COMPLEX ENGINEERING PROBLEM

Title:

CALCULATION
ALCULATION OF COOLING LOAD FOR STUDY HALL

Submitted to:

ENGR. Sir Ashiq

Submitted by:

Muhammad Daniyal Imdad

Roll No:

BMEF18M024

Subject:

Refrigeration and Air conditioning (Th.)

Class:

B.Sc. Mechanical Engineering

Department:

Mechanical Engineering

CET

University of Sargodha
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 Table of Content

Rough Sketch………………………………………………………………………………………………………………........3

Abstract……………………………………………………………………………………………………………………………..4

Introduction……………………………………………………………………………………………………………………….4

Methodology………………………………………………………………………………………………………………………5

Numerical Calculations……………………………………………………………………………………………………….8

Analytical Calculations……………………………………………………………………………………………………….13

Result…………………………………………………………………………………………………………………………………25

Conclusion………………………………………………………………………………………………………………………….25

References…………………………………………………………………………………………………………………………25

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Rough Sketch:
18 ft

Electric Incuba- Do
Kettle tor or
Auto
Wall Facing west
Clave

Wall Facing South Wall Facing North

34 ft

34 ft Oven
W
i
n
d
o
East w

North South
W
i
n
d
West
o
w

Do
or
Wall Facing East

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Abstract:
In the hot countries like Africa, the need of air conditioning is going to increase everyday
because of the global warming. In order to manage the Air conditioning or cooling
temperature for buildings, shopping malls, offices, home, airports, marriage halls and
universities etc, it’s necessary to calculate the cooling load. The main reason behind the
determining of the cooling load is to maintain the temperature in the comfort zone
which ranges from 22 to 27 degree Celsius. The cooling load depends on a number of
factors like environment conditions, humidity, direction, building materials and area of
room etc. In the given document, cooling load of a studying hall will be calculated by
theoretical way and then by analytical method. The hall is located in the University of
Sargodha, Iqbal Hall. The outdoor conditions are different because of sun rays. The
cooling load for each component like walls, windows and room will be calculated
separately.

Introduction:
The cooling load is the amount of heat energy that would need to be removed from a
space (cooling) to maintain the temperature in an acceptable range. Sensible heat
causes the air temperature in the space to increase, while latent heat causes the
moisture content in the space to rise. The cooling load in a building may be influenced
by various heat transfer mechanisms, including the building architecture, internal
equipment, inhabitants, and outdoor weather conditions. The cooling load is
determined in order to choose HVAC equipment with sufficient cooling ability to remove
heat from the zone. A zone is usually characterized as an enclosed space within a
building with the intent of monitoring and controlling the zone's temperature and
humidity with a single sensor, or an area with similar heat gains, similar temperature
and humidity control requirements. The cooling load calculations take heat transfer by
conduction, convection and radiation into account.

The cooling load for hall is determined by calculating;

 Cooling load from heat gain for structure

 Cooling load from heat gain by windows
 Cooling load from heat gain by Doors
 Cooling load for humans
 Cooling load for appliances

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Methodology:
Because the greatest cooling load occurs throughout the day and the outside
circumstances vary greatly during the day due to solar radiation, unstable state
processes must be considered when predicting cooling loads. Furthermore, all internal
sources contribute to cooling loads, and ignoring them would result in an overestimate
of the needed cooling capacity and the risk of failing to maintain the required indoor
conditions. As a result, calculating cooling load is intrinsically more difficult.

For a thorough calculation of the zones and whole-building loads, one of the following
three methods should be employed:

I. Transfer Function Method (TFM): This is the most complex of the methods
proposed by ASHRAE and requires the use of a computer program or advanced
spreadsheet.
II. Cooling Load Temperature Differential/Cooling Load Factors (CLTD/CLF):
This method is derived from the TFM method and uses tabulated data to simplify
the calculation process. The method can be fairly easily transferred into simple
spreadsheet programs but has some limitations due to the use of tabulated data.
III. Total Equivalent Temperature Differential/Time-Averaging (TETD/TA): This
was the preferred method for hand or simple spreadsheet calculation before the
introduction of the CLTD/CLF method.

In this document we will use the second method which is CLTD/CLF method. This
method allows cooling load to be calculated manually by use of simple multiplication
factors.

CLTD is a theoretical temperature difference that accounts for the combined effects of
inside and outside air temp difference, daily temp range, solar radiation and heat
storage in the construction assembly/building mass. It is affected by orientation, tilt,
month, day, hour, latitude, etc. CLTD factors are used for adjustment to conductive heat
gains from walls, roof, floor and glass.

CLF accounts for the fact that all the radiant energy that enters the conditioned space at
a particular time does not become a part of the cooling load instantly. The CLF values for
various surfaces have been calculated as functions of solar time and orientation and are
available in the form of tables in ASHRAE Handbooks. CLF factors are used for

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adjustment to heat gains from internal loads such as lights, occupancy, power
appliances.

SCL factors are used for adjustment to transmission heat gains from glass.

GLF is the glass load factor, and its value depends upon location.

Now, we will write some terminologies and equations that will used in order to calculate
the sensible cooling load of a hall.

As we know that the basic conduction equation for heat gain is; q = U A ∆T

Where,

• q = Heat gain in Btu/hr • U = Thermal Transmittance for roof in Btu/hr.ft².°F

• A = area of roof in ft2 • ∆T = Temperature difference in °F

The heat gain is converted to cooling load using the room transfer functions (sol-air
temperature) for the rooms with light, medium and heavy thermal characteristics. The
equation is modified as

Q = U × A × (CLTDc) …………………………………………………………….. (i)

Where,

• Q = cooling load, Btu/hr

• U = Coefficient of heat transfer roof or wall or glass, Btu/hr.ft².°F

• A = area of roof, ft2

• CLTDc = corrected cooling load temperature difference °F. The values of CLTD and U
are determined from tables available on Wikipedia or other websites or it can be
calculated manually by using formulas.

The equation (i) can be used to determine the cooling load for a roof, window,
appliances, glass, doors or any material in the building. The formula is same but the
value U and CLTD varies according to the area, location or temperature.

The CLTDc is calculated as;

CLTDc = CLTD + LM + (78 – tr) + (ta- 85) ……………………………………… (ii)

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Where LM is the latitude and month, tr and ta are the temperatures of room and
atmosphere (outdoor) in Fahrenheit, CLTD is the cooling load temperature difference,
(78 – tr) is indoor design temperature correction and (ta – 85) is outdoor design
temperature correction.

The value of U and LM depend on temperature and outdoor conditions. That is why they
may vary from time to time.

In order to calculate the cooling load for heat removal from windows we used;

Q = A × GLF ……………………………………………………………………… (iii)

Where A is the area of window and GLF is the glass load factor. The GLF is different for
different materials used for window making. The radiations also affect the GLF.

The cooling load for humans and appliances are calculated by using the formula;

Q= q × No. of persons ………………………………………………………. (iv)

Here, the value of q is taken from the btu/hr table which is available on websites.

Or its value may be found directly from btu/hr table values. It can also be found direct
from Wikipedia.

The above equation it may be as;

Q= No. of persons × time × heat

Here,

 No. of persons = how many people inside

 time = length of time they spend inside each day per person (Hours)
 heat = heat loss per person per hour (Watts)

The heat loss may be checked from the btu/hr table.

The cooling load for infiltration and ventilation is calculated by using;

Q = 1.1 × CFM × TC ……………………………………………………………. (v)

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Where;

TC is the temperature difference and CFM is the cubic feet pr minute.

It is calculated as;

CFM = ACH × V/60 ……………………………………………………………. (vi)

Where ACH is the air changes per hr and V is the volume flow rate.

All above cooling loads are sensible heat. Latent heat is the sum of sensible heat and
latent factor. The latent factor also varies according to the conditions.

Numerical Calculations of Cooling Load:

Before doing numerical calculations we will write some data which is given to us and
some standard data taken from the table and will be used in the calculations.

Name Dimension Name Dimension

Length of Hall (L) 34 ft Width of Hall (w) 18 ft
Height of Hall (L) 21 ft Width of windows (w) 3.2 ft
Height of windows (H) 5.9 ft Outside Temperature (ta) 114.8 °F

Inside Temperature (tr) 76 °F GLF (Glass) 10

Heat loss per person(q) 250 btu/hr Heat loss for tube light (q) 60 btu/hr
U (Roof) 0.47 btu/hrft2 U (Wall, every 3in. block ) 0.06 btu/hrft2
U(Wooden Gates) 0.47 btu/hr ft2 Cooling load Q (incubator) 50 btu/hr
Cooling load Q (Kettle) 150 btu/hr Cooling load Q (oven) 50 btu/hr

I’ll name the above table as TABLE A, for giving referencing easily while calculations.

First of all, we will find out the area of Roof, windows and doors.

AS, Area of a roof = L× w

=34 × 18 = 612ft2

Area of a wall facing east = L× w

= 18 × 21 = 378 ft2

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Area of a wall facing west = 378 ft2

Area of a wall facing north = L× w

=34 × 21 = 714 ft2

Area of a wall facing south = 714 ft2

Area of windows facing North = L× H

In this case, as the windows usually are in pairs so for the pairs we have;

= (3.2 × 5.9) + (3.2 × 5.9) = 37.76 ft 2

Area of a Doors facing North = H× w

=5.9 × 1.9 = 11.21 ft 2

In order to calculate the latent heat we have;

Qt = Qs × LF ………………………………………………………… (A)

Here, LF is the latent heat factor whose value is 1.25 (Psychometric Chart) in our case.
Qs Is the sum of all sensible cooling loads from the appliances, doors, windows, persons
etc.

The outside temperature is equal to ta = 114.8 °F

The inside temperature is equal to tr = 76 °F

Now the area and temperatures are defined, the calculation for sensible cooling load is
given below.

In order to calculate the sensible cooling load we have the formula as;

Q = U × A × (CLTDc) From (i)

Q = sensible load
2
U = 0.47 Btu / (hr ft ) for 12 inches for masonry concrete

The value of CLTD and LM can be obtained from the CLTD chart given in Wikipedia.

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To find the value of CLTDc, we have

CLTDc = CLTD + LM + (78 – tr) + (ta- 85) From (ii)

CLTDC= 14 + 2 + (78 - 76) + (114.8 - 85)

CLTDC = 47.8 °F

Hence;

QRoof = 0.47 × 612 × 47.8 = 13749.192 btu/hr

Similarly sensible cooling load for walls facing west and east are given by;

QWall = QEast + QWest ………………………………………. (B)

U = 0.06 Btu/hr ft2 for every 3 inch brick cement wall.

(CLTDc)East = CLTD + LM + (78 – tr) + (ta- 85) From (ii)

(CLTDC)East = 7+ 0 + (78 - 76) + (114.8 - 85)

(CLTDC)East = 38.8 °F

(CLTDc)West = CLTD + LM + (78 – tr) + (ta- 85) From (ii)

(CLTDC)West = 14+ 1 + (78 - 76) + (114.8 - 85)

(CLTDC)West = 46.8 °F

From B, we have; QWall = QEast + QWest

So, QWall = (U × A × (CLTDc))East + (U × A × (CLTDc))West

QWall = A × U ((CLTDc)East + (CLTDc)West)

QWall = 378 × 0.06 × (38.8 +46.8)

QWall = 1941.408 btu/hr

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Moreover the sensible cooling load for humans, doors, windows and appliance are
given by;

Qperson = no. Of persons × q From (iv)

The value of q for a single person = 250 btu/hr. From TABLE A

Its value is different for different places.

Number persons = 40

Qperson = 40 × 250 = 10000 btu/hr

For machines and appliances, the sensible load is given by;

Q = m × cp × (TC) for autoclave

Here, m is the mass of liquid, Tc temperature difference, and cp is the heat capacity of
liquid in the autoclave. The values for these parameters are taken standard from the
given table in Wikipedia.

Q = 50 × 1.56 × (131 - 125)

Q = 468btu/hr

For tube lights;

Q = q × number of tube lights

The value of q for the tube lights have been taken from the CLTD Table.

Q = 60 × 40 = 2400 btu/hr From TABLE A

The sensible cooling load Q for incubator, oven, and kettle can be determined by using
the formula, but we can take the values from CLTD Table or 2001 ASHRAE Fundamentals
Handbook (SI). The link for the book is given in the reference.

Q for incubator, oven, and kettle = 50 + 34 + 150 = 234btu/hr

Now, the Sensible cooling load for windows is given by;

Q = A × GLF From (iii)

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The area for the two windows is as;

A= 37.76 × 2 (for 2 windows)

2
A = 75.52ft

Location= south,

GLF = 10 for glass From TABLE A

The GLF is taken from the CLTD chart.

Now put the values in the equation iii, we have.

Q = 10 × 75.52

Q= 755.2 btu/hr

Now, the sensible heat of doors are given by;

Q = CLTD × U × A

Location of doors = south

U = 0.47 (for wooden gates) From TABLE A.

The value of U for the doors has been taken from the 2001 ASHRAE Fundamentals
Handbook (SI)

CLTD for relevant direction = 55.8

For 2 Doors we have;

Q = 2(55.8 × 0.47 × 11.21) (2 Doors)

Q= 587.98 btu/hr

Now, the Latent heat load is given as

Qt = Qs × LF From Equation A

Where LF is latent factor = 1.25 (from psychometric chart)

QS = SUM of all sensible loads (that are in blocks)

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QS = (13,749.192+1941.408+10000+468+2400+234+755.2+587.98) btu/hr

QS = 30,135.78 btu/hr

Now, put the value in equation A, we have;

Qt = 30,135.78 × 1.25

Qt = 37,669.725 btu/hr

1 tons of AC = 12000 Btu/ hr

For 37,669.725 btu/hr,

37,669.725/ 12000 = 3.139 tons ≅ 3.14

Analytical Calculations of Cooling Load:

In order to achieve the analytical or the actual value of the cooling load, I used HAP
software which is software for the HVAC calculations. Also I used an HVAC calculator to
take out the tonnage AC. The reason behind using the software and calculator is to
ensure that whether the calculation taken out by hands was correct or there some
issues in the calculations.

The figure shows a view of the Hall whose CEP we have find out. While finding out the
values, I take screenshots of the HAP Window. I’m going to add those screenshots here,
so that may be understandable for the reader.

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Weather Properties:

Design Parameters:

Design Temperature:

The dry and wet bulb temperature are taken as standard for the given location i.e. HALL.

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Design Solar:

The Design solar calculations are also standard by using the latitude and longitude of the
given location i.e. IQBAL HALL. The values are taken from the websites.

Simulation:

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Space Properties:

General Properties:

Internal Properties:

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Walls, windows, Doors Properties:

Roof Properties:

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Infiltration:

Floor Properties:

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Partitions:

Air System Properties:

General properties:

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System components:

If we take fewer factors, the tonnage AC will be of our desire. If we select all the factors,
the selection may be affected and may involve different factors which may fail our
tonnage capacity. So, we will only use some major factors.

Zone Components:

Here, the zone is room. So there is only 1 zone

zone, we used.

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Sizing Data:

Equipment:

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REPORT:

The screenshot patched is of the report generated from the HAP software. The
screenshot shows the latent load and the value of sensible cooling load. By using the
both values, we will find out the value of Tonnage of AC.

From the report, we have;

Qt = 40722.5 btu/hr

1 tons of AC = 12000 Btu/ hr

For 40722.5 btu/hr,

40722.5 / 12000 3.393 tons ≅ 3.40

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The last screenshot shows the graph of sensible zone and zone conditioning and is
generated by the data system in the HAP software.

Now, the screenshots of calculations by using online HVAC calculator are;

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It can be seen that there is a bit difference between the values of HVAC calculator and
HAP software. But, we will use the value of HAP software, as it is more accurate then
the HVAC calculator.

Result:
Here, we can see that the Ton of refrigeration in both, software and HVAC calculator is
same. So, for checking the error between the theoretical and Numerical calculations we
will use the formula given below as;

𝐀𝐜𝐭𝐮𝐚𝐥 (𝐀𝐧𝐚𝐥𝐲𝐭𝐢𝐜𝐚𝐥)𝐕𝐚𝐥𝐮𝐞 –𝐓𝐡𝐞𝐨𝐫𝐞𝐭𝐢𝐜𝐚𝐥 (𝐍𝐮𝐦𝐞𝐫𝐢𝐜𝐚𝐥) 𝐕𝐚𝐥𝐮𝐞

%age Error = × 𝟏𝟎𝟎
𝐀𝐜𝐭𝐮𝐚𝐥(𝐀𝐧𝐚𝐥𝐲𝐭𝐢𝐜𝐚𝐥) 𝐕𝐚𝐥𝐮𝐞

Putting the values we have;

. .
%age Error = × 100
.
.
%age Error = × 100
.

%age Error = 0.0764 × 100

%age Error = 7.64%

Conclusions:
The theoretical and Analytical calculations of cooling load were done in the document. It
was concluded that the error between both the calculations is 7.64% which is round
about 8%. This error may be due to some cooling factors or other various factors like
temperature change, humidification, change in CLTD and the values of U, LM, q, etc. it is
concluded from all the discussion and after checking the results that the AC of 3 tons
will be far enough to deal with the heat for removing it and to provide a comfort zone to
the persons in the study room of IQBAL HALL.

References:
 M. Chandrasekhar, Solar Energy Conversion II, 1981
 https://en.wikipedia.org/wiki/Table_of_specific_heat_capacities
 https://timomarquez.files.wordpress.com/2011/08/calorequipos.pdf

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