ENME 485 HVAC Experiment 1
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School
University of Calgary *
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Course
485
Subject
Aerospace Engineering
Date
Jan 9, 2024
Type
Pages
23
Uploaded by ProfComputer9714
Table of Contents
Letter of Transmittal
Summary/Abstract……………………………………………………………………… Page i
Preface:
List of Figures……………………………………………………………………… Page ii
List of Tables………………………………………………………………………...Page ii
List of Symbols……………………………………………………………………...Page ii
Introduction:
Significance………………………………………………………………… Page 1
Theory & Review…………………………………………………………... Page 2
Objectives……………………………………………………………………Page 3
Apparatus
……………………………………………………………………………….Page 4
Procedure
……………………………………………………………………………….Page 5
Results & Observations………………………………………………………………… Page 6
Data analysis & Computations………………………………………………………Page 7-20
Discussion……………………………………………………………………………...Page 21
Conclusion……………………………………………………………………………...Page 21
References……………………………………………………………………………….Page
Introduction
Significance:
In this experiment we explore the basic principles of air conditioning and a refrigeration cycle. The
HVAC experiment is important as these HVAC systems are prevalent in the lives of many and it is vital to
understand the principles behind them. HVAC systems use the data of change in temperature, relative
humidity and mass flow rates to aid in determining the transfer of energy in air conditioning. We use
thermodynamics and fluid mechanics to ensure controlled environments.
Theory and Review:
To make the analysis simpler, we assume a steady state and that the heating/cooling surfaces are external
when compared to the duct. The system is also a control volume. As a vital portion of this lab is acquiring
the thermal energy transferred to the system, we must use the first law of thermodynamics.
Figure 1: Schematic of HVAC System
Figure 2: Air Flow Diagram
Looking at figure 2 we can use the energy balance equation 1 below:
1
𝑄
˙
?
− ?
˙
?
=
?
˙
?
ℎ
?
− ℎ
𝐷
(
)
+
?
?
2
−?
𝐷
2
2
⎡
⎢
⎣
⎤
⎥
⎦
Where:
ṁ
a
= air mass flow rate (kg/s)
hg = enthalpy at section G (kJ/kg dry air)
hD = enthalpy at section D (kJ/kg dry air)
Vg = air velocity at section G (m/s)
VD = air velocity at section D (m/s)
We can simplify the equation 1 to create equation 2 since
= 0 and V
G
≈ V
D
?
?
˙
𝑄
˙
?
= ?
˙
?
(ℎ
?
− ℎ
𝐷
)
can be defined by equation 3:
?
˙
?
?
˙
?
=
?
𝐷
?
𝐷
𝑣
𝐷
Where:
= cross-sectional area of the duct (m
2
)
?
𝐷
= air velocity at section D (m/s)
?
𝐷
= specific volume of the air at section D (m
3
/kg)
𝑣
𝐷
We can calculate the mass flow rate using equation 4 shown below
?
˙
= 0. 0517
𝑧
𝑣
𝐷
Where:
= mass flow rate (kg/s)
?
˙
= pressure differential orifice (mmH
2
O)
𝑧
= specific volume of the air at section D (m
3
/kg)
𝑣
𝐷
We can calculate the blower work by using the First Law represented in equation 5:
?
˙
= ?
˙
?
(ℎ
𝐷
− ℎ
?
)
2
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Where:
= air mass flow rate (kg/s)
?
˙
= enthalpy at section D (kJ/kg dry air)
ℎ
𝐷
= enthalpy at section A (kJ/kg
dry air)
ℎ
?
In experiment 1 we can find the mass flowrate of steam shown in equation 6
𝑄
˙
?𝑜𝑖???
=
?
˙
??
ℎ
??
(
)
Where:
= Sum of heat transfer to boiler (J/s)
𝑄
?𝑜𝑖???
˙
=
mass flow rate of steam (kg/s)
?
˙
??
= Specific enthalpy of Saturated steam (kJ/kg)
ℎ
??
In order to find the enthalpy of flow we can implement equation 7 below as we are unable to use
the ideal gas law because a mixture of dry air and water vapor is used.
h = 1.005T +
ω
(2501.3 + 1.82T)
Where:
h is enthalpy in Kj/kg
T is temperature in K
ω is the specific humidity
The specific humidity is calculated using equation 8 below:
ω = 0.622(Pv/(P-Pv))
Where: P is the ambient pressure
Pv is the vapor pressure
Experiment 2 follows the same theory as experiment 1 where we determine mass flow rate and apply the
First law to estimate the heat between A and D, and D and G.
Objectives:
For this lab there are two objectives. Objective 1 is to demonstrate the principles of air conditioning by
analyzing the comparison between experimental measured values and theoretical. The second objective is
to demonstrate the principles of a refrigeration cycle.
\
3
Apparatus
The Apparatus used for this experiment is an air-conditioning unit as shown in figure 1 below, the
apparatus consists of two main sections where the first section is the air system. The air system includes a
blower fan that has speed controls, air heaters, a steam injection system and air flow instrumentation. The
second section is the refrigeration system which consists of a compressor-condenser unit, an evaporator,
water extraction tube, controls and instrumentation.
Figure 3 : Air-Conditioning Unit
4
Procedure
Experiment 0:
For Experiment 0 the goal is to determine the blower work. The first step is to measure and record the
local atmospheric pressure and temperature. The second step is to record the data presented on the
computer without flow, this is the zero data. Third step is to turn on the fan to 1 mm and allow it to
stabilize for 1 minute. Step 4 is to collect the temperature and relative humidity data. Step 7 is to repeat
the steps previously until 12 mm is reached with increments of 1 mm each time.
Experiment 1:
Experiment 1 explores the air steam humidification with refrigeration. Step 1 is to turn the fan onto a
speed of 10 mm of water. Step 2 is to turn on all steam heaters. Next, step 3 is to turn the refrigeration
system on and allow it to stabilize. Step 4, we must collect the air and refrigerant temperature, relative
humidity, pressure and mass flow rate. Step 5, turn the fan speed to 1 mm and allow it to stabilize. Step 6,
repeat step 4. Step 7 we must set the fan to 12 mm. Finally for step 8, turn off the steam generator heaters
and refrigeration system.
Experiment 2:
Experiment 2 looks into airstream heating. Step 1, turn on the two 1 kW air preheaters. Then for step 2,
allow the system to run for 5 minutes. Step 3, turn on the two 500 W air heaters. Step 4, allow the system
to stabilize. Step 5, collect the data for air temperatures and relative humidity. Next for step 6 you must
reduce the blower speed to 4 mm of water and allow the system to stabilize. Finally, collect the data
which includes the air temperatures and relative humidity.
5
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Results and Observations
Table 1: Ambient data
Temperature (ᵒC)
Pressure (kPa)
Relative Humidity (%)
20.5
87.3
24.3
Table 2: Initial data
Air Sensors
Refrigeration Section
Inlet (Section
A)
Midsection
(Section D)
Exit (Section
G)
Pre-Throttling
Valve (ᵒC)
Pre-Cooling
Coil (ᵒC)
Post-Cooling
Coil (ᵒC)
Temperature (ᵒC)
19.3
18.8
19.3
18.6
18.7
18.7
Relative Humidity
(%)
20.2
20.7
20
-
-
-
Experiment #0
Table 3: Data for experiment #0 at various fan speeds (mm H2O)
Inlet (Section A)
Midsection (Section D)
Exit (Section G)
fan speeds
()
Temperature (ᵒC)
Relative
Humidity (%)
Temperature (ᵒC)
Relative
Humidity (%)
Temperature (ᵒC)
Relative Humidity (%)
1
19.7
19.6
19.1
19.6
19.7
19.6
2
19.7
19.5
19.1
19.5
19.7
19.6
3
19.8
19.6
19.6
19.6
20.1
19.6
6
4
19.8
19.6
19.7
19.6
20.1
19.6
5
19.7
19.5
19.7
19.4
20.2
19.2
6
19.8
19.6
19.8
19.4
20.3
19.2
7
19.9
19.6
20
19.4
20.4
19.2
8
19.8
19.5
19.9
19.3
20.4
19.2
9
19.7
19.6
19.9
19.4
20.5
19.2
10
19.8
19.6
20
19.4
20.5
19.2
11
19.7
19.5
19.9
19.4
20.5
19.2
12
19.7
19.5
20
19.4
20.5
19.2
Experiment #1
Table 4 : Data for fan speed of 10 (mm H2O) with refrigeration and humidification for experiment #1
Air Sensors
Refrigeration Section
Inlet
(Section
A)
Midsection
(Section D)
Exit
(Section
G)
Pre-Throttli
ng Valve
(ᵒC)
Pre-Cooli
ng Coil
(ᵒC)
Post-Cooli
ng Coil
(ᵒC)
Temperature
(ᵒC)
21.9
23.8
18.2
38.7
5.2
5.6
Relative
Humidity (%)
24.3
52.2
76.1
-
-
-
Coolant Flow (mdot R12) = 33 L/hr
7
Plow= 82 Psi
Phigh= 150 Psi
Rate of water condensation = 0 mL/min
Table 5: Data for fan speed of 1 () with refrigeration and humidification for experiment #1
Air Sensors
Refrigeration Section
Inlet
(Section
A)
Midsection
(Section D)
Exit
(Section
G)
Pre-Throttli
ng Valve
(ᵒC)
Pre-Cooli
ng Coil
(ᵒC)
Post-Cooli
ng Coil
(ᵒC)
Temperature
(ᵒC)
22.5
40.7
21.7
46.8
13.8
15.6
Relative
Humidity (%)
23.1
96.5
96.5
-
-
-
Coolant Flow (
mdot R12
) = 36 L/hr
Plow= 54 Psi
Phigh= 195 Psi
Rate of water condensation = 35 mL/min
8
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Experiment 2:
Table 6: Data for fan speed of 12 (mm H2O) with varying heater settings
Front Heaters with a Value of 2 kW
All Heaters with Values of 2 kW and
1 kW
Inlet
(Section
A)
Midsection
(Section D)
Exit
(Section
G)
Inlet
(Section
A)
Midsection
(Section D)
Exit
(Section
G)
Temperature
(ᵒC)
20.1
33
32
20.2
33.3
37.2
Relative
Humidity
(%)
19.2
10
6.2
19.2
10
1.5
Table 7: Data for fan speed of 4 () with varying heater settings
Front Heaters with a Value of 2 kW
All Heaters with Values of 2 kW and
1 kW
Inlet
(Section
A)
Midsection
(Section D)
Exit
(Section
G)
Inlet
(Section
A)
Midsection
(Section D)
Exit
(Section
G)
Temperature
(ᵒC)
20.6
43.4
40.5
20.8
42.7
49.1
Relative
Humidity
(%)
18.8
8
1.4
18.8
8
1.5
9
Data Analysis/ Computations
Experiment #0
First the offsets must be determined in order to correct the data.
Table 8 : Offsets for Data Collected
Air Sensors
Refrigeration Section
Inlet
(Section
A)
Midsection
(Section D)
Exit
Pre-Throttl
ing Valve
(ᵒC)
Pre-Coolin
g Coil (ᵒC)
Post-Cooli
ng Coil
(ᵒC)
(Section
G)
Temperature
(ᵒC)
-0.16667
0.333333
-0.16667
0.066667
-0.033333
-0.033333
Relative
Humidity
(%)
0.1
-0.4
0.3
-
-
-
Table 9: Corrected Data for Experiment 1
Temperature
Relative Humidity
fan speed
INLET
MID
EXIT
INLET
MID
EXIT
initial
19.13333
19.13333
19.13333
20.3
20.3
20.3
1
19.53333
19.43333
19.53333
19.7
19.2
19.9
2
19.53333
19.43333
19.53333
19.6
19.1
19.9
3
19.63333
19.93333
19.93333
19.7
19.2
19.9
4
19.63333
20.03333
19.93333
19.7
19.2
19.9
5
19.53333
20.03333
20.03333
19.6
19
19.5
6
19.63333
20.13333
20.13333
19.7
19
19.5
7
19.73333
20.33333
20.23333
19.7
19
19.5
8
19.63333
20.23333
20.23333
19.6
18.9
19.5
10
9
19.53333
20.23333
20.33333
19.7
19
19.5
10
19.63333
20.33333
20.33333
19.7
19
19.5
11
19.53333
20.23333
20.33333
19.6
19
19.5
12
19.53333
20.33333
20.33333
19.6
19
19.5
Sample Calculation to determine the offset for temperature and relative humidity for fan speed 1 at the
inlet:
11
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12
13
Table 10: Mass Flow Rate and Work Done at Each Fan Speed
Fan Speed
Blower
mass flow
(dry air)
rate (kW)
rate (m3/kg)
1
-0.01885
0.05257
2
-0.02664
0.074346
3
0.020553
0.091036
4
0.039508
0.105119
5
0.05847
0.117551
6
0.056571
0.128744
7
0.082126
0.139034
8
0.08734
0.148663
9
0.119044
0.157709
10
0.122755
0.166208
11
0.138882
0.174356
12
0.172298
0.182109
14
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Figure 4: Rate of Work vs. Mass Flow Rate at D
15
Experiment 1:
Using the offsets found in experiment 0 we can correct the values for experiment 1 shown below
Table 11 : Offset values found for experiment 0
Temperature
Relative Humidity
Inlet Offset
-0.17
0.1
Mid Offset
0.33
-0.4
Exit Offset
-0.17
0.3
Table 12: Corrected Values for Experiment 1 (10.0 mm of water)
Fan Speed
(mm)
T, inlet
T, mid
T, exit
Relative
Humidity
inlet
Relative
Humidity
mid
Relative
Humidity
exit
10
21.73
24.13
18.03
24.4
51.8
76.4
Table 13: Corrected Values for Experiment 1 (1.0 mm of water)
Fan Speed
(mm)
T, inlet
T, mid
T, exit
Relative
Humidity
inlet
Relative
Humidity
mid
Relative
Humidity
exit
1
22.33
41.03
21.53
23.2
96.1
96.8
16
17
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Table 14: Summary of Calculated Values Including Specific Humidity, Enthalpy and Vapor Pressure
IN (A)
MID (D)
EXIT (G)
z (mm)
ω
h(kJ/kg)
Pv (kPa)
ω
h(kJ/kg)
Pv (kPa)
ω
h(kJ/kg)
Pv (kPa)
1
0.0045
33.8970
0.6280
0.0587
192.6179
7.5296
0.01828
68.0800
2.4926
10
0.0043
32.9400
0.6360
0.0050
36.1623
1.5630
0.0035
54.3829
1.5845
Refrigeration Cycle Analysis:
Sample Calculations for fan speed of 1mm
1.
First we must find the mass flow rate
We can use Equation (6),
ṁ
= 0.0517
= 0.0525kg/s
𝑧
/𝑣 = 0. 0517
1??/0. 978 ?
3
/??
Now we can calculate the mass flow rate of dry air
ṁ
a
using Equation ()
ṁ
a
= ṁ / (1+ ω) = (0.0525 kg/s)/(1 + 0.004506 kg/kg) = 0.0527 kg/s
2.
Next we can calculate the mass flow rate of steam going into the duct
Q
boiler
= m
steam
* h
fg
We can interpolate for the enthalpy using the pressure for 1mm
h
fg
= 2278 kJ/ kg + (86.67-75)kPa*(2257.5 -2278.0)kJ/kg / (100-75) kPa = 2268.4 kJ/kg
ṁ
steam
=
Q
boiler
/h
fg
= 4.72 kW/
2268.4 kJ/kg = 2.08 x 10
-3
kg/s
3.
Next we can compute the heat removed by the refrigeration cycle
Q
̇
R12
= m
R12
(h
1
-h
4
)
m
R12
= 36L/hr x 1kg/L x 1hr/3600s = 0.01 kg/s
We can calculate the enthalpy using the R-134 Tables and interpolating
P1 = 54 psi = 372.317 kPa
P3 = 195 psi = 1344.480 kPa
h3 =h4= 124.172 kJ/kg
h1 = 254.358 kJ/kg
Q
̇
R12
= 0.01 kg/s (254.358-124.172)kJ/kg
Q
̇
R12
= 1.302 kW
18
4.
Water mass balance between sections A and D to estimate
ṁ
steam
ṁ
steam
= ṁ
a
(ωD - ωA)
= 0.0527kg/s (
0.0587
- 0.0045)
= 0.00287 kg/s
5.
Apply the first law between sections A and D to estimate ṁ
steam
We may use the enthalpy values found above to calculate the mass flow rate of steam at 1 mm
ṁ
steam
=
=
= 0.003687 kg/s
??𝑖?(ℎ𝐷 − ℎ?)
ℎ??
0.0527??/?(192.6179?𝐽/?? − 33.897?𝐽/??)
2268.4 ?𝐽/??
6.
Apply the first law between sections A and G to estimate the ṁ
condensed
Q
̇
R12
=
ṁ
a
(
)
ℎ𝐷 − ℎ?
ṁ
condensed
= ṁ
a
(1+ωD)-ṁ
a
(1+ωG) = 0.0527 kg/s(1+0.0050) - 0.0527kg/s(1+0.
01828)
=
0.0006999 kg/s
7.
Apply the first law between sections A and G to estimate the refrigerant load
Q
̇
R12
=
ṁ
a
(
) - ṁ
condensed
ℎ𝐷 − ℎ?
= 0.0527 kg/s (
192.6179- 68.0800) -
0.0006999 kg/s
= 6.56 kW
Table 15: Summary of calculated values for Fan speed 1 mm
Fan Speed 1 mm
1.
ṁ
a
0.0527 kg/s
2.
ṁ
steam
2.08 x 10
-3
kg/s
3.
Q
R12
1.302 kW
4.
ṁ
steam
0.00287 kg/s
5. ṁ
steam
0.003687 kg/s
6. ṁ
condensed
0.0006999 kg/s
7.
Q
̇
R12
6.56 kW
Table 16: Summary of calculated values for Fan speed 10 mm
Fan Speed 10 mm
1.
ṁ
a
0.166 kg/s
2.
ṁ
steam
2.08 x 10
-3
kg/s
19
3.
Q
R12
0.538 kW
4.
ṁ
steam
0.00142 kg/s
5. ṁ
steam
0.00132 kg/s
6. ṁ
condensed
0.0002 kg/s
Experiment 2:
By following the steps previously in experiment 1, we calculate the mass flow rate of dry air with the
same approach and get 0.178 kg/s and 0.101 kg/s for 12 mmH2O and 4 mmH2O.
Calculation of Rate of Heater for sections A-D
After solving for mass flow rate of dry air the rate of heat can be calculated as shown below
ṁ
a
(
-
𝑄
??????
=
˙
ℎ
𝐷
ℎ
?
)
= 0.18(42.6-28.8)
= 2.56 kW
Calculation of Rate of Heater for sections D-G
ṁ
a
(
-
𝑄
??????
=
˙
ℎ
?
ℎ
𝐷
)
= 0.18(37.6-42.6)
= -0.67 kW
Table 17: Calculated Results of Heat Transfer for 2 Air Speeds 12.0 and 4.0
Air Speed (mmH2O)
Sections A-D Heat Transfer
(kj/s)
Sections D-G Heat Transfer
(kj/s)
12.0
2.56
-0.67
4.0
2.82
-1.39
20
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Discussion:
In experiment 0 we analyzed the temperature and relative humidity data and created a mass flow rate vs
blower graph based on calculations. As seen in figure () the graph displays a linear relationship between
mass flow rate and blower work. As there are some outliers present which can be concluded to be
experimental or systematic error, we can disregard those errors and assume a linear relationship. We
assume ideal gas behavior in order to conduct the calculations for mass flow rate and blower work, and
from table () we see that as blower work increases mass flow rate increases. To conclude, the experiment
was a success as it correctly validated our prediction of the linear relationship.
Experiment 1 introduced steam into the HVAC system where the boiler units heated the liquid to vapor in
sections A to D. In the experiment we recorded air temperatures, relative humidity, refrigerant
temperatures, pressures and mass flow rate. The mass flow rates are determined for the depths z= 1 mm
and z=10.0 mm, where we got 0.0522kg/s and 0.166kg/s respectively. As noted in table () and () we can
see that the mass flow rate of steam going into the duct is found to be 2.08 x 10^-3 for both 1 mm and 10
mm as they are assumed to be the same for both fan speeds. We also determined the heat removed
for 1
mm and 10 mm to be
1.302 kW and 0.538 kW respectively. A large factor in our calculations was the
assumption that there was no energy or heat loss in the system.
Experiment 2 had heat going into the system. Two sets of readings, 12mm and 4mm, were read off the
manometer reading. The mass dry air flow rate is 0.178 kg/s and 0.101 kg/s. When looking at the heat
transfer for the 2KW and 1KW heater we see that from A to D we get 2.56 kW for 12.0 mm and -0.67 kW
for D to G. The calculation for D to G makes more sense as heat loss decreases due to the addition of a
second heater. We get a negative value due to the heat loss in the system.
Conclusion:
In conclusion the three experiments were overall a success, experiment 0 correctly showed the predicted
linear relationship between mass flow rate and blower work as we plotted the graph. The corrected values
computed in the beginning of experiment 0 for temperature and relative humidities were accurately found
and used to correct values for experiments 1 and 2. In experiment 1, the refrigeration cycle of the HVAC
system was analyzed, mass flow rates and heat removed were also determined for both z=1 mm and 10
mm. In order to determine the values, a mass balance and energy balance was used to find
the heat
removed for 1 mm and 10 mm to be
1.302 kW and 0.538 kW respectively. In experiment 2 we got 2.56
kW and -0.67 kW for sections A-D and D-G respectively for 12.0 mm, and for 4.0 mm we have 2.82 kW
and -1.39 kW for sections A-D and D-G respectively. Overall all 3 experiments were a success as they
computed reasonable values and validated our theoretical assumptions.
21
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References:
[1] HVAC Video (2022). [online] Available
at:https://d2l.ucalgary.ca/d2l/le/content/425183/viewContent/5178289/View[Accessed 22 Feb. 2022].
[2] HVAC Experiment Data Set. (2022). [online] Available at:
https://d2l.ucalgary.ca/d2l/le/content/425183/viewContent/5178280/View[Accessed 22 Feb. 2022].
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