COOLING LOAD ESTIMATION

COOLING LOAD AND COIL

LOAD CALCULATIONS
 Cooling load calculations for air conditioning system
design are mainly used to determine the volume flow
rate of the air system as well as the coil and
refrigeration load of the equipment and to provide the
inputs to the system for energy use calculations in order
to select optimal design alternatives.
What is BTU?

 BTU refers to British Thermal Unit.

 An Unit of Heat Energy in Imperial System or

I-P System.

1 BTU is the amount of Heat energy required

to raise the temperature of 1 lb of water by
1⁰F.
Sensible Vs Latent Heat

Sensible Heat
60°F 212°F
[15.6°C] [100°C]

Latent Heat
212°F 212°F
[100°C] [100°C]
Important Terms and Definition
associated with Air Conditioning
 Ton of refrigeration
 A ton of refrigeration (TR), also called a refrigeration
ton (RT), is a unit of power used to describe the heat-
extraction capacity of refrigeration and air
conditioning equipment.
 It is defined as the rate of heat transfer that results in the
freezing or melting of 1 short ton (2,000 lb; 907 kg) of
pure ice at 0 °C (32 °F) in 24 hours.
 A refrigeration ton is approximately equivalent to
12,000 BTU/h or 3.5 kW (3516.85 Watts)or 3023.95kcal/hr. Air-
conditioning and Refrigeration equipment capacity is often
specified in “Tons of Refrigeration”.
Important Terms and Definition
associated with Air Conditioning
 Sensible Heat: Sensible heat is literally the heat that can be felt. it
is the heat that can be felt standing near a fire, or standing outside
on a sunny day. When an object is heated, its temperature rises
as heat is added. The increase in heat is called sensible heat.
 Latent Heat:All pure substances in nature are able to change their
state. Solids can become liquids (ice to water) and liquids can
become gases (water to vapor) but changes such as these require
the addition or removal of heat. The heat that causes these change
in phase is called latent heat.
 The latent heat of fusion of ice is 333.55 J/g or 333.55 KJ/kg
 The latent heat of vaporization of water is 2,260 kJ/kg
Important Terms and Definition
associated with Air Conditioning
Specific heat: The quantity of heat required to raise
the temperature of one gram of a substance by one degree Celsius.
For example,
The specific heat of water is 1 calorie (or 4.186 joules) per gram
per degree Celsius.
The specific heat of ice is 0.5 calorie (or 2.090 joules) per gram per
degree Celsius.
The specific heat of steam is o.48 calorie (or 2.010 joules) per gram
per degree Celsius.
Important Terms and Definition
associated with Air Conditioning
 Energy Efficiency Rating (EER): Each air conditioner has an energy
efficiency rating that lists how many BTU's per hour are handled
(moved from the inside of the house and transferd it to the
outside) for each Watt of power the system draws.
 The condition for calculating EER is at an outdoor temperature of
95°F and inside temperature of 80°F with 50% humidity
 Seasonal Energy Efficiency Ratio: For central air conditioners, SEER
is used. Rather than measuring the energy efficiency of an air
conditioner at one operating temperature, SEER is the calculation
of how energy efficient the air conditioner is during the entire
cooling season at varying temperatures between 65 to 104degree F.
Important Terms and Definition
associated with Air Conditioning
 Sensible Heat
 The sensible heat in a heating or cooling process of air can be calculated
as
 Q=m cp dt
 Q= ρ q cp dt
 where
 Q = sensible heat (btu/hr)
 m=mass of air
 cp = specific heat of air (0.24 Btu/lb)
 ρ = density of air (0.075lb/Cu.ft)
 q = air volume flow (cfm)
 dt = temperature difference (oC)
 Q = 1.08 q dt
 Btu/hr=1.08 q dt
Solved Problem

 An air flow of 1 cfm is heated from 32 to 52oF. Using the equation

for the sensible heat, Sensible added to the air can be calculated
as follows
 hs = 1.08 (1 cfm) ((52 oF) - (32 oF))
 = 21.6 (Btu/hr)
Important Terms and Definition
associated with Air Conditioning
 Latent Heat: Latent heat due to the moisture in air can be calculated
as folllows:
 Ql= ρ hwe q dw ,where
 Ql =latent heat (btu/hr)
 ρ = density of air (0.075lb/Cu.ft)
 q = air volume flow (cfm)
 hwe = latent heat vaporization of water (970 btu/lb - in air at
atmospheric pressure )
 dw = humidity ratio difference (grains of water/lb dry air)
 Ql= btu/hr= 0.68 q dw
Solved Problem
 An air flow of 1 cfm is cooled from 52 to 32oF. The relative humidity
of the air is 70% at the start and 100% at the end of the cooling
process.
 From the psychrometric chart, we estimate the water content in
the hot air to be 45 grains water/lb dry air, and the water content
in the cold air to be 27 grains water/lb dry air respectively for dry
bulb temperatures 52and 32 degree F.
 Using (2b) the latent heat removed from the air can be calculated
as
 hl = 0.68 (1 cfm) ((45 grains water/lb dry air) - (27 grains water/lb
dry air))
 = 12.2 (Btu/hr)
Important Terms and Definition
associated with Air Conditioning
 Total Heat - Latent and Sensible Heat: Total heat due to both
temperature and moisture can be expressed in imperial units as:
 ht = ρ q dh ,
 Where ht = total heat (Btu/hr)
 q = air volume flow (cfm, cubic feet per minute)
 ρ = density of air (0.075lb/cu.ft)
 dh = enthalpy difference(btu/lb dry air), this can be estimated from
the Mollier diagram
 ht = 4.5 q dh
 Total heat, thus can be expressed as:
 ht = hs + hl,
 Q =Q + Ql
 = 1.08 q dt + 0.68 q dwgr
Solved Problem

 An air flow of 1 cfm is cooled from 52 to 32oF. The relative

humidity of the air is 70% at the start and 100% at the end of the
cooling process.
 From the psychrometric chart we estimate the water enthalpy in
the hot air to be 19 Btu/lb dry air, and the enthalpy in the cold air
to be 13.5 Btu/lb dry air.
 Using (3b) the total sensible and latent heat removed from the air
can be calculated as
 ht = 4.5 (1 cfm) ((19 Btu/lb dry air) - (13.5 Btu/lb dry air))
 = 24.8 (Btu/hr)
Important Terms and Definition
associated with Air Conditioning
Sensible Heat Ratio:
The dehumidification effectiveness of air conditioning equipment is
commonly characterized by the sensible heat ratio. Which is the ratio of
the sensible cooling capacity to that of total cooling(sensible+latent)
capacity.

1. SHR from 0.95 - 1.00 for Precision air conditioning (computers and data-
centres)

2. SHR from 0.65 - 0.75 for Comfort cooling (people)

3. SHR from 0.50 - 0.60 for Dehumidification (pools and outside air)
 Human Comfort

 Conditions at which most people are likely to feel

comfortable most of the time.

 Also called as Thermal Comfort.

 Temperature: 78⁰F (Summer) – 68⁰F(Winter).

 Relative Humidity: 30 %– 40%.

Factors Affecting Human
Comfort
 Dry-bulb
temperature
 Humidity
 Air movement
 Fresh air
 Clean air
 Noise level
 Adequate lighting
 Proper furniture and
work surfaces
Indoor Design Conditions

Humidity Ratio
A

comfort zone
70°F 80°F
[21.2°C] [26.7°C]

Dry-bulb Temperature
 Determining the Loads on the HVAC System

Establishing HVAC Design Conditions

➢ Design for the worst case conditions
➢ Ensures proper operation in extreme conditions
➢ Add a safety margin to calculations
Effect of Facility Location
➢ Location, Latitude, Elevation
Determining the Loads on the HVAC System

Effect of Site Orientation

➢ Northern exposures vs. Southern exposures
➢ Radiant transfer rates depend on orientation
Selecting Design Weather Conditions
➢ Defined for both summer and winter
➢ Occurrence values
Determining the Loads on the HVAC System

Determining Building Characteristics

➢ Physical dimensions of conditioned space
➢ Construction materials used
➢ Orientation/Dimensions of transfer surfaces
➢ Occupancy levels, use and scheduling
➢ Interior heat sources (lighting, equipment etc.)
➢ Ventilation requirements
Determining the Loads on the HVAC System

Determining Heat Loads

Peak cooling load/heating load calculation
➢ Heat transmissions through walls,
floors, etc.
➢ Transmission through partitions
to conditioned/ unconditioned
spaces
➢ Infiltration of outside air
➢ Ventilation
➢ Transmission from floors
to the ground
Determining the Loads on the HVAC
System

Determining Cooling Loads

➢ Transmission through walls, floors, etc.
➢ Transmissions through partitions from unconditioned
spaces
➢ Solar radiation, ventilation and outside air
➢ Latent and sensible heat losses from people
➢ Lighting, appliances and
equipment heat gains
Cooling Load Estimation
Procedure
Cooling Load Components

roof

lights partition
people wall
infiltration
glass solar
equipment
glass
conduction

exterior
wall
floor
COOLING LOAD AND COIL
LOAD CALCULATIONS
Cooling load usually can be classified into two categories:
➢ External and Internal.
External Cooling Loads.
▪ These loads are formed because of heat gains in the conditioned
space from external sources through the building envelope or
building shell and the partition walls.
 Sources of external loads include the following cooling loads:
 1. Heat gain entering from the exterior walls and roofs
 2. Solar heat gain transmitted through the fenestrations
 3. Conductive heat gain coming through the fenestrations
 4. Heat gain entering from the partition walls and interior doors
 5. Infiltration of outdoor air into the conditioned space Internal
Cooling Loads.
COOLING LOAD AND COIL
LOAD CALCULATIONS
Internal Cooling Loads:
▪ These loads are formed by the release of sensible and
latent heat from the heat sources inside the conditioned
space. These sources contribute internal cooling loads:
 1. People
 2. Electric lights
 3. Equipment and appliances
▪ Out of the above, only infiltrated air, occupants,
equipment, and appliances have both sensible and latent
cooling loads. The remaining components have only
sensible cooling loads.
▪ All sensible heat gains entering the conditioned space
represent radiative heat and convective heat except the
infiltrated air.
Cooling Load Components

cooling load components Sensible latent space coil

Load load load load
Conduction through roof, walls,
windows, and skylights
Solar radiation through windows, skylights
Conduction through ceiling, interior
partition walls, and floor
People
Lights
Equipment/ Appliances
Infiltration
Ventilation
System Heat Gains
Components of Cooling Coil
Load
Components of Cooling Coil Load
➢ Space cooling load Qrc, including sensible and latent
load
➢ Supply system heat gain qss, because of the supply fan
heat gain qsf and supply duct heat gain qsd
➢ Return system heat gain qrs because of heat gains of
recessed electric lights and ceiling plenum qrp, of
return duct qrd , and return fan qrf , if any
➢ Sensible and latent load because of the outdoor
ventilation rates Qo to meet the requirements of the
occupants and others
 Difference between cooling Load and
Cooling Coil Load
For the same air conditioning cycle, note the following:
1. The space cooling load is represented by Qrc, and the cooling coil load is
represented by Qc. Since supply system heat gain qss and return system heat gain
qrs are both instantaneous cooling loads,
then Cooling Coil Load Qc = Qrc + qss + qrs + Qo ,
where Qo = load due to the outdoor ventilation air intake, Btu/h (W).
2. The space cooling load is used to determine the supply volume flow rate ,
whereas the coil load is used to determine the size of the cooling coil in an air-
handling unit or DX coil in a packaged unit.
3. A cooling load component influences both volume flow rate and the size of the
cooling coil, whereas a cooling coil load component may not affect .
4. Infiltration heat gain is an instantaneous cooling load.
It is apparent that the load due to the outdoor ventilation air Qo, sometimes
called the ventilation load, is a coil load. If Qo is considered a cooling load, the
volume flow rate of the air system will be oversized.
 HISTORICAL DEVELOPMENT OF
COOLING LOAD CALCULATIONS
 Because of the need for computerized load and energy calculations, ASHRAE
established a Committee on Energy Consumption in 1965 and named it as
the Task Group on Energy Requirements (TGER) for Heating and Cooling in
1967.
 In the mid-1970s, ASHRAE and the National Bureau of Standards (NBS)
published the computerized calculation of heating and cooling loads in
energy estimating programs. In 1980, the U.S. Department of Energy
sponsored a computer program for energy estimation and load calculation
through hour-by-hour detailed system simulation, called DOE-2, which was
published through Los Alamos National Laboratory and Lawrence Berkeley
Laboratory.
 In this program, a custom weighting factor method for various room
configurations is used for heating and cooling load calculations.
 Many computerized thermal load and energy calculating software programs
had been developed in the 1980s. Since the 1980s, because of the wide
adoption of personal computers, the use of computer-aided HVAC&R design
was rapidly increased and many thermal load and energy analysis programs
were developed in this period.
 COOLING LOAD CALCULATIONS

 Physical
Mode:
 HEAT BALANCE METHOD; The exact method to
calculate the space cooling load is to use heat balance equations to
determine the temperature of the interior surfaces of the building
structure at time ‘t’ simultaneously and then to calculate the space
sensible cooling load, latent load, load due to infiltration. The heat
balance method is more direct and clear in load calculation methodology.
 However, the heat balance method demands laborious work, more
computing time, complicated computer programs, and experienced users.
Only expensive mainframe computers could run computer programs
adopting the heat balance methodology in the 1970s and early 1980s. The
heat balance method is generally used for research and analytical purpose.
COOLING LOAD CALCULATIONS
 TRANSFER FUNCTION METHOD: The transfer function method or weighting
factor method is a simplification of the laborious heat balance method.
 The wide application of the TFM is due to the user-friendliness of the inputs
and outputs of the TFM software and the saving of computing time.
 In the transfer function method, interior surface temperatures and the
space cooling load were first calculated by the exact heat balance method
for many representative constructions.
 The transfer function coefficients (weighting factors) were then calculated
which convert the heat gains to cooling loads. Sometimes, transfer function
coefficients were also developed through test and experiments.
 Today, TFM is the most widely adopted computer-aided load calculation
method in HVAC&R consulting firms.
 Transfer Function and Time Function: The calculation of space cooling load
using the transfer function method consists of two steps. First, heat gains
or heat loss from exterior walls, roofs, and floors is calculated using
response factors or conduction transfer function coefficients; and the solar
and internal heat gains are calcu lated directly for the scheduled hour.
Second, room transfer function coefficients or room weighting factors are
used to convert the heat gains to cooling loads, or the heat losses to
heating loads
COOLING LOAD CALCULATIONS
CLTD/SCL/CLF Method
 The cooling load temperature difference (CLTD)/solar cooling load
(SCL)/cooling load factor (CLF)
 This method first calculates the sensible cooling load based on the TFM.
 The result is divided by the U value, shading coefficient, or sensible heat
gain to generate the CLTD, SCL, or CLF.
 Thus, it provides a direct, one-step space cooling load calculation instead
of a heat gain–cooling load conversion, a two-step calculation in TFM.
 Cooling load calculation using the CLTD/SCL/CLF method can be either
computer-aided or performed manually for a check or rough estimate.
 The CLTD/SCL/CLF method is one of the members of the TFM family.
 In the CLTD/SCL/CLF method, the CLTD is used to calculate the sensible
cooling load for the exterior wall and roofs.
 Recently, an SCL factor has been added which represents the product of
the solar heat gain at that hour and the fraction of heat storage effect due
to various types of room construction and floor coverings.
 CLF is used to calculate internal sensible cooling loads.
 Outdoor design conditions

 Selection of maximum dry and wet bulb temperatures at a particular

location leads to excessively large cooling capacities as the maximum
temperature generally persists for only a few hours in a year.
 Hence it is recommended that the outdoor design conditions for summer
be chosen based on the values of dry bulb and mean coincident wet bulb
temperature that is equaled or exceeded 0.4, 1.0 or 2.0 % of total hours
in an year.
 These values for major locations in the world are available in data
books, such as AHRAE handbooks.
 Whether to choose the 0.4 % value or 1.0 % value or 2.0 % value depends
on specific requirements.
 In the absence of any special requirements, the 1.0% or 2% value may be
considered for summer outdoor design conditions.
 For Trivandrum the outside design conditions based on 0.4 % value, it
is 33.3 deg C DBT & 25.6 def C MWBT
 Outdoor design conditions for winter
 Similar to summer, it is not economical to design a winter air conditioning for
the worst condition on record as this would give rise to very high heating
capacities.
 Hence it is recommended that the outdoor design conditions for winter be
chosen based on the values of dry bulb temperature that is equaled or
exceeded 99.6 or 99.0 % of total hours in an year.
 Similar to summer design conditions, these values for major locations in the
world are available in data books, such as AHRAE handbooks. Generally the
99.0% value is adequate, but if the building is made of light-weight materials,
poorly insulated or has considerable glass or space temperature is critical,
then the 99.6% value is recommended.
 Time of Peak Cooling Load

East-facing
window roof
Heat gain

12 6 12 6 12
mid a.m. noon p.m. mid
Example Office Space (Room
101)
North

Room 101 Room 102

Elevation view (Room 101)

Plan view
 Assumptions
 Orientation of Room 101 -West
 Floor Area 2700Sq.ft
 West facing Wall Area 540Sq.ft
 Exposed window area 160Sq.ft
 Shaded wall area 380Sq.ft
 Sunlit wall area 380Sq.ft
 Orientation of Room102 –East, Other feature similar to Room 101
 Occupancy 18 Nos.
 Type of activity Moderately Active work (SH 250Btu, LH200Btu)
 Equipment Load - Coffee Maker 1500 Watts(SH1050W+LH450W)
 Computers @0.5Watts per Sq.ft = 1350watts
 Lighting @2Watts per Sq.ft = 5400watts
Outdoor Design Conditions

0.4% 1% 2%
DB WB DB WB DB WB

St. Louis, 95°F 76°F 93°F 75°F 90°F 74°F

Missouri [35°C] [25°C] [34°C] [24°C] [32°C] [23°C]
Indoor Design Conditions

 Desired Indoor conditions

 Cooling Season -
 DBT - 78 degree Fahrenheit
 Relative Humidity – Less than 50%

 Heating Season
 DBT - 72 degree Fahrenheit
Heat Conduction through
Surfaces
Conduction through a Shaded
Wall

Q = U  A  T
U – Overall heat transfer coefficient of the surface
A – Area of the surface
 T – Dry bulb temperature difference across the
surface
U-factor

wood studs

insulation

concrete block
gypsum
board
aluminum
siding
 U-factor for Example Wall

thermal resistance (R) 1

U =
Routdoor-air film 0.25 [0.04] Rtotal
Rsiding 0.61 [0.11]
Rconcrete block 2.00 [0.35]
Rinsulation 13.00 [2.29] U = 0.06 Btu/hr•ft2•°F
Rgypsum board 0.45 [0.08]
Rindoor-air film 0.68 [0.12] [ U = 0.33 W/m2•°K ]
Rtotal 16.99 [2.99]
]
 Conduction through a Shaded
Wall

Qwall = 0.06  380  (95 – 78) = 388 Btu/hr

[ Qwall = 0.33  36.3  (35 – 25.6) = 113 W ]

 Sunlit Surfaces

sun solar angle changes throughout the day

rays
 Time Lag
Time
Lag

A B
Solar Effect

12 6 12 6 12
mid a.m. noon p.m. mid
Conduction through Sunlit Surfaces

Q = U  A  CLTD
CLTD : Term used to account for the added
heat transfer due to the sun shining on
exterior walls, roofs, and windows, and the
capacity of the wall and roof to store heat.
 CLTD Factors for West-Facing
Wall
hour
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
CLTD
35 30 25 21 17 14 11 8 7 6 6 7 8 10 12 16 22 30 37 44 48 48 45 41
(°F)

CLTD
19 17 14 12 9 8 6 4 4 3 3 4 4 6 7 9 12 17 21 24 27 27 25 23
(°C)
Conduction through Sunlit Surfaces

Qwall = 0.06  380  22 = 502 Btu/hr

Qroof = 0.057  2700  80 = 12312 Btu/hr

[ Qwall = 0.33  36.3  12 = 144 W ]

[ Qroof = 0.323  250.7  44 = 3563 W ]
U-factors for Windows
fixed frames, vertical installation
aluminum without aluminum with
thermal break thermal break wood/vinyl
single glazing
1/8 in. [3.2 mm] glass 1.13 [6.42] 1.07 [6.07] 0.98 [5.55]
double glazing
1/4 in. [6.4 mm] air space 0.69 [3.94] 0.63 [3.56] 0.56 [3.17]
1/2 in. [12.8 mm] air space 0.64 [3.61] 0.57 [3.22] 0.50 [2.84]
1/4 in. [6.4 mm] argon space 0.66 [3.75] 0.59 [3.37] 0.52 [2.98]
1/2 in. [12.8 mm] argon space 0.61 [3.47] 0.54 [3.08] 0.48 [2.70]
triple glazing
1/4 in. [6.4 mm] air spaces 0.55 [3.10] 0.48 [2.73] 0.41 [2.33]
1/2 in. [12.8 mm] air spaces 0.49 [2.76] 0.42 [2.39] 0.35 [2.01]
1/4 in. [6.4 mm] argon spaces 0.51 [2.90] 0.45 [2.54] 0.38 [2.15]
1/2 in. [12.8 mm] argon spaces 0.47 [2.66] 0.40 [2.30] 0.34 [1.91]
 Conduction through Windows

Qwindows = U  A x CLTD

Qwindows = 0.63  160  13 = 1310 Btu/hr

[ Qwindows = 3.56  14.4  7 = 359 W ]

Solar Radiation through Glass
Solar Heat Gain through Glass

Q = A  SC  SCL
Where,
SC – Shading Coefficient

SCL – Solar Cooling Load Factor

Solar Cooling Load Factor (SCL)
 Direction that the window faces
 Time of day
 Month
 Latitude
 Construction of interior partition walls
 Type of floor covering
 Existence of internal shading devices

SCL: A factor used to estimate the rate at which

solar heat energy radiates directly into the space,
heats up the surfaces and furnishings, and is later
released to the space as a sensible heat gain.
Shading Coefficient (SC) ?

It is an expression used to define how much

of the radiant solar energy, that strikes the
outer surface of the window, is actually
transmitted through the window and into
the space.
 Shading Coefficient (SC)
shading coefficient at normal incidence
aluminum frame other frames
operable fixed operable fixed
uncoated single glazing
1/4 in. [6.4 mm] clear 0.82 0.85 0.69 0.82
1/4 in. [6.4 mm] green 0.59 0.61 0.49 0.59
reflective single glazing
1/4 in. [6.4 mm] SS on clear 0.26 0.28 0.22 0.25
1/4 in. [6.4 mm] SS on green 0.26 0.28 0.22 0.25
uncoated double glazing
1/4 in. [6.4 mm] clear - clear 0.70 0.74 0.60 0.70
1/4 in. [6.4 mm] green - clear 0.48 0.49 0.40 0.47
reflective double glazing
1/4 in. [6.4 mm] SS on clear - clear 0.20 0.18 0.15 0.17
1/4 in. [6.4 mm] SS on green - clear 0.18 0.18 0.15 0.16

SS = stainless-steel reflective coating

Solar Radiation through Windows

Qwindows = 160  0.74  192 = 22733 Btu/hr

[ Qwindows = 14.4  0.74  605 = 6447 W ]

 Internal Heat Gains

Equipment

People

Lights
Appliances
Heat Generated by People

 Metabolism of the human body normally generates

more heat than it needs

 60% heat is transferred by convection and radiation to

the surrounding environment.

 40% is released by perspiration and respiration.

Heat Generated by People (Chart)
Level Of Activity Sensible Heat Latent Heat
Gain Gain
Moderately active work 250 BTU/hr 200 BTU/hr
(Office) (75W) (55W)
Standing, light work, 250 BTU/hr 200 BTU/hr
walking (Store) (75W) (55W)
Light bench work 275 BTU/hr 475BTU/hr
(Factory) (80W) (140W)
Heavy work (Factory) 580BTU/hr(170 870BTU/hr
W) (255W)
Exercise (Gymnasium) 710BTU/hr 1090BTU/hr
(210W) (315W)
 CLF Factors for People

Hours after people enter space

Total hours in
space 1 2 3 4 5 6 7 8 9 10 11 12

2 0.65 0.74 0.16 0.11 0.08 0.06 0.05 0.04 0.03 0.02 0.02 0.01
4 0.65 0.75 0.81 0.85 0.24 0.17 0.13 0.10 0.07 0.06 0.04 0.03
6 0.65 0.75 0.81 0.85 0.89 0.91 0.29 0.20 0.15 0.12 0.09 0.07
8 0.65 0.75 0.81 0.85 0.89 0.91 0.93 0.95 0.31 0.22 0.17 0.13
10 0.65 0.75 0.81 0.85 0.89 0.91 0.93 0.95 0.96 0.97 0.33 0.24

Note: CLF – Cooling Load Factor

Capacity of a space to absorb and store heat.
 Heat Gain from People
QS = No: of people x Sensible heat gain per person x
CLF
Qsensible = 18  250  1.0 = 4500 Btu/hr

QL = No: of people  Latent heat gain/ person

Qlatent = 18  200 = 3600 Btu/hr

[ Qsensible = 18  75  1.0 = 1350 W ]

[ Qlatent = 18  55 = 990 W ]
Heat Gain from Lighting

Q = Btu/hr  Ballast factor  CLF

[ Q = watts  Ballast factor  CLF ]

Ballast factor = 1.2 for fluorescent lights

Ballast factor = 1.0 for incandescent
lights
Heat Gain from Lighting

Qlights = 5400  3.41  1.2  1.0 = 22097 Btu/hr

[ Qlights = 5400  1.2  1.0 = 6480 W ]

Heat generated by equipment

Equipment Sensible Heat Latent Heat Gain

Gain
Coffee maker 3580 BTU/hr 1540 BTU/hr
(1050W) (450W)
Printer 1000 BTU/hr
(292W)
Typewriter 230 BTU/hr
(67W)
Infiltration
Methods of Estimating
Infiltration
 Air change method
 Crack method
 Effective leakage-area method
Infiltration Airflow

Infiltration 32400  0.3

= = 162 CFM
airflow 60

Infiltration 927.6  0.3

= = 0.077 m3/s
airflow 3600
Heat Gain from Infiltration
Qsensible = 1.085  airflow  T
Qlatent = 0.7  airflow  W
[ Qsensible = 1210  airflow  T ]
[ Qlatent = 3010  airflow  W ]
W = (Outdoor Humidity Ratio – Indoor Humidity Ratio)
Air Flow – Quantity of air infiltrating the place
T = (Outdoor D.B.T – Indoor D.B.T)
Density x Specific Heat = 1.085 (1210) Btu•min/hr•ft3•ºF
[J/m3•ºK]
Latent Heat Factor = 0.7 (3010) Btu•min•lb/hr•ft3•gr
[J•kg/m3•g]
 Heat Gain from Infiltration

QS = 1.085  162  (95 – 78) = 2,988 Btu/hr

QL = 0.7  162  (105 – 70) = 3,969 Btu/hr

[ QS = 1,210  0.077  (35 – 25.6) = 876 W ]

[ QL = 3,010  0.077  (15 – 10) = 1,159 W ]
Summary of Space Cooling Loads
space load components sensible load latent load
Btu/hr [W] Btu/hr [W]
conduction through roof 12,312 [3,563]
conduction through exterior wall 502 [144]
conduction through windows 1,310 [359]
solar radiation through windows 22,733 [6,447]
people 4,500 [1,350] 3,600 [990]
lights 22,097 [6,480]
equipment 8,184 [2,404] 1,540 [450]
infiltration 2,988 [876] 3,969 [1,159]
Total space cooling load 74,626 [21,623] 9,109 [2,599]
Ventilation

outdoor-air
supply duct
intake

diffuser
air handler
with fan and
cooling coil
Outdoor Air Requirements

Type of Space Outdoor Air/ person Outdoor Air/ ft2

(m2)
Auditorium 15 CFM (0.008 m3/s)

Class rooms 15 CFM (0.008 m3/s)

Locker rooms 0.5 CFM (0.0025 m3/s)

Office space 20 CFM (0.01 m3/s)

Public restrooms 50 CFM (0.025 m3/s)

Smoking lounge 60 CFM (0.03 m3/s)

 Cooling Load Due to Ventilation

QS = 1.085  360  (95 – 78) = 6640 Btu/hr

QL = 0.7  360  (105 – 70) = 8820 Btu/hr

[ QS = 1210  0.18  (35 – 25.6) = 2047 W ]

[ QL = 3010  0.18  (15 – 10) = 2709 W ]
System Heat Gains

air handler

fan motor
 Components of Fan Heat

blow-through
configuration

draw-through
configuration
Heat Gain in Ductwork
Summary of Cooling Loads
sensible load latent load
Btu/hr [W] Btu/hr [W]
conduction through roof 12,312 [3,563]
conduction through exterior wall 502 [144]
conduction through windows 1,310 [359]
solar radiation through windows 22,733 [6,447]
people 4,500 [1,350] 3,600 [990]
lights 22,097 [6,480]
equipment 8,184 [2,404] 1,540 [450]
infiltration 2,988 [876] 3,969 [1,159]
total space cooling load 74,626 [21,623] 9,109 [2,599]
ventilation 6,640 [2,047] 8,820 [2,709]
total coil cooling load 81,266 [23,670] 17,929 [5,308]
 Psychometric Analysis

© American Standard Inc. 2000 Air Conditioning Clinic TRG-TRC002-EN

Space Load versus Coil Load

space coil
load load
conduction through roof, walls, windows,
and skylights
solar radiation through windows, skylights
conduction through ceiling, interior
partition walls, and floor
people
lights
equipment and appliances
infiltration
ventilation
system heat gains
Space Sensible and Latent Loads

space load components sensible load latent load

Btu/hr [W] Btu/hr [W]
conduction through roof 12,312 [3,563]
conduction through exterior wall 502 [144]
conduction through windows 1,310 [359]
solar radiation through windows 22,733 [6,447]
people 4,500 [1,350] 3,600 [990]
lights 22,097 [6,480]
equipment 8,184 [2,404] 1,540 [450]
infiltration 2,988 [876] 3,969 [1,159]
total space cooling load 74,626 [21,623] 9,109 [2,599]
Sensible Heat Ratio (SHR)

sensible heat gain

SHR =
sensible heat gain + latent heat gain

74,626
SHR = = 0.89
74626 + 9109

21623
SHR = = 0.89
21623 + 2599
Single-Space Analysis

outdoor
supply cooling
coil air
fan
supply
air

return
exhaust
air
air

space
Determine Supply Airflow

supply sensible heat gain

=
airflow 1.085 × (room DB – supply DB)

supply sensible heat gain

=
airflow 1,210 × (room DB – supply DB)
Determine Supply Airflow

supply 74,626
= = 2,990 cfm
airflow 1.085 × (78 – 55)

supply 21,623
= = 1.40 m3/s
airflow 1,210 × (25.6 – 12.8)
Calculate Entering Coil
Conditions
ventilation airflow
% outdoor air =
total supply airflow

360 cfm
%OA = = 0.12
2990 cfm

0.18 m3/s
%OA = = 0.12
1.40 m3/s
 Calculate Entering Coil
Conditions
95°F × 0.12 = 11.4°F
78°F × 0.88 = 68.6°F
mixture = 80.0°F
76°F
[24.4°C]

humidity ratio
35°C × 0.12 = 4.2°C
25.6°C × 0.88 = 22.5°C 66.5°F
B
mixture = 26.7°C [19.2°C]

A C
78°F 80°F 95°F
[25.6°C] [26.7°C] [35°C]

dry-bulb temperature
Determine Supply Air
Temperature

sensible heat ratio

0.4

B 0.6

C 0.8
1.0
D A
59°F
[15°C]

dry-bulb temperature
Recalculate Supply Airflow

supply 74,626
= = 3,620 cfm
airflow 1.085 × (78 – 59)

supply 21,623
= = 1.69 m3/s
airflow 1,210 × (25.6 – 15)
Total Cooling Load on Coil
Room 101
Btu/hr [W]
total space sensible load 74,626 [21,623]
total space latent load 9,109 [2,599]
ventilation 15,460 [4,756]

total coil cooling load 99,195 [28,978]

Multiple-Space Analysis

supply
fan
cooling
coil

Room 101 Room 102

Room 101 (Faces West)
space sensible load 8 a.m. 4 p.m.
components Btu/hr [W] Btu/hr [W]
conduction through roof 2,616 [740] 12,312 [3,563]
conduction through exterior wall 160 [48] 502 [144]
conduction through windows 202 [51] 1,310 [359]
solar radiation through windows 3,552 [1,012] 22,733 [6,447]
people 4,500 [1,350] 4,500 [1,350]
lights 22,097 [6,480] 22,097 [6,480]
equipment 8,184 [2,404] 8,184 [2,404]
infiltration 2,988 [876] 2,988 [876]
total space sensible load 44,299 [12,961] 74,626 [21,623]
Room 102 (Faces East)
space sensible load 8 a.m. 4 p.m.
components Btu/hr [W] Btu/hr [W]
conduction through roof 2,616 [740] 12,312 [3,563]
conduction through exterior wall 160 [48] 844 [252]
conduction through windows 202 [51] 1,310 [359]
solar radiation through windows 21,667 [6,138] 3,078 [874]
people 4,500 [1,350] 4,500 [1,350]
lights 22,097 [6,480] 22,097 [6,480]
equipment 8,184 [2,404] 8,184 [2,404]
infiltration 2,988 [876] 2,988 [876]
total space sensible load 62,414 [18,087] 55,313 [16,158]
“Sum-of-Peaks” versus
“Block”
space sensible load 8 a.m. 4 p.m.
Btu/hr [W] Btu/hr [W]
Room 101 (faces west) 44299 [12961] 74626 [21623]
Room 102 (faces east) 62414 [18087] 55313 [16158]

sum-of-peaks = 74626 + 62414 = 137040 Btu/hr

[21623 + 18087 = 39710 W]

block = 74626 + 55313 = 129939 Btu/hr

[21623 + 16158 = 37781 W]
“Sum-of-Peaks” versus
“Block”
 Sum-of-peaks
supply airflow = 6,648 CFM [3.10 m3/s]

 Block
supply airflow = 6,303 CFM [2.95 m3/s]
“Block” Cooling Load
Room 101 Room 102
loads at 4 p.m. Btu/hr [W] Btu/hr [W]
total space sensible load 74626 [21623] 55313 [16158]
total space latent load 9109 [2599] 9109 [2599]
ventilation 15460 [4756] 15460 [4756]

Total coil cooling load 99195 [28978] 79882 [23513]

Block cooling load = 99195 + 79882 = 179077 Btu/hr

(4 p.m.) [28978 + 23513 = 52491 W]