Explanation Write the expression for entropy balance in a closed system. m ( s 2 − s 1 ) = ∫ 1 2 ( δ Q T ) + σ m ( s 2 − s 1 ) = Q T b + σ (I) Here, the mass of system is m, the specific entropy at state 2 is s 2 , the specific entropy at state 1 is s 1 , the heat transfer through the boundary of system is δ Q , the absolute temperature of system is T , the entropy generation within the system is σ , net heat interaction with surrounding is Q and the surrounding temperature is T b . Write the expression for entropy change between two states for an ideal gas. ( s 2 − s 1 ) = [ s ° ( T 2 ) − s ° ( T 1 ) − R ln ( T 2 T 1 ) ] (II) Substitute [ s ° ( T 2 ) − s ° ( T 1 ) − R ln ( T 2 T 1 ) ] for ( s 2 − s 1 ) in Equation (I). m [ s ° ( T 2 ) − s ° ( T 1 ) − R ln ( T 2 T 1 ) ] = Q T b + σ σ = m [ s ° ( T 2 ) − s ° ( T 1 ) − R ln ( T 2 T 1 ) ] − Q T b (III) Here, the entropy variable depended on temperature T 1 is s ° ( T 1 ) , the entropy variable depended on temperature T 2 is s ° ( T 2 ) , the gas constant is R , the temperature at state 1 is T 1 and the temperature at state 2 is T 2 . Write the expression for energy balance in a system. Δ U + Δ KE + Δ PE = Q − W (IV) Here, the change in internal energy is Δ U , the change in kinetic energy is Δ KE , the change in potential energy is Δ PE , heat supplied to the system is Q and the work done by system is W . Assume that the change in kinetic and potential energy of the system is negligible. Also assume that there is no work done by the rigid tank. Write the expression for change in internal energy. Δ U = − m ( u 2 − u 1 ) Here, mass is m , specific internal energy of state 1 is u 1 and the specific internal energy of state 2 is u 2 . Substitute 0 for ΔKE , 0 for ΔPE , 0 for W and − m ( u 2 − u 1 ) for Δ U in Equation (IV). m ( u 2 − u 1 ) + 0 + 0 = Q − 0 m ( u 2 − u 1 ) = Q (V) Conclusion: Refer to Table A-22E “Ideal Gas Properties of Air” to obtain values of specific internal energy u 1 as 102.34 Btu / lb at 600 ° R and specific internal energy u 2 as 136

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ISBN: 2818440116926

Author: MORAN

Publisher: WILEY CONS

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Chapter 6.13, Problem 53P

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The amount of energy transfer by heat.

The amount of entropy generated.

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Chapter 6.13, Problem 52P

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The amount of energy transfer by heat. The amount of entropy generated.