University of Anbar
College of Engineering
Mechanical Engineering Department
Refrigeration & Air-Conditioning Laboratory
Experiment No.1
Psychometric Process
Cooling with Dehumidification
By
Mohammed Hashim Abdulhassan
Supervisor
Asst. Prof. Dr. Obaid T. Fadhil
Turn in date
9/11/2022
Mechanical Engineering Department
Air Conditioning Report
Abstract
The purpose of the experiment is to better understand how air conditioners work as
well as how much heat and moisture is present in the air. In addition, the
experiment enables students to understand and calculate the necessary heat
exchanges to produce the desired air condition.
Introduction
Dry air (which includes gases like O2, N2, CO2, etc.) and water vapor are
combined to create moist air. A psychrometric chart is a useful tool for researching
and examining the thermodynamic characteristics of wet air. Air conditioning is
necessary to provide thermal comfort cooling in occupied environments. The
methods of air conditioning can include sensible heating or cooling,
dehumidifying, humidifying, etc. depending on the interior requirement and the
indoor temperature. This experiment aims to demonstrate the practical applications
of the psychrometric chart and to investigate some of the air conditioning
processes from the viewpoint of thermodynamics.
Objectives
• To study the components of air conditioning systems.
• To carry out performance test on typical air conditioning processes.
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Air Conditioning Report
Theory and Principles
The cooling and dehumidifying process is one of the fundamental air conditioning
techniques that is widely used in hot, humid locations. Utilizing a cooling coil, this
process is accomplished. Direct expansion (DX) type or chilled water type cooling
coils are both options. To chill and dehumidify moist air, it must pass over a
cooling coil whose surface temperature is lower than the incoming air's dew point
temperature or equal to it.
Two thermometers are installed at each condition in our experimental unit to
measure the temperatures of the dry and wet bulbs. This will make it simpler to
find the states on the map. The air system is set up as a once-through configuration
because the air damper is closed.
Equipment and Instruments
• A660 Air Conditioning Laboratory Unit Fig.1
• Sling psychrometer
Procedure
Turn on the main power switch of the unit and adjust a fan speed around medium
level. Start the refrigeration circuit turning on its switch. Increase the power inputs
of Preheater, Re-heater and Boiler from minimum to maximum for each test.
Collect the data which are seen on the observation sheet while the system is
running.
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Air Conditioning Report
Fig. 1
Fig. 2
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Air Conditioning Report
Calculations and Results
𝑸𝑹𝟏 = 𝒎̇𝒂 (𝒉𝟏 − 𝒉𝟐 ) … … … … . . 𝒆𝒒𝒖. 𝟏
𝑸𝑹𝟐 = 𝒎̇𝒓 (𝒉𝒂 − 𝒉𝒃 ) … … … … . . 𝒆𝒒𝒖. 𝟐
𝒎̇𝒘 = 𝒎̇𝒂 (𝒘𝟏 − 𝒘𝟐 ) … … … … . . 𝒆𝒒𝒖. 𝟑
Where
W1 , W2 : moisture content for outlet and inlet
𝒌𝒈𝒘𝒂𝒕𝒆𝒓
𝒌𝒈𝒂𝒊𝒓
Because of losses Q1 ≠ Q2
𝒎̇𝒂 = 𝟎. 𝟎𝟓𝟎𝟒𝟖√𝐇 ∗ 𝝆𝒂 … … … … . . 𝒆𝒒𝒖. 𝟒
𝝆𝒂 =
1
… … … … . . 𝑒𝑞𝑢. 5
𝑣𝑎
Readings
1
2
Manometer H (mmH2O)
6.6
4.6
db
30
31
wb
19.1
17
db
23
21
wb
14
14.9
Cooling liquid mass flow rate ṁr (g/s)
17
16.8
Cooling liquid pressure after condenser Pc (Kpa)
1260
1240
Cooling liquid pressure after evaporator Pe (Kpa)
295
270
Cooling liquid Temperature before condenser tc (°C)
43
49
Cooling liquid Temperature after evaporator te (°C)
9
9
Inlet air Temperature (°C)
Outlet air Temperature (°C)
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Air Conditioning Report
From psychometric chart for first readings at (tdb ,twb) and crossing the readings:
h2 = 41 kj/kg ,h1 = 53.5 kj/kg ,𝑤1 = 9.4 ∗ 10−3 , 𝑤2 = 7.3 ∗ 10−3
1
1
𝑣
0.848
H = 6.6 mmH2O , 𝜌𝑎 = =
From equ.4
= 1.179
𝑘𝑔
𝑚3
𝑚̇𝑎 = 0.05048 √6.6 ∗ 1.179 = 0.152899
𝑘𝑔
𝑠
∴ 𝑄𝑅1 = 𝑚̇𝑎 (ℎ1 − ℎ2 ) = 0.152899(53.5 − 41) = 1.911 𝑘𝑤
𝑚𝑤 = 𝑚𝑎 (𝑤1 − 𝑤2 ) = 1.911(9.4 − 7.3) ∗ 10−3 = 4.0131 ∗ 10−3
𝑘𝑔
𝑠
And similarly, for second readings
h1 = 48 kj/kg
,h2 = 42 kj/kg ,𝑤1 = 6.4 ∗ 10−3 ,𝑤2 = 8.3 ∗ 10−3
1
1
𝑣
0.848
H = 4.6 mmH2O , 𝜌𝑎 = =
From equ.4
= 1.179
𝑘𝑔
𝑚3
𝑚̇𝑎 = 0.05048 √4.6 ∗ 1.179 = 0.14678
𝑘𝑔
𝑠
∴ 𝑄𝑅1 = 𝑚̇𝑎 (ℎ1 − ℎ2 ) = 0.14678(48 − 42) = 0.88068 𝑘𝑤
𝑚𝑤 = 𝑚𝑎 (𝑤1 − 𝑤2 ) = 0.14678(6.4 − 8.3) ∗ 10−3 = −2.78882 ∗ 10−4
𝑘𝑔
𝑠
From R12 Table after the condenser for first readings at (tdb ,twb)
hb = 242.2 kj/kg ,ha = 356.7 kj/kg
from equ.2 𝑄𝑅2 = 𝑚̇𝑟 (ℎ𝑎 − ℎ𝑏 ) = 17 ∗ 10−3 (356.7 − 242.2) = 1.94 𝑘𝑤
And similarly, for second readings
hb = 220.3 kj/kg ,ha = 356.7 kj/kg
𝑄𝑅2 = 𝑚̇𝑟 (ℎ𝑎 − ℎ𝑏 ) = 16.8 ∗ 10−3 (356.7 − 220.3) = 2.29 𝑘𝑤
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Discussion
The results of the tests demonstrate that there is a considerable difference in the air
qualities before and after the cooling coil, with the enthalpy before the coil being
(53.5 KJ/kg) and after the coil being (41 KJ/kg) for the first measurement, and
when converted into a percentage it becomes
ℎ1 −ℎ2
ℎ1
=
53.5−41
53.5
∗ 100% = 23.36%
For the first reading and the second reading, there is a 23% difference.
ℎ1 − ℎ2
48 − 42
=
∗ 100% = 12.5%
ℎ1
48
The same happens for the second measurement the difference is 12.5% .
We can also see the heat rejected from the air and taken by the cooling liquid,
indicating that there is a difference in heat exchange rate which shows the actual
work of the cooling system.
And that is the expected outcome that the air became cooler and less humid.
Because of various flaws and uncertainties, the results obtained were not reliable.
However, it can be said that the experiment's primary goal was met and that a
complete understanding of the air conditioning system, the refrigeration cycle, and
heat transfer was acquired.
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