11 Energy Conservation

8/9/2019 11 Energy Conservation

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11. ENERGY CONSERVATION


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Energy conservation is the practice of decreasing

the quantity of energy used. It may be achieved

through efficient energy use, in which case energy

use is decreased while achieving a similar outcome,

or by reduced consumption of energy services. Individuals and organizations that are direct

consumers of energy may want to conserve energy

in order to reduce energy costs and promote

economic security. Industrial and commercial users may want to

increase efficiency and thus maximize profit.

WHAT IS ENERGY CONSERVATION?


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Limited fuels available on earth (most of fuels

will be exhausted and we will have to switch

over to alternative sources of energy)

Ever increasing demand of energy

Cost of energy has increased substantially

The product cost has a bearing of energy cost

Energy efficient technology options available

Reduction in green house gases emissions

Over all environment friendly option

NECESSITY OF ENERGY CONSERVATION


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Indian government has passed Energy conservation

act, 2001 for better utilization of energy and itsconservation

25,000 MW capacity creation through energy

efficiency in electricity sector alone has beenestimated in India

23% energy conservation potential

Industrial and agriculture sectors have the

maximum potential

Assessment of potential savings in different sectors

Bureau of Energy Efficiency (BEE) created

ENERGY CONSERVATION IN INDIA


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BEE is an agency of the Government of India,

under the Ministry of Power set up in March2002 under the provisions of the nation's

2001 Energy Conservation Act.

The function of agency is to developprograms which promotes the conservation

and efficient use of energy in India.

The government has proposed to make itmandatory for all appliances in India to have

ratings by the BEE starting in January 2010.

BUREAU OF ENERGY EFFICIENCY (BEE)


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ENERGY CONSERVATION BUILDING CODE

(ECBC)

Mandatory scope covers commercial buildings Applies to new construction only

Building components included

Building Envelope (Walls, Roofs, Windows)

Lighting (Indoor and Outdoor)

Heating Ventilation and Air Conditioning(HVAC) System

Solar Water Heating and Pumping

Electrical Systems (Power Factor,Transformers)


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ENERGY CONSERVATION IN BUILDINGS

Current best practices in building design and

construction result in homes that are

profoundly more energy conserving than

average new homes like

Passive house Super-insulation

Smart ways to construct homes, such that

minimal resources are used for cooling andheating the house in summer and winter

respectively, can significantly reduce energy

costs.


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ENERGY EFFICIENT BUILDING

CONSTRUCTION

Very good insulation of walls, roofsand Basement.

Windows with high quality double

or triple glazing. Air-tight construction to be

checked by blower door test

Avoid cooling demand Sun shading in summer

Natural cooling sources


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CONSERVATION OF ENERGY IN LIGHTING

SYSTEM

Selection of Lighting in Buildings

Two methods are available:

Building area method

Space by space method


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BUILDING AREA METHOD

Interior lighting power allowance (W) by the building

area method is determined in accordance with thefollowing:

a) Determine the allowed lighting power density (LPD)

from Table 1 for each appropriate building area type.

b) Calculate the gross lighted floor area (GLFA) for eachbuilding area type.

c) The interior lighting power allowance (ILPA) is the

sum of the products of the gross lighted floor area of

each building area times the allowed lighting powerdensity for that building area types i.e.

ILPA = (GLFA x LPD)


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TABLE 1: INTERIOR LIGHTING POWER FOR BUILDING

AREA METHOD

Building Area Type LPD (W/m2) Building Area Type LPD (W/m2)

Automotive Facility 9.7 Multifamily Residential 7.5

Convention Center 12.9 Museum 11.8

Dining: Bar Lounge/ Leisure 14.0 Office 10.8

Dining: Cafeteria/ Fast Food 15.1 Parking Garage 3.2

Dining: Family 17.2 Performing Arts Theater 17.2Dormitory/Hostel 10.8 Police/Fire Station 10.8

Gymnasium 11.8 Post Office/ Town Hall 11.8

Healthcare-Clinic 10.8 Religious Building 14.0

Hospital/ Health Care 12.9 Retail/ Mall 16.1

Hotel 10.8 School/ University 12.9

Library 14.0 Sports Arena 11.8

Manufacturing Facility 14.0 Transportation 10.8

Motel 10.8 Warehouse 8.6

Motion Picture Theater 12.9 Workshop 15.1


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SPACE BY SPACE METHOD

Determination of interior lighting power allowance (Watts)

by the space by space method shall be in accordance withthe following:

a) Determine the appropriate building type from Table 2

and the allowed lighting power density.

b) For each space enclosed by partitions 80% or greater ofceiling height, determine the gross interior floor area by

measuring to the center of the partition wall.

c) The interior lighting power allowance is the sum of the

lighting power allowances for all spaces. The lightingpower allowance for a space is the product of the gross

lighted floor area of the space times the allowed lighting

power density for that space.


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TABLE 2: INTERIOR LIGHTING POWER FOR SPACE BY SPACE METHOD

Space Function LPD (W/m2) Space Function LPD (W/m2)

Office-enclosed 11.8 Library 11.8

Office-open plan 11.8 Card File & Cataloging 11.8

Conference/Meeting/Multipurpose 14.0 Stacks 18.3

Classroom/Lecture/Training 15.1 Reading Area 12.9

Lobby 14.0 Hospital

For Hotel 11.8 Emergency 29.1

For Performing Arts Theater 35.5 Recovery 8.6

For Motion Picture Theater 11.8 Nurse Station 10.8

Audience/Seating Area 9.7 Exam Treatment 16.1

For Gymnasium 4.3 Pharmacy 12.9

Patient Room 7.5

For Convention Center 7.5 Operating Room 23.7

For Religious Buildings 18.3 Nursery 6.5

For Sports Arena 4.3 Medical Supply 15.1

For Performing Arts Theater 28.0 Physical Therapy 9.7

For Motion Picture Theater 12.9 Radiology 4.3

For Transportation 5.4 Laundry Washing 6.5

Atrium-first three floors 6.5 Automotive Service Repair 7.5

Atrium-each additional floor 2.2 Manufacturing

Lounge/Recreation 12.9 Low Bay (


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Space Function LPD (W/m2) Space Function LPD (W/m2)

For Hospital 8.6 High Bay (>8m ceiling) 18.3

Dining Area 9.7 Detailed Manufacturing 22.6

For Hotel 14.0 Equipment Room 12.9

For Motel 12.9 Control Room 5.4

For Bar Lounge/Leisure Dining 15.1 Hotel/Motel Guest Rooms 11.8

For Family Dining 22.6 Dormitory Living Quarters 11.8

Food Preparation 12.9

Laboratory 15.1 General Exhibition 10.8

Restrooms 9.7 Restoration 18.3

Dressing/Locker/Fitting Room 6.5 Bank Office Banking Activity Area 16.1

Corridor/Transition 5.4

For Hospital 10.8 Sales Area 18.3

For Manufacturing Facility 5.4 Mall Concourse 18.3

Stairs-active 6.5 Sports Arena

Active Storage 8.6 Ring Sports Area 29.1

For Hospital 9.7 Court Sports Area 24.8

Inactive Storage 3.2 Indoor Field Area 15.1

For Museum 8.6

Electrical/Mechanical 16.1 Fine Material Storage 15.1

Workshop 20.5 Medium/Bulky Material Storage 9.7

Sleeping Quarters 3.2 Parking Garage Garage Area 2.2

Convention Center Exhibit Space 14.0

Airport Concourse 6.5

Air/Train/Bus Baggage Area 10.8

Ticket Counter 16.1

TABLE 2: INTERIOR LIGHTING POWER FOR SPACE BY SPACE METHOD


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ENERGY CONSERVATION IN INDUSTRIES

THROUGH MOTOR

Electric motors converts electrical energy intomechanical energy.

Energy can be saved by using Energy Efficient

Motor (EEM) in place of standard motor.

An EEM produces the same shaft output

power (HP) but uses less input power (kW)

than a standard motor.

C S S O C O


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CASE STUDY ON FINANCIAL EVALUATION

OF EEM

A 60 HP standard AC motor operating at 75% load.At this load efficiency of motor is 82%. The motor

power factor is 0.8. Motor operates 20 hrs a day and

300 days in a year. Per unit energy charge is Rs. 3.10

and per month per kVA demand charge is Rs. 175.Cost of 60 HP standard AC motor is Rs. 1,00,000/-

(Take 1 HP = 750 W)

Rating of standard AC motor

= 60 x 0.75= 45 kW

Input power to motor

= 45 x 0.75 / 0.82 = 41.16 kW



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OF EEM (contd)

kVA demand= 41.16 / 0.8 = 51.45 kVA

kVA charges/year

= 51.45 x 175 x 12 = Rs. 1,08,045/-

Energy (kWh) charges/year

= 41.16 x 3.1 x 20 x 300

= Rs. 7,65,576/-

Total (Demand + Energy) Cost/year

= 1,08,045 + 7,65,576 = Rs. 8,73,621/-



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OF EEM (contd)

A 60 HP Energy Efficient Motor is operating a75% load. At this load, efficiency of the motor

is 87% and the power factor is 0.83. Cost of

60 HP EEM is Rs. 1,20,000/-

Rating of standard AC motor

= 60 x 0.75= 45 kW

Input power to motor

= 45 x 0.75 / 0.87 = 38.79 kW


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OF EEM (contd)

kVA demand= 38.79 / 0.83 = 46.73 kVA

kVA charges/year

= 46.73 x 175 x 12 = Rs. 98,133/-

Energy (kWh) charges/year

= 38.79 x 3.10 x 20 x 300 = Rs. 7,21,494/-

Total (Energy + Demand) charges/year

= 7,21,494 98,133 = Rs. 8,19,627/-


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FINANCIAL EVALUATION

Energy savings achieved by using E.E.M. over

standard motor= 8,73,621 8,19,627 = Rs. 53,994/-

Payback period for replacement of existing standard

motor with E.E.M.= 1,20,000 / 53,994 = 2.3 years

Payback period for purchase of E.E.M. for new

installation= 20,000 / 53,994 = 0.37 yr = 4.5 months

TIPS FOR ENERGY CONSERVATION IN


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DOMESTIC SECTOR

Organized cooking activity can save about 20%Energy.

Use right quantity of water required for cooking and

reduce gas/kerosene usage by 65%.

Cook on low flame as far as possible and save 6 to

10% energy.

The pressure cooker should be loaded 2/3rd of the

foodstuff is solid & hard and if loaded with liquid.Properly used pressure cookers can save upto 50 to

75% of energy as well as time.



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DOMESTIC SECTOR (contd)

Cook the food in solar cookers. Cook anythingexcept roti and save cost of 2 LPG Cylindersannually.

Allow refrigerated foodstuff to come to room

temperature before heating and allow heatedfoodstuff to cool down to normal temperaturebefore placing it in the refrigerator. Avoidfrequent opening and closing of refrigerator &

air-conditioned rooms. Electricity is an expensive as cooking fuel avoid

it


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DOMESTIC SECTOR (contd)

Use your refrigerator & air conditioners efficiently Place the refrigerator about 6 inches away from the

wall to allow air circulation

Air-conditioned room must be leak proof

Instant geysers are considered to be more efficientthan storage type geysers. solar water heaters

operate on solar energy which is available free of

cost.

Switch off lights & fans when not required Stop wastage we cannot afford it

conserve energy

Plant trees


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ENERGY CONSERVATION IN STREET LIGHTS

Normally a tube light of 40W rating with a choke

of 20 W is being utilized for street lights with a

total power of 60 W. Alternately the use of CFLs

of 18 W rating which has an equivalent

luminosity would lead to a power saving to the

extent of 70% ie. 42 W. Also the life of the CFLs

are much longer than that of the tube lights

with a cumulative savings on life and as well asthe energy consumption for the entire life.

CASE STUDY ON ENERGY CONSERVATION


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IN STREET LIGHTS

The experience of such a reported transition

from tube light to CFL in Thiruvallur District is as

follows:

340 tube lights originally deployed were

replaced with 18 W CFL lamps, Consequently,

the annual savings in the cost of energyreported is of Rs.2,12,704/-



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IN STREET LIGHTS (contd)

Parameter40W Tube Light with a 20 W choke

18 W Compact Fluorescent Lamp

Savings

Annual Consumption in units (with an average of 12 hoursper day for 365 days)

With Tube light: 60 x 12 x 365 = 263 kWhWith CFL: 18 x1 2 x 365 = 79 kWh

Energy saving = 263 - 79 = 184 kWh

Annual saving per light point (with cost of energy asRs.3.40 per unit)

= 184 x 3.40 = Rs. 625.60

Total annual saving =Rs.625.60 x 340 =Rs. 2,12,704/-



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AGRICULTURAL PUMP

Selection of proper capacity of pumps according to theirrigation requirement.

Matching of pump set with source of water-canal or

well.

Matching of motor with appropriate size pump. Proper installation of the pump system-shaft alignment,

coupling between motor and pump.

Use of efficient transmission system.

Maintain right tension and alignment of transmissionbelts.

Use of low friction rigid PVC pipes and foot valves.

Avoid use unnecessary bends and throttle valves.

Use bends in place of elbows.



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AGRICULTURAL PUMP (contd)

The suction depth of 6 m is recommended as optimumfor centrifugal pumps. The delivery line should be kept

at minimum required height according to requirement.

Periodically check pump system and carry out corrective

measures - like lubrication, alignment, tuning of enginesand replacement of worn-out parts.

Over irrigation can harm the crops and waste vital water

resource. Irrigate according to established norms for

different crops. Use drip irrigation for specific crops like vegetable,

fruits, tobacco, etc. Drip systems can conserve upto 80%

water and reduce pumping energy requirement.

ENERGY EFFICIENCY IN AGRICULTURAL


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ENERGY EFFICIENCY IN AGRICULTURAL

PUMPING SYSTEM

In centrifugal pumps, the fluid is fed to the

centre of a rotating impeller and is thrown

outward by centrifugal action.

The high speed of rotation of the impeller

imparts high kinetic energy to the fluid. This

kinetic energy when converted into pressure

energy results in pressure difference betweenthe suction and delivery of the pump.

GENERAL PERFORMANCE CHARACTERISTICS


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GENERAL PERFORMANCE CHARACTERISTICS

The hydraulic performance of a centrifugal pump is based on

operating characteristics like:Capacity, Q: expressed in units of volume per unit of time,

such as, m3/s or lps

Head, H : expressed in units of height of liquid column, to

which the liquid is pumped, such as, ft or mPower, P : expressed in units of energy, kW or HP

Efficiency, : expressed in %

The variables influencing the performance of pump are:

Speed, N : expressed speed at which pump runs, in RPM

Diameter, D: expressed as diameter of impeller or wheel,

generally, in mm

PUMP POWER OUTPUT


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PUMP POWER OUTPUT

The power output of a pump is the product of the totaldynamic head and the mass of liquid pumped in a given time.

The power output is given by:Pout = (Q x H x ) / 102Where,

Pout = Pump power output (fluid power), in kWQ = Capacity, in m3/s

H = Total dynamic head, in m of liquid column (LC) = Density of liquid, in kg/m3

The power required for driving the pump is the fluid power(output) divided by the pump efficiency.

The power output of a pump is less than the power input.This is because of internal losses resulting from friction,leakage etc.

PUMP EFFICIENCY


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PUMP EFFICIENCY

The pump efficiency is expressed as:

pump = (Q x H x ) / (102 x kW x motor)

Where

Q = Capacity, in m3/s

H = Dynamic head, in m of LC

= Density of liquid, in kg/m3

kW = Motor input power

motor = Efficiency of motor

The pump efficiency (pump) is the product of three efficiencies:

pump = mvhWhere

m = Mechanical efficiency

v = Volumetric efficiency

h

= Hydraulic efficiency

ENERGY CONSUMPTION


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ENERGY CONSUMPTION

The energy consumption is a factor, which is

affected by three aspects of hydrogeology, namely,the geology of the area, depth of ground water and

the current level of ground water extraction.

For example, the pumps operating in alluvial areas

with low pump density will consume less energy,

while the pumps operating in hard rock areas with

high pump density, will consume more energy.

The energy consumption in the agricultural sector isinfluenced by the cropping pattern also.

ENERGY INDEX


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ENERGY INDEX

Energy index is the ratio of energy consumed by the pumping system per unit

of work done. This index is the inverse of the pumping system efficiency and

is given by:

(i) ELECTRIC PUMP SET

Energy index = (3 VI cos x 102) / (Q x H)

where V = Supply voltage, in V

I = Current measured, in Acos = Power factor (0.70 or 0.80)

Q = Discharge rate, in lps

H = Static head, in m

(ii) DIESEL PUMP SET

Energy index = (27.78 x D) / (Q x H)

where D = Diesel consumed, in lph

Q = Discharge rate, in lps

H = Static head, in m

REDUCING FRICTION ACROSS SUCTION


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REDUCING FRICTION ACROSS SUCTION

PIPING, Hfs

The reduction of Hfscan result in the followingbenefits: Reduction of total system head (H)

Cavitations-free operation

Energy efficient performance The friction loss across pipes is given by:

Hfs = (4 x f x L x v2) / (2 x g x D)

Where,

f = coefficient of friction of pipe

L = equivalent length of pipe, in m

v = velocity of flow, in m/s

D = diameter of pipe, in m

g = acceleration due to gravity, in m/s2

11 Energy Conservation

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