THERMODYNAMIC QUESTION

Question

A heat engine operates between two reservoirs, one at 400^oC and the other at 50^oC. The engine produces 7.5 kW of power and rejects heat into the cold reservoir at the rate of 9.0 kW.

a.        Find the efficiency of the heavy engine.

b.        Is the engine reversible? (Ans: 16.5 kW, no)

Answer 1

a. It is given that the work done by the engine is 7.5 kw per unit time. And the output heat is 9.0 kw per unit time. So from the first law of Thermodynamics, we can say that Heat absorbed = Work done + Heat emitted i.e. Heat absorbed = 7.5 + 9 = 16.5 kw per unit time. Now efficiency of the engine is - η = Work done / Heat absorbed. Thus, η = 7.5/16.5 = 0.45.

b. Reversible process means that if we apply 9 kw heat to the engine and work on the engine by an amount of 7.5 kw, then it should emit a heat of 16.5 kw per unit time. But this is not possible in this case, because it is not a Carnot's engine.