Heat Loss Calculation in Air Conditioning System Unit

This report presents the calculation of heat loss in an air conditioning system unit using temperature sensors and other equipment. The aim of the report was to determine the heat loss between different temperatures at different power levels of the preheater. The results show that heat loss increases with an increase in temperature in the air conditioning cooling system unit. The report also presents the assumptions made during the calculations and the methodology used.

Abstract
The aim of this report has been achieved since from the calculations in test A the heat loss
between temperature T2 and T4 together with the calculation of the heat loss between
temperature T4 and T5 where all determined and found to be 7.484 J and 7.45 J and 7.442 J and
7.461 J at the first 40% of the relative power of preheater and the second 60% of the relative
power of preheater of power of the heater.
The comparison was made between test A and B and was found out through calculation that heat
loss in test B were higher than those that were calculated in test A, because of a higher change in
temperature in test B.
In addition, the calculation of heat loss in test B between temperature T5 and T1 and also
calculation of heat loss between temperature T4 and T5 were determined to be 14.912 J and
14.935 J and 14.924 J and 14.936 J at the first 40% of the relative power of preheater and the
second 60% of the relative power of preheater.

Introduction
Air conditioning system unit comprises of several components that ranges from recirculating
duct, preheater and reheater, velocity meter, a room model, five temperature sensors (T1 to T5),
sensors RH1 to RH5 and three louvres. The components help the system cool the air below its
dew point, the process involves a release of moisture by a humid air, and is drained and collected
on a drain pan and subsequently from the air conditioning system that will help minimizes the
moisture from the room. The process involves mixing hot a humid air from the outdoor with
stale, the mixed air is cooled and the moisture it contains is exhausted by cooling coil. The
supplied air to the room is first reheated electrically, the moist and heat will be collected from the
surroundings of the room and leaves as air in return, in which its nature will be warmer and
more moist as compared when it initially entered. A fraction of this air in return is mixed with
fresh out of door air and further is supplied back to the room, while the other fraction is lost to
the ambient. It is approximated that the air that will leave the cooling coil will be about 12
degree Celsius to 138 degree Celsius and at 100% relative humidity, while the air in supply is
about 16 degree Celsius to 188 degree Celsius at about 50% relative humidity, always supply of
energy is supposed to be applied on reheat coil to help moderate air at its comfort ability level
before the air is supplied to the room and also the energy is applied to the cooling coil to cool
and dehumidify the process.
The loss of cold air is used to cool again and again warm out of door air by applying a recovery
heat coil, that comprises of rotary heat fan that may have or lack desiccants, heat that is being
supplied to the coil for cooling will be reduced by the recovered energy in the heat exchanger,
noting that a coil for precooling will be fixed at the upstream while the coil that is used for
reheating will be fixed at downstream on the system, therefore the air that is precooled will be
hotter as compared to the air that is supplied after is reheated by reheat coil, and hence energy
that is recovered from the precooling will be used to counterbalanced the reheat load, this will
help in lowering the requirement of operating energy.

Methodology
Equipment
Recirculating duct, preheater and reheater, velocity meter, a room model, five temperature
sensors (T1 to T5), sensors RH1 to RH5 and three louvres.
Procedure
The air conditioning is mixed with the hot and moisture air from the room with stale, the moist
air and warm air that is recirculating will return back to a space of air conditioning, the air that is
in a mixed form will be cooled and its moisture is removed by the cooling coil. The air is then
electrically reheated by reheater before it is supplied to the room. The heat and humid is
collected from the environs of the room and leaves as air in return, in which its nature will be
warmer and more moist as compared when it initially entered. A fraction of this air in return is
mixed with fresh out of door air and further is supplied back to the room, while the other fraction
is lost to the ambient, the temperature will be recorded by temperature sensors while the
humidity will be recorded by RH1 and RH5.

Results of calculations
Assumptions
1. The air is assumed to be dry
2. The flow and temperature are taken to be uniform in the duct
3. The ideal gas law will be used to determine air density
4. The specific heat of air will be calculated based on the formula
Cp = 28.11+0.1967∗10−2∗T +0.4802∗10−5∗T 2−0.1966∗10−9∗T 3
28.97
Test A
a) Determination of heat loss between T2 and T4 at first 40% power of the preheater
From the lab results
T2 = 28.50C = 301.5 K
T4 = 280C = 301 K
Change in temperature = T2 – T4 = 28.5 – 28 = 0.5 K
At T4 = 301 K
Cp4 = 28.11+0.1967∗10−2∗301+0.4802∗10−5∗3012 −0.1966∗10−9∗3013
28.97
= 1.00558 kJ/kg.K
At T2 = 301.5
Cp2 = 28.11+0.1967∗10−2∗301.5+0.4802∗10−5∗301.52−0.1966∗10−9∗301.53
28.97
= 1.00567

∆ Q2-4 = ˙m(Cp2*T2 - Cp4*T4)
˙m = ρ*V*A
But,
P/ ρ = RT
ρ = P/RT
T = ∆T = 0.5 K
ρ = 101325/287*0.5
= 706.098 kg/m3
Average velocity = (0.9 + 0.1)/2 = 0.5
˙m = 706.098 * 0.5 * 0.04
= 14.122 kg
∆ Q2-4 = 14.122*(1.00567 *301.5 – 1.00558*301) = 7.484 J
b) Determination of heat loss between T2 and T4 at second 60% power of the preheater
From the lab results
T2 = 30.40C = 303.4 K
T4 = 29.60C = 302.6 K
Change in temperature = T2 – T4 = 30.4 – 29.6 = 0.8 K
At T4 = 302.6 K
Cp4 = 28.11+0.1967∗10−2∗302.6+ 0.4802∗10−5∗302.62−0.1966∗10−9∗302.63
28.97
= 1.00585 kJ/kg.K
At T2 = 303.4 K
Cp2 = 28.11+0.1967∗10−2∗303.4+ 0.4802∗10−5∗303.42−0.1966∗10−9∗303.43
28.97
= 1.00598
∆ Q2-4 = ˙m(Cp2*T2 - Cp4*T4)

˙m = ρ*V*A
But,
P/ ρ = RT
ρ = P/RT
T = ∆T = 0.8 K
ρ = 101325/287*0.8
= 441.31 kg/m3
Average velocity = (0.9 + 0.1)/2 = 0.5
˙m = 441.31 * 0.5 * 0.04
= 8.8262 kg
∆ Q2-4 = 8.8262*(1.00598 *303.4 – 1.00585*302.6) = 7.45 J
c) The calculation results shows that when the temperature T2 is increased in the air
conditioning cooling system unit the heat loss will increase, since it will result to
higher change in temperature and from the formula change in temperature is
directly proportion to heat loss.
d) Determination of heat loss between T5 and T4 at first 40% power of the preheater
From the lab results
T5 = 26.70C = 299.7 K
T4 = 280C = 301 K
Change in temperature = T4 – T5 = 28 – 26.7 = 1.3 K
At T4 = 301 K
Cp4 = 28.11+0.1967∗10−2∗301+0.4802∗10−5∗3012 −0.1966∗10−9∗3013
28.97
= 1.00558 kJ/kg.K
At T2 = 299.7

Cp2 = 28.11+0.1967∗10−2∗299.7+ 0.4802∗10−5∗299.72−0.1966∗10−9∗299.73
28.97
= 1.00537
∆ Q2-4 = ˙m(Cp2*T2 - Cp4*T4)
˙m = ρ*V*A
But,
P/ ρ = RT
ρ = P/RT
T = ∆T = 1.3 K
ρ = 101325/287*1.3
= 271.576 kg/m3
Average velocity = (0.9 + 0.1)/2 = 0.5
˙m = 271.576 * 0.5 * 0.04
= 5.4315 kg
∆ Q2-4 = 5.4315*(1.00558*301 – 1.00537 *299.7) = 7.442 J
e) Determination of heat loss between T5 and T4 at second 60% power of the preheater
From the lab results
T5 = 280C = 301 K
T4 = 29.60C = 302.6 K
Change in temperature = T4 – T5 = 29.6 – 28 = 1.6 K
At T4 = 302.6 K
Cp4 = 28.11+0.1967∗10−2∗302.6+ 0.4802∗10−5∗302.62−0.1966∗10−9∗302.63
28.97
= 1.00585 kJ/kg.K
At T5 = 301 K

Cp5 = 28.11+0.1967∗10−2∗301+0.4802∗10−5∗3012 −0.1966∗10−9∗3013
28.97
= 1.00558 kJ/kg.K
∆ Q5-4 = ˙m(Cp4*T4 - Cp5*T5)
˙m = ρ*V*A
But,
P/ ρ = RT
ρ = P/RT
T = ∆T = 1.6 K
ρ = 101325/287*1.6
= 220.655 kg/m3
Average velocity = (0.9 + 0.1)/2 = 0.5
˙m = 220.655 * 0.5 * 0.04
= 4.4131 kg
∆ Q2-4 = 4.4131*( 1.00585*302.6 – 1.00558 *301) = 7.461 J

Test B
a) Determination of heat loss between T1 and T5 at the first 40% power of the preheater
From the lab results
T1 = 28.70C = 301.7 K
T5 = 29.70C = 302.7 K
Change in temperature = T5 – T1 = 29.7 – 28.7 = 1.0 K
At T1 = 301.7 K
Cp1 = 28.11+0.1967∗10−2∗301.7+ 0.4802∗10−5∗301.72−0.1966∗10−9∗301.73
28.97

= 1.0057 kJ/kg.K
At T5 = 302.7
Cp5 = 28.11+0.1967∗10−2∗302.7+ 0.4802∗10−5∗302.72−0.1966∗10−9∗302.73
28.97
= 1.005866 kJ/kg.k
∆ Q1-5 = ˙m(Cp5*T5 - Cp1*T1)
˙m = ρ*V*A
But,
P/ ρ = RT
ρ = P/RT
T = ∆T = 1.0 K
ρ = 101325/287*1.0
= 353.049 kg/m3
Average velocity = (0.9 + 1.1)/2 = 1.0
˙m = 353.049 * 1.0 * 0.04
= 14.122 kg
∆ Q2-4 = 14.122*(1.005866 *302.7 – 1.0057*301.7) = 14.912 J
b) Determination of heat loss between T1 and T5 at second 60% power of the preheater
From the lab results
T1 = 30.40C = 303.4 K
T5 = 31.70C = 304.7 K
Change in temperature = T5 – T1 = 31.7 – 30.4 = 1.3 K
At T1 = 303.4 K
Cp1 = 28.11+0.1967∗10−2∗303.4+ 0.4802∗10−5∗303.42−0.1966∗10−9∗303.43
28.97
= 1.00598 kJ/kg.K

At T5 = 304.7
Cp5 = 28.11+0.1967∗10−2∗304.7+ 0.4802∗10−5∗304.72−0.1966∗10−9∗304.73
28.97
= 1.0062 kJ/kg.k
∆ Q1-5 = ˙m(Cp5*T5 - Cp1*T1)
˙m = ρ*V*A
But,
P/ρ = RT
ρ = P/RT
T = ∆T = 1.2 K
ρ = 101325/287*1.3
= 271.576 kg/m3
Average velocity = (0.9 + 1.1)/2 = 1
˙m = 271.576 * 1 * 0.04
= 10.863 kg
∆ Q2-4 = 10.863*(1.0062 *304.7 – 1.00598*303.4) = 14.935 J
c) Determination of heat loss between T5 and T4 at first 40% power of the preheater
From the lab results
T4 = 320C = 305 K
T5 = 29.70C = 302.7 K
Change in temperature = T4 – T5 = 32 – 29.7 = 2.3 K
At T4 = 305 K
Cp4 = 28.11+0.1967∗10−2∗305+0.4802∗10−5∗3052−0.1966∗10−9∗3053
28.97
= 1.00625kJ/kg.K
At T5 = 302.7

Cp5 = 28.11+0.1967∗10−2∗302.7+ 0.4802∗10−5∗302.72−0.1966∗10−9∗302.73
28.97
= 1.005866 kJ/kg.k
∆ Q1-5 = ˙m(Cp4*T4 - Cp5*T5)
˙m = ρ*V*A
But,
P/ ρ = RT
ρ = P/RT
T = ∆T = 2.3 K
ρ = 101325/287*2.3
= 153.5 kg/m3
Average velocity = (0.9 + 1.1)/2 = 1.0
˙m = 153.5 * 1.0 * 0.04
= 6.14 kg
∆ Q5-4 = 6.14*(1.00625 *305 – 1.005866*302.7) = 14.924 J
d) Determination of heat loss between T4 and T5 at second 60% power of the preheater
From the lab results
T4 = 34.80C = 307.8 K
T5 = 31.70C = 304.7 K
Change in temperature = T4 – T5 = 34.8 – 31.7 = 3.1 K
At T4 = 307.8 K
Cp4 = 28.11+0.1967∗10−2∗307.8+ 0.4802∗10−5∗307.82−0.1966∗10−9∗307.83
28.97
= 1.006719 kJ/kg.K
At T5 = 304.7

Cp5 = 28.11+0.1967∗10−2∗304.7+ 0.4802∗10−5∗304.72−0.1966∗10−9∗304.73
28.97
= 1.0062 kJ/kg.k
∆ Q4-5 = ˙m(Cp4*T4 - Cp5*T5)
˙m = ρ*V*A
But,
P/ ρ = RT
ρ = P/RT
T = ∆T = 3.1 K
ρ = 101325/287*3.1
= 113.887 kg/m3
Average velocity = (0.9 + 1.1)/2 = 1
˙m = 113.887 * 1 * 0.04
= 4.555 kg
∆ Q5-4 = 4.555*(1.006719 *307.8 – 1.0062*304.7) = 14.936 J
e) The calculation results shows that when the temperature T2 is increased in the air
conditioning cooling system unit the heat loss will increase, since it will result to
higher change in temperature and from the formula change in temperature is
directly proportion to heat loss.
f) Determination of the power supplied by preheater on test A and test B
Test A
1. Determination of heat loss between T1 and T2 at the first 40% power of the preheater
From the lab results
T1 = 23.60C = 296.6 K
T2 = 28.50C = 301.5 K

Change in temperature = T2 – T1 = 28.5 – 23.6 = 4.9 K
At T1 = 296.6 K
Cp1 = 28.11+0.1967∗10−2∗296.6+ 0.4802∗10−5∗296.62−0.1966∗10−9∗296.63
28.97
= 1.00486 kJ/kg.K
At T2 = 301.5
Cp5 = 28.11+0.1967∗10−2∗301.5+0.4802∗10−5∗301.52−0.1966∗10−9∗301.53
28.97
= 1.005667 kJ/kg.k
∆ Q1-5 = ˙m(Cp2*T2 - Cp1*T1)
˙m = ρ*V*A
But,
P/ρ = RT
ρ = P/RT
T = ∆T = 4.9 K
ρ = 101325/287*4.9
= 72.051 kg/m3
Average velocity = (0.9 + 0.1)/2 = 0.5
˙m = 72.051 * 0.5 * 0.04
= 1.44102 kg
∆ Q2-4 = 1.44102*(1.005667 *301.5 – 1.00486*296.6) = 7.446 J
Power = heat loss/time in seconds
Let assume the time taken by the heater to dissipated heat is 1.0 seconds
Power = 7.446/1.0 seconds
= 7.446 W
2. Determination of heat loss between T1 and T2 at second 60% power of the preheater
From the lab results

T1 = 23.50C = 296.5 K
T2 = 30.40C = 303.4 K
Change in temperature = T2 – T1 = 30.4 – 23.5 = 6.9 K
At T1 = 296.5 K
Cp1 = 28.11+0.1967∗10−2∗296.5+ 0.4802∗10−5∗296.52−0.1966∗10−9∗296.53
28.97
= 1.00484 kJ/kg.K
At T2 = 303.4
Cp2 = 28.11+0.1967∗10−2∗303.4+ 0.4802∗10−5∗303.42−0.1966∗10−9∗303.43
28.97
= 1.00598 kJ/kg.k
∆ Q1-2 = ˙m(Cp2*T2 - Cp1*T1)
˙m = ρ*V*A
But,
P/ρ = RT
ρ = P/RT
T = ∆T = 6.9 K
ρ = 101325/287*6.9
= 51.1665 kg/m3
Average velocity = (0.9 + 0.1)/2 = 0.5
˙m = 51.1665 * 0.5 * 0.04
= 1.02333 kg
∆ Q2-1 = 1.02333*(1.00598 *303.4 – 1.00484*296.5) = 7.449 J
Power = heat loss/time in seconds
Let assume the time taken by the heater to dissipated heat is 1.0 seconds
Power = 7.449/1.0 seconds

= 7.449 W
Test B
1. Determination of heat loss between T1 and T2 at the first 40% power of the preheater
From the lab results
T1 = 28.70C = 301.7 K
T2 = 330C = 306 K
Change in temperature = T2 – T1 = 33 – 28.7 = 4.3 K
At T1 = 301.7 K
Cp1 = 28.11+0.1967∗10−2∗301.7+ 0.4802∗10−5∗301.72−0.1966∗10−9∗301.73
28.97
= 1.0057 kJ/kg.K
At T2 = 306
Cp5 = 28.11+ 0.1967∗10−2∗306+ 0.4802∗10−5∗3062−0.1966∗10−9∗3063
28.97
= 1.00642 kJ/kg.k
∆ Q1-5 = ˙m(Cp2*T2 - Cp1*T1)
˙m = ρ*V*A
But,
P/ ρ = RT
ρ = P/RT
T = ∆T = 4.3 K
ρ = 101325/287*4.3
= 82.1044 kg/m3
Average velocity = (0.9 + 1.1)/2 = 1
˙m = 82.1044 * 1 * 0.04
= 3.2842 kg

∆ Q2-4 = 3.2842*(1.00642 *306 – 1.0057*301.7) = 14.926 J
Power = heat loss/time in seconds
Let assume the time taken by the heater to dissipated heat is 1.0 seconds
Power = 14.926/1.0 seconds
= 14.926 W
2. Determination of heat loss between T1 and T2 at second 60% power of the preheater
From the lab results
T1 = 30.40C = 303.4 K
T2 = 36.20C = 309.2 K
Change in temperature = T2 – T1 = 36.2 – 30.4 = 5.8 K
At T1 = 303.4 K
Cp1 = 28.11+0.1967∗10−2∗303.4+ 0.4802∗10−5∗303.42−0.1966∗10−9∗303.43
28.97
= 1.00598 kJ/kg.k
At T2 = 309.2 K
Cp2 = 28.11+ 0.1967∗10−2∗309.2+0.4802∗10−5∗309.22 −0.1966∗10−9∗309.23
28.97
= 1.00695 kJ/kg.k
∆ Q1-2 = ˙m(Cp2*T2 - Cp1*T1)
˙m = ρ*V*A
But,
P/ρ = RT
ρ = P/RT
T = ∆T = 5.8 K
ρ = 101325/287*5.8
= 60.8705 kg/m3
Average velocity = (0.9 + 1.1)/2 = 1.0
˙m = 60.8705 * 1.0 * 0.04

= 2.43482 kg
∆ Q2-1 = 2.43482*(1.00695 *309.2 – 1.00598*303.4) = 14.9367 J
Power = heat loss/time in seconds
Let assume the time taken by the heater to dissipated heat is 1.0 seconds
Power = 14.9367/1.0 seconds
= 14.9367 W
From the calculation the power dissipated by the preheater in test A and test B are different
because of variation on the temperatures T1 and T2.
Conclusion
The results from test A and test B at the percentage of the first 40% of the relative power of
preheater and the second 60% of the relative power of preheater were determined between
temperature T2 and T4 in test A at the first 40% of the relative power of preheater and the
second 60% of the relative power of preheater was 7.484 J and 7.45 J respectively, while
between T4 and T5 of test A at the first 40% of the relative power of preheater and the second
60% of relative power of preheater was 7.442 J and 7.461 J respectively, similarly on test B at
the first 40% of the relative power of preheater and the second 60% of the relative power of
preheater at the temperature between T1 and T5 was 14.912 J and 14.935 J respectively, and
finally for test B at the temperature between T5 and T4 for the first 40% of the relative power of
preheater and the second 60% of the relative power of preheater was 14.924 J and 14.936 J
respectively.
The core reason as to why the was errors in the measurement and calculation of heat loss was
due to the heat that is was being loss on walls of the air conditioning unit and at environs of the
room which could have not been taken in to considerations.
In conclusion it was found out that the heat loss on test B was higher than the heat loss on test A
because the temperature recorded in test B were higher than those in test A, it should be known

that temperature is directly proportion to the heat loss based on the formula ∆ QAB = ˙m(CpB*TB -
CpA*TA)
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