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pav-90 [236]
3 years ago
8

A power cycle operates between hot and cold reservoirs at 1200 K and 300 K, respectively. At steady state the cycle develops a p

ower output of 0.38 MW while receiving energy by heat transfer from the hot reservoir at the rate of 1 MW. Determine the thermal efficiency and the rate at which energy is rejected by heat transfer to the cold reservoir, in MW for the following. (a) The power cycle and conditions above. (b) A reversible power cycle operating between the reservoirs and receiving the same rate of heat transfer from the hot reservoir as in part (a).
Physics
1 answer:
GenaCL600 [577]3 years ago
7 0

Answer:

Explanation:

a ) Thermal efficiency = work output / heat input

= .38 MW / 1 MW = .38

OR 38%

Heat rejected at cold reservoir = heat input - work output

1 MW - .38 MW

= 0.62 MW.

b ) For reversible power output

efficiency = T₂ - T₁ / T₂   ; T₂ is temperature of hot reservoir and T₁ is temperature of cold reservoir.

= 1200 - 300 / 1200 = 900 / 1200

= .75

or 75%

rate at which heat is rejected

= 1 - .75 x 1

= .25 MW .

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<h3>Definition of current</h3>

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