The latent heat of vaporization of water is 2.26 x 109 J/m3. This means that a h
ID: 1969882 • Letter: T
Question
The latent heat of vaporization of water is 2.26 x 109 J/m3. This means that a heat equivalent of 2.26 x 10^9 J is required to convert water at 100 oC to steam at 100 oC; no change in temperature will occur until all water is converted to steam. Assume the following cooling cycle: Water is brought in at 10 oC. Waste heat is transferred to the water heating it to 100 oC (NB the heat capacity of liquid water is given above.) Heat is further transferred to the water until it is completely converted to steam at 100 oC. What volume of water is needed each day to cool a 500 MW powerplant operating at 40% efficiency in converting heat to electrical energy?Explanation / Answer
Water (liquid): cp = 4.1855 *103 j/kg
heat absorbed inconversion from 10 degree water to 100 degree water
= m c T
= m * 4.1855 *103* ( 100-10) = 376.65*103*m j
then water is vapourised to steam at 100 degree celsius
latent heat given is in terms of volume
i.e 2.26*109 j/m3
since density of water is 1000 kg/m3
therefore in terms of mass it is = 2.26*106 j/kg
this means heat absorbed = latent heat *mass = m*2.26*106 j
now, heat generated in the plant = 500 MW - 40 % of 500 MW ( as it has been converted into electrical energy ) = 300 MW = 300*106 W
total energy generated in one day = 300*24*3600 * 106 w
this heat has to be removed by water
therefore , equation becomes
376.65*103*m +m*2.26*106 = 300*24*3600 * 106
(0.37665+ 2.26)m *106= 25920000*106
m = 9830656 kg
volume = 9830656 /1000 = 9830.656 m3
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