Explanation Write the expression for boundary work output and internal energy for adiabatic closed system. W b,out = Δ U = c v ( T 1 − T 2 ) (I) Here, boundary work output is W b,out , change in internal energy is Δ U , specific heat at constant volume is c v , initial temperature is T 1 and final temperature is T 2 . Write the expression for final temperature for the isentropic process. T 2 s = T 1 ( P 2 P 1 ) γ − 1 γ (II) Here, final isentropic temperature is T 2 s , initial temperature is T 1 , initial pressure is P 1 , final pressure is P 2 and specific heat ratio is γ . Write the expression for actual exit temperature. η = T 1 − T 2 T 1 − T 2 s T 2 = T 1 − η ( T 1 − T 2 s ) (III) Here, actual final temperature is T 2 and efficiency of process is η . Write the expression for entropy change of air. Δ s a i r = c p ln ( T 2 T 1 ) − R ln ( P 2 P 1 ) (IV) Here, entropy change of air is Δ s a i r , specific heat at constant pressure is c p and gas constant for air is R . Write the expression for exergy difference between initial and final state. Δ ϕ = c v ( T 1 − T 2 ) + P 0 ( v 1 − v 2 ) − T 0 ( s 1 − s 2 ) v 2 = c v ( T 1 − T 2 ) + P 0 R ( T 1 P 1 − T 2 P 2 ) − T 0 ( s 1 − s 2 ) (V) Here, exergy difference is Δ ϕ , specific heat at constant volume is c v , surrounding temperature is T 0 , surrounding pressure is P 0 , initial state entropy is s 1 and final state entropy is s 2 . Write the expression for useful work. W useful = W b,out − W surr = W b,out − P 0 R ( T 2 P 2 − T 1 P 1 ) (VI) Here, useful work is W useful . Calculate the second-law efficiency. η II = W useful Δ ϕ (VII) Here, second law efficiency is η II . Conclusion: Refer Table A-2, and obtain the specific heat at constant pressure for air as 0.240 Btu/lbm ⋅ R , the specific heat at constant volume for air as 0.171 Btu/lbm ⋅ R , the specific heat ratio as 1.4 and the gas constant for air is 0.06855 Btu/lbm ⋅ R . Substitute 140 ° F for T 1 , 20 psia for P 2 , 180 psia for P 1 and 1.4 for γ in Equation (II). T 2 s = ( 140 ° F ) ( 20 psia 180 psia ) 1.4 − 1 1.4 = ( 140 + 460 ) ° R ( 0.5337 ) = 320.22 ° R Substitute 140 ° F for T 1 , 0.95 for η and 320.22 ° R for T 2 s in Equation (III). T 2 = ( 140 ° F ) − ( 0.95 ) [ ( 140 ° F ) − ( 320.22 ° R ) ] = ( 140 ° F + 460 ) ° R − ( 0.95 ) [ ( 140 ° F + 460 ) ° R − ( 320.22 ° R ) ] = 600 ° R − [ 0.95 × ( 279.78 ) ° R ] = 334.209 ° R Substitute 0.171 Btu/lbm ⋅ R for c v , 140 ° F for T 1 and 334.209 ° R for T 2 in Equation (I). W b,out = ( 0.171 Btu/lbm ⋅ R ) [ ( 140 ° F ) − 334.209 ° R ] = ( 0.171 Btu/lbm ⋅ R ) [ ( 140 ° F + 460 ) ° R − 334.209 ° R ] = ( 0.171 Btu/lbm ⋅ R ) ( 265.791 ° R ) = 45

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Author: CENGEL

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Chapter 8.8, Problem 26P

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The second-law efficiency for air expansion in an adiabatic closed system.

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