CONNECT FOR THERMODYNAMICS: AN ENGINEERI
CONNECT FOR THERMODYNAMICS: AN ENGINEERI
9th Edition
ISBN: 9781260048636
Author: CENGEL
Publisher: MCG
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Chapter 4.5, Problem 55P

Determine the internal energy change Δu of hydrogen, in kJ/kg, as it is heated from 200 to 800 K, using (a) the empirical specific heat equation as a function of temperature (Table A–2c), (b) the cv value at the average temperature (Table A–2b), and (c) the cv value at room temperature (Table A–2a)

(a)

Expert Solution
Check Mark
To determine

The empirical specific heat equation as a function of temperature.

Answer to Problem 55P

The empirical specific heat equation as a function of temperature is 6194kJ/kg_.

Explanation of Solution

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

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)

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

Δu=Δu¯M (IV)

Here, the molar mass is M.

Conclusion:

Substitute 29.11 for a, 8.314 for Ru, 800K for T2, 200K for T1, 0.1961×102 for b, 0.4003×105 for c, and 0.8704×109 for d 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

Substitute 2.016kg/kmol for M and 12487kJ/kmol for Δu¯ in Equation (IV).

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)

Expert Solution
Check Mark
To determine

The cV value at the average temperature.

Answer to Problem 55P

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

Explanation of Solution

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

Δu=cV,avg(T2T1) (V)

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

Conclusion:

Substitute 10.389kJkgK for cV,avg, 800K for T2, and 200K for T1 in Equation (V).

Δu=(10.389kJkgK)(800K200K)=6233kJ/kg

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

(c)

Expert Solution
Check Mark
To determine

The cP value at the room temperature.

Answer to Problem 55P

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

Explanation of Solution

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

Δu=cV,avg(T2T1) (VI)

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

Conclusion:

Substitute 10.183kJkgK for cV,avg, 800K for T2, and 200K for T1 in Equation (VI).

Δu=(10.183kJkgK)(800K200K)=6110kJ/kg

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

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Chapter 4 Solutions

CONNECT FOR THERMODYNAMICS: AN ENGINEERI

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