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 180RQ

a)

To determine

The thermal efficiency of the cycle.

a)

Expert Solution
Check Mark

Explanation of Solution

Given:

Maximum pressure in the cycle (P3) is 10MPa.

Minimum pressure in the cycle (P1) is 30MPa.

Pressure at which steam is reheated for the first time (P4) is 4MPa.

Pressure at which steam is reheated for the second time (P6) is 2MPa.

Net power output of the cycle (W˙net) is 75MW.

Temperature of steam at all three stages of the turbine (T3/T5/T7) is 550°C.

Calculation:

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

Fundamentals of Thermal-Fluid Sciences, Chapter 9, Problem 180RQ

The pressures are constant for the process 4 to 5, process 6 to 7 and process 8 to 1.

  P4=P5P6=P7P8=P1

The entropies are constant for the process 3 to 4, process 5 to 6 and process 7 to 8.

  s3=s4s5=s6s7=s8

Refer Table A-5, “Saturated water-Pressure table”, obtain the specific enthalpy and specific volume at state 1 corresponding to the pressure of 30kPa.

  h1=hf@30kPa=289.18kJ/kgv1=vf@30kPa=0.001022m3/kg

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

  wp,in=v1(P2P1)=(0.001022m3/kg)(10,000kPa30kPa)=(0.001022m3/kg)(10,000kPa30kPa)(1kJ1kPam3)=10.19kJ/kg

Calculate the specific enthalpy at state 2(h2).

  h2=h1+wp,in=289.18kJ/kg+10.19kJ/kg=299.37kJ/kg

Refer Table A-6, “Superheated water”, obtain the specific enthalpy and specific entropy at state 3 corresponding to the pressure of 10MPa and temperature of 550°C.

  h3=3500.9kJ/kgs3=6.7561kJ/kgK

Refer Table A-6, “Superheated water”, obtain the specific enthalpy at state 4 corresponding to the pressure of 4MPa and specific entropy of 6.7561kJ/kgK.

  h4=3204.9kJ/kg

Refer Table A-6, “Superheated water”, obtain the specific enthalpy and specific entropy at state 5 corresponding to the pressure of 4MPa and temperature of 550°C.

  h5=3559.7kJ/kgs5=7.2335kJ/kgK

Refer Table A-6, “Superheated water”, obtain the specific enthalpy at state 6 corresponding to the pressure of 2MPa and specific entropy of 7.2335kJ/kgK.

  h6=3321.1kJ/kg

Refer Table A-6, “Superheated water”, obtain the specific enthalpy and specific entropy at state 7 corresponding to the pressure of 2MPa and temperature of 550°C.

  h7=3578.4kJ/kgs7=7.5706kJ/kgK

Refer Table A-5, “Saturated water-Pressure table”, obtain the following properties corresponding to the pressure of 30kPa and specific entropy of 7.5706kJ/kgK.

  hf=289.27kJ/kghfg=2335.3kJ/kgsf=0.9441kJ/kgKsfg=6.8234kJ/kgK

Calculate the quality of steam at state 8(s8).

  x8=s8sfsfg=7.5706kJ/kgK0.9441kJ/kgK6.8234kJ/kgK=0.9711

Calculate the specific enthalpy at state 8(h8).

  h8=hf+x8hfg=289.27kJ/kg+(0.9711)(2335.3kJ/kg)=2557.1kJ/kg

Calculate the thermal efficiency of the cycle (ηth).

  ηth=wnetqin=qinqoutqin=[(h3h2)+(h5h4)+(h7h8)](h8h1)(h3h2)+(h5h4)+(h7h8)

  =[(3500.9kJ/kg299.37kJ/kg)+(3559.7kJ/kg3204.9kJ/kg)+(3758.4kJ/kg3321.1kJ/kg)](2557.1kJ/kg289.19kJ/kg){(3500.9kJ/kg299.37kJ/kg)+(3559.7kJ/kg3204.9kJ/kg)+(3758.4kJ/kg3321.1kJ/kg)}=1548.8kJ/kg3813.7kJ/kg=0.4053=40.5%

Thus, the thermal efficiency of the cycle is 40.5%.

b)

To determine

The mass flow rate of the steam.

b)

Expert Solution
Check Mark

Explanation of Solution

Calculate the mass flow rate of the steam (m˙).

  m˙=W˙netwnet=W˙netqinqout=W˙net[(h3h2)+(h5h4)+(h7h8)](h8h1)

  =75MW[(3500.9kJ/kg299.37kJ/kg)+(3559.7kJ/kg3204.9kJ/kg)+(3758.4kJ/kg3321.1kJ/kg)](2557.1kJ/kg289.19kJ/kg)=75,000kJ/s[(3500.9kJ/kg299.37kJ/kg)+(3559.7kJ/kg3204.9kJ/kg)+(3758.4kJ/kg3321.1kJ/kg)](2557.1kJ/kg289.19kJ/kg)=75,000kJ/s1548.8kJ/kg=48.5kg/s

Thus, the mass flow rate of the steam is 48.5kg/s.

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Chapter 9 Solutions

Fundamentals of Thermal-Fluid Sciences

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