Question

Lets say that we are testing an engine that produces 17 kW of power. As this engine is operating, it absorbs 95 kW of heat from the surrounding air at 17 C. The air is then exited from a combustor into a room at 20 C.

What is the reversible power?

Group of answer choices

-32.9 kW

32.9 kW

5.882 kW

-5.882 kW

Answer 1

We cannot solve this problem using any formulae because, the formula for the available power (reverisble power) for the heta engine will look like this:

where

• h1 - h2 = enthalpy change = 17 kW (given)
• T(o) = combustor temp = 20 C = 293 K
• S2 - S1 = entropy change of the working medium of the engine (system) =? (NOT GIVEN)
• Q = energy lost by the surroundings = -95 kW
• T(surr) = surrounding temp = 17 C = 290 K

Since S2 - S1 is not given, we cannot calculate the reversible power. But we can guess the correct answer, looking at the options given to us.

• - 32.9 kW and - 5.882 kW are discarded firstly, because these have minus signs, which means that work is being done ON the system. But it is said that we are using an "engine". And an Engine, PRODUCES work or work is done BY the system in an engine. Therefore, we can staright away strike out these two options.
• 5.882 kW can also be discarded because the reversible power is the maximum amount of power possible. If the actual power is given as 17 kW, it is obvious that the "maximum" power CANNOT be less than 17 kW. Therefore, this option is discarded too.
• We are finally left with 32.9 kW. It does not have a negative sign which means that the power is produced BY the system, (which is correct for any "engine"). Also, it is more than the actual power of 17 kW and thus satisfies the criteria for Reversible power.

Thus, the correct answer is => Option B = 32.9 kW

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