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Answer:
Explanation:
Cop of reversible refrigerator = TL / ( TH - TL)
TL = low temperature of freezer = 20 °F
TH = temperature of air around = 75 °F
Heat removal rate QL = 75 Btu/min
W actual, power input = 0.7 hp
conversion on F to kelvin = (T (°F) + 460 ) × 5 / 9
COP ( coefficient of performance) reversible = (20 + 460) × 5/9 / (5/9 ( ( 75 +460) - (20 + 460) ))
COP reversible = 480 / 55 = 8.73
irreversibility expression, I = W actual - W rev
COP r = QL / Wrev
W rev = QL / COP r where 75 Btu/min = 1.76856651 hp where W actual = 0.70 hp
a) W rev = 1.76856651 hp / 8.73 = 0.20258 hp is reversible power
I = W actual - W rev
b) I = 0.7 hp - 0.20258 hp = 0.4974 hp
c) the second-law efficiency of this freezer = W rev / W actual = 0.20258 hp / 0.7 hp = 0.2894 × 100 = 28.94 %
Answer: A) Gradually decrease
Explanation:
The convection value of heat transfer rate are gradually decreasing with the flow of the heat. Flow in a circular pipe, flow direction does not change in the velocity path. The average of the coefficient of heat transfer and the number of pipes are needed and the effects are get neglected so that is why the flow are fully developed.
