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Alex17521 [72]
3 years ago
9

Our aim is to calculate the efficiency of a gas turbine by assuming it operation can be modeled as a Carnot cycle. The kerosene

(jet fuel) combustion is modeled as a hot reservoir at 2000K. The atmosphere is the cold reservoir. Calculate the efficiency of this ideal and reversible engine. Can any real engine operating between the two reservoirs be more efficient than this engine?
Engineering
1 answer:
soldier1979 [14.2K]3 years ago
8 0

Answer:

The efficiency of this ideal and reversible engine is 85 percent.

The efficiency of the Carnot cycle represents the efficiency of a thermal machine with no irreversibilities, hence, it is impossible for any real engine operating between the two reservoirs cannot be more efficient than this engine.

Explanation:

Let assume that the temperature of the atmosphere is 300 K. From Thermodynamics we know that the efficiency of the Carnot's cycle (\eta_{th}), dimensionless, is:

\eta_{th} = 1-\frac{T_{L}}{T_{H}} (1)

Where:

T_{H} - Temperature of the kerosene combustor (hot reservoir), measured in kelvins.

T_{L} - Temperature of the atmosphere (cold reservoir), measured in kelvins.

If we know that T_{L} = 300\,K and T_{H} = 2000\,K, then the efficieny of this ideal and reversible engine is:

\eta_{th} = 1-\frac{300\,K}{2000\,K}

\eta_{th} = 0.85

The efficiency of this ideal and reversible engine is 85 percent.

The efficiency of the Carnot cycle represents the efficiency of a thermal machine with no irreversibilities, hence, it is impossible for any real engine operating between the two reservoirs cannot be more efficient than this engine.

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