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THE HASHIMITE UNIVERSITY FACULTY OF ENGINEERINGDepartment of Mechanical Engineering
Thermal science 1 Lab. - Exp # 1 : Marcet boiler Page 1 of 4
Objective:
To investigate the relationship between (pressure) and (temperature) of a saturated steam, in
equilibrium with (water).
Apparatus: Marcet Boiler, shown in figure 1, is made of steel and fitted with a pressure gauge, a safety valve,
a water cock for testing the water level and a thermo sensor. The boiler is heated by an electrical
immersion heater. To minimize losses and to prevent direct contact to the hot surface, the boiler is
insulated. The temperature is shown on a digital electronic thermometer. An integrated limit switch
prevents the boiler from overheating.
Figure 1: Marcet Boiler.
Exp. # 1
Marcetboiler
THE HASHIMITE UNIVERSITY FACULTY OF ENGINEERINGDepartment of Mechanical Engineering
Thermal science 1 Lab. - Exp # 1 : Marcet boiler Page 2 of 4
Theory:
At a given pressure, the temperature at which a pure substance changes phase is called the
saturation temperature Tsat. Similarly, at a given temperature, the pressure at which a pure substance
changes phase is called the saturation pressure Psat.
This experiment explores the relationship between the saturation temperature and the
corresponding pressure for water.
The water inside the boiler is heated up by the electrical resistance and starts to evaporate. As
more water changes phase from liquid to vapor, more vapor accumulates inside the boiler vessel and
increases the pressure imposed on the water surface. This pressure buildup tends to increase the
resistance faced by liquid molecules as they change into vapor, consequently increasing the saturation
pressure of the remaining liquid.
For a pure substance existing as a mixture of two phases, the Clapeyron relationship relates the
pressure, heat and expansion during a change of phase provided that the two phases are in equilibrium.
The Clapeyron relationship is:
( )fg
g
fg
fg
hTv
hvvT
dPdT
=−
=
where:
vf specific volume of water.
vg specific volume of steam.
hf enthalpy of water.
hg enthalpy of steam.
hfg latent heat of vaporization = hg - hf.
T absolute temperature.
P absolute pressure.
Procedure:
• The boiler was filled with clean water through the filler plug.
• The heater element was connected to a single-phase electrical power supply.
• Switch on the master switch.
• Switch on the heater switch and heat up the boiler.
• Log the boiler pressure and temperature values in increments of approximately 0.5 bars.
• Fill the results in the data sheet table.
THE HASHIMITE UNIVERSITY FACULTY OF ENGINEERINGDepartment of Mechanical Engineering
Thermal science 1 Lab. - Exp # 1 : Marcet boiler Page 3 of 4
Analysis: 1. Fill the table of results below:
S/N Gauge
pressure (bar)
Absolute pressure
(bar)
Steam Temperature
(C°)
Measured slope (dT/dp)
Calculated slope (Tvg/hfg)
1 2 3 4 5 6 7 8 9
10 11 12 13 14 15 16 18 19 20
Atmospheric pressure: . . . . . . . . . . . . . . . . . .
2. Plot the T versus P and measure the slop of each point.
3. To measure the slop ermentaldP
dT
exp⎟⎠⎞
⎜⎝⎛
takes previous & next values of pressure and
temperature for each record.
4. Compare with the theoretical slop calculated using equation (1) and the steam tables or table 2.
5. To measure the slop ltheoriticadP
dT⎟⎠⎞
⎜⎝⎛ take the values of the specific volume & enthalpy for each
corresponding temperature record, use interpolation when required.
6. State what kinds of errors could affect our results in this experiment.
THE HASHIMITE UNIVERSITY FACULTY OF ENGINEERINGDepartment of Mechanical Engineering
Thermal science 1 Lab. - Exp # 1 : Marcet boiler Page 4 of 4
Pressure
(P) bar
Temperature (T) oC
Specific volume (vg)
m3/kg
Latent heat of vaporization
(hfg)
1.0 99.6 1.694 2258
2.0 120.0 0.8856 2202
3.0 133.5 0.6057 2164
4.0 143.6 0.4623 2134
5.0 151.8 0.3748 2109
6.0 158.8 0.3156 2087
7.0 165.0 0.2728 2067
8.0 170.4 0.2403 2048
9.0 175.4 0.2149 2031
10.0 179.9 0.1944 2015
11.0 184.1 0.1774 2000
12.0 188.0 0.1632 1986
13.0 191.6 0.1512 1972
14.0 195.0 0.1408 1960
15.0 198.3 0.1317 1947
16.0 201.4 0.1237 1935
17.0 204.3 0.1167 1923
18.0 207.1 0.1104 1912
19.0 209.8 0.1047 1901
20.0 212.4 0.09957 1890
Table 1: Saturated Water and Steam Tables
THE HASHIMITE UNIVERSITY FACULTY OF ENGINEERING Department of Mechanical Engineering
Thermal science 1 Lab. - Exp # 2 : Gas calorific value Page 1 of 5
Objective:
To determine the calorific value of a gaseous fuel.
Apparatus:
Boys gas calorimeter, with the following parts and instruments:
• Gaseous fuel source.
• Water source and sink.
• Gas control valve.
• Pressure reducing valve.
• “Hyde meter” (Gas meter): To measure the flow rate of the gas, where 2 liter of gas flows
per one revolution.
• The calorimeter with burner.
• Alkaline bath.
• Thermometers.
• Stopwatch.
• Graduated glass vessel.
Exp. # 2 GAS CALORIFIC VALUE
THE HASHIMITE UNIVERSITY FACULTY OF ENGINEERING Department of Mechanical Engineering
Thermal science 1 Lab. - Exp # 2 : Gas calorific value Page 2 of 5
Theory: The calorific value of any fuel is defined as the amount of heat generated by completely burning
(1m3 or 1kg) of that fuel.
In this experiment a given amount of gaseous fuel is burned, and then the generated heat is used
to heat a measured amount of water, so:
Calorific value =f
avgwaterpw
VVGTCM ).()( ×Δ××
.................................. (1)
Where:
Mw : The amount of water collected (L).
Cp (water) : Specific heat of water (4.18 kj/kgK).
ΔTavg. : Average difference of water temperature between inlet and outlet.
G.V : Gas volume factor which can be found from table (3) at atmospheric pressure and
average gas temperature.
Vf : Volume of burned fuel = No. of revolutions ×2 (L).
= No. of revolutions× 2× 10-3 (m3).
Procedure:
1. Turn the gas supply on, light the burner, and adjust gas flow rate using gas control valve to give
one revolution per minute at the Hyde meter.
2. Turn on water to over head funnel, with small over flow to the sink.
3. Lift the coils from the alkali bath (allow to drain for few minutes) and lower into the calorimeter
casing.
4. Allow gas to burn and water run about 45 min. to reach the steady state.
5. Read and record the temperature of the inlet gas by the thermometer on Hyde meter.
6. When the pointer of Hyde meter at (100), turns change over funnel at 300ml beaker to measure the
amount of water.
7. Through a number of a revolutions, record inlet and outlet water temperature (Twi, Two) at each half
revolution.
8. At the completion of the last revolution. Turn change over funnel to sink, then record the values of
the water temperatures, the inlet gas temperature and amount of water collected.
9. Record the atmospheric pressure.
10. Fill the results at tables 1 & 2.
Analysis:
1. Find the average difference of water temperature.
2. At atmospheric pressure and average inlet gas temperature, find the gas volume factor (G.V) from
table 3.1 & 3.2.
3. Calculate the calorific value using equation 1.
THE HASHIMITE UNIVERSITY FACULTY OF ENGINEERING Department of Mechanical Engineering
Thermal science 1 Lab. - Exp # 2 : Gas calorific value Page 3 of 5
Results:
Temperature
(oC)
Number of Revolutions
0.5 1.0 1.5 2.0 2.5 3.0
Twi
Two
Table: 1
(Tg)initial (Tg)final Volume of water (L) Patm. (mbar) G.V
Table: 2
THE HASHIMITE UNIVERSITY FACULTY OF ENGINEERING Department of Mechanical Engineering
Thermal science 1 Lab. - Exp # 2 : Gas calorific value Page 4 of 5
Patm Temperature(oC) Patm.
mm Hg
10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 mm Hg
730 1.095 1.098 1.101 1.105 1.108 1.111 1.116 1.121 1.126 1.132 1.137 1.142 1.148 1.154 1.160 730
732 1.092 1.095 1.098 1.101 1.105 1.108 1.113 1.118 1.123 1.128 1.134 1.139 1.145 1.151 1.157 732
734 1.089 1.092 1.095 1.098 1.102 1.105 1.110 1.115 1.120 1.125 1.131 1.136 1.142 1.148 1.154 734
736 1.085 1.088 1.092 1.095 1.099 1.102 1.107 1.112 1.117 1.122 1.128 1.133 1.139 1.145 1.151 736
738 1.081 1.085 1.088 1.092 1.095 1.099 1.104 1.109 1.114 1.119 1.125 1.130 1.136 1.142 1.148 738
740 1.077 1.081 1.085 1.088 1.092 1.096 1.101 1.106 1.111 1.116 1.122 1.127 1.133 1.139 1.145 740
742 1.074 1.078 1.082 1.085 1.089 1.093 1.098 1.103 1.108 1.113 1.119 1.124 1.130 1.136 1.142 742
744 1.071 1.075 1.079 1.082 1.086 1.090 1.095 1.100 1.105 1.110 1.116 1.121 1.127 1.133 1.139 744
746 1.068 1.072 1.076 1.079 1.083 1.087 1.092 1.097 1.102 1.107 1.113 1.118 1.124 1.130 1.136 746
748 1.065 1.069 1.073 1.076 1.080 1.084 1.089 1.094 1.099 1.104 1.110 1.115 1.121 1.126 1.132 748
750 1.062 1.066 1.070 1.073 1.077 1.081 1.086 1.091 1.096 1.101 1.107 1.112 1.118 1.123 1.129 750
752 1.059 1.063 1.067 1.070 1.074 1.078 1.083 1.088 1.093 1.098 1.104 1.109 1.115 1.120 1.126 752
754 1.056 1.060 1.064 1.067 1.071 1.075 1.080 1.085 1.090 1.095 1.101 1.106 1.112 1.117 1.123 754
756 1.053 1.057 1.061 1.064 1.068 1.072 1.077 1.082 1.087 1.092 1.098 1.103 1.109 1.114 1.120 756
758 1.050 1.054 1.058 1.061 1.065 1.069 1.074 1.079 1.084 1.089 1.095 1.100 1.106 1.111 1.117 758
760 1.047 1.051 1.055 1.058 1.062 1.066 1.071 1.076 1.081 1.086 1.092 1.097 1.103 1.108 1.114 760
762 1.044 1.048 1.052 1.055 1.059 1.063 1.068 1.073 1.078 1.083 1.089 1.094 1.100 1.105 1.111 762
764 1.042 1.046 1.049 1.053 1.056 1.060 1.065 1.070 1.075 1.080 1.086 1.091 1.097 1.102 1.108 764
766 1.039 1.043 1.046 1.050 1.053 1.057 1.062 1.067 1.072 1.077 1.083 1.088 1.094 1.099 1.105 766
768 1.037 1.041 1.044 1.047 1.050 1.054 1.059 1.064 1.069 1.074 1.080 1.085 1.091 1.096 1.102 768
770 1.034 1.038 1.042 1.045 1.048 1.052 1.057 1.062 1.067 1.072 1.078 1.083 1.089 1.094 1.100 770
772 1.031 1.035 1.039 1.042 1.045 1.049 1.054 1.059 1.064 1.069 1.075 1.080 1.086 1.091 1.096 772
774 1.029 1.031 1.035 1.038 1.041 1.046 1.051 1.056 1.061 1.066 1.072 1.077 1.083 1.088 1.094 774
776 1.026 1.029 1.032 1.036 1.039 1.043 1.048 1.053 1.058 1.063 1.069 1.074 1.080 1.085 1.091 776
778 1.024 1.027 1.030 1.033 1.037 1.040 1.045 1.050 1.055 1.060 1.066 1.071 1.077 1.082 1.088 778
780 1.021 1.025 1.028 1.031 1.033 1.038 1.043 1.048 1.053 1.058 1.063 1.068 1.074 1.079 1.085 780
THE HASHIMITE UNIVERSITY FACULTY OF ENGINEERING Department of Mechanical Engineering
Thermal science 1 Lab. - Exp # 2 : Gas calorific value Page 5 of 5
Pamt. Temperature(oC) Patm.
Mm Hg
25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 mm Hg
730 1.166 1.173 1.180 1.187 1.194 1.201 1.208 1.216 1.223 1.231 1.236 1.247 1.256 1.266 1.275 730
732 1.163 1.169 1.176 1.183 1.190 1.197 1.205 1.212 1.220 1.227 1.235 1.244 1.253 1.263 1.272 732
734 1.160 1.166 1.173 1.180 1.187 1.194 1.201 1.209 1.216 1.224 1.231 1.240 1.249 1.259 1.268 734
736 1.157 1.164 1.171 1.178 1.185 1.191 1.198 1.205 1.212 1.220 1.228 1.237 1.246 1.255 1.264 736
738 1.154 1.161 1.168 1.174 1.180 1.187 1.194 1.202 1.209 1.217 1.224 1.233 1.242 1.251 1.260 738
740 1.151 1.158 1.164 1.171 1.178 1.184 1.191 1.199 1.206 1.214 1.221 1.230 1.239 1.247 1.256 740
742 1.148 1.155 1.161 1.168 1.175 1.181 1.188 1.196 1.203 1.210 1.218 1.227 1.235 1.244 1.252 742
744 1.145 1.152 1.158 1.165 1.172 1.178 1.185 1.193 1.200 1.208 1.215 1.223 1.232 1.240 1.249 744
746 1.142 1.148 1.155 1.161 1.168 1.174 1.181 1.189 1.196 1.204 1.211 1.220 1.228 1.237 1.245 746
748 1.138 1.145 1.151 1.158 1.165 1.171 1.178 1.186 1.193 1.201 1.208 1.216 1.225 1.233 1.242 748
750 1.135 1.142 1.148 1.155 1.161 1.167 1.174 1.182 1.169 1.197 1.204 1.212 1.221 1.229 1.238 750
752 1.132 1.139 1.145 1.152 1.158 1.164 1.171 1.179 1.186 1.194 1.204 1.209 1.218 1.226 1.235 752
754 1.129 1.136 1.142 1.149 1.155 1.161 1.168 1.175 1.183 1.190 1.197 1.205 1.214 1.222 1.231 754
756 1.126 1.133 1.139 1.146 1.153 1.158 1.165 1.172 1.180 1.187 1.194 1.202 1.211 1.219 1.228 756
758 1.123 1.130 1.136 1.143 1.149 1.155 1.162 1.169 1.176 1.183 1.190 1.198 1.207 1.215 1.224 758
760 1.120 1.127 1.133 1.140 1.146 1.152 1.159 1.166 1.173 1.180 1.137 1.195 1.204 1.212 1.221 760
762 1.117 1.124 1.130 1.137 1.143 1.149 1.156 1.163 1.170 1.177 1.134 1.191 1.200 1.208 1.217 762
764 1.114 1.120 1.126 1.133 1.139 1.145 1.152 1.159 1.167 1.174 1.131 1.189 1.197 1.206 1.214 764
766 1.111 1.117 1.123 1.129 1.136 1.142 1.149 1.156 1.163 1.170 1.177 1.185 1.194 1.202 1.210 766
768 1.108 1.114 1.120 1.126 1.132 1.139 1.146 1.153 1.160 1.167 1.174 1.182 1.190 1.199 1.207 768
770 1.105 1.111 1.117 1.123 1.130 1.136 1.143 1.150 1.157 1.164 1.171 1.179 1.187 1.196 1.204 770
772 1.102 1.108 1.114 1.120 1.126 1.133 1.140 1.147 1.154 1.161 1.168 1.176 1.184 1.193 1.201 772
774 1.099 1.105 1.111 1.117 1.123 1.130 1.137 1.144 1.151 1.158 1.165 1.173 1.131 1.189 1.197 774
776 1.096 1.102 1.108 1.114 1.120 1.127 1.134 1.141 1.147 1.154 1.161 1.169 1.177 1.186 1.194 776
778 1.093 1.099 1.105 1.111 1.117 1.124 1.131 1.138 1.144 1.151 1.158 1.165 1.173 1.181 1.189 778
780 1.090 1.096 1.102 1.108 1.114 1.121 1.128 1.155 1.141 1.148 1.155 1.162 1.169 1.176 1.183 780
Table 3.1: Gas Volume Factor.
FACULTY OF ENGINEERING THE HASHIMITE UNIVERSITY Department of Mechanical Engineering
Thermodynamics Lab. - Exp # 6 : Nozzle Test Page 1 of 4
Objective:1. To study pressure and velocity distribution along a nozzle.
2. To find critical pressure ratio and efficiency of a nozzle.
Theory:A nozzle is a duct of smoothly varying cross sectional area in which a steadily flowing fluid can be
made. The flow can be accelerated by a pressure drop along the duct. There are many applications in
practice which require a high – velocity stream of fluid such as, gas turbines, jet engines, rockets and flow
measurement.
Consider a stream of fluid at a pressure Pi, enthalpy hi and velocity Vi enters a nozzle. Applying
steady flow steady state energy equation 2 2
( ) ( ) ( )2 2e i
e i e iV Vq w h h g Z Z ......…………………… [ 1 ]
Where Pe, Te, he, Ve, and Ze are exit state or any second state through the nozzle.
Since q, w and Z are equal to zero for nozzles, then
22
22e
ei
iV
hV
h ......…………………… [ 2 ]
Or
2)(2 ieie VhhV ......…………………… [ 3 ]
The continuity equation gives,
VAm.
......…………………… [ 4 ]
Solving for A gives,
VmA.
......…………………… [ 5 ]
Where;
.m is the mass flow rate of air (kg/s).
is the density of air (kg/m3).
A is the cross sectional area of the nozzle (m2).
V is the Velocity of air (m/s).
And TRP orTRP.
, and, iepie TTChh
Let Cp for air equals to 1.005 (kJ/kg K) and = 1.4, and assume the inlet air velocity (Vi) equals to zero.
Assuming air is an ideal gas and the process is isentropic, then
Exp. # 6
Nozzle Test
FACULTY OF ENGINEERING THE HASHIMITE UNIVERSITY Department of Mechanical Engineering
Thermodynamics Lab. - Exp # 6 : Nozzle Test Page 2 of 4
kk
i
e
i
ePP
TT
1
......…………………… [ 6 ]
The nozzle efficiency is defined by the ratio of the actual enthalpy drop (irreversible expansion) to
isentropic enthalpy drop:
)()(
)()(
esi
eai
esip
eaip
esi
eaiTTTT
TTCTTC
hhhh
efficiecyNozzle ......…………………… [ 7 ]
But Ti and Tea are measured values, and then Te can be found from equation 6, thus nozzle
efficiency can be evaluated for any inlet and exit conditions.
For a convergent nozzle expanding into a space, the pressure of which can be varied, while the
inlet pressure remains fixed. When the back pressure (Pb) is equal to inlet pressure (Pi) then no fluid can
flow through the nozzle. As Pb is reduced the mass flow rate through the nozzle increase until the back
pressure reaches the critical value (Pc) after which the mass flow rate remains constant with Pb.
When Pb = Pc then the velocity at exit is sonic and the mass flow through the nozzle is at a
maximum, If Pb < Pc then mass flow rate is maximum and the exit pressure remains at Pc and the fluid
expands violently outside the nozzle down to the back pressure.
When a nozzle operates with the maximum mass flow it is said to be chocked.
Critical pressure ratio is measured by:
112 k
k
i
ckP
P ......…………………… [ 8 ]
Figure 1: h-s diagram of the actual and
isentropic expansion processes of the nozzle.
Figure 2: Pressure distribution along the nozzle for different back
pressures.
FACULTY OF ENGINEERING THE HASHIMITE UNIVERSITY Department of Mechanical Engineering
Thermodynamics Lab. - Exp # 6 : Nozzle Test Page 3 of 4
Apparatus:Figure 1 show the test unit of the nozzle which
consists of the followings:
Three different types of nozzles:
o Convergent – divergent nozzle.
o Convergent nozzle.
o Divergent nozzle.
Pressure gauges for inlet and exit.
8 pressure gauges to measure the pressure
distribution along the nozzle.
Digital thermometer to measure inlet and exit
temperature.
Pressure regulator with filter and control
valves.
Procedure:1. Connect the air supply to the inlet valve.
2. Adjust the inlet air pressure to the required value.
3. Take readings of inlet pressure, exit pressure, inlet temperature, exit temperature, mass flow rate
and the gauge pressures from P1 to P7.
4. Adjust the mass flow rate using the exit valve and record your results.
5. Fill your readings in the table of results.
Analysis:1. Calculate T, V, and A at different sections and compare areas with the given values.
2. Calculate the nozzle efficiency ( ).
3. Plot P, V and T variation along the nozzle.
4. Find the ratio of (Pe/Pi).
5. Plot the mass flow rate m (kg/s) against the ratio (Pe/Pi), then determine the point at which the
chocking phenomena occurs, compare with the value (0.5275).
6. Calculate the speed at the chocking point using previous experiment data.
7. State four applications of nozzles and write the governing equations of one application.
FACULTY OF ENGINEERING THE HASHIMITE UNIVERSITY Department of Mechanical Engineering
Thermodynamics Lab. - Exp # 6 : Nozzle Test Page 4 of 4
Datasheet:Pi
(Bar)Pe
(Bar)P1
(Bar)P2
(Bar)P3
(Bar)P4
(Bar)P5
(Bar)P6
(Bar)P7
(Bar)P8
(Bar)Ti
(oC)Te
(oC)m
(g/s)
Nozzle Efficiency
4 4 %
4 3 %
4 2 %
4 1 %
4 0 %
Table 1: Experimental results.
V1
(m/s) V2
(m/s) V3
(m/s) V4
(m/s) V5
(m/s) V6
(m/s) V7
(m/s) V8
(m/s) A1
(m2)A2
(m2)A3
(m2)A4
(m2)A5
(m2)A6
(m2)A7
(m2)A8
(m2)
Table 2: Calculated results.
THE HASHIMITE UNIVERSITY FACULTY OF ENGINEERING Department of Mechanical Engineering
Thermal science 1 Lab. - Exp # 4: Refrigeration cycle Page 1 of 3
Objective:
To find the coefficient of performance of a refrigeration cycle.
Theory: A refrigerator is a machine whose function is to remove heat from a low temperature region, and
dissipated it to a high temperature region (surroundings). According to the Clausius statement of second
law of thermodynamics, which states that heat will not transfer from a cold body to a hotter one unless
work is added to the system. So the refrigerator will require an external work, which is the compressor
work.
If the function of the system is to use the dissipated heat at high temperature e.g. for space
heating, then the machine is called a heat pump.
The ideal vapor compression refrigeration cycle has four thermodynamic processes which can be
drawn on P-h diagram (Fig 1) where:
Process (1-2)
The compressor increases the pressure and temperature (i.e. enthalpy) of the refrigerant.
Process (2-3)
Condensation through the condenser at a constant pressure and temperature, so at point (2)
refrigerant is saturated liquid.
Process (3-4)
Refrigerant expands from high pressure P2 to low pressure P3 at constant enthalpy.
Process (4-1)
Refrigerant boils and evaporate in the evaporator at a constant pressure and temperature.
Figure 1
h
P
1
23
4
Exp. # 4
Refrigeration Cycle
THE HASHIMITE UNIVERSITY FACULTY OF ENGINEERING Department of Mechanical Engineering
Thermal science 1 Lab. - Exp # 4: Refrigeration cycle Page 2 of 3
Coefficient of performance of a refrigeration cycle is defined as: the amount of heat removed from the cooling space to the work done by the compressor:
WQCOP L
R =
And …
W = QH - QL
QL = mref * (h1 – h4)
QH = mref * (h2 – h3)
mref = ρref * VR Where:
QL heat absorbed at evaporator (kW).
QH heat rejected from condenser (kW).
W compressor work (kW).
h1, h2, h3, and h4 are enthalpies at the given temperature and pressure (experimentally) in (kJ/kg).
mref is the refrigerant flow rate (kg/s).
VR volume flow rate of refrigerant (m3/s).
ρref is the refrigerant density (kg/m3), for R134a take ρref = 1220 kg/m3.
For ideal cycle, take:
P2 = P3
P4 = P1
h4 = h3 …… for throttling process.
Apparatus: Experiment’s rig consists of refrigeration circuit assembled on a metal board. All instruments for reading pressure, temperature, and flow rates are included and installed in place.
1. Compressor. 2. Water connections. 3. Throttle valve. 4. Pressure switch. 5. Variable-area flow meter. 6. Manometer. 7. Dial thermometer. 8. Evaporator. 9. Expansion valve. 10. Filter dryer. 11. Condenser with fan. Figure 2: The refrigeration system rig
THE HASHIMITE UNIVERSITY FACULTY OF ENGINEERING Department of Mechanical Engineering
Thermal science 1 Lab. - Exp # 4: Refrigeration cycle Page 3 of 3
Procedure: 1. Connect water from the tap to the inflow of the circuit and adjust the flow rate to about 10 L/hr.
2. Connect the system to the mains power. Switch on compressor and fan.
3. After steady state reaches take the readings of pressure, temperature and flow rates.
4. Readjust the value of the water flow rate, wait for steady state then take same readings.
Results and analysis:
State 1 State 2 State 3 State 4 mref
P1=PL T1 h1 P2=PH T2 h2 P3=P2 T3 h3 T4 h4=h3
(bar gage) oC (kj/kg) (bar gage) oC (kj/kg) (bar gage) oC (kj/kg) oC (kj/kg) (L/hr)
1. Fill the table of results above.
2. Plot the cycle on the P-h diagram and find the enthalpies.
3. Calculate the COPR.
4. Plot COPR against the condensation temperature, in accordance with your graph state when do
we have better COPR, in summer or in winter?
THE HASHIMITE UNIVERSITY FACULTY OF ENGINEERING Department of Mechanical Engineering
Thermal Science (1) Lab. - Exp # 05 : Gas Turbine Page 1 of 6
Exp. # 5
Gas Turbine Objective:
1- To investigate the overall performance of gas turbine cycle. 2- To investigate the performance of gas generator, power turbine and combustion chamber.
Theory: Brayton cycle - the Ideal Cycle for Gas-turbine Engine
Gas-turbines usually operate on an open cycle, shown on Fig.1.
• A compressor takes in fresh ambient air (state 1), compresses it to a higher temperature and pressure (state 2).
• Fuel and the higher pressure air from compressor are sent to a combustion chamber, where fuel is burned at constant pressure. The resulting high temperature gases are sent to a turbine (state 3).
• The high temperature gases expand to the ambient pressure (state 4) in the turbine and produce power.
• The exhaust gases leave the turbine.
Part of the work generated by the turbine is sent to drive the compressor. The fraction of the turbine work used to drive the compressor is called the back work ratio.
Since fresh air enters the compressor at the beginning and exhaust are thrown out at the end, this cycle is an open cycle.
By utilizing the air-standard assumptions, replacing the combustion process by a constant pressure heat addition process, and replacing the exhaust discharging process by a constant pressure heat rejection process, the open cycle described above can be modeled as a closed cycle, called ideal Brayton cycle Fig.2. The P-v and T-s diagrams of an ideal Brayton cycle are shown on fig.3.
Fig. 1: An Open Gas-Turbine Cycle
Fig. 2: The Ideal Brayton Cycle
Fig. 3: P-v and T-s Diagrams of Ideal Brayton Cycle
THE HASHIMITE UNIVERSITY FACULTY OF ENGINEERING Department of Mechanical Engineering
Thermal Science (1) Lab. - Exp # 05 : Gas Turbine Page 2 of 6
The ideal Brayton cycle is made up of four internally reversible processes.
• 1-2 Isentropic compression (in a compressor)
In this process the following relations are applied: constantkPV = 1
1
2 2
1 1
kkT P
T P
−
⎛ ⎞= ⎜ ⎟⎝ ⎠
2
1
21
1
1
kk
comp pPW mc TP
−⎛ ⎞⎛ ⎞⎜ ⎟= −⎜ ⎟⎜ ⎟⎝ ⎠⎜ ⎟⎝ ⎠
& & 3
Where P is the pressure, T is the temperature, Wcomp is the compressor input work and k is the specific
heat ratio = p
v
cc
⎛ ⎞⎜ ⎟⎝ ⎠
• 2-3 Constant pressure heat addition
( )3 2in pQ mc T T= −& & 4
• 3-4 Isentropic expansion (in a turbine)
In an identical manner to the compression process.1
4 4
3 3
kkT P
T P
−
⎛ ⎞= ⎜ ⎟⎝ ⎠
5
But 3 2 4 1P P and P P= = so the turbine output work is given by: 1
13
2
1
kk
turb pPW mc TP
−⎛ ⎞⎛ ⎞⎜ ⎟= − ⎜ ⎟⎜ ⎟⎝ ⎠⎜ ⎟
⎝ ⎠
& & 6
• 4-1 Constant pressure heat rejection
( )4 1out pQ mc T T= −& & 7
The thermal efficiency of the ideal Brayton cycle under the cold air-standard assumption is given as
( )( )
41
4 1 1,
3 2 32
2
11 1 1
1
pturb in out outth Brayton
in in in p
TTmc T T TW Q Q Q
Q Q Q mc T T TTT
η
⎛ ⎞−⎜ ⎟−− ⎝ ⎠= = = − = − = −
− ⎛ ⎞−⎜ ⎟
⎝ ⎠
& & && &
& & & & 8
Since P2 = P3 and P4 = P1 and Considering equations 2 and 5
( )
1
1 1, 1
2 2
11 1 1
kk
th Brayton kk
p
T PT P r
η
−
−
⎛ ⎞= − = − = −⎜ ⎟
⎝ ⎠ 9
THE HASHIMITE UNIVERSITY FACULTY OF ENGINEERING Department of Mechanical Engineering
Thermal Science (1) Lab. - Exp # 05 : Gas Turbine Page 3 of 6
Where rP = P2/P1 is the pressure ratio. In most designs, the pressure ratio of gas turbines range from
about 11 to 16. Actual Gas-turbine Cycle
The actual gas-turbine cycle is different from the ideal Brayton cycle since there are irreversibilities. Hence, in an actual gas-turbine cycle, the compressor consumes more work and the turbine produces less work than that of the ideal Brayton cycle. The irreversibilities in an actual compressor and an actual turbine can be considered by using the isentropic efficiencies of the compressor and turbine. They are:
( )( )
1
2
2 1 12 1,
22 1 2 1
1
1.
1
kk
p ssc ise
a p
Pmc T T Ph hrev work
Tactual work h h mc T TT
η
−
⎛ ⎞−⎜ ⎟−− ⎝ ⎠= = = =
− − −
&
& 10
Similarly For Turbine
4
3, 1
4
3
1
1
tur ise kk
TT
PP
η −
−=
⎛ ⎞− ⎜ ⎟⎝ ⎠
11
Another difference between the actual Brayton cycle and the ideal cycle is that there are pressure drops in the heat addition and heat rejection processes. Fig 4 shows the T-s diagram for both actual and ideal cycles.
Fig. 4: T-s Diagram of Actual Gas-turbine Cycle
Apparatus: Two Shaft Gas Turbine Unit comprising single shaft compressor/turbine unit combustion chamber for operation on propane, butane or propane/butane mixtures, power turbine, calibrated electrical machine for torque and power measurement, ignition system, oil tank, circulating pump, cooler and filter, five color instrument panel with flow diagram, fitted inlet air flow meter, fuel flow meter, tachometers (2), multi point thermocouple instrument, sensitive pressure gauges (3), manometer, oil pressure gauge and fuel supply pressure gauge. Complete with starting air compressor set and all controls.
THE HASHIMITE UNIVERSITY FACULTY OF ENGINEERING Department of Mechanical Engineering
Thermal Science (1) Lab. - Exp # 05 : Gas Turbine Page 4 of 6
Procedure: 1- Connect cooling water, drain, gas and electric supply to the unit. 2- Set the air inlet control to start position. 3- Close the gas valve and open the bottle valve. 4- Set the dynamometer excitation to maximum. 5- Start the oil pump. 6- Press the reset button. 7- Start the blower. 8- Ste the gas pressure to 2 bar using reducing valve. 9- Press ignition button and open gas valve to give 0.5g/s. 10- Open gas valve slowly to give a gas generator speed to 1000 rps. 11- Turn the air inlet control to run position. 12- Switch off the blower. 13- Take readings and fill the results on table 1. 14- Vary the load on power turbine to get a set of readings.
Analysis:
1. Overall performance (power and plant efficiency) Electrical output power Pelec = Volts × Ampere
Corrected electrical output power P(elec)corrected = Pelec × .
760 288273ambatm TP
×+
Where Patm is the atmospheric pressure in mmHg
Turbine output power Pout(turbine) = ( )
%elec correctedP
Efficiency
Where % efficiency is efficiency figure taken from Fig. 5, taking into account drive, alternator and rectifier
circuit efficiencies.
THE HASHIMITE UNIVERSITY FACULTY OF ENGINEERING Department of Mechanical Engineering
Thermal Science (1) Lab. - Exp # 05 : Gas Turbine Page 5 of 6
The fuel rate ( )f mix corrected readingm Correction factor m= ×& &
Find correction factor From fig. 12 based on Tg
Then the plant efficiency η is given by:
( )f mix corrected
Corrected output power turbinem Cv
η =×&
Cv is the calorific value of the gasses fuel = 46 × 103 KJ/Kg
2. Compressor isentropic efficiency The compressor efficiency is given by:
( )( )
1
2
2 1 12 1,
22 1 2 1
1
1.
1
kk
p ssc ise
a p
Pmc T T Ph hrev work
Tactual work h h mc T TT
η
−
⎛ ⎞−⎜ ⎟−− ⎝ ⎠= = = =
− − −
&
& take P2=P3
Where pressure and temperature are absolute.
1kk−
is taken from fig.1 5 at T1
3. Combustion chamber efficiency
The fuel heat input to the chamber = ( )f correctedm Cv×&
At 2 3
2T T+
, find pCR
from fig. 15
The actual energy added to air = 3 2( )pair
Cm R T T
R× × × −&
The air flow rate am& is given by:
2( / ) 11.09am g s hmmH O=&
3 2
( )
( )pair
CCf corrected
Cm R T T
Rm Cv
η× × × −
=×
&
&
4. Turbine isentropic efficiency The turbine Efficiency is given by:
4
31
4
3
1
1
t kk
TT
PP
η −
−=
⎛ ⎞− ⎜ ⎟⎝ ⎠
where 1k
k−
is from fig. 15 at T3
THE HASHIMITE UNIVERSITY FACULTY OF ENGINEERING Department of Mechanical Engineering
Thermal Science (1) Lab. - Exp # 05 : Gas Turbine Page 6 of 6
5. Heat Rejection QL= ma * (Cp/R) * R *(T5-T1)
Where Cp/R is from fig. 15
6. Flow curves
Plot 3
GGNT
against 3
4
PP
Plot 4
PTNT
against 4
5
PP
Plot PTN against η
Where NGG and NPT is the speed of gas generator and power turbine.
7. State two methods by which the cycle
can be enhanced.
Thermal science 1 Lab. - Exp # 6
TWO STAGE PISTON TYPE AIR COMPRESSOR
1. OBJECTIVE: 1. To determine the polytropic index (n), for the compressor. 2. To calculate the isothermal and polytropic work. 3. To calculate the isothermal efficiency. 2. APPARATUS: Two stage piston type air compressor, operated by an electrical motor coupled by means of pulleys and V-belt.
The compressed air outgoing from the first stage of the compressor passes through a water/air heat exchanger (intercooler), then it is sucked by the second stage. The outlet air from the second stage passes through a second exchanger (after cooler) and it is sent to the storage tank. (See fig. 1).
Fig. 1 Two stage piston type air compressor
• State 1 : inlet conditions to first stage. P1: Atmospheric pressure. T1: Ambient temperature.
• State 2 : outlet from first stage, inlet to intercooler. P2: Pressure of air outlet from first stage. T2: Temperature of air outlet from first stage.
• State 3 : outlet from intercooler, inlet to second stage. P3: Pressure of air outlet from intercooler. (P3 = P2 ) T3: Temperature of air outlet from intercooler.
• State 4 : outlet from second stage, inlet to after cooler. P4: Pressure of air outlet from second stage. T4: Temperature of air outlet from second stage.
• State 5 : outlet from after cooler, inlet to storage tank. P5: Pressure of air outlet from after cooler. (P5 = P4 ) T5: Temperature of air outlet from after cooler. 3. THEORY: 3.1 P-V Diagram for an ideal compressor (fig. 2). Fig. 2 1-2: Compression process: both valves are closed; air is compressed from P1 to P2. 2-3: discharge process: Exit valve is open, air is supplied to the tank at P2. 3-4: Expansion process: both valves are closed; air in clearance volume expands to original state P1. 4-1: intake process: inlet valve is open; air enters the cylinder at state 1, and mixed with air already present in the clearance volume. 3.2 Compression of gases (process 1-2). 1. Isothermal compression: The compression of gases occurs at constant temperature (fig.3) from state 1 to state 2. The equation of path for this process is given by :
V
1
2 3
4
P
P2
P1
1 1 1
2 2 2
T1 T2 T3
P
V
PV = constant........................ (1) Fig. 3 The isothermal work is given by:
PPLnTmR
PPPVLn
VVPVLnW iso
211
21
12 === ................ (2)
Where P: Absolute pressure (bar). V: Specific volume (m3/kg). m: Intake mass flow rate of air (kg/s). Ta: Temperature of inlet air (ambient temperature). R: Gas constant (For air = 287.14 j/kg K ). The subscripts 1,2 denotes for initial and final states. 2. Adiabatic compression (isentropic): The compression is of adiabatic type if it is performed with out thermic exchange with the outside. Of course, in this case the temperature cannot remain constant during the transformation. On P-V field, the equation of path is given by: PVγ = constant Where the exponent γ is the specific heat ratio, and it is a function molecular structure of the gas (for air γ = 1.4) 3. Polytropic compression: Actually, the compression of gases takes place with some kind of thermic exchange with the out side. The transformation is called polytropic, which is intermediate between the isothermal and adiabatic ones. The equation of the polytropic bath and polytropic work is given by: PVn = constant ......................................... (3)
⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛
−
−
−= ⎟
⎠⎞⎜
⎝⎛ 1
12
1
11 PP n
n
TmRn
nW P .......... (4)
Where n is the polytropic index. The polytropic index is a general process, and all other processes is a special case of the polytropic one, so when: (see fig. 4) n = 0 P = constant (isobaric process) W = P (V2-V1) = mR (T2-T1). n = ∞ V = constant (isochoric process) W = 0.
n = 1 T = constant (isothermal process) W = mRT1 Ln 12
PP
.
n = γ S = constant (isentropic process) W =
⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛
−⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛ −
⎟⎠⎞⎜
⎝⎛ 1
12
1
1 PP n
n
TmR .
Fig. 4 The compression produced by a very fast reciprocating compressor is very close to the adiabatic type, since the short period of time in which transformation take place don’t allow an effective thermic exchange with the out side. But for intermediate speed compressors (500-1000 rpm) the compression occurs according to the polytropic process. The work done on air through compression decreases as n decreases, for this reason the compressor cylinder should be cooled. The compression work is minimum when n=1 (i.e.) for isothermal process. The isothermal efficiency is defined as:
ηiso = Isothermal WorkActual indicated Work
...................................................(5)
The P-V diagram for a reversible two stage compressor is shown in fig. 5 , air is compressed from Pa to intermediate pressure P1 , in the low pressure cylinder and then Transferred to the high pressure cylinder for final compression to P2 . Equation 4 indicates that the work required decreases as the inlet temperature decreases, so an inter cooler is fitted between the stages. Cooling the air leaving the first stage to T1 before it enters the second stage, the work required to drive the second is reduced, fig. 5 illustrate this, shaded area represents the saving in work.
4. PROCEDURE: 1. Turn on the power supply to start the compressor. 2. Turn on the cooling water and adjust flow rates to a suitable value. 3. Adjust the compressor speed to 1200 rpm using the potentiometer. 4. Open the exit valve, so that the second stage pressure P2 is 2 bars. 5. Record all temperatures, pressures, voltmeter, ammeter and the manometer readings. 6. Readjust the exit valve, so that P2= 3 bar, and record all values as step 4. 7. Repeat steps 3, 4 and 5 for different values of P2. 8. Fill the results at table 1. 9 . Turn off the compressor, turn off the cooling water and allow air to release, and drain water from cylinder and coolers. 5. RESULTS: Ambient temperature Ta = T1= °C Atmospheric pressure Pa = P1= bar Tes
t Speed Electrical
Power After
stage 1 After inter coole
r
After stage 2
After Second cooler
Air flow rate
No. rpm V volts
I amp
P2 T2 T3 P4 T4 T5
1
2
3
4
5
Table: 1
6. CALCULATIONS: 1. Polytropic index (n) PVn = constant P1V1
n = P2V2n
⎟⎠⎞⎜
⎝⎛=
VV
n
PP
21
12 ............................................................................................... (10)
From gas law P1V1 = mRT1 P2V2 = mRT2
So 1 12 2
12
P VP V
TT
= ............................................................................ (11)
From equations 10 & 11
⎟⎠⎞⎜
⎝⎛ −=
TT n
n
PP
12 1
12
Ln PP
nn
Ln TT
21 1
21
=−
The last equation is a line of form Y = S X where S is the slop. So for the first stage calculate
TTLn
PPLn
12,
12 Then Plot
TTLnAgainst
PPLn
12
12 and find the slop S1 , and from the slop
find the polytropic index n1 . Similar plot
TTLnAgainst
PPLn
34
34 and find n2 .
2. Work a. Isothermal specific work: For first stage (Wiso) 1 = R T1 Ln
PP
21 .
For second stage (Wiso) 2 = R T3 LnPP
43 .
b. Polytropic specific work:
For first stage ( Wp)1 =
⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛
⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛
−
−
− ⎟⎠⎞⎜
⎝⎛ 1
12 1
11
1111
PP n
n
TmRn
n
For second stage ( Wp)2 =
⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛
⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛
−
−
− ⎟⎠⎞⎜
⎝⎛ 1
34 2
12
3122
PP n
n
TmRn
n
3. Indicated isothermal efficiency:
( ) ( )( )
1
11 W p
W isoiso =η for the first stage.
( ) ( )( )
2
22 W p
W isoiso =η for the second stage.
( ) ( )( ) ( )
21
21W pW p
WisoWisoiso +
+=η for the compressor.
THE HASHIMITE UNIVERSITY FACULTY OF ENGINEERING Department of Mechanical Engineering
Thermal science 1 Lab. - Exp # 7: Thermal Resistance of Multilayer Insulation Material Page 1 of 4
Objective:- The purpose of this experiment is to determine thermal resistance of multilayer insulation materials.
Theory:- Several standards have been drawn up to define an acceptable method of thermal conductivity measurements, which are as follows: -
1. steady state method: -
The thermal conductivity is determined from measurements of temperature gradient in the material and the heat input.
2. Transient method: -
The hot wire method is based on transient conditions. The linear heat source is a wire to which is welded a thermocouple. The thermal conductivity is determined from the rate of the thermocouple reading.
3. Heat Flow meter method: -
The specimen under test is placed between a hot plate and the heat flow meter, which is attached to a cold plate. The apparatus is surrounded by insulation. The hot and cold plates are maintained at suitable constant temperatures measured by surface thermocouples. A calibration constant for the individual apparatus is derived from testing a sample of known constant thermal conductivity. By measuring the heat flow meter output and he mean temperature of the test sample, the thermal conductivity is calculated using this calibration constant.
The thermal conductivity is a material property defined by the following equation:-
xTTkAq hc
x Δ−
−=)(
The minus sign is a consequence of the second law of thermodynamics, which requires that heat must flow in the direction of lower temperature. If more than one material is present, as in the multilayer wall shown in the figure 1, the analysis would proceed as follows:
433221 −−− == qqq
Or
cc
bb
aax x
TTAk
xTT
AkxTT
AkqΔ−
=Δ−
=Δ−
=)(
..)(
..)(
.. 433221
Solving these equations gives:
RTT
RRRTT
Akx
Akx
Akx
TTqcba
c
c
b
b
a
a
)(
...
414141 −=
++−
=Δ
+Δ
+Δ
−=
Exp. # (7) Thermal Resistance of Multilayer Insulation Materials
THE HASHIMITE UNIVERSITY FACULTY OF ENGINEERING Department of Mechanical Engineering
Thermal science 1 Lab. - Exp # 7: Thermal Resistance of Multilayer Insulation Material Page 2 of 4
Where:
Ra, Rb, Rc : thermal resistances of each material in oC/W. R : thermal resistance of the multilayer material in oC/W.
Equation (1) can be introduced by:
ResistanceThermaldifference potential ThermalFlowHeat =
Which is quite like Ohm's law in electric circuit theory, therefore we can represent these layers of materials in figure as three electric resistances in series:-
In this experiment the thermal conductivity can be calculated as follows:- _ _ _
21 2 3 4 5 6*[( ( * )) (( ( * )) * ) (( ( * )) * )]sl C C T C C T HFM C C T HFMk
dT+ + + + +
=
Where:-
HFM : Heat flow in mille volts (mV). ls : Specimen thickness (m). C1, C2, C3, C4, C5, C6 : Calibration constants for the apparatus and have
the following values.
6335.20983.04636.5
3
2
1
==−=
CCC
0002.0
0644.00499.0
6
5
4
−===
CCC
Then;
= −
+=
Δ=
1 2_
1 2
dT (T T )
(T T )T2
x RkA
Figure 1
1 2 3 4
q b c
q a Ra Rc Rb
THE HASHIMITE UNIVERSITY FACULTY OF ENGINEERING Department of Mechanical Engineering
Thermal science 1 Lab. - Exp # 7: Thermal Resistance of Multilayer Insulation Material Page 3 of 4
Apparatus:- The thermal conductivity of building and insulating materials unit is shown in figure 1.
Figure (1): Thermal conductivity of building and insulating materials unit.
Procedure:- 1. Switch on the unit at the main switch. 2. Place the specimen then close the lid. 3. Rotate the screw hand wheel anti-clockwise to lower the hot plate assembly down onto the
heat flow meter plate. 4. At the point when the green “Test Position” lamp illuminates stop the turning and note the
dial reading. 5. Multiply this value by 0.25 to give the thickness of the specimen under test in (mm). Analysis:-
• Fill the table(1) below. Table(1)
Readings Time (s) T1 (oC) T2 (oC) HMF (mV)
THE HASHIMITE UNIVERSITY FACULTY OF ENGINEERING Department of Mechanical Engineering
Thermal science 1 Lab. - Exp # 7: Thermal Resistance of Multilayer Insulation Material Page 4 of 4
• Draw T1, T2 versus time and show the steady state region.
• Calculate the equivalent thermal resistance R of the multilayer insulation materials.
• Compare the experimental value of R with the theoretical one, given the table of material
properties below.
K (w/m.K) Thickness (mm) ρ (kg/m3) Cork board 0.043 3 160
Plaster Board 0.182 10 720 Plaster Gypsum 0.170 11 800
• Mention other different ways of measuring the thermal conductivity k
THE HASHIMITE UNIVERSITY FACULTY OF ENGINEERING Department of Mechanical Engineering
Thermal science 1 Lab. - Exp # 8: Double pipe concentric tube heat exchanger Page 1 of 5
Objective:- To study the performance and the characteristics of double pipe, water to water,
concentric tube heat exchanger in both parallel and counter flow. Theory:- One of the most common, conductive-convective, heat exchanger types is the concentric tube heat exchanger. These exchangers are built of coaxial tubes placed the ones inside the others. When both the fluids enter from the same side and flow through the same direction we have the parallel flow (cocurrent flow), otherwise, if the fluids enter from opposite sides and flow through the contrary direction we have the countercurrent flow. Usually the countercurrent flow is more efficient from the heat transfer point of view. This type of heat exchangers can also be built with the internal tube made with longitudinal fins which could be placed either in its internal surface or in its external one or both. This configuration is useful mainly if one of the fluids is a gas or a liquid with a very high viscosity and it's very difficult to have a good thermal convection coefficient. The heat transfer from the hot fluid to the cold fluid is given by the following equation:
LMTDAUq ××= Where: U is the overall heat transfer coefficient. A is the internal exchange surface area between the two fluids.
LMTD is a log mean temperature difference, and it's given by 1 2
1 2ln( / )T T
T TΔ −ΔΔ Δ
ΔT1=T hot in- T cold in ΔT2=T hot out- T cold out for the parallel flow exchanger. ΔT1=T hot in- T cold out ΔT2=T hot out- T cold in for the counter flow exchanger.
Counter flow
Figure(1): Temperature distribution for counter flow heat exchangers
Exp. # (8) Double pipe concentric tube heat exchanger
THE HASHIMITE UNIVERSITY FACULTY OF ENGINEERING Department of Mechanical Engineering
Thermal science 1 Lab. - Exp # 8: Double pipe concentric tube heat exchanger Page 2 of 5
Parallel flow
Figure(2): Temperature distribution for parallel flow heat exchangers.
Apparatus:-
The apparatus is a double – pipe, water to water heat exchanger test unit with 4m concentric pipes. The built in heater includes a series of resistors with fixed and variable heating capacity.
Figure (3): double – pipe heat exchanger (Photo)
THE HASHIMITE UNIVERSITY FACULTY OF ENGINEERING Department of Mechanical Engineering
Thermal science 1 Lab. - Exp # 8: Double pipe concentric tube heat exchanger Page 3 of 5
Figure (): double – pipe heat exchanger (layout)
Procedure:-
1. Adjust V5 and V6 valves to get the parallel flow circuit. 2. Adjust the hot circuit valve V3 so as to obtain the required flow rate m hot with turbulent
rate. 3. Adjust the cold circuit valve V4 so as to obtain the required flow rate m cold with
turbulent rate. 4. Wait until the stationary heat flow between the two fluids is obtained and measure the
values of inlet, intermediate and outlet temperature of the two circuits 5. Keep the hot flow rate m hot at a constant level; increase the cold flow rate, wait for
steady state then repeat the temperature reading. 6. Repeat same procedure for the counter flow circuit.
THE HASHIMITE UNIVERSITY FACULTY OF ENGINEERING Department of Mechanical Engineering
Thermal science 1 Lab. - Exp # 8: Double pipe concentric tube heat exchanger Page 4 of 5
Results:-
Flow meters Temperatures (οC)
LMTD (οC)
U
(W/m2K)Hot
water (L/hr)
Cold water (L/hr)
Hot water Cold water
Inlet T1
Middle T2
Outlet T3
Outlet T4
Middle T5
Inlet T6
320
100
50
320
150
50
320
200
50
320
250
50
Table(1): Parallel flow results
Flow meters Temperatures (οC)
LMTD (οC)
U
(W/m2K)Hot
water (L/hr)
Cold water (L/hr)
Hot water Cold water
Inlet T1
Middle T2
Outlet T3
Inlet T4
Middle T5
Outlet T6
320
100
50
320
150
50
320
200
50
320
250
50
Table (2): Counter flow results
THE HASHIMITE UNIVERSITY FACULTY OF ENGINEERING Department of Mechanical Engineering
Thermal science 1 Lab. - Exp # 8: Double pipe concentric tube heat exchanger Page 5 of 5
Analysis:- NOTE: The following analysis should be performed for both parallel and counter flow heat exchangers.
1. Characteristic curve: • Plot ΔThot versus cold water flow rate.
2. Temperature distribution in heat exchanger. • Plot the average inlet, intermediate and outlet temperatures of the two fluids as a
function of the length of the heat exchanger. 3. Overall heat transfer coefficient U
• LMTDAUQ ××=
• LMTDA
QU hot
×=
• hotpwhothot TCmQ Δ××= • coldpwcoldcold TCmQ Δ××= • coldhotloss QQQ −= • Plot U versus cold water flow rate. • Plot Qloss versus cold water flow rate
Where: ΔThot = Thot water inlet – Thot water outlet ΔTcold = Tcold water outlet – Tcold water inlet Cpw = 4.18 kj/kgK. A = Internal exchange surface area between the two fluids = 0.226 m2
THE HASHIMITE UNIVERSITY FACULTY OF ENGINEERING Department of Mechanical Engineering
Thermal science 1 Lab. - Exp # 09: Cross flow heat exchanger Page 1 of 16
NOTE: The following theory and apparatus description are served for both experiments (4) &(5) Theory:
• Introduction In order to transfer heat between two fluids many forms of heat exchanger have been
devised. In one of the most common arrangements, heat is transferred between a fluid flowing through a bundle of tubes and another fluid flowing transversely over the outside of the tubes. This configuration is known as a Cross Flow Heat Exchanger and is shown schematically in Fig.1
Figure(1): Cross flow heat exchanger
Various tube layouts have been devised in order to improve the efficiency of the cross
flow heat exchanger and thereby reduce the physical size for a given heat transfer rate. However, the objective of all of the arrangements is to promote turbulence in the fluid flowing across the tube bundle.
The reason for this lies in the fact that the overall heat transfer coefficient for a cross flow heat exchanger is made up of three components. Firstly the surface heat transfer coefficient for the fluid is flowing through the tubes, secondly the thermal conductivity and thickness of the tube material, and thirdly the surface heat transfer coefficient for the fluid flowing over the external surface of the tubes.
Enhancement of the first two components may be achieved by increasing flow velocity in the tubes and reducing the tube wall thickness, or using a material of higher thermal conductivity.
The third component may be increased by raising the stream velocity, thereby increasing the external Reynolds Number of each individual tube. Alternatively, the tube layout may be changed in order to maximize turbulence. This is achieved by ensuring that each row of tubes is positioned such that turbulence induced by the preceding row is incident upon the next row. Hence a cascade effect is produced such that the degree of turbulence increases with the depth of the tube bundle.
The effect of turbulence is to enhance the surface heat transfer coefficient beyond the level achieved by increased Reynolds Number alone.
If the fluid flowing over the outside of the tubes is a gas, then the effective heat transfer coefficient may be further increased by the use of extended surfaces, e.g. fins.
Exp. # (9)
Cross flow heat exchanger
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Thermal science 1 Lab. - Exp # 09: Cross flow heat exchanger Page 2 of 16
As cross flow heat exchangers occur in many varied forms throughout industry, it is essential that engineers and technologists should be aware of the performance of such units.
• Theoretical background:
The simplest form of cross flow heat exchanger may be regarded as a series of identical heat transfer surfaces in a transverse stream that each has an influence on, and is in turn influenced by, its neighbor. Therefore, in order to obtain a prediction for the heat transfer rate to or from a bundle of surfaces in cross flow it is usual to initially consider a single surface in isolation as a basis for correlation.
A. Isolated Cylinder In Cross Flow
Two distinct types of convective heat transfer exist, these being laminar and turbulent. In the case of laminar flow the fluid flows in filaments, or stream lines that do not mix.
Hence heat transfer from a surface in laminar flow must occur by conduction through the fluid itself. Therefore the rate of heat transfer will be low and highly dependent upon the thermal conductivity of the fluid.
In the case of turbulent flow mixing of the fluid occurs. Hence a “packet” of fluid may at one instant be close to the heated surface and then rapidly transfer and dissipate in the stream, thus transferring heat very quickly to the bulk of the fluid. Hence the higher the degree of turbulence, the higher the rates of heat transfer.
For laminar flow it is possible to devise expressions for the mean surface heat transfer coefficient in particular cases of geometry. For example, laminar flow in pipes and laminar flow over flat plates. However, for external flow over cylinders this is not generally possible and empirical methods must be used.
Similarly, except for special cases, turbulent flow conditions do not lend themselves to simple theoretical analysis and therefore alternative methods are required in order to evaluate surface heat transfer coefficients for general flow conditions.
One such method is to apply the principle of dynamic similarity. This, along with certain assumptions, proves that the following statements are valid for
both laminar and turbulent flow: 1. The velocity distribution within two boundaries will be similar when the Reynolds
Numbers μ
ρUL are the same for both fields.
2. The temperature distribution within two boundaries will be similar when in addition to
(1) the Prandtl Numbers (k
C pμ ) are the same for both fluids.
3. When (1) and (2) are satisfied, then the Nusselt Numbers (k
hL ) for corresponding surface
elements will be the same and hence the average Nusselt Numbers will be the same for both surfaces. These conditions may be summarized by writing:
Pr)(Re,fNu = ………………………………………………………………….(1) It follows, therefore, that empirical data obtained for a certain set of conditions on
perhaps a scale model heat exchanger may be equally applied to a full scale unit providing that the geometry, Reynolds and Prandtl Numbers are equal.
In order to reduce equation (1) to a usable form, dimensional analysis may be used and this results in the following general relationship,
nmCNu PrRe= ……………………………………………………….……….(2)
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Thermal science 1 Lab. - Exp # 09: Cross flow heat exchanger Page 3 of 16
Generally in the case of gases the Prandtl Number varies little. For the variations in temperature and pressure normally encountered, and the Prandtl Number factor may be assumed part of the constant C.
Therefore, by carrying out a series of tests on apparatus of a particular geometry at varying Reynolds Numbers, it is possible to obtain values for the constants C and m.
For the case of an isolated cylinder in turbulent cross flow conditions, the following relationship is generally accepted for Reynolds Numbers (based on cylinder diameter) between 4000 and 40,000.
618.0Re174.0=Nu ………………………………………………………...…(3)
B. Tube Bundles in Cross flow: In the case of an isolated cylinder in cross flow, the velocity, used to calculate the
Reynolds Number of the flow is that of the stream approaching the cylinder. However, for the case of a tube bundle obstructing the duct, it can be readily appreciated
that the velocity of the flow approaching the bundle will be far lower than the velocity between the rows of tubes, the duct area having been reduced by some function of the transverse plan area of the tubes.
A characteristic reference velocity for a particular tube bundle is therefore taken, and an accepted value is the stream velocity at the minimum free area.
Hence, if the empty duct has a cross sectional area of Ad and the minimum inter tube area is At the velocity through the beat exchanger will be,
t
dAA
UU ×=′ ……………………………………………………..……………(4)
It is the velocity that is used to calculate the Reynolds Numbers used in the correlations. As in the case of a single tube in cross flow, determination of a correlation for the mean
convective heat transfer coefficient for the tubes forming a cross flow heat exchanger must be carried out experimentally.
The tube position within the bundle adds a further variable to the general turbulent flow equation (2) and this then has the form,
FnCNu nm PrRe= …………………………………………………………….(5) Where Fn is a function of the number of tube rows crossed by the transverse stream. Fn = 0.95 for six tube rows. An accepted form of the generalized equation (5) is,
FnNu 34.0635.0 PrRe273.0= …………………………………...…………….(6) The Nusselt Number obtained from this correlation is a mean value for all of the tubes
within a bundle. Hence for design purposes a prediction may be obtained for the overall heat transfer rate of a cross flow heat exchanger of a particular size and number of rows.
The above equation is applicable to a staggered arrangement of tubes shown in fig.1 for Reynolds Numbers between 300 and 200,000.
Similar correlations exist for the various other geometries possible and these are generally available in textbooks, or from references. Apparatus: The cross flow heat exchanger shown in fig. 2 with the following specifications:
• Air Duct: Vertically mounted glass reinforced plastic duct of 65 x 150mm cross section with bell mouth intake at its upper end.
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Thermal science 1 Lab. - Exp # 09: Cross flow heat exchanger Page 4 of 16
Front cover of opaque plastic with a central opening of 200mm length to receive one of two standard tube plates.
• Fan: Three phase centrifugal blower of 1.l KW power input, mounted on an epoxy coated welded steel frame. Air duct is directly mounted on the frame and fan intake.
• Fan Starter: Three phase contactor with current operated overload and On - Off buttons. Mounted on fan frame.
• Air Flow Control: A lever operated iris damper mounted on the fan exhaust. • Single Tube Plate: A clear plastic plate with a centrally drilled hole to accept the single
active element. Plate dimensions such that it snugly fits the 200mm opening in the air duct.
• Multi Tube Plate: A clear plastic plate with 27 fixed plastic tubes of 16mm nominal diameter arranged on an equilateral triangular pitch of 32mm between centers. Tubes form six rows. Near the centre of each row is a dummy tube that may be removed and replaced with the active element.
• Active element: Electrically heated (maximum 70V) thick copper cylinder of nominally 15.8mm diameter and 50mm length. Heated surface area = 2.482 x 10-3 m2. Extreme ends are insulated to reduce errors due to wall effects. Integral thermocouple senses surface temperature.
• Console: All electronic instrumentation and control is housed in a plastic coated steel console with brushed aluminum front and rear panels.
• Temperature: Digital electronic thermometer with 0.1°C resolution. Indicates element surface temperature and, via a biased switch, the duct air temperature.
• Voltage: Analogue voltmeter indicating the voltage across the active element heater. Range: 0 to 70V.
• Active Element Control: Rotary variable transformer regulates voltage across active element heater between 0 and 70V.
• Pressure Measurement: 1 Duct mounted inclined manometer recording intake depression. Range: 0 to 70 mm H2O. I Duct mounted inclined manometer. Range: 0 to 30 mm H2O.
• Voltage Switch: Controls maximum voltage from rotary voltage transformer. 70V maximum supply for 5 pin heater and thermocouple plug. 35V for 7 pin accessories socket.
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Thermal science 1 Lab. - Exp # 09: Cross flow heat exchanger Page 5 of 16
Figure (2): Cross flow heat exchanger apparatus
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Thermal science 1 Lab. - Exp # 09: Cross flow heat exchanger Page 6 of 16
Experiment No (4): Cylinder in cross flow Objective: To study the steady state heat transfer, and to determine the surface heat transfer coefficient for a single tube in a transverse flow air stream. Procedure:
1. Ensure the instrument console main switch is in the off position. Ensure the fan is switched off.
2. If the single tube plate is not in position, remove the four knurled brass nuts retaining the clear plastic tube plate. Remove the existing tube plate and replace with the single tube plate. Replace and retighten the brass nuts.
3. Insert the active element into the hole in the single tube plate and plug the lead into the instrument console.
4. Connect the duct pressure tapping to the right hand tube of the lower manometer with the grey hose provided.
5. Close the Iris damper on the fan discharge to position number 9 and press the fan start button.
6. Adjust the iris damper in conjunction with the intake manometer to obtain a low velocity air flow through the duct (a depression H of approximately 4mm H2O).
7. Switch the voltage switch to 70V. Depress the main switch on the instrument console and adjust the heater control to give an indicated active element surface temperature Ts, of approximately 95°C. At low air velocities the heat transfer rate is low and it is advisable to adjust the heater control in increments, allowing time between each adjustment for the system to stabilize.
8. When stable conditions occur indicated by a constant active element surface temperature record the values of Ts, Ta, H and V.
9. Adjust the iris damper on the fan exhaust to increase the indicated air depression H and hence the duct air velocity.
10. Adjust the heater control to give approximately the original active element surface temperature Ts.
11. Again when stable, record Ts, Ta, H and V. 12. Repeat the above procedure for increasing air velocities up to the maximum (iris damper
fully open).
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Thermal science 1 Lab. - Exp # 09: Cross flow heat exchanger Page 7 of 16
Results: Atmospheric pressure Pa: Pa Heater element resistance R: 66.7 Ohms
Test No. 1 2 3 4 5 6 Active element surface
temperature Ts (oC)
Duct air temperature Ta (oC)
Intake air depression H (mm H2O) 4 11 20 32.5 44 54
Active element heater voltage V (volts)
Tube row Single tube
Table(1): Experimental results for single tube.
Heat transfer rate Q (W)
Heat flux Ф (W/m2)
Active element surface to air temperature
difference Ta-Ts (K)
Mean surface heat transfer coefficient
h (W/m2K)
Duct air velocity U (m/s)
Effective air velocity U' (m/s)
Reynolds number (Re)
Nusselt number (Nu)
Table (2): Derived results for single tube.
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Thermal science 1 Lab. - Exp # 09: Cross flow heat exchanger Page 8 of 16
Analysis: Calculate the followings:
• Heat transfer rate from the active element.
RVQ
2
=
• Heat flux.
AQ
=Φ Where A is the area of heat transfer surface = 2.482 × 10-3 m2
• Mean surface heat transfer coefficient.
)( as TTh
−Φ
=
• Duct air velocity.
a
a
PHT
U×
= 294.74
• Reynolds number.
νUd
=Re Whereν is the kinematic viscosity of air at Ta
On the same graph paper, plot h versus Re for the data obtained from the experiment, and the data given by the following correlation:
618.0
618.0
Re174.0
Re174.0
×=
==
dkh
khdNu
Where k is the thermal conductivity of the air at Ta.
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Thermal science 1 Lab. - Exp # 09: Cross flow heat exchanger Page 9 of 16
Experiment No (5): Tube bundles in cross flow Objective: To determine the steady state mean surface heat transfer coefficient for tubes in the 1st, 2nd, 3rd, 4th, 5th, and 6th rows of a cross flow over a tube bundle heat exchanger. Procedure:
1. Ensure the instrument console main switch is in the off position. Ensure the fan is switched off.
2. If the multi-tube plate is not in position, remove the four knurled brass nuts retaining the clear plastic tube plate. Remove the existing tube plate and replace with the multi-tube plate. Replace and retighten the brass nuts.
3. Insert the active element into the top open hole in the tube plate and plug the lead into the instrument console. Ensure that the five remaining dummy tubes are in position in the lower holes.
4. Connect the duct pressure tapping to the right hand tube of the lower manometer with the grey hose provided.
5. Close the Iris damper on the fan discharge to position number 9 and press the fan start button.
6. Adjust the iris damper in conjunction with the intake manometer to obtain a low velocity air flow through the duct (a depression H of approximately 1.5mm H2O).
7. Switch the voltage switch to 70V. Depress the main switch on the instrument console and adjust the heater control to give an indicated active element surface temperature Ts, of approximately 95°C. At low air velocities the heat transfer rate is also low and it is advisable to adjust the heater control in increments, allowing time between each adjustment for the system to stabilize.
8. When stable conditions occur indicated by a constant active element surface temperature record the values of Ts, Ta, H and V.
9. Adjust the iris damper on the fan exhaust to increase the indicated air depression H and hence the duct air velocity.
10. Adjust the heater control to give approximately the original active element surface temperature Ts.
11. Again when stable, record Ts, Ta, H and V. 12. Repeat the above procedure for increasing air velocities up to the maximum (iris damper
fully open). 13. Turn the heater control to minimum and allow the active element to cool. 14. Place the active element in the second row hole and place the dummy tube from this hole
in the first row hole. 15. Repeat the experiment for a similar range of tunnel intake depressions. 16. Repeat the entire procedure with the active element in rows 3, 4, 5 and 6.
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Thermal science 1 Lab. - Exp # 09: Cross flow heat exchanger Page 10 of 16
Results: Atmospheric pressure: Pa Heater element resistance R:66.7 Ohms(Ω) Test No. 1 2 3 4 Active element surface temperature Ts (oC)
Duct air temperature Ta (oC)
Intake air depression H (mm H2O) 2.5 4.5 6.5 8.5
Active element heater voltage V (volts)
Tube row 1 1 1 1
Table (1): Experimental results for tube No (1)
Heat transfer rate Q (W)
Heat flux Ф (W/m2)
Active element surface to air temperature
difference Ta-Ts (K)
Mean surface heat transfer coefficient
h (W/m2K)
Duct air velocity U (m/s)
Effective air velocity U' (m/s)
Reynolds number (Re)
Nusselt number (Nu)
Table(2): Derived results for tube No (1)
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Thermal science 1 Lab. - Exp # 09: Cross flow heat exchanger Page 11 of 16
Test No. 1 2 3 4 Active element surface
temperature Ts (oC)
Duct air temperature Ta (oC)
Intake air depression H (mm H2O) 2.5 4.5 6.5 8.5
Active element heater voltage V (volts)
Tube row 2 2 2 2
Table (3): Experimental results for tube No (2)
Heat transfer rate Q (W)
Heat flux Ф (W/m2)
Active element surface to air temperature
difference Ta-Ts (K)
Mean surface heat transfer coefficient
h (W/m2K)
Duct air velocity U (m/s)
Effective air velocity U' (m/s)
Reynolds number (Re)
Nusselt number (Nu)
Table(4): Derived results for tube No (2)
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Thermal science 1 Lab. - Exp # 09: Cross flow heat exchanger Page 12 of 16
Test No. 1 2 3 4 Active element surface
temperature Ts (oC)
Duct air temperature Ta (oC)
Intake air depression H (mm H2O) 2.5 4.5 6.5 8.5
Active element heater voltage V (volts)
Tube row 3 3 3 3
Table(5): Experimental results for tube No (3)
Heat transfer rate Q (W)
Heat flux Ф (W/m2)
Active element surface to air temperature
difference Ta-Ts (K)
Mean surface heat transfer coefficient
h (W/m2K)
Duct air velocity U (m/s)
Effective air velocity U' (m/s)
Reynolds number (Re)
Nusselt number (Nu)
Table(6): Derived results for tube No (3)
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Thermal science 1 Lab. - Exp # 09: Cross flow heat exchanger Page 13 of 16
Test No. 1 2 3 4 Active element surface
temperature Ts (oC)
Duct air temperature Ta (oC)
Intake air depression H (mm H2O) 2.5 4.5 6.5 8.5
Active element heater voltage V (volts)
Tube row 4 4 4 4
Table (7): Experimental results for tube No (4)
Heat transfer rate Q (W)
Heat flux Ф (W/m2)
Active element surface to air temperature
difference Ta-Ts (K)
Mean surface heat transfer coefficient
h (W/m2K)
Duct air velocity U (m/s)
Effective air velocity U' (m/s)
Reynolds number (Re)
Nusselt number (Nu)
Table (8): Derived results for tube No (4)
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Thermal science 1 Lab. - Exp # 09: Cross flow heat exchanger Page 14 of 16
Test No. 1 2 3 4 Active element surface
temperature Ts (oC)
Duct air temperature Ta (oC)
Intake air depression H (mm H2O) 2.5 4.5 6.5 8.5
Active element heater voltage V (volts)
Tube row 5 5 5 5
Table (9): Experimental results for tube No (5)
Heat transfer rate Q (W)
Heat flux Ф (W/m2)
Active element surface to air temperature
difference Ta-Ts (K)
Mean surface heat transfer coefficient
h (W/m2K)
Duct air velocity U (m/s)
Effective air velocity U' (m/s)
Reynolds number (Re)
Nusselt number (Nu)
Table (10): Derived results for tube No (5)
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Thermal science 1 Lab. - Exp # 09: Cross flow heat exchanger Page 15 of 16
Test No. 1 2 3 4 Active element surface
temperature Ts (oC)
Duct air temperature Ta (oC)
Intake air depression H (mm H2O) 2.5 4.5 6.5 8.5
Active element heater voltage V (volts)
Tube row 6 6 6 6
Table (11): Experimental results for tube No (6)
Heat transfer rate Q (W)
Heat flux Ф (W/m2)
Active element surface to air temperature
difference Ta-Ts (K)
Mean surface heat transfer coefficient
h (W/m2K)
Duct air velocity U (m/s)
Effective air velocity U' (m/s)
Reynolds number (Re)
Nusselt number (Nu)
Table (12): Derived results for tube No (6)
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Thermal science 1 Lab. - Exp # 09: Cross flow heat exchanger Page 16 of 16
Analysis: Calculate the followings:
• Heat transfer rate from the active element.
RVQ
2
=
• Heat flux.
AQ
=Φ Where A is the area of heat transfer surface = 2.482 × 10-3 m2
• Mean surface heat transfer coefficient.
)( as TTh
−Φ
=
• Duct air velocity.
a
a
PHT
U×
= 294.74
• Effective air velocity U' = U×2.343
• Reynolds number.
νUd
=Re Whereν is the kinematic viscosity of air at Ta
On the same graph paper, plot h versus Re for tubes with the active element in the rows from 1 to 6.
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Thermal science 1 Lab. - Exp # 10: Combined convection &radiation Page 1 of 4
Objective:- 1. To determine the combined heat transfer (radiation convection) from a horizontal cylinder in
natural convection over a wide range of power inputs and corresponding surface temperatures.
2. To demonstrate the relationship between power input and surface temperature in
Free convection.
Apparatus:-
Figure(1):Combined convection and radiation apparatus
Exp. # (10) Combined convection &radiation
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Thermal science 1 Lab. - Exp # 10: Combined convection &radiation Page 2 of 4
Theory:- If a surface, at a temperature above that of its surroundings, is located in stationary air at the
same temperature as the surroundings then heat will be transferred from the surface to the air and
surroundings. This transfer of heat will be a combination of natural convection to the air (air
heated by contact with the surface) and radiation to the surroundings. A horizontal cylinder is
used in this experiment to provide a simple shape from which the heat transfer can be calculated.
In the case of natural (free) convection the mean heat transfer coefficient (Hcm) can be calculated
using the following steps.
1. Grashof number calculation
2
3)(υ
β DTaTsgGrD−
=
Where:-
g = Acceleration due to gravity = 9.81 (m/s2)
β = Volume expansion coefficient (K-1)
ν = Dynamic viscosity of air (m2/s)
The volumetric expansion coefficient (β) = 1/ Tf
Where Tf is the film temperature which equal (Ts+Ta)/2
2.Raleigh number Ra
Pr)(Pr 2
3
υ
β DTaTsgGrRa DD−
==
Where Pr is the prandtl number
3. Nusselt number
nDm RacNu )(=
Where c and n are obtained from the table below
RaD C n 10-9to10-2 0.675 0.058 10-2to102 1.02 0.148102to104 0.850 0.188 104to107 0.480 0.250 107to1012 0.125 0.333
Table (1): listing constant c and exponent n for natural convection on a horizontal cylinder
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Thermal science 1 Lab. - Exp # 10: Combined convection &radiation Page 3 of 4
4. Mean heat transfer coefficient (Hcm)
DKNu
Hc mm
)(=
Where:-
Hcm is the mean heat transfer coefficient for natural convection (W/m2K).
K is thermal conductivity of air (W/Mk).
Note The physical properties of air K, υ,and Pr are take at film temperature (Tf).
Also the heat transfer coefficient for free convection may be calculated using the following
simplified equation. 25.0)(32.1 ⎥⎦
⎤⎢⎣⎡ −
=D
TaTsHcm
Then the heat loss due to natural convection (Qc) can be calculated using the following relation.
)( TaTsAsHcQc m −= (W)
Where AS is the heat transfer area (surface area).
In the case of radiation the mean heat transfer coefficient (Hrm) can be calculated using the
following relationship.
TaTsTaTsFHrm −
−=
)( 44σξ
Where:-
σ is Stefan Boltzman constant = 5.67 x 10-8 ( W/m2K4).
ξ is the emissivity of surface = 0.95.
F is the view factor = 1.
Then the heat loss due to radiation (Qr) can be calculated using the following relationship.
)( TaTsAsHrQr m −= (W)
The total heat loss from the cylinder (Qtot ) = Qc + Qr
Procedure:- 1. Set the heater voltage to 5 Volt (adjust the voltage control potentiometer to give reading
of 5 Volt on the top panel meter with the selector switch set to position V).
2. Allow the surface temperature of the cylinder T10 to stabilize using the lower selector
switch/meter
3. When the temperatures are stable record T9, T10, V, and I in the table below.
4. Repeat steps 2&3 for 8, 12, and 15 Voltage.
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Thermal science 1 Lab. - Exp # 10: Combined convection &radiation Page 4 of 4
Diameter of cylinder (D) = 10mm.
Heated length of cylinder (L) = 70mm.
Test No Voltage (V) Current(I) Air
temperature T9(°C)
Surface temperature
T10 (°C) 1 5
2 8
3 12
4 15
Analysis & Results:-
No Power Qin(W) Hcm(W/m2K) Hrm(W/m2K) Qc(W) Qr(W) Qtot(W)
1
2
3
4
• Compare the theoretical values for Qtot with the measured values for Qin and explain
any differences in values.
• Compare the calculated heat transferred due to Convection Qc and radiation Qr.
• Compare the value for Hcm obtained using the simplified and full empirical equations and
comment on any difference.
• Plot a graph of surface temperature T1O against power input Qin and observe the
relationship.
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Thermal science 1 Lab. - Exp # 11: Force convection & radiation Page 1 of 5
Objective:- 1. To determine the effect of force convection on heat transfer from the surface of a cylinder at
varying air velocities and surface temperatures.
2. To demonstrate the relationship between air velocity and surface temperature for a cylinder
subject to forced convection.
Apparatus:-
Figure(1):Combined convection and radiation apparatus
Exp. # (11)
Force convection & radiation
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Thermal science 1 Lab. - Exp # 11: Force convection & radiation Page 2 of 5
Theory:- In free (natural) convection the heat transfer rate from a surface is limited by the movement of
air which are generated by change in the density of the air as the air is heated by the surface. In
force convection the air movement can be greatly increased resulting in improved heat transfer
rate from a surface. Therefore a surface subjected to force convection will have a lower surface
temperature than the same surface subjected to free convection, for the same power input
. If a surface, at a temperature above that of its surroundings, is located in moving air at the same
temperature as the surroundings then heat will be transferred from the surface to the air and
surroundings. This transfer of heat will be a combination of force convection to the air (heat is
transferred to the air passing the surface) and radiation to the surroundings. A horizontal cylinder
is used in this experiment to provide a simple shape from which the heat transfer can be
calculated.
The heat transfer coefficient Hfm due to force convection and Hrm due to radiation can be
calculated using the following relationships:
• Calculation of heat transfer coefficient for radiation
TaTsTaTsFHrm −
−=
)( 44σξ
Where:-
σ is Stefan Boltzman constant = 5.67 x 10-8 ( W/m2K4).
ξ is the emissivity of surface = 0.95.
F is the view factor = 1.
Ts is surface temperature of the cylinder (K).
Ta is the ambient temperature.
Then the heat loss due to radiation (Qr) can be calculated using the following relationship.
)( TaTsAsHrQr m −= (W)
Where AS is the heat transfer area (surface area).
• Calculation of heat transfer coefficient for force convection
Where:
k is the thermal conductivity of the air (W/m2K).
D is the diameter of the cylinder. (m).
Num is the average Nusselt number.
mm NuDkHf =
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Thermal science 1 Lab. - Exp # 11: Force convection & radiation Page 3 of 5
An empirical formula can be used to calculate the value for Num as follows:
Where;
Re is the Reynolds number = UcD/υ
Pr is the Prandtl number for air.
Uc is the corrected air velocity (m/s).
Corrected air velocity = 1.22Ua(m/s)
Note The physical properties of air K, υ,and Pr are take at film temperature (Tf).
Then the heat loss due to force convection (Qf) can be calculated using the following relation.
)( TaTsAsHfQf m −= (W)
Where AS is the heat transfer area (surface area).
The total heat loss from the cylinder (Qtot ) = Qf + Qr
Procedure:- 1. Start the centrifugal fan by pressing the switch on the connection box.
2. Open the throttle plate on the front of the fan by rotating the knob at the center to give a
reading of 0.5m/s on the upper panel meter.
3. Set the heater voltage to 20 Volt (adjust the voltage control potentiometer to give reading
of 20 Volt on the top panel meter with the selector switch set to position V).
4. Allow the surface temperature of the cylinder T10 to stabilize using the lower selector
switch/meter
5. When the temperatures are stable record Ua, T9, T10, V, and I in the table below.
6. Adjust the throttle plate to give a velocity of 1.0 m/s (stop selector switch set to position
Ua).
7. Allow the temperature stabilize then repeat the above reading.
8. Repeat the above procedure changing the air velocity in steps of 1.0 m/s until the air
velocity is 7.0 m/s.
( )⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎠⎞
⎜⎝⎛+
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎠⎞
⎜⎝⎛+
+=5.0
25.066.0
33.05.0
282000Re1
Pr4.01
PrRe62.03.0Num
THE HASHIMITE UNIVERSITY FACULTY OF ENGINEERING Department of Mechanical Engineering
Thermal science 1 Lab. - Exp # 11: Force convection & radiation Page 4 of 5
Diameter of cylinder (D) = 10mm.
Heated length of cylinder (L) = 70mm.
Teat No
Velocity of
air
Ua
(m/s)
Voltage
V
(V)
Current
I
(A)
Air
temperature
T9(°C)
Surface
temperature
T10 (°C)
1 0.5
2 1
3 2
4 3
5 4
6 5
7 6
8 7
Analysis & Results:-
Test
No
Power
Qin(W)
Corrected
Air velocity
Uc (m/s)
Hfm
(W/m2K)
Hrm
(W/m2K) Qc(W) Qr(W) Qtot(W)
1
2
3
4
5
6
7
8
• Compare the theoretical values for Qtot with the measured values for Qin and explain
any difference in the two value values.
• Compare the calculated heat transferred due to force Convection Qf and radiation Qr.
• Plot a graph of surface temperature T1O against corrected air velocity.
THE HASHIMITE UNIVERSITY FACULTY OF ENGINEERING Department of Mechanical Engineering
Thermal science 1 Lab. - Exp # 11: Force convection & radiation Page 5 of 5
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