HEAT LOAD CALCULATION (manual)
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
Project
Location
Space usage
Orientation
Weather properties
Solar gain through glass
Transmission gains (wall and roof)
Partition gains
Internal heat
Infiltration
Sensible heat
Latent heat
Tonnage of refrigeration
Coil ADP
Dehumidified air supply
Room Dimension
Length
Width
Height
Area
Volume
Window size
= 15 ft
= 10 ft
= 10 ft
= 150 ft
= 1500 ft³
= 3*3 ft
1. WEATHER
Location- Trivandrum
From data book PAGE NO: 20 - 21 we get,
Latitude
= 8.3˚N
Daily Range
= 15˚F
(Maximum temperature difference acquiring a day during a year)
Dry bulb (D.B) = 92.1˚F
Wet bulb (W.B) = 79˚F
Relative humidity (R.H) = 56.6 %
Specific humidity (W) =129.4 gr/lb
{gram/pound }
We get all this details from both CARRIER’S PSYCHOMETRIC CHART SOFTWERE by entering DB&WB,
and HAP by entering the location.
1.1 Recommended inside design condition for summer and winter
D.B
R.H
W.B
w
= 75˚F
= 50%
= 62.5˚F
= 64.9 gr/lb
1.2 Difference between Required and actual condition
D.B
W.B
R.H
w
= ( 92.1 – 75 ) = 17.1˚F
= ( 79 – 62.5 ) = 16.5˚F
= ( 56.6 – 50 ) = 6.6%
= ( 129.4 – 64.5 ) = 64.5 gr/lb
(it is used for finding correction factor for Δt in section-6 and 7)
2. SPACE
From the above drawing
Length
= 15 ft
Width
= 10 ft
Floor area
= 150 ft²
Avg ceiling height = 10 ft
3. OA VENTILATION REQUIREMENT
From data book PAGE NO : 10 we get,
Occupancy
category
People outdoor air
rates
cfm/person
Reception
areas
Area outdoor
air rate
Notes
Occupant
density
#/1000ft²
Or
#/100m²
30
L/s
Cfm/ft² L/sm²
person
3
2.5
for reception area 1000 ft²
for 150 ft²
0.06
0.3
Default values
Combined outdoor air rate
cfm/person
L/s person
7
3.5
Air
class
1
= 30 people
= x people
30
x
= 150 ×
x
= 5 person
1000
= 4.5
cfm/person for reception area = 7/person
no. of people = 5
Total cfm
= 5*7
= 35 cfm
4. INTERNAL HEAT
Sensible heat
4.1 People
For normal person, Sensible
Latent
= 250 Btu/hr
= 200 Btu/hr
No of person = 5
From data book PAGE NO: 62 we get,
Moderately active office work = 250 Btu/hr
Heat gain from people = 5*250
= 1250 Btu/hr
4.2 Power
1 HP
1 RT (Refrigerant Tonnage) = 12,000 BTU/HR=3.51KW
1 WATT =3.41 BTU/HR
= 2545 Btu/hr
Btu – British thermal unit – unit of energy
4.3 Lights
Heat gain through Lights = Power*3.41*1.2
1 watt = 3.41 Btu/hr
3.41 is a constant for converting Watts electrical power unit into mechanical.
1.2 is Ballast factor (such as choke), we can take 1 to 1.25
If,
No of lights
=5
Power of a single light = 40
Then,
Total power
= 5*40
= 200 watts
Heat gain through lights = 200*3.41*1.2
= 818.4 Btu/hr
4.3 Appliances
Heat gain = Power*3.41*Usage factor
1 watt = 3.41 Btu/hr
3.41 is a constant for converting Watts electrical power unit into mechanical.
Usage factor is 1 for 24 hours working.
If,
Power = 500 watts
Then,
Heat gain = 500*3.41*1
= 1705 Btu/hr
TOTAL SENSIBLE INTERNAL HEAT = People + Lights + Appliances
= 1250 + 818.4 + 1705
= 3773.4 Btu/hr
5. HEAT GAIN THROUGH GLASS
Orientation = West (from figure)
Area of 1 window
= 3*3 = 9 ft²
Total area of windows in west side = 2*9 = 18 ft²
Heat gain = Area*Δt*U
A = Area of window
Δt = Equivalent temperature difference / Solar factor
U = Overall value
Method of finding Δt
Δt is get From data book PAGE NO: 42 (41 to 46) by considering LATITUDE ( in this case, it is 8.3˚N
from section 1. )
Available charts for finding solar heat gain through ordinary glass is, LATITUDE 0˚,10˚,20˚,30˚,40˚,50˚.
 For this case we can choose chart of 10˚ LATITUDE.
 Heat gain is maximum on summer season, so take “JULY 23 & MAY 21” in “TIME OF YEAR”
section
 Window is in west direction (in this case), so take peak value ( ie, in peak time) in WEST
direction
 From above procedure we get Δt =158 Btu/hr
Method of finding U value
U value for Ordinary glass = 0.86 Btu/hr ft²
ie,
Heat gain
= A*Δt*U
= 18*158*0.86
= 2445.86 Btu/hr
Similarly, we can find heat gain through glass in south direction
Orientation = South (from figure)
Area of 1 window
= 3*3 = 9 ft²
Total area of windows in west side = 1*9 = 9 ft²

From above procedure we get Δt =14 Btu/hr
ie,
Heat gain
= A*Δt*U
= 9*14*0.86
= 108.36 Btu/hr
TOTAL HEAT GAIN THROUGH GLASS = HEAT GAIN THROUGH WEST + HEAT GAIN THROUGH SOUTH
= 2445.86 + 108.36
= 2554.22 Btu/hr
6. TRANSMISSION GAIN THROUGH ROOF
Heat gain = Area*Δt*U
A = Area of window
Δt = Equivalent temperature + Correction factor
U = Overall value
Roof Area
= 10*15 = 150 ft²
Roof Material = Construction type, 6” thickness concrete and 1” thickness plaster
Method of finding U value
U vale finding is by considering ROOF MATERIAL, from data book PAGE NO: 54 & 55





Choose CONCRETE, SEND & GREVEL AGG (because, here construction type is concrete) in
“TYPE OF DECK” section.
Choose 4, 6, 8, [47], [70], [93] (because, here concrete of thickness is 6” is used) in
“THICKNESS OF DESK (INCHES) AND WEIGHT (LB PER SQ FT)” section.
(where the values in rectangular brackets [] are WEIGHT OF ROOF corresponding to the
THICKNESS OF DESK, it is required for the calculation of equivalent temperature. ie, for 4
inch, weight of roof is 47 lb/ft²)
Choose SUSPENDED PLASTER {6} (here plaster is used) in “CEILING” section.
Take values for 1” (ie, in this case plaster thickness is 1”) in “INSULATION ON TOP OF DECK,
INCHES” section.
From above procedure we get U = 0.16 Btu/hr ft².
Method of finding Δt
Δt = Equivalent Temperature + correction factor
 Finding of equivalent temperature is with respect to weight of roof. Weight of room is get
from the table used to find U value.
 From data book PAGE NO: 54 table for finding u value, we get the WEIGHT OF ROOF
corresponding to the THICKNESS OF DESK. (here thickness of desk is 6 inch, ie weight of roof
is 70 lb/ft²)
 From data book PAGE NO: 49 take the peak values for 60 lb/ft² (ie 70 lb/ft² is not available)
by considering CONDITION as EXPOSED TO SUN
 From above procedure We get equivalent temperature = 39˚F
To get correction factor, we need difference between Outdoor temperature and room temperature.
That we find in section 1.2 as difference between required and actual condition, and Daily range in
section 1 as weather.
ΔDB
= 17.1˚F
Daily range
= 15˚F
From data book PAGE NO:49, TABLE NO: 6 `by matching “OUTDOOR DESIGN FOR MONTH AT 3PM
MINUS ROOM TEMPERATURE” as 20˚F (higher than 17.1˚F) and “DAILY RANGE” as 16˚F (15 is not
available, 16 is next higher value), we get CORRECTION FACTOR as 7.
Δt
= Equivalent temperature + Correction factor
= 39 + 7
Δt
= 46˚F
Heat gain
= Area*Δt*U
= 150*46*0.16
= 1104 Btu/hr
7. TRANSMISSION GAIN THROUGH WALL
Heat gain = Area*Δt*U
A = Effective wall area
Δt = Equivalent temperature + Correction factor
U = Overall value
Roof Area
Wall Material
= 10*15 = 150 ft²
= Construction type, 8” thickness stone and 3/4” thickness gypsum board plaster
WEST WALL AREA
Total wall area
Area of windows in west
Effective area of west wall
= length*height = 10*10 = 100 ft²
= 2 window = 18 ft²
= 100 – 18 = 82 ft²
SOUTH WALL AREA
Total wall area
Area of windows in south
Effective area of south wall
= 15*10 = 150 ft²
= 1 window = 9 ft²
= 150 – 9 = 141 ft²
Method of finding U value
U vale finding is by considering WALL MATERIAL, from data book PAGE NO:50 (50 to 53)




Choose STONE (because, here stone is used for construction) in “EXTERUIR FINISH” section.
Choose 8 (100) (because, here thickness of stone is 8”) in “THICKNESS (INCHES) AND
WEIGHT (LB PER SQ FT)” section.
(where the values in brackets () are WEIGHT OF WALL corresponding to the THICKNESS OF
DESK, it is required for the calculation of equivalent temperature. ie, for 8-inch thickness
stone, weight of wall is 100 lb/ft²)
Take value corresponding to “¾” GYPSUM BOARD (PLASTER BOARD)” in INTERIOR FINISH
section.
From above procedure we get U = 0.55 Btu/hr ft
Method of finding Δt
Δt = Equivalent Temperature + correction factor
 Finding of equivalent temperature is with respect to weight of wall and Exposure (direction)
of wall. Weight of room is get from the table used to find U value.

From data book PAGE NO: 50 table for finding u value, we get the WEIGHT OF WALL
corresponding to the THICKNESS. (here thickness of stone is 8 inch, ie weight of wall is 100
lb/ft²)
WEST WALL
 From data book PAGE NO: 48 take the values for EXPOSURE as WEST and WEIGHT OF WALL
as 100 lb/ft² at 6pm
 From above procedure We get equivalent temperature = 20˚F
Correction factor is same as that of roof.
To get correction factor, we need difference between Outdoor temperature and room temperature.
That we find in section 1.2 as difference between required and actual condition, and Daily range in
section 1 as weather.
ΔDB
= 17.1˚F
Daily range
= 15˚F
From data book PAGE NO:49, TABLE NO: 6 `by matching “OUTDOOR DESIGN FOR MONTH AT 3PM
MINUS ROOM TEMPERATURE” as 20˚F (higher than 17.1˚F) and “DAILY RANGE” as 16˚F (15 is not
available, 16 is next higher value), we get CORRECTION FACTOR as 7.
Δt
Δt
= Equivalent temperature + Correction factor
= 20 + 7
= 27˚F
SOUTH WALL
 From data book PAGE NO: 48 take the values for EXPOSURE as SOUTH and WEIGHT OF WALL
as 100 lb/ft² at 6pm
 From above procedure We get equivalent temperature = 18˚F
Correction factor is same as that of roof.
To get correction factor, we need difference between Outdoor temperature and room temperature.
That we find in section 1.2 as difference between required and actual condition, and Daily range in
section 1 as weather.
ΔDB
= 17.1˚F
Daily range
= 15˚F
From data book PAGE NO:49, TABLE NO: 6 `by matching “OUTDOOR DESIGN FOR MONTH AT 3PM
MINUS ROOM TEMPERATURE” as 20˚F (higher than 17.1˚F) and “DAILY RANGE” as 16˚F (15 is not
available, 16 is next higher value), we get CORRECTION FACTOR as 7.
Δt
Δt
= Equivalent temperature + Correction factor
= 18 + 7
= 25˚F
Heat gain through west wall
= Area of west wall * Δt of west * U
= 82 * 27 * 0.55
= 1217.7 Btu/hr
Heat gain through south wall
Total heat gains through wall
= Area of south wall * Δt of south * U
= 141 * 25 * 0.55
= 1938.75 Btu/hr
= 1217.7 + 1938.75
= 3156.45 Btu/hr
8. HEAT GAIN THROUGH THE UNCONDITIONED SPACE
Heat gain = Area*Δt*U
A = Transmission area
Δt = (Outside temperature - 5˚F) – Inside temperature
Where, from (Outside temperature - 5˚F) is mean by the temperature of
nearest unconditioned space, it’s not same as that of outside air.
U = Overall value
Inside wall material = Hollow concrete block, 4” thickness, both side finished, 3/8” gypsum
board plastering.
ΔT
= (Outside temperature - 5˚F) – Inside temperature
= (92.1 – 5) – 75
= 12.1˚F
Total area
Door area
Transmission area
= 10*10=100 ft²
= 3*7 = 21 ft²
= 100 – 21 = 79 ft²
Method of finding U value
U vale finding is by considering INSIDE WALL MATERIAL, from data book PAGE NO: 53





Choose HOLLOW CONCRETE BLOCK (because, here construction type is hollow concrete) in
“RACKING” section.
Choose 4 (20) (because, thickness is 4”) in “THICKNESS (INCHES) AND WEIGHT (LB PER SQ
FT)” section.
(where the values in brackets () are WEIGHT OF INSIDE WALL corresponding to the
THICKNESS.)
Choose NO. OF SIDES FINISHED as Both
Take value corresponding to “3/8” GYPSUM BOARD (PLASTER BOARD)” in FINISH section.
From above procedure we get U = 0.32 Btu/hr ft².
Heat gain
= Area * Δt * U
= 79 * 12.1 * 0.32
= 305.89 Btu/hr
9. INFILTRATION
.DSASADA
VOLUME (ft2 ) × ACH (Air change per hour)
CFM =
60
ACH (Air change per hour) = 0.5 to 1.5
Cfm =
1500 ×1
60
Specific heat formula
Qs = 1.08*CFM*ΔT
Where, Qs= sensible heat
gain.
1.08=sensible heat value.
CFM=dry air
ΔT= difference of required
and actual temperature.
= 25 cfm
INFILTRATION HEAT GAIN = CFM * ΔT * 1.08
CFM – Infiltration
ΔT = (Outside temperature - 5˚F) – Inside temperature
1.08 = Sensible heat value (for more details refer last page)
Infiltration Heat gain
10.
Q=m*c*ΔT
Where, Q= heat added/
subtracted to that
substance/ control volume.
m= mass flow into that
substance/ control volume.
ΔT= difference of required
and actual temperature.
= CFM*ΔT*1.08
= 25 * [(92.1 – 5) – 75] * 1.08
= 326.7 Btu/hr
BYPASSED OUT SIDE AIR
HEAT GAIN BY BYPASSED AIR = CFM * ΔT * 1.08 * B.F
CFM – OA VENTILATION REQUIREMENT (from section 3)
ΔT = (Outside temperature – Inside required temperature)
1.08 = Sensible heat value
B.F = Bypass factor, normally we take 0.12
Heat gain by bypassed air
11.
= CFM * ΔT * 1.08 * B.F
= 35 * (92.1 – 75) * 1.08 * 0.12
= 70.92 Btu/hr
TOTAL SENSIBLE HEAT
TOTAL SENSIBLE HEAT = TOTAL SENSIBLE INTERNAL HEAT
SOLAR GAIN THROUGH GLASS
TRANSMISSION GAIN THROUGH ROOF
TRANSMISSION GAIN THROUGH WALL
TRANSMISSION GAIN THROUGH
UNCONDITIONED SPACE
INFILTRATION
BYPASSED OUTSIDE AIR
+
+
+
+
+
+
TOTAL SENSIBLE HEAT
Adding 10% extra for safety
= 3773.4 + 2554.22 + 1104 + 3156.45 + 305.89 + 326.7 + 70.92
= 11,291.58 Btu/hr
= 1.1 * 11,291.58
= 12,420.738 Btu/hr
EFFECTIVE ROOM SENSIBLE HEAT [ERSH]
12.
= 12,420.738 Btu/hr
TOTAL LATENT HEAT
TOTAL SENSIBLE HEAT = LATENT HEAT DUE TO PEOPLE
+
LATENT HEAT DUE TO BYPASSED OA +
LATENT HEAT DUE TO INFILTRATION
12.1
Latent heat due to people
Heat gain
= No. of people * Latent heat/person
For normal person, latent heat/person = 200 Btu/hr
Heat gain = No. of people * Latent heat/person
= 5 * 200
= 1000 Btu/hr
12.2
Latent heat due to Bypassed outside air
Bypassed OA = CFM * Δw * B.F * 0.68
CFM - OA VENTILATION REQUIREMENT (from section 3)
Δw = Grain size difference, or specific humidity difference (from section 1.2)
B.F = Bypass factor, normally we take 0.12
0.68 = Latent heat value
Bypassed OA
= CFM * Δw * B.F * 0.68
= 35 * 64.5 * 0.12 * 0.68
= 184.212 Btu/hr
12.3
Latent heat due to infiltration
Infiltration
= CFM * Δw * 0.68
CFM – Infiltration Air (from section 9)
Δw = Grain size difference, or specific humidity difference (from section 1.2)
0.68 = Latent heat value
Infiltration
= CFM * Δw * 0.68
= 25 * 64.5 * 0.68
= 1096.5 Btu/hr
TOTAL SENSIBLE HEAT
Adding 10% extra for safety
= LATENT HEAT DUE TO PEOPLE
+
LATENT HEAT DUE TO BYPASSED OA +
LATENT HEAT DUE TO INFILTRATION
= 1000 + 184.212 + 1096.5
= 2,280.712 Btu/hr
= 1.1 * 2280.71
= 2508.78 Btu/hr
EFFECTIVE ROOM LATENT HEAT [ERLH]
13.
= 2508.78 Btu/hr
EFFECTIVE ROOM TOTAL HEAT [ERTH]
EFFECTIVE ROOM TOTAL HEAT = EFFECTIVE ROOM SENSIBLE HEAT + EFFECTIVE ROOM LATENT HEAT
ERTH = ERSH + ERLH
ERTH = 12,420.738 + 2508.78
= 14,929.518 Btu/hr
14.
OUTSIDE AIR HEAT (EXCLUDING BYPASS FACTOR)
TOTAL HEAT
= ΔHS + ΔHL
Where, ΔHs = Sensible heat
ΔHL = Latent heat
14.1
Sensible heat
ΔHS = Q * ΔT * (1 – B.F) * 1.08
Q = O.A VENTILATION REQUIRED (from section 3), or fresh air required
ΔT = (Outside temperature – Inside required temperature)
1.08 = Sensible heat value
(1 - B.F) = Excluding bypass factor
Q
ΔT
= 35 cfm
= (92.1 – 75)
= 17.1˚F
B. F = 0.12
ΔHS = Q * ΔT * (1 – B.F) * 1.08
= 35 * 17.1 * (1 – 0.12) * 1.08
= 568.81 Btu/hr
14.2
Latent heat
ΔHL = Q * Δw * (1 – B.F) * 0.68
Q = O.A VENTILATION REQUIRED (from section 3), or fresh air required
ΔT = Specific humidity (from section 1.2)
0.68 = Latent heat value
(1 - B.F) = Excluding bypass factor
Q
= 35 cfm
ΔT = 64.5 gr/lb
B. F = 0.12
ΔHS = Q * Δw * (1 – B.F) * 0.68
= 35 * 64.5 * (1 – 0.12) * 0.68
= 2145.53 Btu/hr
TOTAL HEAT = ΔHS + ΔHL
= 568.81 + 2145.53
= 2714.34 Btu/hr
Adding 5 % extra for safety
= 1.05 * 2714.34
= 2850.06 Btu/hr
15.
GRAIN TOTAL
GRAIN TOTAL = EFFECTIVE ROOM TOTAL HEAR [ERTH] + TOTLA OUTSIDE AIR HEAT
= 14,929.518 + 2850.06
= 17,779.578 Btu/hr
16.
TONS OF REFRIGERATION
1 TON = 12,000 Btu/hr
Tons of refrigeration =
17,779.578
12,000
= 1.48 TR
17.
COIL APPARATUS DEW POINT (COIL ADP)
To get coil ADP from data book, we need
i.
Inside room temperature (from section 1.1)
ii.
R.H (from section 1.1)
iii.
Effective room sensible heat factor (ESHF)
ESHF
ESHF
=
=
ERSH
ERTH
12,420.738
14,929.518
= 0.8316
= 83.16%
From data book PAGE NO: 67 By using this data’s we get coil ADP = 52˚F
18.
SUPPLY AIR
DEHUMIDIFIED AIR RISE
DEHUMIDIFIED AIR RISE = (1- B.F) * (ROOM D.B – COIL ADP)
Dehumidified air rise = (1 – B.F) * (room D.B – coil ADP)
= (1 – 0.12) * (75 – 52)
= 20.24˚F
DEHUMIDIFIED AIR QUANTITY (SUPPLY AIR)
VOLUME OF SUPPLY
VOLUME OF SUPPLY
=
=
=
ERSH
1.08×DEHUMIDIFIED AIR RISE
ERSH
1.08×DEHUMIDIFIED AIR RISE
12,420.738
1.08×20.24
= 568.22 cfm
19.
RESULT
Tonnage
Supply air
Fresh air
Coil ADP
= 1.4 TR
= 568.22 cfm
= 35 cfm
= 52˚F
Specific heat formula
Qs = 1.08*CFM*ΔT
Where, Qs
1.08
CFM
ΔT
Where, Q
M
ΔT
= sensible heat gain.
=sensible heat value.
=dry air
= difference of required and actual temperature.
Q
= m*c*ΔT
= heat added/ subtracted to that substance/ control volume.
= mass flow into that substance/ control volume.
= difference of required and actual temperature.
From a chart “Ideal gas specific heat of various common gases at 80˚F)
We get, for
Air, cp, a
= 0.24 Btu/lb ˚F
moist air/steam, cp, v
= 0.445 Btu/lb ˚F
Abbreviation used,
Cp = specific heat at constant pressure
Cp, a = spec. heat at const. press. of air
Cp, v = spec. heat at const. press. of vapour
ma = mass of air
mv = mass of vapour
Q̇ = heat flow
ṁ a = mass flow of air (lb/hr)
ṁ v = mass flow of vapour (lb/hr)
Q
= ma cp, a (ΔT) + mv cp, v (ΔT)
By dividing both side by t (time), we get its rate,
Q̇
= ṁ a cp, a (ΔT) + ṁ v cp, v (ΔT)
Substitute the values for cp, a & cp, v
Q̇
= ṁ a (0.24) (ΔT) + ṁ v (0.445) (ΔT)
For typical air-conditioning process
Q̇
ṁv
ṁa
= 0.01
= ṁ a (0.24) (ΔT) + (0.01) ṁ a (0.445) (ΔT)
By rearranging
Q̇
= ṁ a (ΔT) [0.24 + 0.0045]
In many case, we can neglect the value 0.0045, it won’t make any difference in final answer.
Q̇
= ṁ a (ΔT) (0.24)
Conversion of ṁ a from lb/hr to cfm.
ṁ a
lb
hr
=
𝑓𝑡²
𝑚𝑖𝑛
×
60 𝑚𝑖𝑛
1 ℎ𝑟
×
1 𝑙𝑏
13.3 𝑓𝑡³
Q̇
= (4.5 CFM) (ΔT) (0.24)
Q̇
= 1.08 * CFM * ΔT
= 4.5 𝐶𝐹𝑀