air Delivery

In the above example the measured flow is 1407.6 m3/hr and actual power consumption is100 kW.

Specific power requirement = 1001407.6

= 0.071 kW/m3/hr

8.8 Measurement of FAD by Pump Up Method

(Note: The following section is a repeat of material provided in the chapter-3 on CompressedAir System in Book-3.)



Another way of determining the Free Air Delivery of the compressor is by Pump UpMethod - also known as receiver filling method. Although this is less accurate, this can beadopted where the elaborate nozzle method is difficult to be deployed.

Simple method of Capacity Assessment in Shop floor

• Isolate the compressor along with its individual receiver being taken for test from maincompressed air system by tightly closing the isolation valve or blanking it, thus closingthe receiver outlet.

• Open water drain valve and drain out water fully and empty the receiver and the pipeline.Make sure that water trap line is tightly closed once again to start the test.

• Start the compressor and activate the stopwatch.• Note the time taken to attain the normal operational pressure P2 (in the receiver) from ini-

tial pressure P1.• Calculate the capacity as per the formulae given below:

Actual Free air discharge

P2 – P1 VQ = X Nm3/Minute

P0 T

Where P2 = Final pressure after filling (kg/cm2 a)P1 = Initial pressure (kg/cm2a) after bleedingP0 = Atmospheric Pressure (kg/cm2 a)V = Storage volume in m3 which includes receiver,

after cooler, and delivery pipingT = Time take to build up pressure to P2 in minutes

The above equation is relevant where the compressed air temperature is same as the ambi-ent air temperature, i.e., perfect isothermal compression. In case the actual compressed air tem-perature at discharge, say t2°C is higher than ambient air temperature say t1°C (as is usual case),the FAD is to be corrected by a factor (273 + t1) / (273 + t2).

EXAMPLE

An instrument air compressor capacity test gave the following results (assume the final com-pressed air temperature is same as the ambient temperature) - Comment?

Piston displacement : 16.88 m3/minuteTheoretical compressor capacity : 14.75 m3/minute @ 7 kg/cm2

Compressor rated rpm 750 : Motor rated rpm : 1445Receiver Volume : 7.79 m3

Additional hold up volume,i.e., pipe / water cooler, etc., is : 0.4974 m3

Total volume : 8.322 m3



Initial pressure P1 : 0.5 kg/cm2

Final pressure P2 : 7.03 kg/cm2

Atmospheric pressure P0 : 1.026 kg/cm2,aTime taken to buildup pressure from P1 to P2 : 4.021 minutes

(P2 – P1) × Total VolumeCompressor output m3/minute :

Atm. Pressure × Pumpup time(7.03 – 0.5) × 8.322

: = 13.17 m3/minute1.026 × 4.021

Capacity shortfall with respect to 14.75 m3/minute rating is 1.577 m3/minute i.e., 10.69 %, which indicates compressor performance needs to be investigated further.





QUESTIONS

1) What is meant by Free Air Delivery?

2) Describe the method of estimating flow by nozzle method.

3) Describe the method of estimating flow by pump up method.

4) Define the term isothermal efficiency and explain its significance.

5) Define the term volumetric efficiency and explain its significance.

6) How is specific power requirement calculated?

REFERENCES 1. IS 10431:1994: Measurement of airflow of compressors and exhausters by nozzles.2. IS 5456:1985 code of practice for testing of positive displacement type air compressors

and exhausters3. Compressor performance – Aerodynamics for the user by M Theodore Gresh-

Butterworth Heinemann.

9. ENERGY PERFORMANCE ASSESSMENTOF HVAC SYSTEMS


9.1 Introduction

Air conditioning and refrigeration consume significant amount of energy in buildings and inprocess industries. The energy consumed in air conditioning and refrigeration systems is sensi-tive to load changes, seasonal variations, operation and maintenance, ambient conditions etc.Hence the performance evaluation will have to take into account to the extent possible all thesefactors.


The purpose of performance assessment is to verify the performance of a refrigeration systemby using field measurements. The test will measure net cooling capacity (tons of refrigeration)and energy requirements, at the actual operating conditions. The objective of the test is to esti-mate the energy consumption at actual load vis-à-vis design conditions.


Tons of refrigeration (TR): One ton of refrigeration is the amount of cooling obtained by oneton of ice melting in one day: 3024 kCal/h, 12,000 Btu/h or 3.516 thermal kW.

Net Refrigerating Capacity. A quantity defined as the mass flow rate of the evaporator watermultiplied by the difference in enthalpy of water entering and leaving the cooler, expressed inkCal/h, tons of Refrigeration.

kW/ton rating: Commonly referred to as efficiency, but actually power input to compressormotor divided by tons of cooling produced, or kilowatts per ton (kW/ton). Lower kW/ton indi-cates higher efficiency.

Coefficient of Performance (COP): Chiller efficiency measured in Btu output (cooling) divid-ed by Btu input (electric power).

Energy Efficiency Ratio (EER): Performance of smaller chillers and rooftop units is fre-quently measured in EER rather than kW/ton. EER is calculated by dividing a chiller's coolingcapacity (in Btu/h) by its power input (in watts) at full-load conditions. The higher the EER, themore efficient the unit.

9.4 Preparatory for Measurements

After establishing that steady-state conditions, three sets of data shall be taken, at a minimumof five-minute intervals. To minimize the effects of transient conditions, test readings should betaken as nearly simultaneously.

9.5 Procedure

9.5.1 To determine the net refrigeration capacity

The test shall include a measurement of the net heat removed from the water as it passesthrough the evaporator by determination of the following:

a. Water flow rateb. Temperature difference between entering and leaving water

The heat removed from the chilled water is equal to the product of the chilled water flowrate, the water temperature difference, and the specific heat of the water is defined asfollows

The net refrigeration capacity in tons shall be obtained by the following equation:

9. Energy Performance Assessment of HVAC Systems


The accurate temperature measurement is very vital in refrigeration and air conditioning andleast count should be at least one decimal.

Methods of measuring the flow

In the absence of an on-line flow meter the chilled water flow can be measured by the follow-ing methods

• In case where hot well and cold well are available, the flow can be measured from the tanklevel dip or rise by switching off the secondary pump.

• Non invasive method would require a well calibrated ultrasonic flow meter using whichthe flow can be measured without disturbing the system

• If the waterside pressure drops are close to the design values, it can be assumed that thewater flow of pump is same as the design rated flow.

9.5.2 Measurement of compressor power

The compressor power can be measured by a portable power analyser which would give read-ing directly in kW.

If not, the ampere has to be measured by the available on-line ammeter or by using a tongtester. The power can then be calculated by assuming a power factor of 0.9

Power (kW) = √3 x V x I x cosφ

9.5.3 Performance calculations

The energy efficiency of a chiller is commonly expressed in one of the three following ratios:



First calculate the kW/ton rating from the measured parameters.

a) kW/ton rating = Measured compressor power, kWNet refrigeration Capacity (TR)

Use this data to calculate other energy efficiency parameters with the following relations

COP = 0.293 EER EER = 3.413 COP

kW/Ton = 12 / EER EER = 12 / (kW/Ton)

kW/Ton = 3.516 / COP COP = 3.516 / (kW/Ton)

b) Coefficient of performance (COP) = 3.516kW/ton rating

c) Energy Efficiency Ratio (EER) = 12kW/ton rating

9.5.4 Performance evaluation of air conditioning systems

For centralized air conditioning systems the air flow at the air handling unit (AHU) can be mea-sured with an anemometer. The dry bulb and wet bulb temperatures can be measured at theAHU inlet and outlet. The data can be used along with a psychrometric chart (Figure 9.1) todetermine the enthalpy (heat content of air at the AHU inelt and outlet)

* Source : American Refrigeration Institute

Heat load (TR) = m x (hin – hout)4.18 x 3024

m – mass flow rate of air, kg/hrhin – enthalpy of inlet air at AHU, kJ/kghout – enthalpy of outlet air at AHU, kJ/kg

Heat load can also be calculated theoretically by estimating the various heat loads, both sen-sible and latent, in the air-conditioned room (refer standard air conditioning handbooks). Thedifference between these two indicates the losses by way of leakages, unwanted loads, heatingress etc.

9.6 Measurements to be Recorded During the Test

All instruments, including gauges and thermometers shall be calibrated over the range of testreadings for the measurement of following parameters.

Evaporator

a. Temperature of water entering evaporator b. Temperature of water leaving evaporatorc. Chilled water flow ratesd. Evaporator water pressure drop (inlet to outlet)

Compressor

e. Power input to compressor electrical power, kW

9.7 Example

In a brewery chilling system, ethylene glycol is used a secondary refrigerant. The designedcapacity is 40 TR. A test was conducted to find out the operating capacity and energy perfor-mance ratios. The flow was measured by switching off the secondary pump and measuring thetank level difference in hot well.

Measurements data:

Temperature of ethylene glycol entering evaporator = (-) 1°C Temperature of ethylene glycol leaving evaporator = (-) 4°CEthylene glycol flow rates = 13200 kg/hrEvaporator ethylene glycol pressure drop (inlet to outlet) = 0.7 kg/cm2

Power input to compressor electrical power, kW = 39.5 kWSpecific heat capacity of ethylene glycol = 2.34 kCal/kg°C





QUESTIONS

1) What is meant by a ton of refrigeration?

2) Define the terms net refrigeration capacity, COP, energy efficiency ratio.

3) What is the relation between COP and kW/ton of refrigeration?

4) How would you calculate the heat load for a room to be air-conditioned?

5) If the power consumed by a refrigerating unit / ton of refrigeration is 2 kW then findenergy efficiency ratio?

REFERENCES 1. Refrigeration and Air Conditioning by Richard C.Jordan & Gayle B.Priester - Prentice

Hall of India pvt.ltd.2. Modern Air Conditioning Practice by Norman C.Harris - McGraw-Hill International

Edition.

10. ENERGY PERFORMANCE ASSESSMENT OF LIGHTINGSYSTEMS


10.1 Introduction

Lighting is provided in industries, commercial buildings, indoor and outdoor forproviding comfortable working environment. The primary objective is to provide therequired lighting effect for the lowest installed load i.e highest lighting at lowest powerconsumption.


Most interior lighting requirements are for meeting average illuminance on a horizontal plane,either throughout the interior, or in specific areas within the interior combined with generallighting of lower value.

The purpose of performance test is to calculate the installed efficacy in terms oflux/watt/m² (existing or design) for general lighting installation. The calculated value canbe compared with the norms for specific types of interior installations for assessingimprovement options.

The installed load efficacy of an existing (or design) lighting installation can be assessedby carrying out a survey as indicated in the following pages.


Lumen is a unit of light flow or luminous flux. The lumenrating of a lamp is a measure of the total light output of thelamp. The most common measurement of light output (orluminous flux) is the lumen. Light sources are labeled withan output rating in lumens.

Lux is the metric unit of measure for illuminance of a sur-face. One lux is equal to one lumen per square meter.

Circuit Watts is the total power drawn by lamps and ballasts in a lighting circuit underassessment.

Installed Load Efficacy is the average maintained illuminance provided on a horizontal work-ing plane per circuit watt with general lighting of an interior. Unit: lux per watt per squaremetre (lux/W/m²)

Lamp Circuit Efficacy is the amount of light (lumens) emitted by a lamp for each watt ofpower consumed by the lamp circuit, i.e. including control gear losses. This is a moremeaningful measure for those lamps that require control gear. Unit: lumens per circuitwatt (lm/W)

Installed Power Density. The installed power density per 100 lux is the power needed persquare metre of floor area to achieve 100 lux of average maintained illuminance on a horizon-

10. Energy Performance Assessment of Lighting Systems


tal working plane with general lighting of an interior. Unit: watts per square metre per 100 lux(W/m²/100 lux)

100Installed power density (W/m²/100 lux) = —————————————–

Installed load efficacy (lux/W/m²)

Installed Load Efficacy Ratio (ILER)= Actual Lux/W/m² Target W/m²/100lux

——————— or ————————Target Lux/W/m² Actual W/m²/100lux

Average maintained illuminance is the average of lux levels measured at various points in adefined area.

Color Rendering Index (CRI) is a measure of the effect of light on the perceived color of objects.To determine the CRI of a lamp, the color appearances of a set of standard color chips are measuredwith special equipment under a reference light source with the same correlated color temperature asthe lamp being evaluated. If the lamp renders the color of the chips identical to the reference lightsource, its CRI is 100. If the color rendering differs from the reference light source, the CRI is lessthan 100. A low CRI indicates that some colors may appear unnatural when illuminated by the lamp.

10.4 Preparation (before Measurements)

Before starting the measurements, the following care should be taken:

• All lamps should be operating and no luminaires should be dirty or stained.

• There should be no significant obstructions to the flow of light throughout the interior,especially at the measuring points.

• Accuracies of readings should be ensured by– Using accurate illuminance meters for measurements– Sufficient number and arrangement of measurement points within the interior– Proper positioning of illuminance meter– Ensuring that no obstructions /reflections from surfaces affect measurement.

• Other precautions– If the illuminance meter is relatively old and has not been checked recently, it

should be compared with one that has been checked over a range of illuminances,e.g. 100 to 600 lux, to establish if a correction factor should be applied.

– that the number and arrangement of measurement points are sufficient andsuitable to obtain a reasonably accurate assessment of the average illuminancethroughout an interior. The procedure recommended in the CIBSE Code forsuch site measurements is as follows:

The interior is divided into a number of equal areas, which should be as square as possible.The illuminance at the centre of each area is measured and the mean value calculated. Thisgives an estimate of the average illuminance on the horizontal working plane.



10.5 Procedure for Assessment of Lighting Systems

10.5.1 To Determine the Minimum Number and Positions of Measurement Points

Calculate the Room Index: RI = L x W————–Hm(L + W)

Where L = length of interior; W = width of interior; Hm = the mounting height, whichis the height of the lighting fittings above the horizontal working plane. The working planeis usually assumed to be 0.75m above the floor in offices and at 0.85m above floor level inmanufacturing areas.

It does not matter whether these dimensions are in metres, yards or feet as long as thesame unit is used throughout. Ascertain the minimum number of measurement points fromTable10.1.

TABLE 10.1 DETERMINATION OF

MEASUREMENT POINTS

Room Index Minimum number ofmeasurement points

Below 1 9

1 and below 2 16

2 and below 3 25

3 and above 36

To obtain an approximately "square array", i.e. the spacing between the points oneach axis to be approximately the same, it may be necessary to increase the number ofpoints.

For example, the dimensions of an interior are:

Length = 9m, Width = 5m, Height of luminaires above working plane (Hm) = 2m

Calculate RI = 9 x 5 = 1.6072(9 + 5)

From Table 10.1 the minimum number of measurement points is 16

As it is not possible to approximate a "square array" of 16 points within such a rectan-gle it is necessary to increase the number of points to say 18, i.e. 6 x 3. These should bespaced as shown below:



Therefore in this example the spacing between points along rows along the length of theinterior = 9 ÷ 6 = 1.5m and the distance of the 'end' points from the wall = 1.5 ÷ 2 = 0.75m.

Similarly the distance between points across the width of the interior = 5 ÷ 3 = 1.67m withhalf this value, 0.83m, between the 'end' points and the walls.

If the grid of the measurement points coincides with that of the lighting fittings, large errorsare possible and the number of measurement points should be increased to avoid such anoccurrence.

STEP 1 Measure the floor area of the interior: Area = -------------------- m²

STEP 2 Calculate the Room Index RI = -----------------------

STEP 3 Determine the total circuit watts of the installation by a power Total circuit watts = --------meter if a separate feeder for lighting is available. If the actualvalue is not known a reasonable approximation can be obtained bytotaling up the lamp wattages including the ballasts:

STEP 4 Calculate Watts per square metre, Value of step 3 ÷ value of step 1 W/m² = ----------------------

STEP 5 Ascertain the average maintained illuminance by using lux meter, Eav. Maintained Eav.maint. = ----------------

STEP 6 Divide 5 by 4 to calculate lux per watt per square Metre Lux/W/m² = ---------------

STEP 7 Obtain target Lux/W/m² lux for type of the type ofinterior/application and RI (2): Target Lux/W/m² =

STEP 8 Calculate Installed Load Efficacy Ratio ( 6 ÷ 7 ). ILER =

10.5.2 Calculation of the Installed Load Efficacy and Installed Load Efficacy Ratio of aGeneral Lighting Installation in an Interior



The principal difference between the targets for Commercial and Industrial Ra: 40-85(Cols.2 & 3) of Table 10.2 is the provision for a slightly lower maintenance factor for the lat-ter. The targets for very clean industrial applications, with Ra: of 40 -85, are as column 2.

10.5.3 ILER Assessment

Compare the calculated ILER with the information in Table 10.3.

Ra : Colour rendering index

TABLE 10.3 INDICATORS OF

PERFORMANCE

ILER Assessment

0.75 or over Satisfactory to Good

0.51 – 0.74 Review suggested

0.5 or less Urgent action required

TABLE 10.2 Target lux/W/m² (W/m²/100lux) values for

maintained illuminance on horizontal

plane for all room indices and applica-

tions:

ILER Ratios of 0.75 or more may be considered to be satisfactory. Existing installationswith ratios of 0.51 - 0.74 certainly merit investigation to see if improvements are possible. Ofcourse there can be good reasons for a low ratio, such as having to use lower efficacy lamps orless efficient luminaires in order to achieve the required lighting result -but it is essential tocheck whether there is a scope for a more efficient alternative. Existing installations with anILER of 0.5 or less certainly justify close inspection to identify options for converting theinstallation to use more efficient lighting equipment.



Having derived the ILER for an existing lighting installation, then the difference betweenthe actual ILER and the best possible (1.0) can be used to estimate the energy wastage. For agiven installation:

Annual energy wastage (in kWh)= (1.0 - ILER) x Total load (kW) x annual operating hours (h)

This process of comparing the installed load efficacy (ILE) with the target value for theRoom Index and type of application can also be used to assess the efficiency of designs for newor replacement general lighting installations. If, when doing so, the calculated ILE (lux/W/m²)is less than the target value then it is advisable to ascertain the reasons. It may be that therequirements dictate a type of luminaire that is not as efficient as the best, or the surfacereflectances are less than the normal maxima, or the environment is dirty, etc., Whatever thereasons, they should be checked to see if a more efficient solution is possible.

10.6 Example of ILER Calculation (for the room as mentioned in paragraph 10.5.1)

STEP 1 Measure the floor area of the interior: Area = 45 m²

STEP 2 Calculate the Room Index RI = 1.93

STEP 3 Determine the total circuit watts of the installation by a powermeter if a separate feeder for lighting is available. If the actualvalue is not known a reasonable approximation can be obtained bytotaling up the lamp wattages including the ballasts:

STEP 4 Calculate Watts per square metre, 3 ÷1 : W/m² = 22

STEP 5 Ascertain the average maintained illuminance, Eav. Maintained(average lux levels measured at 18 points) Eav.maint. = 700

STEP 6 Divide 5 by 4 to calculate the actual lux per watt per square Metre Lux/W/m² = 31.8

STEP 7 Obtain target Lux/W/m² lux for type of the type of interior/application and RI (2):(Refer Table 10.2) Target Lux/W/m² = 46

STEP 8 Calculate Installed Load Efficacy Ratio ( 6 ÷ 7 ). ILER = 0.7

Total circuit watts = 990 W

Referring to table 3, ILER of 0.7 means that there is scope for review of the lighting system. Annual energy wastage = (1 - ILER) x watts x no. of operating hours

= (1 - 0.7) x 990 x 8 hrs/day x 300 days= 712 kWh/annum

10.7 Areas for Improvement

• Look for natural lighting opportunities through windows and other openings

• In the case of industrial lighting, explore the scope for introducing translucent sheets

• Assess scope for more energy efficient lamps and luminaries

• Assess the scope for rearrangement of lighting fixtures



10.8 Other Useful Information

10.8.1 IES - Recommendations

The Illuminating Engineering Society (IES) has published illuminance recommendations forvarious activities. These tables cover both generic tasks (reading, writing etc), and 100's ofvery specific tasks and activities (such as drafting, parking, milking cows, blowing glass andbaking bread).

All tasks fall into 1 of 9 illuminance categories, covering from 20 to 20,000 lux, (2 to 2000foot candles). The categories are known as A - I, and each provide a range of 3 iluminance val-ues (low, mid and high). See Table 10.4.

ACTIVITY CATEGORY LUX FOOTCANDLES

Public spaces with dark surroundings A 20-30-50 2-3-5

Simple orientation for short temporary visits B 50-75-100 5-7.5-10

Working spaces where visual tasks are only C 100-150-200 10-15-20occasionally performed

Performance of visual tasks of high contrast D 200-300-500 20-30-50or large size

Performance of visual tasks of medium E 500-750-1000 50-75-100contrast or small size

Performance of visual tasks of low contrast F 1000-1500-2000 100-150-200or very small size

Performance of visual tasks of low contrast G 2000-3000-5000 200-300-500or very small size over a prolonged period

Performance of very prolonged and exacting H 5000-7500-10000 500-750-1000visual tasks

Performance of very special visual tasks of I 10000-15000-20000 1000-1500-2000extremely low contrast

TABLE 10.4 IES ILLUMINANCE CATEGORIES AND VALUES - FOR GENERIC INDOOR

ACTIVITIES

A-C for illuminances over a large area (i.e. lobby space)

D-F for localized tasks

G-I for extremely difficult visual tasks

10.8.2 Example Using IES Recommendations

Let us determine the appropriate light level for a card file area in a library.

Step 1: The visual task is reading card files in a library. A number of tasks are accomplished inthe room. In such a cases, a category is chosen based on the generic descriptions in the IESIlluminance Category and Illuminance table discussed in step 3. For example, offices will usu-ally require Category E: 500-750-1000 lux.



Step 2: More detailed task descriptions are given in the recommended illuminance level tablesin the IES Handbook. (For an intensive lighting survey) Under the task category "Libraries,"subheading "Card files," the illuminance category is E.

Step 3: From the IES Illuminance Category and Ranges table, find category E and choose 500-750-1000 lux for the range of illuminance recommended. The first column in the table is illuminancevalues in units of lux, the metric version of footcandle. Notice that categories A through C are forgeneral illumination throughout the area, but D through I are for illuminance on the task. CategoriesG through I would require a combination of general lighting and task lighting.

Step 4: Use the weighting factors to decide which of the values in the illuminance range to use.Since libraries are public facilities, there may be many individuals over 55 years of age so selectthe category 'Over 55' for a weighting factor of +1.

Next, decide whether the demand for speed and accuracy is not important, important or crit-ical. Filing of cards correctly is not a critical activity, so the weighting factor of zero (0) isselected. An example of critical might be drafting work. The task background reflectance forblack type on a white page is 85%. So choose "greater than 70 percent" for a weighting factorof -1. The total weighting factor is 0. So use the middle recommended illuminance, or 750 lux.

For more detailed information on this the IES handbook may be referred.

10.8.3 Characteristics of Different Types of Lamps

Type Lamp Lumens Lamp Efficiency Choke Life of Capacitor Colorof Wattage (Lumens/Watt) Rating Lamp Rating Rendering

Lamp (Watts) (Watts) (Hours) Required Index(Microfarads)

HPSV 70 5600 80 13 15000 - 0.2 - 0.39 1220000

HPSV 150 14000 93 20 15000 - 0.2 - 0.39 2020000

HPSV 250 25000 100 20 15000 - 0.2 - 0.39 3220000

HPSV 400 47000 118 40 15000 - 0.2 - 0.39 4520000

HPSV 70 --- --- --- --- --- ---Super

HPSV 100 9500 95 18 15000 - 0.2 - 0.39 ---Super 20000

HPSV 150 15500 103 20 15000 - 0.2 - 0.39 ---Super 20000

HPSV 250 30000 120 25 15000 - 0.2 - 0.39 ---Super 20000

HPSV 400 54000 129 40 15000 - 0.2 - 0.39 ---



Super 20000

HPSV 600 --- --- --- --- --- ---Super

HPMV 80 3400 43 9 4000 - 0.6 - 0.69 85000

HPMV 125 6300 50 12 4000 - 0.6 - 0.69 105000

HPMV 250 13000 52 16 4000 - 0.6 - 0.69 185000

HPMV 400 22000 55 25 4000 - 0.6 - 0.69 185000

Metal 70 4200 84 26 10000 0.9 - 0.93 ---Halide

Metal 150 10500 70 20 10000 0.9 - 0.93 ---Halide

Metal 250 19000 76 25 10000 0.9 - 0.93 ---Halide

Metal 400 31000 76 60 10000 0.9 - 0.93 ---Halide

Metal 1000 80000 80 65 10000 0.9 - 0.93 ---Halide

FTL 40 2400 60 15 4400 0.8 - 0.89 3.2 - 3.8

FTL 36 3250 90 5 14000 0.8 - 0.89 3.2 - 3.8Super



QUESTIONS

1) What is circuit watts?

2) Define ILER and its significance.

3) Distinguish between lux and lumens.

4) What do you understand by the term colour rendering index?

5) Define room index?

6) For a room of length 10 m and width 20 m, calculate room index?

7) For a room of 9 x 6 m, determine the appropriate number of measuring points for luxlevels?

8) What possible improvement measures you would look for in a general lighting sys-tem?

9) Which of the following lamps has the maximum lamp efficiency?(lumes/Watt) a) Metal Hallide b) Fluorescent c) Incandescent d) HPSV

REFERENCES 1. Illumination engineering for energy efficient luminous environments by Ronald N.

Helms, Prentice-Hall, Inc. 2. The 'LIGHTSWITCH' programme, Energy Saving Trust, UK

11. PERFORMING FINANCIAL ANALYSIS


11.1 Introduction

When planning an energy efficiency or energy management project, the costs involvedshould always be considered. Therefore, as with any other type of investment, energy man-agement proposals should show the likely return on any capital that is invested. Consider thecase of an energy auditor who advises the senior management of an organisation that capi-tal should be invested in new boiler plant. Inevitably, the management of the organisationwould ask:

• How much will the proposal cost?• How much money will be saved by the proposal?

These are, of course, not unreasonable questions, since within any organisation there aremany worthy causes, each of which requires funding and it is the job of senior management toinvest in capital where it is going to obtain the greatest return. In order to make a decision aboutany course of action, management needs to be able to appraise all the costs involved in a pro-ject and determine the potential returns.

This however, is not quite as simple as it might first appear. The capital value of plantor equipment usually decreases with time and it often requires more maintenance as it getsolder. If money is borrowed from a bank to finance a project, then interest will have to bepaid on the loan. Inflation too will influence the value of any future energy savings thatmight be achieved. It is therefore important that the cost appraisal process allows for allthese factors, with the aim of determining which investments should be undertaken, and ofoptimising the benefits achieved. To this end a number of accounting and financial appraisaltechniques have been developed which help energy managers and auditors make correct andobjective decisions.

The financial issues associated with capital investment in energy saving projects are inves-tigated in this chapter. In particular, the discounted cash flow techniques of net present valueand internal rate of return are discussed in detail.

11.2 Fixed and Variable Costs

When appraising the potential costs involved in a project it is important to understand the dif-ference between fixed and variable costs. Variable costs are those which vary directly with theoutput of a particular plant or production process, such as fuel costs. Fixed costs are those costs,which are not dependent on plant or process output, such as site-rent and insurance. The totalcost of any project is therefore the sum of the fixed and variable costs. Example 1 illustrateshow both fixed and variable costs combine to make the total operating cost.

Example 1

The capital cost of the DG set is Rs.9,00,000, the annual output is 219 MWh, and the mainte-nance cost is Rs.30,000 per annum. The cost of producing each unit of electricity is 3.50Rs./kWh. The total cost of a diesel generator operating over a 5-year period, taking into con-sideration both fixed and variable cost is:

From Example 1, it can be seen that the fixed costs represent 21.5% of the total cost. In fact,the annual electricity output of 219 MWh assumes that the plant is operating with an averageoutput of 50 kW. If this output were increased to an average of 70 kW, then the fuel cost wouldbecome Rs. 53,65,500, with the result that the fixed costs would drop to 16.37% of the total.Thus the average unit cost of production decreases as output increases.

The concept of fixed and variable costs can be used to determine the break-even pointfor a proposed project. The break-even point can be determined by using the followingequation.

11. Performing Financial Analysis


Item Type of cost Calculation Cost

Capital cost of generator Fixed - 9,00,000

Annual maintenance Fixed 30,000 x 5 (years) 1,50,000

Fuel cost Variable 219,000 x 3.50 x 5 3,83,2500

Total cost 4,88,2500

Example 2

If the electricity bought from a utility company costs an average of Rs.4.5/kWh, the break-even point for the generator described in Example 1, when the average output is 50 kW isgiven by:

4.5 x 50 x n = (9,00,000 + 150000) + (3.5 x 50 x n)

n = 21000 hours

If the average output is 70 kW, the break-even point is given by:

4.5 x 70 x n = (9,00,000 + 150000) + (3.50 x 70 x n)

n = 15000 hours

Thus, increasing the average output of the generator significantly reduces the break-eventime for the project. This is because the capital investment (i.e. the generator) is being betterutilised.



11.3 Interest Charges

In order to finance projects, organizations often borrow money from banks or other leadingorganizations. Projects financed in this way cost more than similar projects financed fromorganisation's own funds, because interest charges must be paid on the loan. It is thereforeimportant to understand how interest charges are calculated. Interest charges can be calculatedby lending organization in two different ways: simple interest and compound interest.

(i) Simple interest: If simple interest is applied, then charges are calculated as a fixed per-centage of the capital that is borrowed. A fixed interest percentage is applied to each year ofthe loan and repayments are calculated using the equation.

(ii) Compound interest: Compound interest is usually calculated annually (although this isnot necessarily the case). The interest charged is calculated as a percentage of the outstandingloan at the end of each time period. It is termed 'compound' because the outstanding loan isthe sum of the unpaid capital and the interest charges up to that point. The value of the totalrepayment can be calculated using the equation.

Example 3

A company borrows Rs.3,00,00,00 to finance a new boiler installation. If the interest rate is10% per annum and the repayment period is 5 years, let us calculate the value of the totalrepayment and the monthly repayment value, assuming (i) simple interest and (ii) compoundinterest.

(i) Assuming simple interest:

Total repayment = 30,00,000 + (10/100 x 30,00,000 x 5) = Rs.45,00,000Monthly repayment = 45,00,000 / (5 x 12) = Rs.75,000

(ii) Assuming compound interest

Repayment at end of year 1 = 30,00,000 + (10/100 x 30,00,000) = Rs.33,00,000

Repayment at end of year 2 = 33,00,000 + (10/100 x 33,00,000) = Rs.36,30,000

Similarly, the repayments at the end of years 3, 4 and 5 can be calculated:

Repayment at end of year 3 = Rs. 39,93,000

Repayment at end of year 4 = Rs. 43,92,300

Repayment at end of year 5 = Rs. 48,31530

Alternatively, the following equation can be used to determine the compound interest repay-ment value.

Total repayment value = 30,00,000 x (1 + 10 / 100)5 = Rs.48,31,530

4831530Monthly repayment = = Rs.80,525

5 x 12

It can be seen that by using compound interest, the lender recoups an additional Rs.33,1530.It is not surprisingly lenders usually charge compound interest on loans.

11.4 Simple Payback Period

This is the simplest technique that can be used to appraise a proposal. The simple payback peri-od can be defined as 'the length of time required for the running total of net savings beforedepreciation to equal the capital cost of the project'. In theory, once the payback period hasended, all the project capital costs will have been recouped and any additional cost savingsachieved can be seen as clear 'profit'. Obviously, the shorter the payback period, the moreattractive the project becomes. The length of the maximum permissible payback period gener-ally varies with the business culture concerned. In some companies, payback periods in excessof 3 years are considered acceptable.

The payback period can be calculated using the equation



The annual net cost saving (AS) is the least savings achieved after all the operational costs havebeen met. Simple payback period is illustrated in Example 4.



Example 4

A new small cogeneration plant installation is expected to reduce a company's annual energybill by Rs.4,86,000. If the capital cost of the new boiler installation is Rs.22,20,000 and theannual maintenance and operating costs are Rs. 42,000, the expected payback period for theproject can be worked out as.

Solution

PB = 22,20,000 / (4,86,000 – 42,000) = 5.0 years

11. 5 Discounted Cash Flow Methods

The payback method is a simple technique, which can easily be used to provide a quick evalu-ation of a proposal. However, it has a number of major weaknesses:

• The payback method does not consider savings that are accrued after the payback periodhas finished.

• The payback method does not consider the fact that money, which is invested, shouldaccrue interest as time passes. In simple terms there is a 'time value' component to cashflows. Thus Rs.1000 today is more valuable than Rs.1000 in 10 years' time.

In order to overcome these weaknesses a number of discounted cash flow techniques havebeen developed, which are based on the fact that money invested in a bank will accrue annualinterest. The two most commonly used techniques are the 'net present value' and the 'internalrate of return' methods.

Net Present Value Method

The net present value method considers the fact that a cash saving (often referred to as a'cash flow') of Rs.1000 in year 10 of a project will be worth less than a cash flow of Rs.1000in year 2. The net present value method achieves this by quantifying the impact of time onany particular future cash flow. This is done by equating each future cash flow to its currentvalue today, in other words determining the present value of any future cash flow. The pre-sent value (PV) is determined by using an assumed interest rate, usually referred to as a dis-count rate. Discounting is the opposite process to compounding. Compounding determinesthe future value of present cash flows, where" discounting determines the present value offuture cash flows.

In order to understand the concept of present vale, consider the case described in Example 3.If instead of installing a new cogeneration system, the company invested Rs.22,20,000 in a

bank at an annual interest rate of 8%, then:

The value of the sum at the end of year 1 = 22,20,000 + (0.08 x 22,20,000) = Rs.23,97,600

The value of the sum at the end of year 2 = 23,97,600 + (0.08 x 23,97,600) = Rs.25,89,408

The value of the investment would grow as compound interest is added, until after n yearsthe value of the sum would be:

Example :

The future value of the investment made at present, after 5 years will be:

FV = 22,20,000 x (1 + 8/100)5 = Rs.32,61,908.4

So in 5 years the initial investment of 22,20,000 will accrue Rs.10,41,908.4 in interest and willbe worth Rs.32,61,908.4. Alternatively, it could equally be said that Rs.32,61908.4 in 5 yearstime is worth Rs.22,20,000 now (assuming an annual interest rate of 8%). In other words thepresent value of Rs.32,61,908.40 in 5 years time is Rs.22,00,000 now.

The present value of an amount of money at any specified time in the future can be deter-mined by the following equation.



The net present value method calculates the present value of all the yearly cash flows (i.e.capital costs and net savings) incurred or accrued throughout the life of a project, and summatesthem. Costs are represented as a negative value and savings as a positive value. The sum of allthe present values is known as the net present value (NPV). The higher the net present value,the more attractive the proposed project.

The present value of a future cash flow can be determined using the equation above.However, it is common practice to use a discount factor (DF) when calculating present value.The discount factor is based on an assumed discount rate (i.e. interest rate) and can be deter-mined by using equation.

DF = (1 + IR/100)–n

The product of a particular cash flow and the discount factor is the present value.

PV = S x DF

The values of various discount factors computed for a range of discount rates (i.e. interest rates)are shown in Table 11.1. The Example 5 illustrates the process involved in a net present valueanalysis.

Example 5

Using the net present value analysis technique, let us evaluate the financial merits of the proposedprojects shown in the Table below. Assume an annual discount rate of 8% for each project.



TABLE 11.1COMPUTED DISCOUNT FACTORS

Discount rate % (or interest rate %)

Year 2 4 6 8 10 12 14 16

0 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000

1 0.980 0.962 0.943 0.926 0.909 0.893 0.877 0.862

2 0.961 0.825 0.890 0.857 0.826 0.797 0.769 0.743

3 0.942 0.889 0.840 0.794 0.751 0.712 0.675 0.641

4 0.924 0.855 0.792 0.735 0.683 0.636 0.592 0.552

5 0.906 0.822 0.747 0.681 0.621 0.567 0.519 0.476

6 0.888 0.790 0.705 0.630 0.564 0.507 0.456 0.410

7 0.871 0.760 0.665 0.583 0.513 0.452 0.400 0.354

8 0.853 0.731 0.627 0.540 0.467 0.404 0.351 0.305

9 0.837 0.703 0.592 0.500 0.424 0.361 0.308 0.263

10 0.820 0.676 0.558 0.463 0.386 0.322 0.270 0.227

11 0.804 0.650 0.527 0.429 0.350 0.287 0.237 0.195

12 0.788 0.625 0.497 0.397 0.319 0.257 0.208 0.168

13 0.773 0.601 0.469 0.368 0.290 0.229 0.182 0.145

14 0.758 0.577 0.442 0.340 0.263 0.205 0.160 0.125

15 0.743 0.555 0.417 0.315 0.239 0.183 0.140 0.108

16 0.728 0.534 0.394 0.292 0.218 0.163 0.123 0.093

17 0.714 0.513 0.371 0.270 0.198 0.146 0.108 0.080

18 0.700 0.494 0.350 0.250 0.180 0.130 0.095 0.069

19 0.686 0.475 0.331 0.232 0.164 0.116 0.083 0.060

20 0.673 0.456 0.312 0.215 0.149 0.104 0.073 0.051

Project – 1 Project – 2

Capital cost (Rs.) 30 000.00 30 000.00

Year Net annual saving (Rs.) Net annual saving (Rs.)

1 +6 000.00 +6 600.00

2 +6 000.00 +6 600.00

3 +6 000.00 +6 300.00

Solution

The annual cash flows should be multiplied by the annual discount factors for a rate of 8% todetermine the annual present values, as shown in the Table below:



4 +6 000.00 +6 300.00

5 +6 000.00 +6 000.00

6 +6 000.00 +6 000.00

7 +6 000.00 +5 700.00

8 +6 000.00 +5 700.00

9 +6 000.00 +5 400.00

10 +6 000.00 +5 400.00

Total net saving at +60 000.00 + 60 000.00end of year 10

Year Discount Project 1 Project 2Factor for Net Present Net Present

8% savings value (Rs.) savings value (Rs.)(a) (Rs.) (a x b) (Rs.) (a x c)

(b) (c)

0 1.000 –30 000.00 –30 000.00 –30 000.00 –30 000.00

1 0.926 +6 000.00 +5 556.00 +6 600.00 +6 111.60

2 0.857 +6 000.00 +5 142.00 +6 600.00 +5 656.20

3 0.794 +6 000.00 +4 764.00 +6 300.00 +5 002.20

4 0.735 +6 000.00 +4 410.00 +6 300.00 +4 630.50

5 0.681 +6 000.00 +4 086.00 +6 000.00 +4 086.00

6 0.630 +6 000.00 +3 780.00 +6 000.00 +3 780.00

7 0.583 +6 000.00 +3 498.00 +5 700.00 +3323.10

8 0.540 +6 000.00 +3 240.00 +5 700.00 +3 078.00

9 0.500 +6 000.00 +3 000.00 +5 400.00 +2 700.00

10 0.463 +6 000.00 +2 778.00 +5 400.00 +2 500.20

NPV = +10 254.00 NPV = +10 867.80

It can be seen that over a 10 year life-span the net present value for Project 1 isRs.10,254.00, while for Project 2 it is Rs.10,867.80. Therefore Project 2 is the preferentialproposal.

The whole credibility of the net present value method depends on a realistic prediction offuture interest rates, which can often be unpredictable. It is prudent therefore to set the discountrate slightly above the interest rate at which the capital for the project is borrowed. This willensure that the overall analysis is slightly pessimistic, thus acting against the inherent uncertainties in predicting future savings.

Internal rate of return method

It can be seen from Example 5 that both projects returned a positive net present value over 10years, at a discount rate of 8%. However, if the discount rate were reduced there would come apoint when the net present value would become zero. It is clear that the discount rate whichmust be applied, in order to achieve a net present value of zero, will be higher for Project 2 thanfor Project 1. This means that the average rate of return for Project 2 is higher than for Project1, with the result that Project 2 is the better proposition.



Example 6 illustrates how an internal rate of return analysis is performed.

Example 6

A proposed project requires an initial capital investment of Rs.20 000. The cash flows generat-ed by the project are shown in the table below:

Year Cash flow (Rs.)

0 –20,000.00

1 +6000.00

2 +5500.00

3 +5000.00

4 +4500.00

5 +4000.00

6 +4000.00

Given the above cash flow data, let us find out the internal rate of return for the project.

It can clearly be seen that the discount rate which results in the net present value being zerolies somewhere between 12% and 16%.

For12% discount rate, NPV is positive; for 16% discount rate, NPV is negative. Thusfor some discount rate between 12 and 16 percent, present value benefits are equated topresent value costs. To find the value exactly, one can interpolate between the two ratesas follows:

459.5Internal rate of return = 0.12 + (0.16 – 0.12) x

(459.5 – (–1508.5))

459.5Internal rate of return = 0.12 + (0.16 – 0.12) x = 12.93%

(459.5 + 1508.5)

Thus the internal rate of return for the project is 12.93 %. At first sight both the net presentvalue and internal rate of return methods look very similar, and in some respects are. Yet thereis an important difference between the two. The net present value method is essentially a com-parison tool, which enables a number of projects to be compared, while the internal rate ofreturn method is designed to assess whether or not a single project will achieve a target rate ofreturn.

Profitability index

Another technique, which can be used to evaluate the financial viability of projects, is the prof-itability index. The profitability index can be defined as:



Cash 8% discount rate 12% discount rate 16% discount rateflow Discount Present Discount Present Discount Present(Rs.) factor value factor value factor value(Rs.) (Rs.) (Rs.)

0 –20000 1.000 –20000 1.000 –20000 1.000 –20000

1 6000 0.926 5556 0.893 5358 0.862 5172

2 5500 0.857 4713.5 0.797 4383.5 0.743 4086.5

3 5000 0.794 3970 0.712 3560 0.641 3205

4 4500 0.735 3307.5 0.636 3862 0.552 2484

5 4000 0.681 2724 0.567 2268 0.476 1904

6 4000 0.630 2520 0.507 2028 0.410 1640

NPV = 2791 NPV = 459.5 NPV = –1508.5

Solution

x 100

x 100

The application of profitability index is illustrated in Example 7.

Example 7

Determine the profitability index for the projects outlined in Example 5

10254For Project 1: Profitability index = = 0.342

30,000

10867For Project 2: Profitability index = = 0.362

30,000

Project 2 is therefore a better proposal than Project 1.

11.6 Factors Affecting Analysis

Although the Examples 5 and 6 illustrate the basic principles associated with the financialanalysis of projects, they do not allow for the following important considerations:

• The capital value of plant and equipment generally depreciates over time• General inflation reduces the value of savings as time progresses. For example, Rs.1000

saved in 1 year's time will be worth more than Rs.1000 saved in 10 years time.

The capital depreciation of an item of equipment can be considered in terms of its salvagevalue at the end of the analysis period. The Example 8 illustrates the point.

Example 8

It is proposed to install a heat recovery equipment in a factory. The capital cost of installing theequipment is Rs.20,000 and after 5 years its salvage value is Rs.1500. If the savings accrued bythe heat recovery device are as shown below, we have to find out the net present value after5 years. Discount rate is assumed to be 8%.



Data

Year 1 2 3 4 5

7000 6000 6000 5000 5000

Solution



Year Discount Capital Net PresentFactor for Investment Savings Value

8% (Rs.) (Rs.) (Rs.)(a) (b) (c) (a) x (b + c)

0 1,000 –20,000.00 –20,000.00

1 0.926 +7000.00 +6482.00

2 0.857 +6000.00 +5142.00

3 0.794 +6000.00 +4764.00

4 0.735 +6000.00 +3675.00

5 0.681 +1,500.00 +5000.00 +4426.50

NPV = +4489.50

It is evident that over a 5-year life span the net present value of the project is Rs.4489.50.Had the salvage value of the equipment not been considered, the net present value of the pro-ject would have been only Rs.3468.00.

Real value

Inflation can be defined as the rate of increase in the average price of goods and services. Insome countries, inflation is expressed in terms of the retail price index (RPI), which is deter-mined centrally and reflects average inflation over a range of commodities. Because of infla-tion, the real value of cash flow decreases with time. The real value of sum of money (S)realised in n years time can be determined using the equation.

RV = S x (1 + R/100)–n

Where RV is the real value of S realized in n years time. S is the value of cash flow in nyears time and R is the inflation rate (%).

As with the discount factor it is common practice to use an inflation factor when assessingthe impact of inflation on a project. The inflation factor can be determined using the equation.

IF = (1 + R/100)–n

The product of a particular cash flow and inflation factor is the real value of the cash flow.

RV = S x IF

The application of inflation factors is considered in Example 9.

Example 9

Recalculate the net present value of the energy recovery scheme in Example 8, assuming thediscount rate remains at 8% and that the rate of inflation is 5%.

Solution

Because of inflation; Real interest rate = Discount rate – Rate of inflationTherefore Real interest rate = 8 – 5 = 3%



Year Capital Net real Inflation Net real Real PresentInvestment Savings Factor Savings Discount Value

(Rs.) (Rs.) For 5% (Rs.) Factor (Rs.)For 3%

0 –20,000.00 1.000 –20,000.00 1.000 –20,000.00

1 +7000.00 0.952 +6664.00 0.971 +6470.74

2 +6000.00 0.907 +5442.00 0.943 +5131.81

3 +6000.00 0.864 +5184.00 0.915 +4743.36

4 +5000.00 0.823 +4145.00 0.888 +3654.12

5 +1500.00 +5000.00 0.784 +5096.00 0.863 +4397.85

NPV = +4397.88

The Example 9 shows that when inflation is assumed to be 5%, the net present value of theproject reduces from Rs.4489.50 to Rs.4397.88. This is to be expected, because general infla-tion will always erode the value of future 'profits' accrued by a project.



QUESTIONS

1. Why fresh investments are needed for energy conservation in industry ?

2. Cost of an heat exchanger is Rs.1.00 lakhs. Calculate simple pay back period consid-ering annual saving potential of Rs.60,000/- and annual operating cost ofRs.15,000/-.

3. What is the main draw back of simple pay back method?

4. Calculate simple pay back period for a boiler that cost Rs.75.00 lakhs to purchaseand Rs.5 lakhs per year on an average to operate and maintain and is expected tosave Rs.30 lakhs.

5. What are the advantages of simple pay back method?

6. What do you understand by the term " present value of money"?

7. Define ROI.

8. What is the objective of carrying out sensitivity analysis?

9. You are investing Rs.100 in a bank. The bank gives 10% interest per year for twoyears. What is the present value and what is the future value?

REFERENCES 1. Energy Management, Supply and Conservation, Dr. Clive Beggs, .Butterworth

Heinemann

12. APPLICATION OF NON-CONVENTIONAL & RENEWABLE ENERGY SOURCES


12.1 Concept of Renewable Energy

Renewable energy sources also called non-conventional energy, are sources that are continu-ously replenished by natural processes. For example, solar energy, wind energy, bio-energy -bio-fuels grown sustain ably), hydropower etc., are some of the examples of renewable energysources

A renewable energy system converts the energy found in sunlight, wind, falling-water, sea-waves, geothermal heat, or biomass into a form, we can use such as heat or electricity. Most ofthe renewable energy comes either directly or indirectly from sun and wind and can never beexhausted, and therefore they are called renewable.

However, most of the world's energy sources are derived from conventional sources-fossilfuels such as coal, oil, and natural gases. These fuels are often termed non-renewable energysources. Although, the available quantity of these fuels are extremely large, they are neverthe-less finite and so will in principle 'run out' at some time in the future

Renewable energy sources are essentially flows of energy, whereas the fossil and nuclearfuels are, in essence, stocks of energy

Various forms of renewable energy

Solar energyWind energyBio energyHydro energyGeothermal energyWave and tidal energy

This chapter focuses on application potential of commercially viable renewable energysources such as solar, wind, bio and hydro energy in India.

12.2 Solar Energy

Solar energy is the most readily available and freesource of energy since prehistoric times. It is esti-mated that solar energy equivalent to over 15,000times the world's annual commercial energy con-sumption reaches the earth every year.

India receives solar energy in the region of 5 to7 kWh/m2 for 300 to 330 days in a year. This ener-gy is sufficient to set up 20 MW solar power plant per square kilometre land area.

Solar energy can be utilised through two different routes, as solar thermal route and solarelectric (solar photovoltaic) routes. Solar thermal route uses the sun's heat to produce hot wateror air, cook food, drying materials etc. Solar photovoltaic uses sun's heat to produce electricityfor lighting home and building, running motors, pumps, electric appliances, and lighting.

Solar Thermal Energy Application

In solar thermal route, solar energy can be converted into thermal energy with the help of solarcollectors and receivers known as solar thermal devices.

The Solar-Thermal devices can be classified into three categories:

Low-Grade Heating Devices - up to the temperature of 100°C. Medium-Grade Heating Devices -up to the temperature of 100°-300°C High-Grade Heating Devices -above temperature of 300°C

Low-grade solar thermal devices are used in solar water heaters, air-heaters, solar cookersand solar dryers for domestic and industrial applications.

Solar water heaters Most solar water heating systems have two main parts: asolar collector and a storage tank. The most common col-lector is called a flat-plate collector (see Figure 12.1). Itconsists of a thin, flat, rectangular box with a transparentcover that faces the sun, mounted on the roof of buildingor home. Small tubes run through the box and carry thefluid - either water or other fluid, such as an antifreezesolution – to be heated. The tubes are attached to anabsorber plate, which is painted with special coatings toabsorb the heat. The heat builds up in the collector, whichis passed to the fluid passing through the tubes.

An insulated storage tank holds the hot water. It issimilar to water heater, but larger is size. In case of systems that use fluids, heat is passed fromhot fluid to the water stored in the tank through a coil of tubes.

Solar water heating systems can be either active or passive systems. The active system,which are most common, rely on pumps to move the liquid between the collector and the storage tank. The passive systems rely on gravity and the tendency for water to naturally circulate as it is heated. A few industrial application of solar water heaters are list-ed below:

❑ Hotels: Bathing, kitchen, washing, laundry applications❑ Dairies: Ghee (clarified butter) production, cleaning and sterilizing, pasteurization ❑ Textiles: Bleaching, boiling, printing, dyeing, curing, ageing and finishing ❑ Breweries & Distilleries: Bottle washing, wort preparation, boiler feed heating ❑ Chemical /Bulk drugs units: Fermentation of mixes, boiler feed applications ❑ Electroplating/galvanizing units: Heating of plating baths, cleaning, degreasing applica-

tions ❑ Pulp and paper industries: Boiler feed applications, soaking of pulp.

Solar Cooker

Solar cooker is a device, which uses solar energy for cooking, and thus saving fossil fuels, fuelwood and electrical energy to a large extent. However, it can only supplement the cooking fuel,and not replace it totally. It is a simple cooking unit, ideal for domestic cooking during most ofthe year except during the monsoon season, cloudy days and winter months

12. Application of Non-Conventional & Renewable Energy Sources


Figure 12.1 Solar Flat plate collector

Box type solar cookers: The box type solar cookers with asingle reflecting mirror are the most popular in India.These cookers have proved immensely popular in ruralareas where women spend considerable time for collectingfirewood. A family size solar cooker is sufficient for 4 to5 members and saves about 3 to 4 cylinders of LPG everyyear. The life of this cooker is upto 15 years. This cookercosts around Rs.1000 after allowing for subsidy. Solarcookers.(Figure 12.2) are widely available in the market.

Parabolic concentrating solar cooker:

A parabolic solar concentrator comprises of sturdy FibreReinforced Plastic (FRP) shell lined with Stainless Steel(SS) reflector foil or aluminised polyester film. It can accom-modate a cooking vessel at its focal point. This cooker isdesigned to direct the solar heat to a secondary reflectorinside the kitchen, which focuses the heat to the bottom of acooking pot. It is also possible to actually fry, bake and roastfood. This system generates 500 kg of steam, which isenough to cook two meals for 500 people (see Figure 12.3).This cooker costs upward of Rs.50,000.

Positioning of solar panels or collectors can greatlyinfluence the system output, efficiency and payback. Tiltingmechanisms provided to the collectors need to be adjustedaccording to seasons (summer and winter) to maximise thecollector efficiency.

The period four to five hours in late morning and earlyafternoon (between 9 am to 3pm) is commonly called the"Solar Window". During this time, 80% of the total collectable energy for the day falls ona solar collector. Therefore, the collector should be free from shade during this solar win-dow throughout the year - Shading, may arise from buildings or trees to the south of thelocation.

Solar Electricity Generation

Solar Photovoltaic (PV): Photovoltaic is the technicalterm for solar electric. Photo means "light" and voltaicmeans "electric". PV cells are usually made of silicon,an element that naturally releases electrons whenexposed to light. Amount of electrons released fromsilicon cells depend upon intensity of light incident on it. The silicon cell is covered with a grid of metal that directs the electrons to flow in a path to create an electric current. This current is guided into a wire that is connected to a battery or DC appli-ance. Typically, one cell produces about 1.5 watts ofpower. Individual cells are connected together to form a



Figure 12.2 Box Type Solar Collector

Figure 12.3 Parabolic Collector

Figure 12.4 Solar Photovoltaic Array

solar panel or module, capable of producing 3 to 110 Watts power. Panels can be connectedtogether in series and parallel to make a solar array (see Figure 12.4), which can produceany amount of Wattage as space will allow. Modules are usually designed to supplyelectricity at 12 Volts. PV modules are rated by their peak Watt output at solar noon on aclear day.

Some applications for PV systems are lighting for commercial buildings, outdoor (street)lighting (see Figure 12.5),rural and village lightingetc. Solar electric powersystems can offer indepen-dence from the utility gridand offer protection duringextended power failures.Solar PV systems arefound to be economicalespecially in the hilly andfar flung areas where con-ventional grid power supply will be expensive to reach.

PV tracking systems is an alternative to the fixed, stationary PV panels. PV tracking sys-tems are mounted and provided with tracking mechanisms to follow the sun as it moves throughthe sky. These tracking systems run entirely on their own power and can increase output by40%.

Back-up systems are necessary since PV systems only generate electricity when the sun isshining. The two most common methods of backing up solar electric systems are connecting thesystem to the utility grid or storing excess electricity in batteries for use at night or on cloudydays.

Performance

The performance of a solar cell is measured in terms of its efficiency at converting sunlight intoelectricity. Only sunlight of certain energy will work efficiently to create electricity, and muchof it is reflected or absorbed by the material that make up the cell. Because of this, a typicalcommercial solar cell has an efficiency of 15%—only about one-sixth of the sunlight strikingthe cell generates electricity. Low efficiencies mean that larger arrays are needed, and higherinvestment costs. It should be noted that the first solar cells, built in the 1950s, had efficienciesof less than 4%.

Solar Water Pumps

In solar water pumping system, the pump is driven by motor run by solar electricity insteadof conventional electricity drawn from utility grid. A SPV water pumping system consists ofa photovoltaic array mounted on a stand and a motor-pump set compatible with the photo-voltaic array. It converts the solar energy into electricity, which is used for running the motorpump set. The pumping system draws water from the open well, bore well, stream, pond,canal etc



Figure 12.5 Photovoltaic Domestic and Streetlights

Case Example:

Under the Solar PhotovolaticWater Pumping Programmeof the Ministry of Non-conventional Energy Sourcesduring 2000-01 the PunjabEnergy Development Agency(PEDA) has completedinstallation of 500 solarpumps in Punjab for agricul-tural uses.

Under this project, 1800watt PV array was coupledwith a 2 HP DC motor pumpset. The system is capable ofdelivering about 140,000 litres water every day from a depth of about 6 – 7 metres. This quan-tity of water is considered adequate for irrigating about 5 – 8 acres land holding for most of thecrops. Refer Figure 12.6.

12.3 Wind Energy

Wind energy is basically harnessing of wind power toproduce electricity. The kinetic energy of the wind isconverted to electrical energy. When solar radiation entersthe earth's atmosphere, different regions of the atmosphereare heated to different degrees because of earth curvature.This heating is higher at the equator and lowest at the poles.Since air tends to flow from warmer to cooler regions, this causes what we call winds, and it is these airflows thatare harnessed in windmills and wind turbines to producepower.

Wind power is not a new development as this power, in theform of traditional windmills -for grinding corn, pumpingwater, sailing ships - have been used for centuries. Now windpower is harnessed to generate electricity in a larger scale withbetter technology.

Wind Energy Technology

The basic wind energy conversion device is the wind turbine. Although various designs andconfigurations exist, these turbines are generally grouped into two types:

1. Vertical-axis wind turbines, in which the axis of rotation is vertical with respect to theground (and roughly perpendicular to the wind stream),

2. Horizontal-axis turbines, in which the axis of rotation is horizontal with respect to theground (and roughly parallel to the wind stream.)



Figure 12.6 Photovoltaic Water Pumping

The Figure 12.7 illustrates the two types of turbines and typical subsystems for an electric-ity generation application. The subsystems include a blade or rotor, which converts the energyin the wind to rotational shaft energy; a drive train, usually including a gearbox and a genera-tor, a tower that supports the rotor and drive train, and other equipment, including controls, elec-trical cables, ground support equipment, and interconnection equipment.

Wind electric generators (WEG)

Wind electric generator converts kinetic energy available in wind to electrical energy by usingrotor, gear box and generator. There are a large number of manufacturers for wind electric gen-erators in India who have foreign collaboration with different manufacturers of Denmark,Germany, Netherlands, Belgium, USA, Austria, Sweden, Spain, and U.K. etc. At present,WEGs of rating ranging from 225 kW to 1000 kW are being installed in our country.

Evaluating Wind Mill Performance

Wind turbines are rated at a certain wind speed and annual energy output

Annual Energy Output = Power x Time

Example: For a 100 kW turbine producing 20 kW at an average wind speed of 25 km/h, the cal-culation would be:

100 kW x 0.20 (CF) = 20 kW x 8760 hours = 175,200 kWh

The Capacity Factor (CF) is simply the wind turbine's actual energy output for the yeardivided by the energy output if the machine operated at its rated power output for the entireyear. A reasonable capacity factor would be 0.25 to 0.30 and a very good capacity factor wouldbe around 0.40. It is important to select a site with good capacity factor, as economic viabilityof wind power projects is extremely sensitive to the capacity factor.

Wind Potential

In order for a wind energy system to be feasible there must be an adequate wind supply. A windenergy system usually requires an average annual wind speed of at least 15 km/h. The followingtable represents a guideline of different wind speeds and their potential in producing electricity.



Figure 12.7 Wind Turbine Configuration



Average Wind Speed Suitabilitykm/h (mph)

Up to 15 (9.5) No good

18 (11.25) Poor

22 (13.75) Moderate

25 (15.5) Good

29 (18) Excellent

A wind generator will produce lesser power in summer than in winter at the same windspeed as air has lower density in summer than in winter.

Similarly, a wind generator will produce lesser power in higher altitudes - as air pressure aswell as density is lower -than at lower altitudes.

The wind speed is the most important factor influencing the amount of energy a wind tur-bine can produce. Increasing wind velocity increases the amount of air passing the rotor, whichincreases the output of the wind system.

In order for a wind system to be effective, a relatively consistent wind flow is required.Obstructions such as trees or hills can interfere with the wind supply to the rotors. To avoidthis, rotors are placed on top of towers to take advantage of the strong winds available high above the ground. The towers are generally placed 100 metres away from the nearestobstacle. The middle of the rotor is placed 10 metres above any obstacle that is within 100 metres.

Wind Energy in India

India has been rated as one of the most promising countries for wind power development, withan estimated potential of 20,000 MW. Total installed capacity of wind electric generators in theworld as on Sept. 2001 is 23270 MW. Germany 8100 MW, Spain- 3175 MW, USA 4240 MW,Denmark 2417 MW, and India - 1426 MW top the list of countries. Thus, India ranks fifth inthe world in Wind power generation.

There are 39 wind potential stations in Tamil Nadu, 36 in Gujarat, 30 in Andhra Pradesh, 27in Maharashtra, 26 in Karnataka, 16 in Kerala, 8 in Lakshadweep, 8 Rajasthan, 7 in MadhyaPradesh, 7 in Orissa, 2 in West Bengal, 1 in Andaman Nicobar and 1 in Uttar Pradesh. Out of208 suitable stations 7 stations have shown wind power density more than 500 Watts/ m2.

Central Govt. Assistance and Incentives

The following financial and technical assistance are provided to promote,support and accelerate the development of wind energy in India:

Five years tax holiday 100% depreciation in the first year Facilities by SEB's for grid connection Energy banking and wheeling and energy buy back Industry status and capital subsidy Electricity tax exemptionSales tax exemption

Applications

• Utility interconnected wind turbines generate power which is synchronous with the grid andare used to reduce utility bills by displacing the utility power used in the household and byselling the excess power back to the electric company.

• Wind turbines for remote homes (off the grid) generate DC current for battery charging. • Wind turbines for remote water pumping generate 3 phase AC current suitable for driving

an electrical submersible pump directly. Wind turbines suitable for residential or villagescale wind power range from 500 Watts to 50 kilowatts.

12.4 Bio Energy

Biomass is a renewable energy resource derived from the car-bonaceous waste of various human and natural activities. It isderived from numerous sources, including the by-products fromthe wood industry, agricultural crops, raw material from the for-est, household wastes etc.

Biomass does not add carbon dioxide to the atmosphere as it absorbs the same amount ofcarbon in growing as it releases when consumed as a fuel. Its advantage is that it can be usedto generate electricity with the same equipment that is now being used for burning fossil fuels.Biomass is an important source of energy and the most important fuel worldwide after coal, oiland natural gas. Bio-energy, in the form of biogas, which is derived from biomass, is expectedto become one of the key energy resources for global sustainable development. Biomass offershigher energy efficiency through form of Biogas than by direct burning (see chart below).

Application



Biogas Plants

Biogas is a clean and efficient fuel, generated from cow-dung,human waste or any kind of biological materials derived throughanaerobic fermentation process. The biogas consists of 60%methane with rest mainly carbon-di-oxide. Biogas is a safe fuelfor cooking and lighting. By-product is usable as high-grademanure.

A typical biogas plant has the following components: A digester in which the slurry (dungmixed with water) is fermented, an inlet tank - for mixing the feed and letting it into thedigester, gas holder/dome in which the generated gas is collected, outlet tank to remove the

spent slurry, distribution pipeline(s) to transport the gas into the kitchen, and a manure pit,where the spent slurry is stored.

Biomass fuels account for about one-third of the total fuel used in the country. It is the mostimportant fuel used in over 90% of the rural households and about 15% of the urban households.Using only local resources, namely cattle waste and other organic wastes, energy and manure arederived. Thus the biogas plants are the cheap sources of energy in rural areas. The types of biogasplant designs popular are: floating drum type, fixed dome-type and bag-type portable digester.

Biomass Briquetting

The process of densifying loose agro-waste into asolidified biomass of high density, which can beconveniently used as a fuel, is called BiomassBriquetting (see Figure 12.8). Briquette is alsotermed as "Bio-coal". It is pollution free and eco-friendly. Some of the agricultural and forestryresidues can be briquetted after suitable pre-treat-ment. A list of commonly used biomass materialsthat can be briquetted are given below:

CornCob, JuteStick, Sawdust, PineNeedle,Bagasse, CoffeeSpent, Tamarind, CoffeeHusk,AlmondShell, Groundnutshells, CoirPith,BagaseePith, Barleystraw, Tobaccodust, RiceHusk, Deoiled Bran

Advantages

Some of advantages of biomass briquetting are high calorific value with low ash content,absence of polluting gases like sulphur, phosphorus fumes and fly ash- which eliminate the needfor pollution control equipment, complete combustion, ease of handling, transportation & stor-age - because of uniform size and convenient lengths.

Application

Biomass briquettes can replace almost all conventional fuels like coal, firewood and lignite inalmost all general applications like heating, steam generation etc. It can be used directly as fuelinstead of coal in the traditional chulhas and furnaces or in the gasifier. Gasifier converts solidfuel into a more convenient-to-use gaseous form of fuel called producer gas.

Biomass Gasifiers

Biomass gasifiers (see Figure 12.9) convert thesolid biomass (basically wood waste, agricul-tural residues etc.) into a combustible gas mix-ture normally called as producer gas. The con-version efficiency of the gasification process isin the range of 60%–70%. The producer gasconsists of mainly carbon-monoxide, hydro-gen, nitrogen gas and methane, and has a lowercalorific value (1000–1200 kcal/Nm3).



Figure 12.8 Biomass Briquetting

Figure 12.9 Biomass Gasifiers

Gasification of biomass and using it in place of conventional direct burning devices willresult in savings of atleast 50% in fuel consumption. The gas has been found suitable for com-bustion in the internal combustion engines for the production of power.

Applications:

Water pumping and Electricity generation: Using biomass gas, it possible to operate a dieselengine on dual fuel mode-part diesel and part biomass gas. Diesel substitution of the order of75 to 80% can be obtained at nominal loads. The mechanical energy thus derived can be usedeither for energizing a water pump set for irrigational purpose or for coupling with an alterna-tor for electrical power generation - 3.5 KW - 10 MW

Heat generation: A few of the devices, to which gasifier could be retrofitted, are dryers- fordrying tea, flower, spices, kilns for baking tiles or potteries, furnaces for melting non-ferrousmetals, boilers for process steam, etc.

Direct combustion of biomass has been recognized as an important route for generation ofpower by utilization of vast amounts of agricultural residues, agro-industrial residues and for-est wastes. Gasifiers can be used for power generation and available up to a capacity 500 kW.The Government of India through MNES and IREDA is implementing power-generating sys-tem based on biomass combustion as well as biomass gasification

High Efficiency Wood Burning Stoves

These stoves save more than 50% fuel wood consumption. They reduce drudgery of womensaving time in cooking and fuel collection and consequent health hazards. They also help in sav-ing firewood leading to conservation of forests. They also create employment opportunities forpeople in the rural areas.

Bio fuels

Unlike other renewable energy sources, biomass can be converteddirectly into liquid fuels— biofuels— for our transportation needs(cars, trucks, buses, airplanes, and trains). The two most commontypes of biofuels are ethanol and biodiesel. See Figure 12.10.

Ethanol is an alcohol, similar to that used in beer and wine. Itis made by fermenting any biomass high in carbohydrates (starch-es, sugars, or celluloses) through a process similar to brewing beer.Ethanol is mostly used as a fuel additive to cut down a vehicle'scarbon monoxide and other smog-causing emissions. Flexible-fuelvehicles, which run on mixtures of gasoline and up to 85%ethanol, are now available.

Biodiesel, produced by plants such as rapeseed (canola), sunflowers and soybeans, can beextracted and refined into fuel, which can be burned in diesel engines and buses. Biodiesel canalso made by combining alcohol with vegetable oil, or recycled cooking greases. It can be usedas an additive to reduce vehicle emissions (typically 20%) or in its pure form as a renewablealternative fuel for diesel engines.



Figure 12.10 BiodieselDriven Bus

Biopower

Biopower, or biomass power, is the use of biomass to generate electricity. There are sixmajor types of biopower systems: direct-fired, cofiring, gasification, anaerobic digestion,pyrolysis, and small - modular.

Most of the biopower plants in the world use direct-fired systems. They burn bioenergyfeedstocks directly in boiler to produce steam. This steam drives the turbo-generator. In someindustries, the steam is also used in manufacturing processes or to heat buildings. These areknown as combined heat and power facilities. For example, wood waste is often used to pro-duce both electricity and steam at paper mills.

Many coal-fired power plants use cofiring systems to significantly reduce emissions, espe-cially sulfur dioxide emissions. Cofiring involves using bio energy feedstock as a supplemen-tary fuel source in high efficiency boilers.

Gasification systems use high temperatures and an oxygen-starved environment to convertbiomass into a gas (a mixture of hydrogen, carbon monoxide, and methane). The gas fuels a gasturbine, which runs an electric generator for producing power.

The decay of biomass produces methane gas, which can be used as an energy source.Methane can be produced from biomass through a process called anaerobic digestion.Anaerobic digestion involves using bacteria to decompose organic matter in the absence of oxy-gen. In landfills -scientific waste disposal site - wells can be drilled to release the methane fromthe decaying organic matter. The pipes from each well carry the gas to a central point where itis filtered and cleaned before burning. Methane can be used as an energy source in many ways.Most facilities burn it in a boiler to produce steam for electricity generation or for industrialprocesses. Two new ways include the use of microturbines and fuel cells. Microturbines haveoutputs of 25 to 500 kilowatts. About the size of a refrigerator, they can be used where there arespace limitations for power production. Methane can also be used as the "fuel" in a fuel cell.Fuel cells work much like batteries, but never need recharging, producing electricity as long asthere is fuel.

In addition to gas, liquid fuels can be produced from biomass through a process calledpyrolysis. Pyrolysis occurs when biomass is heated in the absence of oxygen. The biomass thenturns into liquid called pyrolysis oil, which can be burned like petroleum to generate electrici-ty. A biopower system that uses pyrolysis oil is being commercialized.

Several biopower technologies can be used in small, modular systems. A small, modularsystem generates electricity at a capacity of 5 megawatts or less. This system is designed for use at the small town level or even at the consumer level. For example, some farmers use the waste from their livestock to provide their farms with electricity. Not only do these systems provide renewable energy, they also help farmers meet environmentalregulations.

Biomass Cogeneration

Cogeneration improves viability and profitability of sugar industries. Indian sugar mills arerapidly turning to bagasse, the leftover of cane after it is crushed and its juice extracted, to gen-erate electricity. This is mainly being done to clean up the environment, cut down power costsand earn additional revenue. According to current estimates, about 3500 MW of power can begenerated from bagasse in the existing 430 sugar mills in the country. Around 270 MW ofpower has already been commissioned and more is under construction.



12.5 Hydro Energy

The potential energy of fallingwater, captured and converted tomechanical energy by water-wheels, powered the start of theindustrial revolution.

Wherever sufficient head, orchange in elevation, could befound, rivers and streams weredammed and mills were built.Water under pressure flowsthrough a turbine causing it tospin. The Turbine is connected toa generator, which produceselectricity (see Figure 12.11). Inorder to produce enough elec-tricity, a hydroelectric systemrequires a location with the fol-lowing features:

Change in elevation or head: 20 feet @ 100 gal/min = 200 Watts.100 feet head @ 20 gal/min gives the same output.

In India the potential of small hydro power is estimated about 10,000 MW. A total of 183.45MW small Hydro project have been installed in India by the end of March 1999. Small HydroPower projects of 3 MW capacity have been also installed individually and 148 MW project isunder construction.

Small Hydro

Small Hydro Power is a reliable, mature and proven technology. Itis non-polluting, and does not involve setting up of large dams orproblems of deforestation, submergence and rehabilitation. Indiahas an estimated potential of 10,000 MW

Micro Hydel

Hilly regions of India, particularly the Himalayan belts, are endowed withrich hydel resources with tremendous potential. The MNES has launcheda promotional scheme for portable micro hydel sets for these areas. Thesesets are small, compact and light weight. They have almost zero mainte-nance cost and can provide electricity/power to small cluster of villages.They are ideal substitutes for diesel sets run in those areas at high genera-tion cost.

Micro (upto 100kW) mini hydro (101-1000 kW) schemes canprovide power for farms, hotels, schools and rural communities, and help create local industry.



Figure 12.11 Hydro Power Plant

12.6 Tidal and Ocean Energy

Tidal Energy

Tidal electricity generation involves the construction of a bar-rage across an estuary to block the incoming and outgoing tide.The head of water is then used to drive turbines to generateelectricity from the elevated water in the basin as in hydro-electric dams.

Barrages can be designed to generate electricity on the ebbside, or flood side, or both. Tidal range may vary over a widerange (4.5-12.4 m) from site to site. A tidal range of at least 7m is required for economical operation and for sufficient headof water for the turbines.

Ocean Energy

Oceans cover more than 70% of Earth's surface, making them the world's largest solar collec-tors. Ocean energy draws on the energy of ocean waves, tides, or on the thermal energy (heat)stored in the ocean. The sun warms the surface water a lot more than the deep ocean water, andthis temperature difference stores thermal energy.

The ocean contains two types of energy: thermal energy from the sun's heat, and mechan-ical energy from the tides and waves.

Ocean thermal energy is used for many applications, including electricity generation. Thereare three types of electricity conversion systems: closed-cycle, open cycle, and hybrid. Closedcycle systems use the ocean's warm surface water to vaporize a working fluid, which has a lowboiling point, such as ammonia. The vapour expands and turns a turbine. The turbine then acti-vates a generator to produce electricity. Open-cycle systems actually boil the seawater by oper-ating at low pressures. This produces steam that passes through a turbine / generator. The hybridsystems combine both closed-cycle and open-cycle systems.

Ocean mechanical energy is quite different from ocean thermal energy. Even though the sun affects all ocean activity, tides are driven primarily by the gravitational pull of themoon, and waves are driven primarily by the winds. A barrage (dam) is typically used to convert tidal energy into electricity by forcing the water through turbines, activating agenerator.

India has the World's largest programmes for renewable energy. Several renewable energytechnologies have been developed and deployed in villages and cities of India. A Ministry ofNon-Conventional Energy Sources (MNES) created in 1992 for all matters relating to Non-Conventional / Renewable Energy. Government of India also created Renewable EnergyDevelopment Agency Limited (IREDA) to assist and provide financial assistance in the formof subsidy and low interest loan for renewable energy projects.

IREDA covers a wide spectrum of financing activities including those that are connected toenergy conservation and energy efficiency. At present, IREDA's lending is mainly in the fol-lowing areas: -



• Solar energy technologies, utilization of solar thermal and solar photo voltaic systems• Wind energy setting up grid connected Wind farm projects• Small hydro setting up small, mini and micro hydel projects• Bio-energy technologies, biomass based co-generation projects, biomass gasification, ener-

gy from waste and briquetting projects• Hybrid systems• Energy efficiency and conservation

The estimated potential of various Renewable Energy technologies in India by IREDA aregiven below.

Energy source estimated potential

Solar Energy 20 MW / sq. kmWind Energy 20,000 MWSmall Hydro 10,000 MWOcean Thermal Power 50,000 MWSea Wave Power 20,000 MWTidal Power 10,000 MWBio energy 17,000 MWDraught Animal Power 30,000 MWEnergy from MSW 1,000 MWBiogas Plants 12 Million PlantsImproved Wood Burning Stoves 120 Million StovesBagasse-based cogeneration 3500 MW

Cumulative achievements in renewable energy sector (As on 31.03.2000)

Sources / Technologies Unit Upto31.03.2000Wind Power MW 1167Small Hydro MW 217Biomass Power & Co-generation MW 222Solar PV Power MW / Sq. km 42Urban & MSW MW 15.21Solar Heater m2. Area 480000Solar Cookers No. 481112Biogas Plants Nos. in Million 2.95Biomas Gasifier MW 34Improved Chulhas Nos. in Million 31.9





QUESTIONS

1 What do you mean by renewable energy

2 Why is solar energy potential high in India?

3. Explain working of solar water heater?

4. List few applications of low temperature water heaters in domestic and industrial use

5. What are the two methods by which energy can be recovered from solar radiation

6. How can the performance of solar collectors be improved?

7. Explain any two applications of concentrated solar energy?

8. What do you mean by photovoltaic?

9. Explain the terms cell, module and array as applicable to photovoltaic.

10. What are the typical applications of photovoltaic power?

11. Name the few states with high wind energy potential in India.

12. What are the criteria for selection of wind mill installation?

13. What ere the incentives available for wind mill installation?

14. Explain the bio-energy potential in India and its applications.

15. What are the various methods by which power can be generated from biomass?

16. What is the role of IREDA in renewable energy sector

17. India has recorded good growth in wind energy sector. Do you agree? What are thefactors responsible for such a high growth?

REFERENCES 1. Alternate Energy Sources by T H Taylor.Adam Hilger Ltd, Bristol2. Renewable Energy Sources for rural areas in Asia and Pacific, APO, Tokyo, 20003. www.ireda.org4. www.windenergy.com

13. WASTE MINIMISATION AND RESOURCECONSERVATION


13.1 Introduction

Traditionally, waste is viewed as an unnecessary element arising from the activities of anyindustry. In reality, waste is a misplaced resource, existing at a wrong place at a wrong time.

Waste is also the inefficient use of utilities such as electricity, water, and fuel, which areoften considered unavoidable overheads. The costs of these wastes are generally underestimat-ed by managers. It is important to realise that the cost of waste is not only the cost of waste dis-posal, but also other costs such as:

� Disposal cost� Inefficient energy use cost� Purchase cost of wasted raw material� Production cost for the waste material� Management time spent on waste material� Lost revenue for what could have been a product instead of waste � Potential liabilities due to waste.

What is waste minimisation?

Waste minimisation can be defined as "systematically reducing waste at source". It means:

• Prevention and/or reduction of waste generated• Efficient use of raw materials and packaging • Efficient use of fuel, electricity and water• Improving the quality of waste generated to facilitate recycling and/or reduce hazard• Encouraging re-use, recycling and recovery.

Waste minimisation is also known by other terms such as waste reduction, pollution preven-tion, source reduction and cleaner technology. It makes use of managerial and/or technical inter-ventions to make industrial operations inherently pollution free

It should be also clearly understood that waste minimization, however attractive, is not apanacea for all environmental problems and may have to be supported by conventional treat-ment/disposal solutions.

Waste minimization is best practiced by reducing the generation of waste at the source itself.After exhausting the source reduction opportunities, attempts should be made to recycle the

TABLE 13.1 WASTES AND POSSIBLE RESOURCES

Wastes Resources

Fly ash from power plant Raw material for cement or brickmanufacture

Bagasse wastes from sugar manufacture Fuel for boiler

CO2 release from ammonia plant Raw material for Urea manufacture

waste within the unit. Finally, modification or reformulation of products so as to manufactureit with least waste generation should be considered. Few wastes and possible resources are indi-cated in the Table 13.1

13.2 Classification of Waste Minimization (WM) Techniques

The waste minimization is based on different techniques. These techniques are classified ashereunder.

13. Waste Minimisation and Resource Conservation


Source Reduction

Under this category, four techniques of WM are briefly discussed below:

a) Good Housekeeping- Systems to prevent leakages & spillages through preventive main-tenance schedules and routine equipment inspections. Also, well-written working instructions,supervision, awareness and regular training of workforce would facilitate good housekeeping.

b) Process Change: Under this head, four CP techniques are covered:

(i) Input Material Change - Substitution of input materials by eco-friendly (non-toxic or less toxic than existing and renewable) material preferably having longerservice time.

(ii) Better Process Control - Modifications of the working procedures, machine-oper-ating instructions and process record keeping in order to run the processes at higherefficiency and with lower waste generation and emissions.

(iii) Equipment Modification - Modification of existing production equipment andutilities, for instance, by the addition of measuring and controlling devices, in orderto run the processes at higher efficiency and lower waste and emission generationrates.

(iv) Technology change - Replacement of the technology, processing sequenceand/or synthesis route, in order to minimise waste and emission generation duringproduction.

c) Recycling

i) On-site Recovery and Reuse - Reuse of wasted materials in the same process orfor another useful application within the industry.

ii) Production of Useful by-product - Modification of the waste generation processin order to transform the wasted material into a material that can be reused or recy-cled for another application within or outside the company.

d) Product Modification

Characteristics of the product can be modified to minimise the environmental impactsof its production or those of the product itself during or after its use (disposal).

13.3 Waste Minimization Methodology

For an effective Waste Minimization programme, it is essential to bring together variousgroups in the industry to ensure implementation. How formalised the programme wouldbe depends upon the size and composition of the industry and its waste and emission prob-lems. The programme should be flexible enough so that it can adapt itself to changing cir-cumstances. A methodical step-by-step procedure ensures exploitation of maximum wasteminimization opportunities. The steps in a typical waste minimization progamme areillustrated below:



Step 1: Getting Started

Form a Waste Minimization Team

Waste Minimization (WM) concept encompasses all departments and sections in an industry.Activities involved in WM audits require assistance and cooperation not only from the person-nel belonging to concerned department, but also from other related departments. Hence, making



an inter-disciplinary and inter-departmental team is a prerequisite for successful conduct of aWM audit. In special cases, it would be advantageous to have external experts in order to havean unbiased, professional approach.

List Process Steps / Unit Operations

The WM team should familiarize itself with the manufacturing processes including utilities,waste treatment and disposal facilities, and list all the process steps. Preparation of sketches ofprocess layout drainage system, vents and material-loss areas would be useful. This helps inestablishing cause-effect relationships and ensuring that important areas are not overlooked.Periodic, intermittent and continuous discharge streams should be appropriately labeled.

Identify and Select Wasteful Process Steps

In multi-process type industries, it may be difficult to start detailed Waste minimization activi-ties covering the complete unit. In such cases, it is advisable to start with fewer process stepsto begin with. The selected step(s) could be the most wasteful and / or one with very high wasteminimization potential.

This activity could also be considered a preliminary prioritization activity. All the variouswasteful steps identified in 1.2 should be broadly assessed in terms of volume of waste, sever-ity of impact on the environment, Waste minimization opportunities, estimated benefits (spe-cially cost savings), cost of implementation etc. Such assessment would help in focusing on theprocess steps / areas for detailed analysis.

Step 2: Analysing Process Steps

Prepare Process Flow Charts

This activity follows the activity described at 1.2. Flow charts are diagrammatic / schemat-ic representation of production, with the purpose of identifying process steps and thesource of waste streams and emissions. A flow chart should list, and characterize the inputand output streams, as well as recycle streams. Even the so called free or less costly inputslike water, air, sand, etc should be taken into account as these often end up in being themajor cause of wastes. Wherever required, the process flow diagram should be supple-mented with chemical equations to facilitate understanding of the process. Also the mate-rials which are used occasionally and / or which do not appear in output streams (for exam-ple catalysts, coolant oil) should be specified. The periodic / batch / continuous stepsshould also be appropriately highlighted. Preparation of a detailed and correct process flowdiagram is a key step in the entire analysis and forms the basis for compilation of materi-als and energy balance.

Make material and Energy Balances

Material and Energy balances are important for any Waste minimization programme since theymake it possible to identify and quantify, previously unknown losses or emissions. These bal-ances are also useful for monitoring the progress achieved in a prevention programme, andevaluating the costs and benefits. Typical components of a material balance and energy balanceare given below (see Figures 13.1 & 13.2):



It is not advisable to spent more time and money to make a perfect material balance. Evena rough / preliminary material balance throws open Waste Minimization opportunities whichcan be profitably exploited.

On the other hand, the precision of analytical data and flow measurements is important asit is not possible to obtain a reliable estimate of the waste stream by subtracting the materialsin the product from those in the raw materials. In such cases, a direct monitoring and analysisof waste streams should be carried out.

Assign Costs To Waste Stream

In order to assess the profit potential of waste streams, a basic requirement would be to assigncosts to them. This cost essentially reflects the monetary loss arising from waste. Apparently, awaste stream does not appear to have any quantifiable cost attached to it, except where directraw material / product loss is associated with it. However, a deeper analysis would show sev-eral direct and indirect cost components associated with waste streams such as:

� Cost of raw materials in waste.� Manufacturing cost of material in waste� Cost of product in waste� Cost of treatment of waste to comply with regulatory requirements� Cost of waste disposal� Cost of waste transportation



� Cost of maintaining required work environment� Cost due to waste cess.

Based on this, for each waste stream, total cost per unit of waste (Rs/KL or Rs/Kg)should be worked out. This figure would be useful in working out the feasibility of the wasteminimization measures. The result can also be used to categorize the waste streams forpriority action.

Review of Process

Through the material and energy balances, it is possible to carry out a cause analysis to locateand pinpoint the causes of waste generation. These causes would subsequently become the toolsfor evolving Waste Minisation measures. There could be a wide variety of causes for wastegeneration ranging from simple lapses of housekeeping to complex technological reasons asindicated below.

Typical Causes Of Waste

Poor housekeeping ;

� Leaking taps / valves / flanges� Spillages� Overflowing tanks� Worn out material transfer belts



Operational and maintenance negligence

� Unchecked water / air consumption� Unnecessary running of equipment� Sub optimal loading� Lack of preventive maintenance� Sub-optimal maintenance of process condition� Ritualistic operation

Poor raw material quality

� Use of substandard cheap raw material� Lack of quality specification� Improper purchase management system� Improper storage

Poor process / equipment design

� Mismatched capacity of equipment� Wrong material selection� Maintenance prone design� Adoption of avoidable process steps� Lack of information / design capability

Poor layout

� Unplanned / adhoc expansion� Poor space utilization plan� Bad material movement plan

Bad technology

� Continuation of obsolete technology� Despite product / raw material change� High cost of better technology� Lack of availability of trained manpower� Small plant size� Lack of information

Inadequately trained personnel

� Increased dependence on casual / contract labour� Lack of formalized training system� Lack of training facilities� Job insecurity� Fear of losing trade secrets� Lack of availability of personnel� Understaffing hence work over pressure



Employee Demotivation

� Lack of recognition� Absence of reward � Emphasis only on production, not on people� Lack of commitment and attention by top management

Step 3: Generating Waste Minimization Opportunities

Develop Waste Minimization Opportunities

Once the origin and causes of waste and emissions are known, the process enters the creativephase. The WM team should now start looking for possible opportunities for reducing waste.Finding potential options depends on the knowledge and creativity of your team members. Thepotential sources of help in finding Waste Minimization Opportunities are:

� Other personnel from the same or similar plant elsewhere� Trade associations� Consultancy organizations� Equipment suppliers� Consultants

The process of finding Waste Minimization opportunities should take place in anenvironment, which stimulates creativity and independent thinking. It would be benefi-cial to move away from the routine working environment for better results. Variousanalysis tools and techniques like "brainstorming", "group discussions" etc would be use-ful in this step.

Select Workable Opportunities

The Waste Minimization opportunities developed should be screened and those, which areimpractical, should be discarded.

The discarding process should be simple, and straightforward and may often be onlyqualitative. There should be no ambiguity or bias. Unnecessary effort in doing detailed fea-sibility analysis of opportunities, which are impractical or non-feasible, should be avoided.The remaining Waste Minimization opportunities are then subjected to a more detailedfeasibility analysis.

Step 4: Selecting Waste Minimization Solutions

The selection of a Waste Minimization solution for implementation requires that it should notonly be techno-economically viable, but also environmentally desirable.

Assess Technical Feasibility

The technical evaluation determines whether a proposed Waste Minimization option will workfor the specific application. The impact of the proposed measure on product production rateshould be evaluated first. In case of significant deviation from the present process practices, lab-oratory testing trial runs might be required to assess the technical feasibility.



A typical checklist for technical evaluation could be as follows:

� Availability of equipment� Availability of operating skills� Space availability� Effect on production� Effect on product quality� Safety aspects� Maintenance requirements� Effect on operational flexibility� Shut down requirements for implementation

Assess Economic Viability

Economic viability would often be the key parameter to promote or discourage WasteMinimization. For a smooth take-off, it is therefore essential that the first few WasteMinimization measures should be economically very attractive. Such a strategy helps in creat-ing more interest and sustains commitment.

Options requiring little investment, but involving more of procedural changes (housekeep-ing measures, measures, operational improvements) do not need an intensive economic analy-sis and simple methods like pay back period could be used. However as Waste Minimizationmeasures become more involved and capital intensive, other profitability analysis methods viz.Return on Investment (IRR) or Net Present Value (NPV) are recommended to get the totalpicture.

While doing the economic investment, the costs may include fixed capital i.e. cost of equip-ment, working capital cost, shutdown cost, O & M costs etc. The savings could be direct sav-ings of input materials / energy, increased production levels and hence lower specific input cost,lower O & M cost, value of by products, reduction in environmental cost i.e. waste treatmenttransportation and disposal cost etc.

Evaluate Environmental Aspects

The options for Waste Minimization with respect to their impact on the environmentshould be assessed. In many cases the environmental advantage will be obvious if thereis a net reduction in the toxicity and / or quantity of waste. Other impacts could bechanges in treatment of the wastes. In the initial stages, environmental aspects may notappear to be as compelling as economic aspects. In future as in developed countries,environmental aspects would become the most important criteria irrespective of the eco-nomic viability.

Select Solutions for Implementation

After technical, economic and environmental assessment, Waste Minimization measures shouldbe selected for implementation. Understandably, the most attractive ones would be those hav-ing best financial benefits, provided technical feasibility is favourable. However, in a growingnumber of cases especially when active pressure groups are present, environmental prioritiesmay become the final deciding factor.



The work done so far should be documented. Apart from becoming a reference documentfor seeking approvals in implementation, the document would also be useful in obtainingfinances from external finance institutions, reporting status to other agencies, and establishingbase levels for performance evaluation and review.

Step 5: Implementing Waste Minimization Solutions

Prepare for Implementation

The selected solutions could be taken for implementation. Apart from simple housekeepingmeasures several others would require a systematic plan of implementation.

The Waste Minimization team should be well prepared to take up the job of implementa-tion. The preparation would include arranging finances, establishing linkages in case of multi-department solutions, technical preparations, etc.

Implement Solutions

The task comprises layout and drawing preparation equipment fabrication / procurement, trans-portation to site, installation and commissioning. Whenever required, simultaneous training ofmanpower should be taken up as many excellent measures have failed miserably because ofnon-availability of adequately trained people.

Monitor and Evaluate Results

The WM solutions should be monitored for performance. The results obtained should bematched with those estimated / worked out during technical evaluation to establish causes fordeviation, if any. The implementation job is considered to be over, only after successful com-missioning and sustained stable performance over a reasonable length of time.

Step 6: Sustain Waste Minimization

The biggest challenges in Waste Minimization lies in sustaining Waste Minimization. Theenthusiasm of the Waste Minimization team wanes off with time. Such tragic ends should beavoided. Backing out from commitment, predominance of production at any cost, absence ofrewards and appreciation, and shifting of priorities are some of the commonly encountered rea-sons, which one should check and avoid.

Also monitoring and review of the implemented measures should be communicated to allemployees in the industry so that it fans the desires for minimizing wastes. Involvement of aslarge a number of employees as possible and rewarding the deserving ones, will help long termsustenance of Waste Minimization.

Having implemented Waste Minimization solutions in the area under study, the WasteMinimization team should go back to Step-2 i.e. analysing the process steps and identifyingand selecting the next wasteful area. In this way, the cycle continues, till all the steps areexhausted.

In a nutshell, a philosophy of minimizing waste must be developed within the compa-ny. This means that Waste Minimization should become an integral part of company'sactivities. All successful Waste Minimization programmes, till date, have been founded onthis philosophy.



13.4 Case Study

Maximising Cullet Recovery Reduces Batch Costs

At a Lead Crystal Glass Works, glass was produced by melting a charge of blended raw mate-rials together with process cullet. However, only about 30% of the cullet produced at the glass-works was of a size appropriate for remelting. Concern about the significant amounts of valu-able raw materials lost in this cullet and being sent for waste disposal, led to the installation ofa crushing unit to reduce the cullet to an optimum size for recovery. Operation of the crushingunit was subsequently enhanced by the addition of a vibrating screen and cullet washingsystem.

The ideal size for cullet pieces, to produce a high quality melt of uniform composition andavoid faults in the blown glass, is 12 – 20 mm. Most of the heavy cullet at the company waspresent in large pieces that cannot be easily broken up manually to the optimum size. Lighterpieces such as those from wine glasses, were also difficult to recycle because they have to bebroken up manually to obtain a satisfactory charge weight. This generates a lot of fine materi-al, which was unsuitable because it tends to result in air bubbles being trapped in the glass gath-ered from the furnace pot by the glass blower.

Before waste minimization programme, about 560 tonnes of cullet were disposed for wastedisposal each year, costing the company considerably in terms of lost raw materials. The com-pany therefore decided to install a crushing plant capable of producing a consistent output of asize suitable for remelting and with minimum generation of fine material. Such a plant wouldallow more cullet to be recycled, leading to a reduction in the cost of both primary raw materi-als and cullet disposal.

Following discussions with suppliers of crushing plant, the company installed a rotary ham-mer mill. This resulted in recycling of 74% of process cullet as against 30% previously. Alsoalternative uses avoiding waste disposal have been found for the crusher fines and other formsof waste glass. Crushing has also increased the bulk density of the cullet by a factor of three andimproved its size distribution.

The benefits of maximising inhouse cullet recovery include:

• Cost savings• Reduction of 37% in the purchase of primary raw materials• Improved yield of first quality glass• Payback period of three weeks

Associated Waste Minimization Measures

In addition to installing the cullet crusher, the company had initiated a number of other associatedwaste minimization measures such as segregation by source of inhouse cullet, segregating stonesfrom cullet, lead recovery from reject cullet, crusher fines and waste glass prior to disposal.





QUESTIONS

1. Explain the concept of waste minimization with suitable examples.

2. "Waste is a misplaced resource" Explain.

3. What are the 3R's in waste minimization techniques?

4. Which would you prefer between recycling and source reduction? Justify.

5. List down few housekeeping measures by which wastes can be reduced.

6. Explain how modifying a product can help minimize the wastes with few examples.

7. For a coal-fired boiler, draw a block diagram and indicate various material and ener-gy inputs, outputs and wastes.

8. Can employee be a factor in reducing wastes? Explain.

REFERENCES 1. From Waste to Profits, Guidelines for Waste Minimization by National Productivity

Council, New Delhi2. Waste Minimization Manual for Textile Processing by National Productivity Council,

Chennai.

35894243 Energy Efficiency Assessment Book

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