Question

A reversible power cycle whose thermal efficiency is 40% receives 50 kJ by heat transfer from a hot reservoir at 600 K and rejects energy by heat transfer to a cold reservoir at temperature TC.
1. Determine the energy rejected, in kJ, and TC, in K.

Answer 1

Answer:

$-Q_{c}= 30 KJ\\-T_{c} = 360 K$

Explanation:

For this question, we take into consideration, the following:

• The values are constant average
• The power cycle is reversible

We calculate the efficiency of the cycle using the following equation:

$\eta = 1 - \frac{T_{c}}{T_{H}}$

where $T_{c}$ is the temperature of the cold reservoir and $T_{H}$ is the temperature of the hot reservoir.

To calculate, $T_{c}$, we rearrange the above equation.

$T_{c}=T_{H} (1 - \eta)$

⇒ 600 ( 1 - 0.4 )

⇒ 600 × 0.6

⇒ 360 K

The efficiency of the cycle is then calculated using the following equation:

$\eta = \frac{Q_H - Q_c}{Q_H}$

Using this equation, we find the amount of heat added to the cold reservoir:

$Q_c = Q_H (1 - \eta)$

⇒ 50 ( 1 - 0.4)

⇒ 50 × 0.6

⇒ 30 KJ