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 8, Problem 135P
To determine

The rate of entropy generation during the process.

Expert Solution & Answer
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Explanation of Solution

Given:

The inlet pressure of the water (P1) is 20 psia.

The inlet temperature of the water (T1) is 50°F.

The mass flow rate of water (m˙1) is 300lbm/min.

The inlet pressure of the steam (P2) is 20 psia.

The inlet temperature of the steam (T2) is 240°F.

The heat lost to the surroundings (Q˙out) is 180Btu/min.

The surrounding temperature of the air (Tsurr) is 70°F.

Calculation:

Refer to Table A-4E, “Saturated water—Temperature table”, obtain the below properties at the pressure of 20psia and temperature of 50°F.

The fluid entropy (sf=s1) is 0.03609Btu/lbmR.

The fluid enthalpy (hf=h1) is 18.07Btu/lbm.

Refer to Table A-4E, “Saturated water—Temperature table”, obtain the below properties at the pressure of 20psia and temperature of 240°F.

The entropy at inlet 2 (s2) is 1.7406Btu/lbmR.

The enthalpy at inlet 2 (h2) is 1162.3Btu/lbm.

Refer to Table A-4E, “Saturated water—Temperature table”, obtain the below properties at the pressure of 20psia and temperature of 130°F.

The entropy at exit 3 (s3) is 0.18174Btu/lbmR.

The enthalpy at exit 3 (h3) is 97.99Btu/lbm.

Write the expression for the energy balance equation for closed system.

  E˙inE˙out=ΔE˙system                              …… (I).

Here, rate of net energy transfer in to the control volume is E˙in, rate of net energy transfer exit from the control volume is E˙out and rate of change in internal energy of system is ΔE˙system.

Substitute 0 for ΔE˙system in Equation (I).

  E˙inE˙out=0E˙in=E˙outm˙1h1+m˙2h2=Q˙out+m˙3h3Q˙out=m˙1h1+m˙2h2m˙3h3                           …… (II).

Here, mass flow rate at inlet 1 is m˙1, mass flow rate at inlet 2 is m˙2, mass flow rate at exit is m˙3.

Substitute m˙3=m˙1+m˙2 in Equation (II).

  Q˙out=m˙1h1+m˙2h2(m˙1+m˙2)h3=m˙1(h1h3)+m˙2(h2h3)

  180Btu/min=[300lbm/min(18.07Btu/lbm97.99Btu/lbm)+m˙2(1162.3Btu/lbm97.99Btu/lbm)]180Btu/min=(23976Btu/min)+m˙2(1064.31Btu/lbm)m˙2=22.7lbm/min

The rate of change in internal energy of the system is zero at steady state,

Write the expression for the mass balance of the system.

  m˙inm˙out=Δm˙system                    …… (III).

Here, inlet mass flow rate is m˙in and outlet mass flow rate is m˙out and change in mass flow rate is Δm˙system.

Substitute m˙in=m˙1+m˙2, m˙out=m˙3 and Δm˙system=0 in Equation (III)

  m˙1+m˙2m˙3=0m˙1+m˙2=m˙3

  m˙3=(300lbm/min)+(22.7lbm/min)=322.7lbm/min

Write the expression for the entropy balance during the process.

  S˙inS˙out+S˙gen=ΔS˙system                …… (IV).

Here, rate of net input entropy is S˙in, rate of net output entropy is S˙out, rate of entropy generation is S˙gen, and rate of change of entropy of the system is ΔS˙system.

Substitute S˙in=m˙1s1+m˙2s2, S˙out=m˙3s3+Q˙outTsurr and ΔS˙system=0 in Equation (IV).

  m˙1s1+m˙2s2m˙3s3Q˙outTsurr+S˙gen=0S˙gen=m˙3s3m˙1s1m˙2s2+Q˙outTsurr

  S˙gen=[(322.7lbm/min)(0.18174Btu/lbmR)(300lbm/min)(0.03609Btu/lbmR)(22.7lbm/min)(1.7406Btu/lbmR)+180Btu/min70°F]=[(322.7lbm/min)(0.18174Btu/lbmR)(300lbm/min)(0.03609Btu/lbmR)(22.7lbm/min)(1.7406Btu/lbmR)+180Btu/min(70+460)R]=8.65Btu/minR

Thus, the rate of entropy generation during the process is 8.65Btu/minR.

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Chapter 8 Solutions

Fundamentals of Thermal-Fluid Sciences

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