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 127P

a)

To determine

The pressure at which the reheating takes place.

a)

Expert Solution
Check Mark

Explanation of Solution

Given:

Pressure of steam at state 3(P3) is 800psia.

Pressure of steam at state 1(P1) is 1psia.

Temperature of steam at the state 3(T3) is 900°F.

Temperature of steam at the state 5(T5) is 800°F.

Temperature of steam at the condenser (Tcondenser) is 45°F.

Rate of heat input (Q˙in) is 6×104Btu/s.

Calculation:

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

Fundamentals of Thermal-Fluid Sciences, Chapter 9, Problem 127P

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

  s3=s4s5=s6

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

  h1=hsat@1psia=69.72Btu/lbmv1=vsat@1psia=0.016124ft3/lbmT1=Tsat@1psia=101.69°F

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

  wp,in=v1(P2P1)=(0.016124ft3/lbm)(800psia1psia)=(0.016124ft3/lbm)(800psia1psia)(1Btu5.4039psiaft3)=2.39Btu/lbm

Calculate the specific enthalpy at state 2(h2).

  h2=h1+wp,in=69.72Btu/lbm+2.39Btu/lbm=72.11Btu/lbm

Refer Table A-6E, “Superheated water”, obtain the specific enthalpy and specific entropy at state 3 corresponding to the pressure of 800psia and temperature of 900°F.

  h3=1456Btu/lbms3=1.6413Btu/lbmR

Refer Table A-5E, “Saturated water-Pressure table”, obtain the pressure at state 4 corresponding to the specific entropy of 1.6413Btu/lbmR.

  h4=hg@sg=s4=1178.5Btu/lbmP4=Psat@sg=s4=62.23psia

Thus, the pressure at which the reheating takes place is 62.23psia.

b)

To determine

The thermal efficiency of the cycle.

b)

Expert Solution
Check Mark

Explanation of Solution

Calculation:

Refer Table A-6E, “Superheated water”, obtain the specific enthalpy and specific entropy at state 5 corresponding to the pressure of 62.23psia and temperature of 800°F.

  h5=1431.4Btu/lbms5=1.8985Btu/lbmR

Refer Table A-5E, “Saturated water-Pressure table”, obtain the following properties corresponding to the pressure of 1psia and specific entropy of 1.8985Btu/lbmR.

  hf=69.72Btu/lbmhfg=1035.7Btu/lbmsf=0.13262Btu/lbmRsfg=1.84495Btu/lbmR

Calculate the quality of steam at state 6(x6).

  x6=s6sfsfg=1.8985Btu/lbmR0.13262Btu/lbmR1.84495Btu/lbmR=0.9572

Calculate the specific enthalpy at state 6(h6).

  h6=hf+x6hfg=69.72Btu/lbm+(0.9572)(1035.7Btu/lbm)=1061.0Btu/lbm

Calculate the thermal efficiency of the cycle (ηth).

  ηth=1qoutqin=1h6h1(h3h2)+(h5h4)=11061.0Btu/lbm69.72Btu/lbm{(1456Btu/lbm72.11Btu/lbm)+(1431.4Btu/lbm1178.5Btu/lbm)}

  =0.394=39.4%

Thus, the thermal efficiency of the cycle is 39.4%.

c)

To determine

The minimum mass flow rate of the cooling water.

c)

Expert Solution
Check Mark

Explanation of Solution

Calculation:

Calculate the rate of heat output from the cycle (Q˙out).

  Q˙out=Q˙inW˙net=(1ηth)Q˙in=(10.394)(6×104Btu/s)=3.634×104Btu/s

Calculate the minimum mass flow rate of the cooling water (m˙cool).

  m˙cool=Q˙outcΔT=Q˙inW˙net=3.634×104Btu/s(1Btu/lbm°F)(101.69°F45°F)=641lbm/s

Thus, the minimum mass flow rate of the cooling water is 641lbm/s.

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Chapter 9 Solutions

Fundamentals of Thermal-Fluid Sciences

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