Answer:
[tex]\dot Q _{L} = 511.111 MW[/tex]. Heat transfer can be higher if themal efficiency is lower.
Explanation:
The heat transfer rate to the river water is calculated by this expression:
[tex]\dot Q_{L} = \dot Q_{H} - \dot W[/tex]
[tex]\dot Q_{L} = (\frac{1}{\eta_{th}}-1 )\cdot \dot W\\\dot Q_{L} = (\frac{1}{0.54}-1)\cdot (600 MW)\\\dot Q _{L} = 511.111 MW[/tex]
The actual heat transfer can be higher if the steam power plant reports an thermal efficiency lower than expected.
Following are the solution to heat transfer rate:
Work output,[tex]W = 600 \ MW[/tex]
Thermal efficiency, [tex]\eta_{th} , = 54\% = 0.54[/tex]
Using the following equation, control the rate of heat supplied to the power plant:
[tex]\eta =\frac{W}{Q_{H}} \\\\0.54=\frac{600}{Q_H}\\\\Q_H =1,111.11\ MW[/tex]
Using the following equation, calculate the rate of heat transfer to river water:
[tex]W=Q_H-Q_L\\\\600 =1,111.11-Q_L\\\\Q_L=511.11\ MW[/tex]
In actuality, the rate of heat transfer to river water will be smaller than the calculated number because some heat would be lost to the surrounding.
Learn more:
brainly.com/question/13963175
