Question

A power cycle operates between hot and cold reservoirs at 500 K and 310 K, respectively. At steady state the cycle develops a power output of 0.1 MW. Determine the minimum theoretical rate at which energy is rejected by heat transfer to the cold reservoir, in MW.

Answer 1

Answer:

$Q_{out} = 0.16 MW$

Explanation:

As we know that efficiency of theoretical engine is given as

$\eta = 1 - \frac{T_2}{T_1}$

so we will have

$T_2 = 310 K$

$T_1 = 500 K$

so here we will have

$\eta = 1 - \frac{310}{500}$

$eta = 0.38$

so we know that

$\frac{W}{Q_{in}} = 0.38$

$\frac{0.1}{Q_{in}} = 0.38$

$Q_{in} = 0.26 MW$

So heat rejected by the engine to cold reservoir is given as

$Q_{out} = Q_{in} - W$

$Q_{out} = 0.26 - 0.1$

$Q_{out} = 0.16 MW$

Answer 2

The minimum theoretical rate at which energy is rejected is mathematically given as

Qr = 0.16 MW

What is the minimum theoretical rate at which energy is rejected by heat transfer to the cold reservoir?

Question Parameter(s):

A power cycle operates between hot and cold reservoirs at 500 K and 310 K,

At steady state the cycle develops a power output of 0.1 MW.

Generally, the equation for the efficiency  is mathematically given as

$n = 1 - \frac{T_2}{T_1}$

Therefore

$n = 1 - \frac{310}{500}$

n=0.38

Where

W/Q = 0.38

Hence

0.1/Q = 0.38

Q=0.26MW

In conclusion,heat rejected

Qr = Q - W

Qr = 0.26 - 0.1

Qr = 0.16 MW

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