Fundamentals of Thermal-Fluid Sciences
Fundamentals of Thermal-Fluid Sciences
5th Edition
ISBN: 9780078027680
Author: Yunus A. Cengel Dr., Robert H. Turner, John M. Cimbala
Publisher: McGraw-Hill Education
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Chapter 9, Problem 179RQ
To determine

Compare the thermal efficiency of the cycle when it is operated such that the saturated liquid enters the pump against that when the sub-cooled liquid enters the pump.

Expert Solution & Answer
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Explanation of Solution

Given:

Pressure of steam at the boiler (P3) is 6000kPa.

Pressure of steam at the condenser (P1) is 50kPa.

Temperature of steam at the turbine inlet (T3) is 600°C.

Temperature of sub-cooled liquid (T) is 11.3°C.

Calculation:

Draw the Ts diagram of the cycle as in Figure (1).

Fundamentals of Thermal-Fluid Sciences, Chapter 9, Problem 179RQ

The pressures are constant for the process 2 to 3 and process 4 to 1.

  P2=P3P4=P1

The entropies are constant for the process 1 to 2 and process 3 to 4.

  s1=s2s3=s4

The thermal efficiency of the cycle when it is operated such that when the saturated liquid enters the pump:

Refer Table A-5, “Saturated water-Pressure table”, obtain the specific enthalpy and specific volume at state 1 corresponding to the pressure of 50kPa.

  h1=hf@50kPa=340.54kJ/kgv1=vf@50kPa=0.001030m3/kg

Calculate the work done by the pump during process 1-2(wp,in).

  wp,in=v1(P2P1)=(0.001030m3/kg)(6000kPa50kPa)=(0.001030m3/kg)(6000kPa50kPa)(1kJ1kPam3)=6.13kJ/kg

Calculate the specific enthalpy at state 2(h2).

  h2=h1+wp,in=340.54kJ/kg+6.13kJ/kg=346.67kJ/kg

Refer Table A-6, “Superheated water”, obtain the specific enthalpy and temperature at state 3 corresponding to the pressure of 6MPa and temperature of 600°C.

  h3=3658.8kJ/kgs3=7.1693kJ/kgK

Refer Table A-5, “Saturated water-Pressure table”, obtain the following properties corresponding to the pressure of 50kPa and specific entropy of 7.1693kJ/kgK.

  hf=340.54kJ/kghfg=2304.7kJ/kgsf=1.0912kJ/kgKsfg=6.5019kJ/kgK

Calculate the quality of steam at state 4(s4).

  x4=s4sfsfg=7.1693kJ/kgK1.0912kJ/kgK6.5019kJ/kgK=0.9348

Calculate the specific enthalpy at state 4(h4).

  h4=hf+x4hfg=340.54kJ/kg+(0.9348)(2304.7kJ/kg)=2495kJ/kg

Calculate the thermal efficiency of the cycle (ηth).

  ηth=1qoutqin=1h4h1h3h2

  =2495kJ/kg340.64kJ/kg3658.8kJ/kg346.67kJ/kg=2154.5kJ/kg3312.1kJ/kg=0.3495=34.95%

The thermal efficiency of the cycle when it is operated such that when the sub-cooled liquid enters the pump:

Refer Table A-5, “Saturated water-Pressure table”, obtain the saturation temperature corresponding to the pressure of 50kPa.

  Tsat@50kPa=81.3°C

Calculate the temperature at state 1(T1) .

  T1=Tsat@50kPaT=81.3°C11.3°C=70°C

Refer Table A-4, “Saturated water-Temperature table”, obtain the specific enthalpy and specific volume at state 1 corresponding to the temperature of 70°C.

  h1=hf@70°C=293.07kJ/kgv1=vf@70°C=0.001023m3/kg

Calculate the work done by the pump during process 1-2(wp,in).

  wp,in=v1(P2P1)=(0.001023m3/kg)(6000kPa50kPa)=(0.001023m3/kg)(6000kPa50kPa)(1kJ1kPam3)=6.09kJ/kg

Calculate the enthalpy at state 2(h2).

  h2=h1+wp,in=293.07kJ/kg+6.09kJ/kg=299.16kJ/kg

Calculate the thermal efficiency of the cycle (ηth).

  ηth=1qoutqin=1h4h1h3h2

  =2495kJ/kg293.09kJ/kg3658.8kJ/kg299.16kJ/kg=2201.9kJ/kg3359.6kJ/kg=0.3446=34.46%

By comparing the thermal efficiencies of the cycle, it is found that the thermal efficiency slightly reduces because of the sub-cooling at the inlet of the pump.

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Chapter 9 Solutions

Fundamentals of Thermal-Fluid Sciences

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