SPSolution27-29

SPSolution27-29 - Naik, Sameer V. Solution for Special...

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Naik, Sameer V. Fall 2009 1000 Solution for Special Problem 27 Given: - Ice at 0 ° C mixed with liquid water at 30 ° C Find: - Final equilibrium temperature ( ° C) - Entropy production (kJ/K) during the process System: Assumptions: - Water is incompressible liquid - Negligible changes in KE and PE Basic Equation(s): ln final water initial T Sm C T Δ= ice water ice Q dh vdP ds S TT T =− Solution: (a) Considering the energy balance, we can write: 03 0 final final ice water water C waterT water C waterT QQ Q →→ += DD () ( ) ,, ice water ice water final initial ice water water initial water final Q m CT T m T +− = kJ kJ kJ 1 kg 335 1 kg 4.18 273 K 7 kg 4.18 303 K kg kg-K kg-K final final × + ×× = 289.2 K final T = i.e. 16.2 C final T = D (b) Entropy production during the process will be: 0 final final ice water water C waterT water C waterT SS S σ kJ 1 kg 335 kJ 289.2 K kJ 289.2 K kg 1 kg 4.18 ln 7 kg 4.18 ln 273 K kg-K 273 K kg-K 303 K × ⎛⎞ =+ × + × ⎜⎟ ⎝⎠ kJ kJ kJ 1.2272 0.2410 1.3640 KKK =+− kJ 0.1042 K irreversible process
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Naik, Sameer V. Fall 2009 1000 Solution for Special Problem 28 Given: - Insulated rigid tank containing CO 2 Find: Assuming constant specific heats - Final temperature ( ° C) for CO 2 - Entropy change (kJ/K) for CO 2 Considering variable specific heats - Final temperature ( ° C) for CO 2 - Entropy change (kJ/K) for CO 2 System: W in = 200 kJ CO 2 m = 2.7 kg P 1
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This note was uploaded on 05/15/2010 for the course ME 200 taught by Professor Gal during the Spring '08 term at Purdue.

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SPSolution27-29 - Naik, Sameer V. Solution for Special...

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