R & AC Load Calculations In refririgration system

REFRIGERATION AND AIR
CONDITIONING
Dr. Muzaffar Ali
Lecture # 12

Cooling and Heating Load Calculations
2

Human Thermal Comfort
3
Operative temperature. 20 to 26°C
Relative Humidity. 30-60%
A dew-point temperature of 2 to 17°C
Average air velocity. Up to 0.25 m/s

Manual Calculation Methods
Method Best For Accuracy Complexity
CLTD/SCL/CLF (Cooling
Load Temperature
Difference / Solar Cooling
Load / Cooling Load Factor)
Cooling load
calculations
High Medium
ETD/ETR (Equivalent
Temperature Difference /
Equivalent Temperature
Radiance)
Insulated
buildings
Moderate Medium
Degree-Day
Annual load
estimation
Low Easy
Heat Balance
Precise load
calculation
Very High High
Bin Method
Seasonal energy
estimation
Moderate Medium

Brief Overview of Manual Calculation
Methods
• CLTD/SCL/CLF Method: Uses empirical data from ASHRAE to calculate
cooling loads, accounting for conduction, solar heat gain, and internal heat
sources. Suitable for quick manual HVAC load calculations.
• ETD/ETR Method: Considers equivalent temperature differences for walls
and roofs to estimate heat gain in insulated buildings. Commonly used in
envelope heat gain calculations.
• Degree-Day Method: Estimates annual heating and cooling energy
requirements using historical temperature data and degree-day factors.
Suitable for broad energy consumption analysis.
• Heat Balance Method: The most precise approach, solving energy
balance equations for each building surface. Requires detailed input data
and is mainly used in computational models.
• Bin Method: Categorizes outdoor temperature ranges into bins and
calculates load contributions accordingly. Useful for seasonal energy
estimations and HVAC system performance analysis.

Software-Based Methods
Software Methodology Best For Key Features
HAP Transfer Function
Method
HVAC load
calculations
Used for HVAC load calculations,
widely used in industry
EnergyPlus Heat Balance
Method
Detailed energy
modeling
Detailed energy modeling with
advanced physics-based
calculations
TRNSYS Component-
Based Simulation
Research &
transient analysis
Research-focused, excellent for
transient system analysis
IES VE Heat Balance
Method
Green buildings
& LEED
Ideal for green buildings and
LEED certification modeling
eQUEST DOE-2 Simplified energy
modeling
Free, simplified, and widely used
for quick energy analysis
DesignBuilder EnergyPlus GUI Easy energy
modeling
User-friendly, integrates
EnergyPlus for detailed
simulations

Best Method Based on Use Case
Use Case Recommended Method
Manual HVAC Load
Calculation
CLTD/SCL/CLF Method
High-Accuracy Energy
Simulation
EnergyPlus or IES VE
Research (CSP, HVAC, Solar) TRNSYS
LEED Certification IES VE
Quick Energy Analysis eQUEST
User-Friendly Energy
Modeling
DesignBuilder

Recommendations
 For Manual Calculations: CLTD/CLF method is most
practical.
 For Software-Based Analysis: IES VE for green
buildings, TRNSYS/EnergyPlus for detailed
simulations.
 For HVAC System Design: HAP is widely used.
 Choose the right tool based on accuracy needs,
complexity, and project requirements.

Design Conditions
 Indoor  outdoor
11
• For heating operation an indoor temperature of 20 to 22°C is generally
assumed,
• For cooling operation 24 to 26°C is typical.
• A minimum relative humidity of 30% in the winter and a maximum of 60%
in the summer is also assumed.
• For heating operation the 97 5 percent value of the outside temperature is-
usually chosen.
• This means that on a long-term basis the outside dry-bulb temperature equals
or exceeds this value for 97.5 percent of the hours during the coldest months
of the year.
• At the 97.5 percent outdoor temperature the air is assumed to be saturated.

Categories of heating and cooling
loads.
 Internal Load
 External Load…
Solar
15
Solar. Heat gain due to transmission of solar energy through a transparent building
component or absorption by an opaque building component
Infiltration. Heat loss or heat gain due to the infiltration of outside air into a
conditioned space
Internal. Heat gain due to the release of energy within a space (lights, people,
equipment, etc.)

Cooling Load Components
roof
lights
equipment
floor
exterior
wall
glass solar
glass
conduction
infiltration
people
partitio
wall

Cooling Load Components
Sensible
Load
latent
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
space
load
coil
load
cooling load components

Building Envelope Factors
 Heat transfer through a
building envelope is
influenced by:
 the materials used;
 by geometric factors
such as size, shape, and
orientation;
 by the existence of
internal heat sources;
and
 by climatic factors.
18

External Loads
 Solar Radiation by windows
 Heat Conduction through:
 Walls
 Roof
 Floor
 Doors
19

External Loads- Building Data
 Location (lat. and long.)
 2D/3D map with dims
 Materials of walls, roof, windows,
floor, door etc.
 Size of all above
 Application/Utility of building
 Orientation of building ( E, W, N, S)
 Locations of windows, doors
 No. of windows and doors
 No. of people, gender, age, and
occupancy rate
 Nature of clothing
 Others
20

Architectural Layout
Understanding Architectural layout and Client's requirement is the key parameter in HVAC
design

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
gypsum
board
concrete block
aluminum
siding
• aluminum siding (R = 0.11 m2•ºK/W)
• 200 mm lightweight concrete block (R = 0.35]
• 3.5 in. [90 mm] of fiberglass insulation (R = 2.29)
• 12.7 mm gypsum board (R = 0.08)
• a film of air on the outside surface of the wall (R
= 0.044, assuming air moving at 12 km/hr
• a film of air on the inside surface of the wall (R =
0.12, assuming still air

U-factor for Example Wall
thermal resistance (R)
Routdoor-air film 0.25 [0.04]
Rsiding 0.61 [0.11]
Rconcrete block 2.00 [0.35]
Rinsulation 13.00 [2.29]
Rgypsum board 0.45 [0.08]
Rindoor-air film 0.68 [0.12]
Rtotal 16.99 [2.99]
]
U =
Rtotal
1
U = 0.06 Btu/hr•ft2•°F
[ U = 0.33 W/m2•°K ]
•100 mm concrete with 90 mm insulation and steel decking
U-factorroof = 0.057 Btu/hr•ft2•°F [0.323 W/m2•°K]

Sunlit Surfaces
sun
rays
solar angle changes throughout the day

Time Lag
Solar
Effect
12 6 12 6 12
noon
a.m. p.m. mid
mid
A B
Time Lag

Q = U × A × CLTD
Conduction through Sunlit Surfaces
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
21 17 14 11 8 7 6 6 7
CLTD
(°F)
35 30 25
6 7 8 9 10 11 12
1 2 3 4 5 13 14 15 16 18 19
17 20 21 22 23 24
CLTD
(°C)
8 10 12 16 22 30 37 44 48
48 45 41
12 9 8 6 4 4 3 3 4
19 17 14 4 6 7 9 12 17 21 24 27
27 25 23
• 78ºF [25.6ºC] indoor air
• 95ºF [35ºC] maximum outdoor air
• Average outdoor daily temperature
range of 21ºF [11.7ºC]
• 21st day of July
• 40º north latitude
• Dark-colored surface
• Wall Type 9. At 4 p.m. (Hour 17)
∆T = 35 – 25.6 = 9.4ºC
“effective temperature difference”
West wall at 5 p.m. CLTDhour=17 = 12ºC
Roof at 5 p.m. CLTDhour=17 = 44ºC

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 ]
Conduction heat gain through the roof:
•Area of roof = 45 ft x 60 ft = 2,700 ft2 [13.7 m x 18.3 m = 250.7 m2]

Q=U*A*(CLTD)corr
Where:
(CLTD)corr =(CLTD + LM) K + (25.5 – Ti )+ (To – 29.4)
Where :
CLTD: cooling load factor
K:color factor: K=1 dark color
K=0.83 medium color
K=0.5 light color
For walls :

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

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 Heat Gain through Glass
Q = A × SC × (SCL)
Where,
SC – Shading Coefficient
SCL – Solar Cooling Load Factor

For glass :
Heat transmitted through glass
Q=A*(SHG)*(SC)*(CLF)
SHG: solar heat gain
SC: shading coefficient
CLF: cooling load factor

Solar Heat Gain Factor (SHGF/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
uncoated single glazing
1/4 in. [6.4 mm] clear
1/4 in. [6.4 mm] green
reflective single glazing
1/4 in. [6.4 mm] SS on clear
1/4 in. [6.4 mm] SS on green
uncoated double glazing
1/4 in. [6.4 mm] clear - clear
1/4 in. [6.4 mm] green - clear
reflective double glazing
1/4 in. [6.4 mm] SS on clear - clear
1/4 in. [6.4 mm] SS on green - clear
0.82
aluminum frame other frames
operable fixed
0.85 0.69 0.82
0.59 0.61 0.49 0.59
0.26 0.28 0.22 0.25
0.26 0.28 0.22 0.25
0.70 0.74 0.60 0.70
0.48 0.49 0.40 0.47
0.18 0.18 0.15 0.16
0.20 0.18 0.15 0.17
SS = stainless-steel reflective coating
operable fixed

Solar Radiation through Windows
Qwindows = 160 × 0.74 × 192 = 22733 Btu/hr
[ Qwindows = 14.4 × 0.74 × 605 = 6447 W ]

Internal Heat Gains
People
Equipment
Appliances
Lights

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
Gain
Latent Heat
Gain
Moderately active work
(Office)
250 BTU/hr
(75W)
200 BTU/hr
(55W)
Standing, light work,
walking (Store)
250 BTU/hr
(75W)
200 BTU/hr
(55W)
Light bench work (Factory) 275 BTU/hr
(80W)
475BTU/hr
(140W)
Heavy work (Factory) 580BTU/hr(170W
)
870BTU/hr
(255W)
Exercise (Gymnasium) 710BTU/hr
(210W)
1090BTU/hr
(315W)

CLF Factors for People
Hours after people enter space
0.11 0.08 0.06 0.05 0.04 0.03 0.02 0.02 0.01
0.65 0.74 0.16
6 7 8 9 10 11 12
1 2 3 4 5
Total hours in
space
2
4
6
8
10
0.65
0.65
0.65
0.65
0.85 0.24 0.17 0.13 0.10 0.07 0.06 0.04 0.03
0.75 0.81
0.85 0.89 0.91 0.29 0.20 0.15 0.12 0.09 0.07
0.75 0.81
0.85 0.89 0.91 0.93 0.95 0.31 0.22 0.17 0.13
0.81
0.75
0.85 0.89 0.91 0.93 0.95 0.96 0.97 0.33 0.24
0.81
0.75
Note: CLF – Cooling Load Factor
Capacity of a space to absorb and store heat.
If the space is not maintained at a constant temperature during the 24-hour period,
however, the CLF is assumed to equal 1.0.

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 ]
Internal heat gain from lighting:
• Amount of lighting in space = 2 W/ft2 [21.5 W/m2]
• Floor area = 45 ft x 60 ft = 2,700 ft2 [13.7 m x 18.3 m = 250.7 m2]
• Total lighting energy = 21.5 W/m2 x 250.7 m2 = 5,400 W
• Ballast factor = 1.2 (fluorescent lights)
• CLF = 1.0 (because the space temperature set point is increased at night)

Heat generated by equipment
Equipment Sensible Heat Gain Latent Heat Gain
Coffee maker 3580 BTU/hr
(1050W)
1540 BTU/hr
(450W)
Printer 1000 BTU/hr
(292W)
Typewriter 230 BTU/hr
(67W)

Methods of Estimating Infiltration
 Air change method
the quantity of infiltration air is estimated using the equation:
infiltration airflow = (volume of space x air change rate) ÷ 60 min/hr
[infiltration airflow = (volume of space x air change rate) ÷ 3,600 s/hr]
where,
• Infiltration airflow = quantity of air infiltrating into the space, cfm [m3/s]
• Volume of space = length x width x height of space, ft3 [m3]
• Air change rate = air changes per hour (ACH, volume per hr)
•Crack method is a little more complex and is based upon the average
quantity of air known to enter through cracks around windows and doors
when the wind velocity is constant
•Effective leakage-area method takes wind speed, shielding, and
“stack effect” into account, and requires a very detailed calculation.

Infiltration Airflow
Infiltration
airflow
32400 × 0.3
60
= = 162 CFM
Infiltration
airflow
927.6 × 0.3
3600
= = 0.077 m3/s
Assuming that the space is of average construction and kept at a positive pressure relative to the
outdoors, we estimate 0.3 air changes/hr of infiltration.
volume of space = 45 ft x 60 ft x 12 ft = 32,400 ft3
[13.7 m x 18.3 m x 3.7 m = 927.6 m3]

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]
air at “standard” conditions (69°F [21°C] dry air at sea level)
• Density = 0.075 lb/ft3 [1.2 kg/m3]
• Specific heat = 0.24 Btu/lb•°F [1,004 J/kg•°K]
• Latent heat of water vapor = 1,076 Btu/lb [2,503 kJ/kg]
0.075 x 0.24 x 60 min/hr = 1.085 [1.2 x 1,004 = 1,210]

Heat Gain from Infiltration
QS = 1.085 × 162 × (95 – 78) = 2,988 Btu/hr
[ QS = 1,210 × 0.077 × (35 – 25.6) = 876 W ]
QL = 0.7 × 162 × (105 – 70) = 3,969 Btu/hr
[ QL = 3,010 × 0.077 × (15 – 10) = 1,159 W ]
Heat gain from infiltration:
• Infiltration airflow = 162 cfm [0.077 m3/s]
• Outdoor conditions: 95ºF [35ºC] dry bulb and 76ºF [25ºC] wet bulb results in Wo = 105 grains of
water/lb dry air [15 grams of water/kg dry air]
• Indoor conditions: 78ºF [25.6ºC] dry bulb and 50% relative humidity results in Wi = 70 grains of
water/lb dry air [10 grams of water/kg dry air]

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

Ventilation
air handler
with fan and
cooling coil
supply duct
diffuser
outdoor-air
intake

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)
• Ventilation airflow = 18 people × 20 CFM/person = 360 cfm [0.18 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 ]
Cooling load due to the conditioning of ventilation air:
• Ventilation airflow = 360 cfm [0.18 m3/s]
• Outdoor conditions: To = 95ºF [35ºC], Wo = 105 grains of water/lb dry air [15 grams of
water/kg dry air]
• Indoor conditions: Ti = 78ºF [25.6ºC], Wi = 70 grains of water/lb dry air [10 grams of water/kg
dry air]

System Heat Gains
air handler
fan motor
Heat gain from a fan is associated with three energy conversion losses.
• fan motor heat gain = power input to motor × (1 – motor efficiency)
• fan blade heat gain = power input to fan × (1 – fan efficiency)
• duct friction heat gain = power input to fan × fan efficiency

Components of Fan Heat
blow-through
configuration
draw-through
configuration
the fan heat causes an increase in the
temperature of the air entering the coil
the fan heat causes an increase in the
temperature of the air supplied to the space

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

Space Load versus Coil Load
space
load
coil
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
sensible load
Btu/hr [W]
people
lights
equipment
infiltration
12,312 [3,563]
4,500 [1,350]
22,097 [6,480]
8,184 [2,404]
2,988 [876]
74,626 [21,623]
total space cooling load
3,600 [990]
1,540 [450]
3,969 [1,159]
latent load
Btu/hr [W]
9,109 [2,599]
space load components
502 [144]
1,310 [359]
22,733 [6,447]

Sensible Heat Ratio (SHR)
SHR
sensible heat gain
sensible heat gain + latent heat gain
=
= 0.89
74,626
74626 + 9109
SHR =
= 0.89
21623
21623 + 2599
SHR =

Single-Space Analysis
space
supply
fan
cooling
coil
outdoor
air
return
air
supply
air
exhaust
air

Determine Supply Airflow
sensible heat gain
supply
airflow
=
1.085 × (room DB – supply DB)
sensible heat gain
supply
airflow =
1,210 × (room DB – supply DB)

Determine Supply Airflow
74,626
2,990 cfm
=
1.085 × (78 – 55)
supply
airflow =
21,623
1.40 m3/s
=
1,210 × (25.6 – 12.8)
supply
airflow =
Assume the supply air dry-bulb temperature is 55ºF [12.8ºC].

Calculate Entering Coil Conditions
ventilation airflow
% outdoor air =
total supply airflow
360 cfm
%OA =
2990 cfm
= 0.12
0.18 m3/s
%OA =
1.40 m3/s
= 0.12

Calculate Entering Coil Conditions
B
A C
95°F × 0.12 = 11.4°F
78°F × 0.88 = 68.6°F
mixture = 80.0°F
35°C × 0.12 = 4.2°C
25.6°C × 0.88 = 22.5°C
mixture = 26.7°C
dry-bulb temperature
humidity
ratio
95°F
[35°C]
76°F
[24.4°C]
80°F
[26.7°C]
78°F
[25.6°C]
66.5°F
[19.2°C]

Determine Supply Air Temperature
dry-bulb temperature
sensible
heat
ratio
D
59°F
[15°C]
B
A
1.0
0.8
0.6
0.4
C
Assume supply air at 90-95%RH depending cooling coil type.
This supply-air condition is 59°F dry bulb, 57.4°F wet bulb
[15°C dry bulb, 14.1°C wet bulb].

Recalculate Supply Airflow
21,623
1.69 m3/s
=
1,210 × (25.6 – 15)
supply
airflow =
74,626
3,620 cfm
=
1.085 × (78 – 59)
supply
airflow =
The supply air temperature is 59ºF [15ºC],
different from assumed as 55ºF [12.8ºC].

Room 101
Btu/hr [W]
total coil cooling load 99,195 [28,978]
ventilation 15,460 [4,756]
Total Cooling Load on Coil
total space sensible load
9,109 [2,599]
74,626 [21,623]
total space latent load
8.3 refrigeration tons

Multiple-Space Analysis
supply
fan
Room 101 Room 102
cooling
coil
* Constant volume (CV) system, the fan is sized by summing
the peak sensible loads for each of the spaces it serves
* Variable-air-volume (VAV) system, the fan delivers a varying amoun
of air to the system, the fan is sized based on the one-time, worst-case
airflow requirement

Room 101 (Faces West)
8 a.m.
Btu/hr [W]
people
lights
equipment
infiltration
conduction through exterior wall 160 [48]
2,616 [740]
202 [51]
3,552 [1,012]
4,500 [1,350]
22,097 [6,480]
8,184 [2,404]
2,988 [876]
44,299 [12,961]
4 p.m.
Btu/hr [W]
4,500 [1,350]
22,097 [6,480]
8,184 [2,404]
2,988 [876]
space sensible load
components
74,626 [21,623]
502 [144]
1,310 [359]
22,733 [6,447]
12,312 [3,563]

Room 102 (Faces East)
8 a.m.
Btu/hr [W]
people
lights
equipment
infiltration
conduction through exterior wall 160 [48]
21,667 [6,138]
4,500 [1,350]
22,097 [6,480]
2,988 [876]
62,414 [18,087]
4 p.m.
Btu/hr [W]
844 [252]
1,310 [359]
3,078 [874]
4,500 [1,350]
22,097 [6,480]
2,988 [876]
55,313 [16,158]
202 [51]
space sensible load
components
8,184 [2,404] 8,184 [2,404]
2,616 [740] 12,312 [3,563]

“Sum-of-Peaks” versus “Block”
Room 101 (faces west)
Room 102 (faces east)
space sensible load
sum-of-peaks = 74626 + 62414 = 137040 Btu/hr
[21623 + 18087 = 39710 W]
block = 74626 + 55313 = 129939 Btu/hr
[21623 + 16158 = 37781 W]
8 a.m.
Btu/hr [W]
4 p.m.
Btu/hr [W]
44299 [12961] 74626 [21623]
62414 [18087] 55313 [16158]
CV
VAV

“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]
CV
VAV

Room 101
Btu/hr [W]
Total coil cooling load
Room 102
Btu/hr [W]
99195 [28978]
ventilation 15460 [4756]
“Block” Cooling Load
79882 [23513]
15460 [4756]
9109 [2599]
74626 [21623]
total space latent load
55313 [16158]
9109 [2599]
Block cooling load
(4 p.m.)
= 99195 + 79882 = 179077 Btu/hr
[28978 + 23513 = 52491 W]
loads at 4 p.m.
14.9 refrigeration tons

80
 Solution of Exercise problems:
 Chapter 03,
 Chapter # 04 and
 Chapter # 05
Assignment # 03

R & AC Load Calculations In refririgration system

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