Question

A system that operates in a reversible cycle receives energy by heat transfer from a 15.7oC reservoir at a rate of 26.9 kW, and delivers energy by heat transfer to a 49.8oC reservoir. Determine the rate of heat transfer (kW) to the high temperature reservoir.

Answer 1

figure shows a reversible cycle system which receives energy from lower temperature reservoir and delivers to higher temperature reservoir hence it is either refrigerator or heat pump.

to calulate the rate of heat transfer Q₁ (kW) to the high temperature reservoir. we have given values

Q₂ = 26.9 kW

T₁ =49.8⁰ C = 49.8 +273 = 322.8 K

T₂ =15.7⁰ C = 15.7 +273 = 288.7 K

now for a reversible cycle,

we know that,

COP = T₂ / (T₁ - T₂ ) = Q₂ / (Q₁ - Q₂)

so, taking

T₂ / (T₁ - T₂ ) = Q₂ / (Q₁ - Q₂)

put all given values in above equation

288.7 / (322.8 - 288.7) = 26.9 / ( Q₁ - 26.9)

by calculating this equation we have,

Q₁ = 30.07 kW