301 Test_3_Solution

301 Test_3_Solution - SOLUTIONS 1 MAE 301: Engineering...

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Unformatted text preview: SOLUTIONS 1 MAE 301: Engineering Thermodynamics I Spring 2010 Test 3 (15% of final grade) April 22, 2010 Closed books & notes Problem 1. (25 points) A refrigeration cycle operating between two reservoirs receives energy Q L from a cold reservoir at T L = 7 ˚C and rejects energy Q H to a hot reservoirs at T H = 320 ˚C. For each of the following cases determine whether the cycle operates reversibly, irreversible, or is impossible: (a) Q L = 1500 kJ, W cycle = 150 kJ. (b) Q L = 1400 kJ, Q H = 1600 kJ. (c) Q H = 1600 kJ, W cycle = 400 kJ. (d) COP R = 5. Solution: 7 280 320 280 , = − = − = L H L rev R T T T COP , based on this: (a) 10 150 1500 = = = cycle L R W Q COP- impossible (b) 7 1400 1600 1400 = − = − = L H L R Q Q Q COP- reversible (c) 3 400 1200 = = − = = cycle cycle H cycle L R W W Q W Q COP- irreversible (d) 5 = R COP- irreversible Problem 2. (25 points) At steady state, a refrigeration cycle operating between hot and cold reservoirs at 300 K and 270 K, respectively, removes energy by heat transfer from the cold reservoir...
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This note was uploaded on 05/19/2010 for the course MAE 301 taught by Professor Hassan during the Spring '08 term at N.C. State.

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301 Test_3_Solution - SOLUTIONS 1 MAE 301: Engineering...

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