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 91P

a)

To determine

The power delivered by the plant assuming constant specific heats for air.

a)

Expert Solution
Check Mark

Explanation of Solution

Given:

Temperature of the air at the compressor inlet (T1) is 290K.

Pressure of the air at the compressor inlet (P1) is 95kPa.

Temperature of the air at the turbine inlet (T2) is 1100K.

Pressure of the air at the turbine inlet (P2) is 880kPa.

Rate of heat transferred to air (Q˙in) is 30,000kJ/s.

Effectiveness for the regenerator (ε) is 0.100.

Calculation:

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

Fundamentals of Thermal-Fluid Sciences, Chapter 9, Problem 91P

Refer Table A-2, “Ideal-gas specific heats of various common gases”, obtain the properties of the nitrogen.

  R=0.2968kJ/kgKcv=0.743kJ/kgKcp=1.039kJ/kgKk=1.4

Calculate the pressure ratio (rP).

  rP=P2P1=880kPa95kPa=9.263

Calculate the temperature at state 2(T2).

  T2=T1(P2P1)k1k=T1(rP)k1k=(290K)(9.263)1.411.4=547.8K

Calculate the temperature at state 4(T4).

  T4=T3(P4P3)k1k=T3(1rP)k1k=(1100K)(19.263)1.411.4=582.3K

For 100% effectiveness of the regenerator, the temperatures at states 5 and 4 are equal and the temperatures at states 6 and 2 are equal.

  T5=T4=582.3 KT6=T2=547.8 K

Calculate the thermal efficiency of the cycle (ηth).

  ηth=1qoutqin=1cP(T6T1)cP(T3T5)=1T6T1T3T5

  =1547.8K290K1100K582.3K=0.5020=50.2%

Calculate the power developed by the plant (W˙net).

  W˙net=ηth(Q˙in)=(0.5020)(30,000kJ/s)=(0.5020)(30,000kW)=15,060kW

Thus, the power delivered by the plant assuming constant specific heats for air is 15,060kW.

b)

To determine

The power delivered by the plant accounting for the variation of specific heats with temperature.

b)

Expert Solution
Check Mark

Explanation of Solution

Calculation:

Refer table A-21, “Ideal gas properties of the air”, obtain the enthalpy and relative pressure of air at the temperature of 290K.

  h1=290.16kJ/kgPr1=1.2311

Calculate the relative pressure at state 2(Pr2).

  Pr2=P2P1Pr1

  Pr2=(rP)Pr1=(9.263)(1.2311)=11.40

Refer table A-21, “Ideal gas properties of the air”, obtain the enthalpy of air at the relative pressure of 11.40.

  h2=548.85kJ/kg

Refer table A-21, “Ideal gas properties of the air”, obtain the enthalpy and relative pressure of air at the temperature of 1100K.

  h3=1161.07kJ/kgPr3=167.1

Calculate the relative pressure at state 4(Pr4).

  Pr4=P4P3Pr3

  Pr4=(1rP)Pr3=(19.263)(167.1)=18.04

Refer table A-21, “Ideal gas properties of the air”, obtain the enthalpy of air at the relative pressure of 18.04.

  h4=624.89kJ/kg

For 100% effectiveness of the regenerator, the enthalpies at states 5 and 4 are equal and enthalpies at states 6 and 2 are equal.

  h5=h4=624.89kJ/kgh6=h2=548.85kJ/kg

Calculate the thermal efficiency of the cycle (ηth).

  ηth=1qoutqin=1h6h1h3h5

  =1548.85kJ/kg290.16kJ/kg1161.07kJ/kg624.89kJ/kg=0.5175=51.75%

Calculate the power developed by the plant (W˙net).

  W˙net=ηth(Q˙in)=(0.5175)(30,000kJ/s)=(0.5175)(30,000kW)=15,525kW

Thus, the power delivered by the plant accounting for the variation of specific heats with temperature is 15,525kW.

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Chapter 9 Solutions

Fundamentals of Thermal-Fluid Sciences

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