The temperature of the heat reservoirs for a Carnot cycle (reversible) engine ar

Question

The temperature of the heat reservoirs for a Carnot cycle (reversible) engine are T h = 1200 K and T c = 300 K. (a) Calcula te the efficiency of this engine to two significan figures. (b) If w = - 100 kJ (note that the sign means work is done by the system) calculate q 1 and q 2 in kJ. Enter them in that order to two significant figures. (c) The same engine can be operated in reverse, i.e. as a heat pump which is an engine that turns work into heat. If w = +100 kJ, calculate q 1 and q 2 in kJ (note that the signs change). (d) Suppose it were possible to have an engine with a higher efficiency of 0.80. If for this engine w = - 100 kJ, calculate q 1 and q 2 in kJ. (e) Now let’s use the engine of part (d) to drive the heat pump in part (c). The total heat transferred is given by Q 1 = sum of q 1 for parts (c) and (d) and Q 2 = sum of q 2 . Enter Q 1 and Q 2 in kJ and in that order. Note that the signs indicate the net effect of combining the engines is a “spontaneous” transfer of heat from a colder reservoir to a hotter on e. F rom experience this should not happen. This type of argument is used to show that all reversible engines operating between the same two reservoirs have the same efficien cy

Explanation / Answer

Efficiency of Carnot Enginer= 1-TC/TH= 1-300/1200=1-0.25=0.75

but n= W/QH

W= 100Kj   and QH= W/0.75= 100/0.75=133.3 KJ( QH= heat added in the boiler)

QH= W+QC ( Qc= heat removed in the condenser)

QC= 133.3-100 =33.3 KJ

b) If its work as a heat pump

W= 100Kj

W= QC*(TH-TC)/TC

100= QC* (1200-300)/300= 3QC

QC= 100/3= 33.3

QC= 33.3+33.3 =66.7 Kj

QH= 133.3+133.3=266.6 KJ

100= QH-33.3

QH =133.3 Kj

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