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THE HASHIMITE UNIVERSITY FACULTY OF ENGINEERINGDepartment of Mechanical Engineering

Thermal science 1 Lab. - Exp # 1 : Marcet boiler 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



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.



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.



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



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.



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



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



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



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.



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.



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.


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



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



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?


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 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



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.



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 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



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.


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



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



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)



• 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


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



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)



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.



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



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


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



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)



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.



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.



Figure (2): Cross flow heat exchanger apparatus



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).



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.



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.



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.



Results: Atmospheric pressure: Pa Heater element resistance R:66.7 Ohms(Ω) Test No. 1 2 3 4 Active element surface temperature Ts (oC)


Intake air depression H (mm H2O) 2.5 4.5 6.5 8.5


Tube row 1 1 1 1

Table (1): Experimental results for tube No (1)


Heat flux Ф (W/m2)




h (W/m2K)




Nusselt number (Nu)

Table(2): Derived results for tube No (1)



Test No. 1 2 3 4 Active element surface

temperature Ts (oC)




Tube row 2 2 2 2



Heat flux Ф (W/m2)




h (W/m2K)




Nusselt number (Nu)





temperature Ts (oC)




Tube row 3 3 3 3

Table(5): Experimental results for tube No (3)


Heat flux Ф (W/m2)




h (W/m2K)




Nusselt number (Nu)





temperature Ts (oC)




Tube row 4 4 4 4



Heat flux Ф (W/m2)




h (W/m2K)




Nusselt number (Nu)

Table (8): Derived results for tube No (4)




temperature Ts (oC)




Tube row 5 5 5 5



Heat flux Ф (W/m2)




h (W/m2K)




Nusselt number (Nu)





temperature Ts (oC)




Tube row 6 6 6 6



Heat flux Ф (W/m2)




h (W/m2K)




Nusselt number (Nu)




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.


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



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



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.



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.


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



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)


• 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 =



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)


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



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.

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