Answer:
A The heat transfer to the river would be 900 MW
B. The actual heat transfer would be lower than the theoretical heat transfer.
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 = [tex]Q_{H}[/tex] - [tex]Q_{c}[/tex] ......................1
where W is the work output of the steam power plant
[tex]Q_{H}[/tex] is the heat transfer at the hot reservoir
[tex]Q_{c}[/tex] 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 ([tex]Q_{H}[/tex]). The efficiency can be obtained with equation 2;
e = [tex]\frac{W}{Q_{H} }[/tex]
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
[tex]Q_{H}[/tex] is the heat transfer at the hot reservoir
Rearranging equation 2 to make [tex]Q_{H}[/tex] the subject formula we have;
[tex]Q_{H}[/tex] = W/e
substituting the values we have;
[tex]Q_{H}[/tex] = 600 MW/ 0.4
[tex]Q_{H}[/tex] = 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;
[tex]Q_{c}[/tex] = [tex]Q_{H}[/tex] - W
[tex]Q_{c}[/tex] = 1500 MW - 600 MW
[tex]Q_{c}[/tex] = 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.
The actual heat transfer rate is higher than the power delivered by the steam power plant.
The heat transfer rate released to the river ([tex]\dot Q_{L}[/tex]), in megawatts, is derived from the definition of energy efficiency ([tex]\eta[/tex]), no unit, that is to say:
[tex]\eta = \frac{\dot W}{\dot Q_{L}+\dot {W}}[/tex]
[tex]\eta \cdot \dot Q_{L} + \eta \cdot \dot W = \dot W[/tex]
[tex](1-\eta)\cdot \dot W = \eta \cdot \dot Q_{L}[/tex]
[tex]\dot Q_{L} = \left(\frac{1}{\eta}-1 \right)\cdot \dot W[/tex] (1)
Where [tex]\dot W[/tex] is the power delivered by the steam power plant, in megawatts.
If we know that [tex]\eta = 0.4[/tex] and [tex]\dot W = 600\,MW[/tex], then the heat transfer rate released to the river is:
[tex]\dot Q_{L} = \left(\frac{1}{0.4}-1 \right)\cdot (600\,MW)[/tex]
[tex]\dot Q_{L} = 900\,MW[/tex]
Thus, the actual heat transfer rate is higher than the power delivered by the steam power plant.
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