The heat capacity of liquid water is 4.2 x 106 J/m3/K. This means that 1 m3 of w
ID: 1969881 • Letter: T
Question
The heat capacity of liquid water is 4.2 x 106 J/m3/K. This means that 1 m3 of water will be increased in temperature by 1 K when it absorbs 4.2 x 10^6 J.a. Assuming a discharge temperature allowing a 3K change, what volume of water is needed each day to cool a 500 MW powerplant operating at 40% efficiency in converting heat to electrical energy? Hint: If the plant is 40% efficient and it must generate 500MW, how much heat must be put into the system (it will be more than 500 MW.) The notes on thermodynamics will help.
Explanation / Answer
as 40% is 500 mw so 60% heat into the system is lost for cooling the plant
750 Mw of cooling is done by heat capacity of liquid water
amount of water neede per second is found first
cool load =750 * 10^6 watts ---> j/sec
Tmax for water is 3 klevin
max heat extracted by 1m3 of liquid water = Tmax *4.2 * 10^6 =12.6 * 10^6 j
cool load/heat extracted = 750/12.6 = 59.524 m3/sec
86400 sec a day
59.524 * 86400 = 5.143 * 10^6 m3/day
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