During the operation of the heat engine for some time, the working fluid received from the heater

| April 26, 2021 | education

During the operation of the heat engine for some time, the working fluid received from the heater an amount of heat Q = 1.5 * 10 ^ 6 J, transferred to the refrigerator Q = -1, 2 * 10 ^ 6 J. Calculate the efficiency of the machine and compare it with the maximum possible Efficiency if the temperatures of the heater and refrigerator are 250 ° C and 30 ° C, respectively.

Task data: Qн (heat received) = 1.5 * 10 ^ 6 J; Qx (given off heat) = 1.2 * 10 ^ 6 J; Tн (heater temperature) = 523 K (250 ºС); Тх (refrigerator temperature) = 303 К (30 ºС).

1) Actual efficiency: ηд = 1 – Qх / Qн = 1 – 1.2 * 10 ^ 6 / (1.5 * 10 ^ 6) = 0.2 = 20%.

2) The maximum possible efficiency: ηmax = 1 – Тх / Тн = 1 – 303/523 = 0.42065 ≈ 42%.

3) Comparison of efficiency: k = ηmax / ηd = 42/20 = 2.1 p.

Answer: The efficiency of the machine is 2.1 times less than the maximum possible and is 20%.



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