301 HW_2_Solution

301 HW_2_Solution - Answer: (d) 3600 kJ Problem 3. Problem...

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1 MAE 301: Engineering Thermodynamics Spring 2010 HOMEWORK #2 ( Solutions ) Problem 1. Problem 2.119 from the book. Solution: The unit cost of each Btu of useful energy supplied to the house by each system can be determined from Unit cost of useful energy Unit cost of energy supplied Conversion efficiency = Substituting, Natural gas heater : kJ / 10 5 . 13 $ kJ 105,500 therm 1 0.87 m $1.24/ther energy useful of cost Unit 6 × = = Heating oil heater : kJ / 10 4 . 10 $ kJ 138,500 gal 1 0.87 $1.25/gal energy useful of cost Unit 6 × = = Electric heater : kJ / 10 0 . 25 $ kJ 3600 kWh 1 1.0 $0.09/kWh) energy useful of cost Unit 6 × = = Therefore, the system with the lowest energy cost for heating the house is the heating oil heater . Problem 2. Problem 2.132 from the book.
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Unformatted text preview: Answer: (d) 3600 kJ Problem 3. Problem 2.135 from the book. Answer: (e) 37 kW Problem 4. Problem 3.1C from the book. Answer: Yes, since the chemical composition throughout the tank remain the same Problem 5. Problem 3.8C from the book. Answer: Yes. Problem 6. Problem 3.10C from the book. 2 Answer: At supercritical pressures, there is no distinct phase change process. The liquid uniformly and gradually expands into a vapor. At subcritical pressures, there is always a distinct surface between the phases. Problem 7. Problem 3.19C from the book. Answer: Yes. It decreases with increasing pressure and becomes zero at the critical pressure....
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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 HW_2_Solution - Answer: (d) 3600 kJ Problem 3. Problem...

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