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 113P
To determine

The mass flow rate through the boiler, the power produced by the turbine, the rate of heat supply in the boiler and the thermal efficiency of the cycle.

Expert Solution & Answer
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Explanation of Solution

Given:

Pressure of steam at the condenser (P1) is 2psia.

Pressure of steam at the turbine (P3) is 1500psia.

Temperature of steam at the turbine (T3) is 800°F.

Net power produced by the cycle (W˙net) is 2500kW.

Isentropic efficiency of the turbine (ηT) is 0.90.

Calculation:

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

Fundamentals of Thermal-Fluid Sciences, Chapter 9, Problem 113P

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

Refer Table A-5E, “Saturated water-Pressure table”, obtain the enthalpy and specific volume at state 1 corresponding to the pressure of 2psia.

  h1=hf@2psia=94.02Btu/lbmv1=vf@2psia=0.016230ft3/lbm

Refer Table A-5E, “Saturated water-Pressure table”, obtain the following properties at state 1 corresponding to the pressure of 5psia.

  hf=94.02Btu/lbmhfg=1021.7Btu/lbmsf=0.1750Btu/lbmRsfg=1.7444Btu/lbmR

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

  wp,in=v1(P2P1)=(0.016230ft3/lbm)(1500psia2psia)=(0.016230ft3/lbm)(1500psia2psia)(1Btu5.404psiaft3)=4.50Btu/lbm

Calculate the enthalpy at state 2(h2).

  h2=h1+wp,in=94.02Btu/lbm+4.50Btu/lbm=98.52Btu/lbm

Refer Table A-6E, “Superheated water”, obtain the enthalpy and entropy at state 3 corresponding to the pressure of 1500psia and temperature of 800°F.

  h3=1363.1Btu/lbms3=1.5064Btu/lbmR

Calculate the quality of water at state 4(x4s).

  x4s=s4sfsfg=s3sfsfg=1.5064Btu/lbmR0.1750Btu/lbmR1.7444Btu/lbmR=0.7633

Calculate the enthalpy at state 4s (h4s).

  h4s=hf+x4shfg=94.02Btu/lbm+(0.7633)(1021.7Btu/lbm)=873.86Btu/lbm

Calculate the enthalpy at state 4(h4).

  ηT=h3h4h3h4s

  h4=h3(ηT)(h3h4s)=1363.1Btu/lbm(0.90)(1363.1Btu/lbm873.86Btu/lbm)=922.79Btu/lbm

Calculate the heat supplied in the boiler (qin).

  qin=h3h2=1363.1Btu/lbm98.52Btu/lbm=1264.6Btu/lbm

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

  m˙=W˙netwnet=W˙netqinqout=W˙netqin(h4h1)

  =2500kW1264.6Btu/lbm(922.79Btu/lbm94.02Btu/lbm)=2500kJ/s(0.94782Btu1kJ)435.8Btu/lbm=5.437lbm/s

Thus, the mass flow rate through the boiler is 5.437lbm/s.

Calculate the power output from the turbine (W˙T,out).

  W˙T,out=m˙(h3h4)=(5.437lbm/s)(1363.1Btu/lbm922.79Btu/lbm)=(5.437lbm/s)(1363.1Btu/lbm922.79Btu/lbm)(1kJ0.94782Btu)=2526kW

Thus, the power produced by the turbine is 2526kW.

Calculate the rate of heat addition (Q˙in).

  Q˙in=m˙qin=(5.437lbm/s)(1264.6Btu/lbm)=6876Btu/s

Thus, the rate of heat supply in the boiler is 2526kW.

Calculate the thermal efficiency of the cycle (ηth).

  ηth=W˙netQ˙in=2500kJ/s(0.94782Btu1kJ)6876Btu/s=0.3446=34.5%

Thus, the thermal efficiency of the cycle is 34.5%.

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Chapter 9 Solutions

Fundamentals of Thermal-Fluid Sciences

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