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 161RQ

a)

To determine

The final temperature of the helium

a)

Expert Solution
Check Mark

Explanation of Solution

Given:

The volume of the nitrogen is 0.2 m3.

The mass of the helium (m) is 0.1 kg.

The initial temperature of the gases (T1) is 20°C.

The initial pressure of the gases (P1) is 95 kPa.

The final temperature of the nitrogen is 500°C.

The final pressure of the helium (P2) is 120 kPa.

Calculation:

Refer the Table A-2, “Ideal-gas specific heats of various common gases table”,

The specific heat ratio (k) for Helium gas is 1.667.

Write the relation between the pressure and temperature in the isentropic process.

  T2T1=(P2P1)k1kT2=T1(P2P1)k1kT1=20°(120 kPa95 kPa)1.66711.667=(20°+273) K(120 kPa95 kPa)1.66711.667=321.7 K

b)

To determine

The final volume of nitrogen

b)

Expert Solution
Check Mark

Explanation of Solution

Refer the Table A-2, “Ideal-gas specific heats of various common gases table”,

The gas constant for the helium gas is 2.0769 kPam3kgK.

Calculate the initial volume of helium (VHe,1).

  VHe,1=mRT1P1VHe,1=0.1 kg×2.0769 kPam3kgK×(20°C)95kPa=0.1 kg×2.0769 kPam3kgK×(20+273)95kPa=0.6406 m3

Calculate the final volume of helium (VHe,2).

  VHe,2=mRT2P2VHe,1=0.1 kg×2.0769 kPam3kgK×(321.7 K)120kPa=0.5568 m3

Calculate the final volume of the nitrogen (VN2)2.

  (VN2)2=(VN2)1+(VHe)1+(VHe)2(VN2)2=0.2 m3+0.6406 m3+0.5568m3=0.2838 m3

Thus, the final volume of nitrogen is 0.2838m3_.

c)

To determine

The heat transferred to the nitrogen.

c)

Expert Solution
Check Mark

Explanation of Solution

Refer the Table A-2, “Ideal-gas specific heats of various common gases table”,

The gas constant for the helium gas (R)He is 2.0769 kPam3kgK

The specific heat constant the helium gas (CV)He is 3.1156kJkgK.

The gas constant for the helium gas (R)N2 is 0.2968kPam3kgK

The specific heat constant the helium gas (CV)N2 is 0.743kJkgK.

Calculate the mass of the nitrogen (mN2).

  mN2=P1V1RT1mN2=95kPa×0.2 m32.0769 kPam3kgK×(20°C)=95kPa×0.2 m32.0769 kPam3kgK×(20+273)K=0.2815 kg

Calculate the temperature of the nitrogen TN2,2.

  TN2,2=P2V2mRTN2,2=120kPa×0.2838 m30.2185kg×0.2968 kPam3kgK=120kPa×0.2838 m30.2185kg×0.2968kPam3kgK=525.1 K

Calculate the change in internal energy of the nitrogen.

  ΔUN2=mcVN2(T2T1)ΔUN2=0.2185 kg×0.743kJkgK(521.5 K20°C)=0.2185 kg×0.743kJkg×K(521.5 K(20+273) K)=37.1 kJ

Calculate the change in internal energy of the Helium.

  ΔUHe=mcV(T2T1)ΔUHe=0.1 kg×3.1156kJkgK(321.7 K20°C)=0.1 kg×3.1156kJkg×K(321.7 K(20+273) K)=8.942 kJ

Calculate the heat transferred to the nitrogen using the energy balance equation for the system.

  Qin=ΔUN2+ΔUHeQin=37.1 kJ+8.942 kJ=46.042 kJ

Thus, the heat transferred to the nitrogen is 46.042 kJ_.

d)

To determine

The entropy generation during the process

d)

Expert Solution
Check Mark

Explanation of Solution

Write the expression for the entropy generation (Sgen) for the isentropic process:

  Sgen=ΔSN2+ΔSsurr

Sgen=mN2[cPN2ln(T2T1)Rln(P2P1)]+QinTRSgen=0.2185 kg×[1.039 kJkg×K×ln(521.5 K20°C)0.296kJ/kgK120 kPa95 kPa]+[[46.042 kJ500°C]]=0.2185 kg×[1.039 kJkg×K×ln(521.5 K(20+273 K))0.296kJ/kgK120 kPa95 kPa]+[46.042 kJ(500+273) K]=0.057 kJ/K

Thus, the entropy generation during the process is 0.057 kJ/K_.

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Chapter 8 Solutions

Fundamentals of Thermal-Fluid Sciences

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