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Unformatted text preview: Department of Mechanical Science and Engineering ME 300 Thermodynamics Section D Fall 2010 Homework #3 - Solution 1 Problem 3.1 Known: The specifications of a heat pump. Find: The energy the heat pump provides to the apartment per hour. The coefficient of performance. The required amount of energy transfer from the surroundings to the heat pump. Annual cost of running the heat pump. Comparison of the annual cost between heat pump and electrical heater. Engr. Model/Assumptions: The heat transfer rates and power of the heat pump are constant. The work input into the electric heater is equal to the energy released into the room. Analysis: (a) To determine the energy the heat pump provides to the room we need to integrate the heat transfer rate over time: Q out = Z Q out dt = 4 kJ s 3600 s = 14400 kJ (b) Intergrating the power over time gives as the energy per hour the heat pump consumes: W cycle = Z W cycle dt = 1 . 6 kJ s 3600 s = 5760 kJ The coefficient of performance is then: = Q out W cycle = 14400 kJ 5760 kJ = 2 . 5 (c) To determine the necessary heat transfer from the surroundings to the heat pump when Q out = 14400 kJ we rearrange the coefficient of performance equation to solve for Q in : = Q out Q out- Q in Q in = Q out ( - 1) = 14400 kJ (2 . 5- 1) 2 . 5 = 8640 kJ (d) The cost for running the heat pump 1500 hours a year is: cost = 1 . 6 kW 1500 h year $0 . 06 kW h = $144 . 00 year (e) The cost for running the electric heater 1500 hours a year is: cost = 4 kW 1500 h year $0 . 06 kW h = $360 . 00 year The annual savings by using the heat pump instead of an electrical heater is $216.00 perThe annual savings by using the heat pump instead of an electrical heater is $216....
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This note was uploaded on 05/29/2011 for the course ME 300 taught by Professor Staff during the Spring '08 term at University of Illinois, Urbana Champaign.
- Spring '08