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 5, Problem 53P

(a)

To determine

The empirical specific heat equation as a function of temperature.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given:

The initial temperature (T1) is 200 K.

The final temperature (T2) is 200 K.

Calculation:

From Appendix Table A-2 (c) “Ideal-gas specific heats of various common gases”.

  a=29.11b=0.1916×102c=0.4003×105d=0.8704×109

Write the expression for the empirical relation between c¯p(T) and c¯V(T).

  c¯V(T)=c¯PRu=(aRu)+bT+cT2+dT3        (I)

Here, the universal gas constant is Ru and the temperature is T.

Write the expression for the change in internal energy.

  Δu¯=12c¯V(T)dT        (II)

Substitute (aRu)+bT+cT2+dT3 for c¯V(T) in Equation (II).

  Δu¯=12((aRu)+bT+cT2+dT3)dT=(aRu)(T)T1T2+b(T22)T1T2+c(T33)T1T3+d(T44)T1T2=(aRu)(T2T1)+1/2b(T22T12)+1/3c(T23T13)+1/4d(T24T14)        (III)

Substitute the table constants (a, b, c, and d), T2=800K, and T1=200K in Equation (III)

  Δu¯=[(29.118.314J/molK)(800K200k)12(0.1961×102)(8002K2002K)+13(0.4003×105)(8003K2003K)14(0.8704×109)(8004K2004K)]=12487kJkmol

Refer table A-1, “Molar mass, gas constant, and critical-point properties”.

The molar mass of the hydrogen (M) is 2.016kg/kmol.

Write the expression for internal energy of empirical specific heat equation.

  Δu=Δu¯M

  Δu=(12487kJ/kmol)(2.016kg/kmol)=6193.948kJ/kg6194kJ/kg

Thus, the empirical specific heat equation as a function of temperature is 6194kJ/kg_.

(b)

To determine

The cV value at the average temperature.

(b)

Expert Solution
Check Mark

Explanation of Solution

Determine the average temperature for the cV value.

  Tavg=200+8002=500K

From Table A-2b, write the value of ideal gas specific heat of various gases at various temperatures at 500 K average temperature.

  cV,avg=cV@500K=10.389kJkgK

Write the expression for internal energy of cV value at the average temperature.

  Δu=cV,avg(T2T1)

  Δu=(10.389kJkgK)(800K200K)=6233kJ/kg

Thus, the cV value at the average temperature is 6233kJ/kg_.

(c)

To determine

The cP value at the room temperature.

(c)

Expert Solution
Check Mark

Explanation of Solution

Determine the room temperature for the cP value.

  Troom=27°C=27+273K=300K

From Table A-2b, write the value of ideal gas specific heat of various gases at various temperatures at 300 K room temperature.

  cV,avg=cV@300K=10.183kJkgK

Write the expression for internal energy of cP value at the room temperature.

  Δu=cV,avg(T2T1)

  Δu=(10.183kJkgK)(800K200K)=6110kJ/kg

Thus, the cP value at the room temperature is 6110kJ/kg_.

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Chapter 5 Solutions

Fundamentals of Thermal-Fluid Sciences

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