Thermodynamics: An Engineering Approach
Thermodynamics: An Engineering Approach
8th Edition
ISBN: 9780073398174
Author: Yunus A. Cengel Dr., Michael A. Boles
Publisher: McGraw-Hill Education
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Chapter 7.13, Problem 176RP
To determine

The work and heat transfer for each process.

Expert Solution & Answer
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Answer to Problem 176RP

The heat transfer for the isothermal process 1–2 is 117.7kJ.

The work done during the process 1-2 is 117.7kJ.

The work done during the isentropic compression process 2-3 is 97.1kJ.

The heat transfer for the isentropic process 2–3 is 0kJ.

The work done during constant pressure compression process 3-1 is 37kJ.

The heat transfer during constant pressure compression process 3-1 is 135.8kJ.

Explanation of Solution

Write the expression to calculate the enthalpy change in process 1-2.

ΔS12=mRlnP2P1        (I)

Here, pressure at process 1 is P1 , mass of the air is m, gas constant of air is R and pressure at process 2 is P2.

Write the expression to calculate the ideal gas equation, to find mass of the air.

m=P1ν1RT1        (II)

Here, mass of the air is m , volume at process 1 is ν1 and temperature at process1 is T1

Write the expression to calculate the heat transfer for the isothermal process 1–2.

Qin,12=T1ΔS12             (III)

Here, enthalpy change in process 1-2 is ΔS12, heat transfer for the isothermal process 1–2 is Qin,12 and temperature at 1 stage is T1,

Write the expression to calculate the work done during the process 1-2 (Wout,12).

Wout,12=Qin,12      (IV)

Write the expression to calculate the work done during the isentropic compression process 2-3 (Win,23).

Win,23=m(u3u2)        (V)

Here, mass of the air is m, internal energy at process 3 is u3 and internal energy at process 2 is u2.

Write the expression to calculate the relative pressure at process 3 (Pr3).

Pr3=P3P2Pr2        (VI)

Here, relative pressure at process 2 is Pr2.

Write the expression to calculate the volume at process 3(ν3)

ν3=mRT3P3      (VII)

Write the expression to calculate the work done during constant pressure compression process 3-1 (Win,31).

Win,31=P3(ν3ν1)      (VIII)

Here, volume at process 3 is ν3 and volume at process 1 is ν1.

Write the expression to calculate the heat transfer during constant pressure compression process 3-1(Qout,31).

Qout,31=Win,31m(u1u3)    (IX)

Here, heat transfer during constant pressure compression process 3-1 is Qout,31 and internal energy at process 1 is u1

Conclusion:

From Table A-1 “the molar mass, gas constant and critical point properties table”, obtain the gas constant (R) of air as 0.287kJ/kgK.

Substitute 400kPa for P1, 0.3m3 for ν1, 0.287kJ/kgK for R and 300 K for T1 in Equation (II).

m=(400 kPa)(0.3m3)(0.287kJ/kgK)(300K)=1.394kg

Substitute 1.394 kg for m, 0.287kJ/kgK for R, 400kPa for P1 and 150kPa for P2 in Equation (I).

ΔS12=(1.394kg)(0.287kJ/kgK)ln150kPa400kPa=0.3924kJ/K

Substitute 27°C for T1 and 0.3924kJ/K for ΔS12 in Equation (III).

Qin,12=27°C(0.3924kJ/K)=(27+273)K(0.3924kJ/K)=117.7kJ

Thus, the heat transfer for the isothermal process 1–2 is 117.7kJ.

Substitute 117.7kJ for Qin,12 in Equation (IV).

Wout,12=117.7kJ

Thus, the work done during the process 1-2 is 117.7kJ.

From Table A-17, “Ideal-gas properties of air”, obtain the internal energy (u1) or (u2), entropy (s1) or (s2), and relative pressure (Pr1) or (Pr2), at temperature of 300K as

214.07kJ/kg, 1.70203kJ/kgK and 1.3860 respectively.

Substitute 400kPa for P3, 150kPa for P2 and 1.3860 for Pr2 in Equation (VI).

Pr3=400kPa150kPa(1.3860)=3.696

Refer to Table A-17, “Ideal-gas properties of air”.

Obtain the select the internal energy (u3) and temperature (T3) at the relative pressure of 3.696 by using interpolation method.

Write the formula of interpolation method of two variables.

y2=(x2x1)(y3y1)(x3x1)+y1 (X)

Here, the variables denoted by x and y are relative pressure and internal energy.

Show relative pressure and internal energy values from the Table A-17.

Relative pressure (Pr3)Internal energy (u3), in kJ/kg
3.481278.93
3.696?
3.806286.16

Substitute 3.481 for x1, 3.696 for x2, 3.806 for x3, 278.93 for y1, and 286.16 for y3 in Equation (X).

y2=(3.6963.481)(286.16278.931)(3.8063.481)+278.93=283.71

The value of internal energy process 1 (u3) at the relative pressure of 3.696 is 283.71kJ/kg.

Show temperature and initial internal energy values from the Table A-17.

Temperature (T3)Internal energy (u3), in kJ/kg
3.481390
3.696?
3.806400

Substitute 3.481 for x1, 3.696 for x2, 3.806 for x3, 390 for y1, and 400 for y3 in Equation (X).

y2=(3.6963.481)(400390)(3.8063.481)+390=396.615

The value of Temperature (T3) at the relative pressure of 3.696 is 396.615K.

Substitute 1.394 kg for m, 283.71kJ/kg for u3 and 214.07kJ/kg for u2 in Equation (V).

Win,23=(1.394kg)(283.71kJ/kg214.07kJ/kg)=97.1kJ

Thus, the work done during the isentropic compression process 2-3 is 97.1kJ.

The heat transfer for the isentropic process 2–3 is zero when entropy change remains unchanged for the isentropic compression process.

Thus, the heat transfer for the isentropic process 2–3 is 0kJ.

Substitute 1.394 kg for m, 0.287kJ/kgK for R, 396.6 K for T3 and 400 kPa for P3 in Equation (VII).

ν3=(1.394 kg)(0.287kJ/kgK)(396.6K)(400kPa)=0.3967m3

Substitute 400kPa for P3, 0.3967m3 for v3 and 0.3m3 for v1 in Equation (VIII).

Win,31=(400 kPa)(0.3967m30.3m3)=37kJ

Thus, the work done during constant pressure compression process 3-1 is 37kJ.

Substitute 37 kJ for Win,31, 1.394 kg for m, 214.07kJ/kg for u1 and 283.71kJ/kg for u3 in Equation (IX).

Qout,31=37kJ(1.394 kg)(214.07kJ/kg283.71kJ/kg)=135.8kJ

Thus, the heat transfer during constant pressure compression process 3-1 is 135.8kJ.

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