Chemical and Engineering Thermodynamics Solutions Manual by Y.V.C Rao

By Y.V.C Rao

This publication is a really helpful reference that includes worked-out recommendations for all of the workout difficulties within the ebook Chemical Engineering Thermodynamics by means of a similar writer. step by step strategies to all workout difficulties are supplied and ideas are defined with distinct and vast illustrations. it is going to come in useful for all academics and clients of Chemical Engineering Thermodynamics.

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18 J mol Stage 1 is as in the previous problem. Stage 2 Following the same calculation as above. 957 × 10-9 Pa = Stage 3 allowed pressure. J = Stage 3 work mol Question for the student: Why is the calculated work the same for each stage? 19 The mass, energy and entropy balances are dM & +M & = 0, M& = − M& =M 1 2 2 1 dt dU & H$ + M & H$ + Q& + W& ; M & H$ − H$ + W& = 0; =0= M 1 1 2 2 s 1 1 2 s dt & H$ − H$ W&s = + M 1 2 1 c c h h dS Q& = 0 = M& 1S$1 + M& 2 S$2 + + S&gen = M& 1 S$1 − S$2 + S&gen = 0 dt T $ $ & & Sgen = M1 S2 − S1 c c h h 300° C, 5 bar = 05 .

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WSrev = 9334 . 14K. 464 ⋅ 10−9 T 3 . 44K. 92K. So there is a significant difference between the results for the constant heat capacity and variable heat capacity cases. 092 × 106 Pa. T f = 800K Wrev = z × 104 . 18 J mol Stage 1 is as in the previous problem. Stage 2 Following the same calculation as above. 957 × 10-9 Pa = Stage 3 allowed pressure. J = Stage 3 work mol Question for the student: Why is the calculated work the same for each stage? 19 The mass, energy and entropy balances are dM & +M & = 0, M& = − M& =M 1 2 2 1 dt dU & H$ + M & H$ + Q& + W& ; M & H$ − H$ + W& = 0; =0= M 1 1 2 2 s 1 1 2 s dt & H$ − H$ W&s = + M 1 2 1 c c h h dS Q& = 0 = M& 1S$1 + M& 2 S$2 + + S&gen = M& 1 S$1 − S$2 + S&gen = 0 dt T $ $ & & Sgen = M1 S2 − S1 c c h h 300° C, 5 bar = 05 .

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