A 600-MW steam power plant, which is cooled by a nearby river, has a thermal efficiency of 40 percent. Determine the rate of hea

Question

natima [27]

3 years ago

10

A 600-MW steam power plant, which is cooled by a nearby river, has a thermal efficiency of 40 percent. Determine the rate of hea

t transfer to the river water. Will the actual heat transfer rate be higher or lower than this value

Engineering

2 answers:

Serhud [2] · Answer 1 · 2022-05-20T04:16:06+03:00

Answer:

<em>A The heat transfer to the river would be 900 MW</em>

<em>B. The actual heat transfer would be lower than the theoretical heat transfer.</em>

Explanation:

Part A

For a steam power plant, the process is continuous and the law of thermodynamic is conserved. The steam power plant consists of the cold and hot reservoir and the rate of heat transfer in the steam power plant can be gotten with the expression below.

W = $Q_{H}$ - $Q_{c}$ ......................1

where W is the work output of the steam power plant

$Q_{H}$ is the heat transfer at the hot reservoir

$Q_{c}$ is the heat transfer at the cold reservoir( heat transfer to the nearby river)

The efficiency of the steam power plant would be used to obtain the heat transfer at the hot reservoir ( $Q_{H}$ ). The efficiency can be obtained with equation 2;

e = $\frac{W}{Q_{H} }$

e is the thermal efficiency of the steam power plant = 40 % = 0.4

W is the work output of the steam power plant = 600 MW

$Q_{H}$ is the heat transfer at the hot reservoir

Rearranging equation 2 to make $Q_{H}$ the subject formula we have;

$Q_{H}$ = W/e

substituting the values we have;

$Q_{H}$ = 600 MW/ 0.4

$Q_{H}$ = 1500 MW

The heat transfer at the hot reservoir is 1500 MW and putting it into equation 1 to find the heat transfer to the river;

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

$Q_{c}$ = 1500 MW - 600 MW

$Q_{c}$ = 900 MW

Therefore the heat transfer to the river would be 900 MW

Part B

The actual heat transfer would be lower than the theoretical heat transfer. Losses dues to friction, head losses and heat loss by evaporation to the surroundings account for the lower value.