Answer
We can use the following equations to solve this problem:
(a) The rate of energy extraction from the hot reservoir is given by:
$Q_H = P_{out}/\eta$
where $P_{out}$ is the mechanical power output and $\eta$ is the efficiency of the engine.
$Q_H = 48.4 \text{ kW}/(1-41.7/48.4) = 81.7 \text{ kW}$
(b) The efficiency of the engine is given by:
$\eta = 1 - Q_C/Q_H$
where $Q_C$ is the rate at which the engine rejects heat to the ambient environment.
$\eta = 1 - 41.7 \text{ kW}/81.7 \text{ kW} = 0.489$ or 48.9%
(c) The temperature of the ambient environment is given by:
$\frac{Q_C}{Q_H} = \frac{T_C}{T_H}$
where $T_C$ and $T_H$ are the temperatures of the ambient environment and the hot reservoir, respectively.
$T_C = T_H \times Q_C/Q_H = 625 \text{ K} \times 41.7 \text{ kW}/81.7 \text{ kW} = 320 \text{ K}$
Therefore, the temperature of the ambient environment is 320 K (46.8°C).
Work Step by Step
We can use the following equations to solve this problem:
(a) The rate of energy extraction from the hot reservoir is given by:
$Q_H = P_{out}/\eta$
where $P_{out}$ is the mechanical power output and $\eta$ is the efficiency of the engine.
$Q_H = 48.4 \text{ kW}/(1-41.7/48.4) = 81.7 \text{ kW}$
(b) The efficiency of the engine is given by:
$\eta = 1 - Q_C/Q_H$
where $Q_C$ is the rate at which the engine rejects heat to the ambient environment.
$\eta = 1 - 41.7 \text{ kW}/81.7 \text{ kW} = 0.489$ or 48.9%
(c) The temperature of the ambient environment is given by:
$\frac{Q_C}{Q_H} = \frac{T_C}{T_H}$
where $T_C$ and $T_H$ are the temperatures of the ambient environment and the hot reservoir, respectively.
$T_C = T_H \times Q_C/Q_H = 625 \text{ K} \times 41.7 \text{ kW}/81.7 \text{ kW} = 320 \text{ K}$
Therefore, the temperature of the ambient environment is 320 K (46.8°C).
