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eat at a rate of 300 I hich is at 7°C, serves vo reservoirs and det

eat at a rate of 300 I hich is at 7°C, serves vo reservoirs and det

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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I just need clarification what are TL and TH? Where do we get the values from?
A completely reversible heat pump produces heat at a rate of 300 kW to warm a house maintained at 24°C. The exterior air, which is at 7°C, serves as the source. Calculate the rate of entropy change of the two reservoirs and determine if this heat pump satisfies the second law according to the increase of entropy principle.
Transcribed Image Text:A completely reversible heat pump produces heat at a rate of 300 kW to warm a house maintained at 24°C. The exterior air, which is at 7°C, serves as the source. Calculate the rate of entropy change of the two reservoirs and determine if this heat pump satisfies the second law according to the increase of entropy principle.
COPHP,rev 1 1 = 17.47 net 1-T, /T, 1-(280 K)/(297 K) The power required to drive this heat pump, according to the coefficient of performance, is then 7°C 300 kW %3D net, in = 17.17 kW %3D COP HP.rev 17.47 According to the first law, the rate at which heat is removed from the low-temperature energy reservoir is 300 kW -17.17 kW = 282.8 kW %3D net, in The rate at which the entropy of the high temperature reservoir changes, according to the definition of the entropy, is 300 kW 1.01kW/K TH 297 K and that of the low-temperature reservoir is - 282.8 kW =-1.01kW/K AS, T, 280 K The net rate of entropy change of everything in this system is AS AS, + AS, = 1.01 – 1.01 = 0 kW/K total as it must be since the heat pump is completely reversible.
Transcribed Image Text:COPHP,rev 1 1 = 17.47 net 1-T, /T, 1-(280 K)/(297 K) The power required to drive this heat pump, according to the coefficient of performance, is then 7°C 300 kW %3D net, in = 17.17 kW %3D COP HP.rev 17.47 According to the first law, the rate at which heat is removed from the low-temperature energy reservoir is 300 kW -17.17 kW = 282.8 kW %3D net, in The rate at which the entropy of the high temperature reservoir changes, according to the definition of the entropy, is 300 kW 1.01kW/K TH 297 K and that of the low-temperature reservoir is - 282.8 kW =-1.01kW/K AS, T, 280 K The net rate of entropy change of everything in this system is AS AS, + AS, = 1.01 – 1.01 = 0 kW/K total as it must be since the heat pump is completely reversible.
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