The table below indicates recommended air change rates (air changes per hour) in some common types of rooms and buildings:

Building / RoomAir Change Rates

- n - (1/hr)

All spaces in general min 4

Attic spaces for cooling 12 - 15

Auditoriums 8 - 15

Bakeries 20

Banks 4 - 10

Barber Shops 6 - 10

Bars 20 - 30

Beauty Shops 6 - 10

Boiler rooms 15 - 20

Bowling Alleys 10 - 15

Cafeterias 12 - 15

Churches 8 - 15

Club rooms 12

Clubhouses 20 - 30

Cocktail Lounges 20 - 30

Computer Rooms 15 - 20

Court Houses 4 - 10

Dental Centers 8 - 12

Department Stores 6 - 10

Dining Halls 12 -15

Dining rooms (restaurants) 12

Dress Shops 6 - 10

Drug Shops 6 - 10

Engine rooms 4 - 6

Factory buildings, ordinary 2 - 4

Factory buildings, fumes and moisture

10 - 15

Fire Stations 4 - 10

Foundries 15 - 20

Galvanizing plants 20 - 30

Garages repair 20 - 30

Garages storage 4 - 6

Homes, night cooling 10 - 18

Jewelry shops 6 - 10

Kitchens 15 - 60

Laundries 10 - 15

Libraries, public 4

Lunch Rooms 12 -15

Luncheonettes 12 -15

Nightclubs 20 - 30

Malls 6 - 10

Medical Centers 8 - 12

Medical Clinics 8 - 12

Medical Offices 8 - 12

Mills, paper 15 - 20

Mills, textile general buildings

4

Mills, textile dye houses 15 - 20

Municipal Buildings 4 - 10

Museums 12 -15

Offices, public 3

Offices, private 4

Police Stations 4 - 10

Post Offices 4 - 10

Precision Manufacturing 10 - 50

Pump rooms 5

Restaurants 8 - 12

Retail 6 - 10

School Classrooms 4 - 12

Shoe Shops 6 - 10

Shopping Centers 6 - 10

Shops, machine 5

Shops, paint 15 - 20

Shops, woodworking 5

Substation, electric 5 - 10

Supermarkets 4 - 10

Town Halls 4 - 10

Taverns 20 - 30

Theaters 8 - 15

Turbine rooms, electric 5 - 10

Warehouses 2

Waiting rooms, public 4

Note that in many cases local regulations and codes will govern the ventilation requirements.

The fresh air supply to a room can be calculated as

q = n V         (1)

where

q = fresh air supply (ft3/h, m3/h)

n = air change rate (1/n)

V = volume of room (ft3, m3)

Example - Fresh Air Supply to a Public Library

The fresh air supply to a public library with volume 1000 m3 can be calculated as

Q = (4 1/h) (1000 m3)

    = 4000 m3/h

THE COOLING COIL HAS RELATION SHIP TO AIRVELOCITY SO THAT THERE IS NO CONDENSATE CARRY OVER AND THE RELATION SHIP IS A SIMPLE CALCULATION.

CFM=12525 AND FACE AREA OF COIL IS CFM/VELOCITY= 12525/500= 25 SQFEET

HENCE THE SIZE CAN BE 5FEET*5FEET = SQUARE COIL( hEIGHT *LENGTH)

2.5FEET* 10FEET = RECTANGULAR COIL ( hEIGHT *LENGTH)

aS REGARDS CAPACITY USE THE FOLLOWING FORMULA

CAPACITY= 1.08*DELTA T(F)* CFM

NOW TR=58 SO IN BTU/H= 58*12000= 696000

SO 696000/12525CFM*1.08=DELTA T (F) =51.4

SO ITS RECOMENDED TO USE 6 OR EVEN 8 ROW COIL TO ACHIEVE THIS LOAD REQUIREMNT

HOPE THE ABOVE IS HELPFUL FOR YOU

http://www.cedengineering.com/upload/Cooling%20Load%20Calculations%20and%20Principles.pdf

Determining Air Flow in Cubic Ft./MinPg. 1Rev. 10/31/06

To calculate Air Flow in Cubic Feet per Minute (CFM), determine the Flow Velocity in feet perminute, then multiply this figure by the Duct Cross Sectional Area.Air Flow in CFM (Q) = Flow Velocity in Feet Per Minute (V) x Duct Cross Sectional Area (A)The easiest way to determine Flow Velocity is to measure the Velocity Pressure in the ductwith a Pitot Tube Assembly connected to a differential pressure sensor. The Pitot Tube Assemblyincludes a Static Pressure Probe and a Total Pressure Probe.A Total Pressure Probe, aligned into the airflow, senses the duct velocity pressure and the staticpressure, which equals the total pressure. A Static Pressure Probe aligned at a right angle to theairflow senses only the static pressure. The difference between the total pressure reading and thestatic pressure reading is the Velocity Pressure.If you connect the Total Pressure Probe to the HIGH port on a differential pressure sensor andthe Static Pressure Probe to the LOW port on the differential pressure sensor, then the sensor’soutput will be the Velocity Pressure, as shown in the figures below.Static Pressure ProbeTotal Pressure ProbeFig 2: BAPI Differential ZonePressure Sensor (ZPS)measuring Velocity PressureThe Flow Velocity is then determined with the following equation:V = 4005 x Sqrt(P)V = Flow Velocity in feet per minute.Sqrt() = Square root of the number in the parenthesis.P = The Velocity Pressure measured by the pressure sensorExample: Measuring a Velocity Pressure of .75” W.C. equals a Flow Velocity of 3,468 Ft/Min.V = 4005 x Sqrt(0.75)Sqrt(0.75) = 0.866 • 4005 x 0.866 = 3,468 • Flow Velocity = 3,468 Ft/MinContinued on next page....Fig 1: BAPI Pitot Tube Assembly,includes Static and Total PressureProbe Assemblies (ZPS-ACC12)Determining Flow VelocityApplication NoteBuilding Automation Products, Inc., 750 North Royal Avenue, Gays Mills, WI 54631 USATel: +1-608-735-4800 • Fax: +1-608-735-4804 • E-mail:sales@bapihvac.com • Web:www.bapihvac.com

Determining Air Flow in Cubic Ft./MinPg. 2After obtaining the Flow Velocity from the previous procedure, that figure is now multiplied bythe Duct Cross Sectional Area to determine the Air Flow in CFM.There are two different equations for determining the Duct Cross Sectional Area, one forround ducts and one for square or rectangular ducts.The equation for a round duct is:A = x r2

A = Duct Cross Sectional Area= 3.14159r = radius of duct in feetExample: An 18” diameter round duct has a Duct Cross Sectional Area of 1.77 Ft2

A = x r2 or A = 3.14158 x .562518” diameter is 1.5 feet, therefore the radius is .75 feet • r2 = 0.752 = 0.5265 • = 3.14159A = 3.14159 x 0.5625 = 1.77 Ft2

After obtaining the Flow Velocity and the Duct Cross Sectional Area from the previous twoprocedures, the Air Flow in CFM is determined by multiplying the two:Air Flow in CFM (Q) = Flow Velocity in Feet Per Minute (V) x Duct Cross Sectional Area (A)ExampleAn 18” diameter round duct with a Velocity Pressure of .75” W.C. has an Air Flow of 6,128CFMThe Flow Velocity is 3,468 Ft/Min.V = 4005 x Sqrt(P)V = 4005 x Sqrt(0.75)Sqrt(0.75) = 0.866 • 4005 x 0.866 = 3,468 • Flow Velocity = 3,468 Ft/MinThe Duct Cross Sectional Area is 1.65 Ft2

A = x r2

= 3.14159 • r2 = 0.752 = 0.5625Duct Cross Sectional Area (A) = 3.14159 x 0.5625 = 1.77 Ft2

The Air Flow in CFM is 6,128 Ft3/MinAir Flow in CFM (Q) = Flow Velocity in Feet Per Minute (V) x Duct Cross Sectional Area (A)Air Flow in CFM (Q) = 3,468 Ft/Min x 1.77 Ft2 = 6,128 CFMIf you have any questions about this procedure, please call your BAPI representative.

Selected questions Chapter 4.9: HVAC System

Short type questions1. Define ‘tons of refrigeration’?

Ans. One ton of refrigeration is the amount of cooling obtained by one ton of ice melting in one day. It is equivalent to 3024 kcal/h, 12,000 Btu/h or 3.516 thermal kW.

2.Where is performance evaluation of air conditioning unit term ‘energy efficiency ratio (EER)’ is used?

Energy Efficiency Ratio mainly used in air condition systems. Performance of smaller units and roof top units are frequently measured in EER rather than kW/ton.

3.To assess the refrigeration capacity, list the basic parameters required?

Ans For assessment of refrigeration load, basic parameters required are:

a) Water flow rate through evaporatorb) Temperature difference between entering and leaving water

4.Write the relation between COP, EER and kW/tonne, which are the different energy efficiency performance indicators of Refrigeration systems?

Ans The relation between different energy efficiency performances of chillers are:

COP - 0.293 EER

kW/ton - 12/EER

kW/ton - 3.516/CoP

5.Define the term ‘refrigeration’?

Refrigeration is defined as an art of producing and maintaining the temperature in a space below atmospheric temperature.

6.Which is the popular low energy cost comfort cooling system in tropical climate (dry regions)?

Evaporative cooling systems are popular in tropical climate. These systems control humidity up to 50% for human comfort.

7.Find the dew point temperature (°C) of air for following parameters from psychrometric chart?

Ans Dry bulb temp (°C) = 25 °C, Wet bulb temperature (°C) = 16 °C

Dew point temperature of above air is 10.0

8.How cooling tower should be designed for refrigeration systems?

Cooling water inlet temperature to refrigeration system should be as low as possible. In general, cooling towers will be design with approach WBT + 3 °C for refrigeration systems.

9.What will be the additional refrigeration (TR) required, if a coffee machine of 3 kWh consumption, added in conditioned space?

Ans Additional heat inside the space = 3 x 860 = 2580 kcal/h

Equivalent refrigeration load (TR) = 2580/3024 = 0.85 TR

10.Write the coefficient of performance (CoP) for an ideal refrigeration system (Carnot cycle) ?

Ans COP Carnot = Te/Tc-Te

Te=Evaporator temperature; Tc-

condenser temperature

9. ENERGY PERFORMANCE ASSESSMENT OF HVAC SYSTEMS

9.1 Introduction

Air conditioning and refrigeration consume significant amount of energy in buildings and in process industries. The energy consumed in air conditioning and refrigeration systems is sensitive 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 these factors.

9.2 Purpose of the Performance Test

The purpose of performance assessment is to verify the performance of a refrigeration system by 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 estimate the energy consumption at actual load vis-à-vis design conditions.

9.3 Performance Terms and Definitions

Tons of refrigeration (TR): One ton of refrigeration is the amount of cooling obtained by one ton 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 water multiplied by the difference in enthalpy of water entering and leaving the cooler, expressed in kCal/h, tons of Refrigeration.

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

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

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

9.4 Preparatory for Measurements

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

Bureau of Energy Efficiency 115 9. Energy Performance Assessment of HVAC Systems

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 passes through the evaporator by determination of the following:

a. Water flow rate

b. Temperature difference between entering and leaving water

The heat removed from the chilled water is equal to the product of the chilled water flow rate, the water temperature difference, and the specific heat of the water is defined as follows

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

3024)(Cxm (TR)Capacity ion refrigeratNet poutinttx−=

Where m – mass flow rate of chilled water, kg/hr

cp - Specific heat, kcal/kg oC

tin – Chilled water temperature at evaporator inlet oC

tout - Chilled water temperature at evaporator outlet oC

The accurate temperature measurement is very vital in refrigeration and air conditioning and least 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 following methods

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

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

• If the waterside pressure drops are close to the design values, it can be assumed that the water 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 reading directly in kW.

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

Power (kW) = √3 x V x I x cosφ

Bureau of Energy Efficiency 1169. Energy Performance Assessment of HVAC Systems

9. ENERGY PERFORMANCE ASSESSMENT OF HVAC SYSTEMS

9.1 Introduction

Air conditioning and refrigeration consume significant amount of energy in buildings and in process industries. The energy consumed in air conditioning and refrigeration systems is sensitive 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 these factors.

9.2 Purpose of the Performance Test

The purpose of performance assessment is to verify the performance of a refrigeration system by 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 estimate the energy consumption at actual load vis-à-vis design conditions.

9.3 Performance Terms and Definitions

Tons of refrigeration (TR): One ton of refrigeration is the amount of cooling obtained by one ton 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 water multiplied by the difference in enthalpy of water entering and leaving the cooler, expressed in kCal/h, tons of Refrigeration.

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

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

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

9.4 Preparatory for Measurements

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

Bureau of Energy Efficiency 115 9. Energy Performance Assessment of HVAC Systems

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 passes through the evaporator by determination of the following:

a. Water flow rate

b. Temperature difference between entering and leaving water

The heat removed from the chilled water is equal to the product of the chilled water flow rate, the water temperature difference, and the specific heat of the water is defined as follows

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

3024)(Cxm (TR)Capacity ion refrigeratNet poutinttx−=

Where m – mass flow rate of chilled water, kg/hr

cp - Specific heat, kcal/kg oC

tin – Chilled water temperature at evaporator inlet oC

tout - Chilled water temperature at evaporator outlet oC

The accurate temperature measurement is very vital in refrigeration and air conditioning and least 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 following methods

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

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

• If the waterside pressure drops are close to the design values, it can be assumed that the water 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 reading directly in kW.

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

Power (kW) = √3 x V x I x cosφ

Bureau of Energy Efficiency 1169. Energy Performance Assessment of HVAC Systems

9.5.3 Performance calculations

The energy efficiency of a chiller is commonly expressed in one of the three following ratios: 1.Coefficient of performance

COP=

kW refrigeration effectkW input

2. Energy efficiency ratio

EER=

Btu/h refrigeration effectWatt input

3. Power per Ton

kW/Ton=

kW inputTons refrigeration effect

Energy Use Standards●The available norms of energy use in hotel industry arefollows:-- Power - 60 – 75 kWh/day/room- Fuel - 4 – 5 Liters/day/room- Water - 1.15 – 1.3 KL/day/roomBEE has recently announced the Building Codes.

I assumed that you have an engineering background. Designing a simple centralized A/C system is a complicated job. Its not simply deviding the the total air volume flow to the number of rooms you want to supply air. You need to know the exact A/C load for each room, the contaminants present, the amount of fresh air to supply to ventilate the rooms, the orientation of the rooms, and you need to have a stong knowledge of air psychrometry. Anyway if you really insist in knowing the equation for air flow in a circular duct, the equation is Q = f x Ai x V, where Q is in CFM, Ai (inside cross sectional area of duct, V is the velocity of air in FPM, and f is the flow factor. In ducts the ideal V is 1700 FPM w/ f = 0.75. For air diffusers V is 400 FPM w/ f = 0.85 to .90. Let me remind you theat the air flow from the outlet of the fan to the farthest air outlet will experience a pressure drop . Your calculations should consider this factor. You may assume an equal pressure drop to each outlet. Goodluck to your design!

9.5.3 Performance calculations

The energy efficiency of a chiller is commonly expressed in one of the three following ratios: 1.Coefficient of performance

COP=

kW refrigeration effectkW input

2. Energy efficiency ratio

EER=

Btu/h refrigeration effectWatt input

3. Power per Ton

kW/Ton=

kW inputTons refrigeration effect