Question

A Carnot engine operates between 1350 °F and 125 °F. If it rejects 55 Btu as heat, determine the work output.

Answer 1

Answer:

C. $W = 115.12\,Btu$

Explanation:

Thermodynamically speaking, a Carnot engine represents an entirely reversible thermal process and its energy efficiency represents the maximum theoretical efficiency that thermal machines can reach. The efficiency of the ideal thermal process ($\eta$), no unit, is:

$\eta = \left(1-\frac{T_{L}}{T_{H}} \right)$ (1)

Where:

$T_{L}$ - Temperature of the cold reservoir, measured in Rankine.

$T_{H}$ - Temperature of the hot reservoir, measured in Rankine.

If we know that $T_{H} = 1809.67\,R$ and $T_{L} = 584.67\,R$, then the energy efficiency of the ideal thermal process is:

$\eta = 0.678$

By First Law of Thermodynamics, we calculate the work output:

$W = Q_{H}-Q_{L}$

$W = \frac{W}{\eta} -Q_{L}$ (By definition of efficiency)

$Q_{L} = \frac{W}{\eta}-W$

$Q_{L} = \left(\frac{1}{\eta}-1 \right)\cdot W$(2)

Where:

$Q_{H}$ - Heat received by the engine, measured in Btu.

$Q_{L}$ - Heat rejected by the engine, measured in Btu.

$W$ - Work output, measured in Btu.

If we know that $\eta = 0.678$ and $Q_{L} = 55\,Btu$, then the work output of the Carnot engine is:

$W = \frac{Q_{L}}{\frac{1}{\eta}-1 }$

$W = 115.807\,Btu$

The work output of the Carnot engine is 115.807 Btu. (Answer: C)