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Byrgg
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Ok, I don't really need to know because it's something I've been told to ignore in the text, but I'm curious about it anyway, here goes:
At a nuclear power plant, water at 8 deg. celsius from a nearby lake is used in a heat exchanger to condense spent steam, at a temperature of 120 deg. celsius, to water, at 85 deg. celsius before it is recycled into the reactor. If the cooling water returns to the lake at a temperature of 19.deg celsius, how many kg of water are needed per kg of steam? Ignore the pressure changes on the steam.
Here's how I solved it, could you tell me if I'm right?
Q_gained(1) = -Q_lost(2)
Q_1 = mcT (for the water)
Q_2 = mcT + ml + mcT (for the steam)
specific heat capacity of water is 4200
specific heat capacity of steam is 2000
latent heat of vapourization of water is 2255000
mcT = -(mcT + ml + mcT)
m(4200)(19-8) = -(m(2000)(100-120) + m(2255000) + m(2000)(85-100))
m46200 = -(m(-40000) + m(2255000) + m(-3000))
m46200 = m4000 - m2255000 + m3000
divide both sides by m2... (mass of the steam)
m1(46200)/m2 = 4000 - 2255000 + 3000
divide both sides by 462000...
m1/m2 = (4000 - 2255000 + 3000)/46200
m1/m2 = 48.66
Therefore the ratio of water to steam is 48.66, the book says 50, if I followed the significant figures than I would've acheived 50 I'm pretty sure so, I'm confident I did that right does it look ok? But now here's what I want to know:
I want to know exactly how the pressure changes in the steam would affect the answer(considering it told me to ignore them).
Thanks in advance, someone please respond soon
At a nuclear power plant, water at 8 deg. celsius from a nearby lake is used in a heat exchanger to condense spent steam, at a temperature of 120 deg. celsius, to water, at 85 deg. celsius before it is recycled into the reactor. If the cooling water returns to the lake at a temperature of 19.deg celsius, how many kg of water are needed per kg of steam? Ignore the pressure changes on the steam.
Here's how I solved it, could you tell me if I'm right?
Q_gained(1) = -Q_lost(2)
Q_1 = mcT (for the water)
Q_2 = mcT + ml + mcT (for the steam)
specific heat capacity of water is 4200
specific heat capacity of steam is 2000
latent heat of vapourization of water is 2255000
mcT = -(mcT + ml + mcT)
m(4200)(19-8) = -(m(2000)(100-120) + m(2255000) + m(2000)(85-100))
m46200 = -(m(-40000) + m(2255000) + m(-3000))
m46200 = m4000 - m2255000 + m3000
divide both sides by m2... (mass of the steam)
m1(46200)/m2 = 4000 - 2255000 + 3000
divide both sides by 462000...
m1/m2 = (4000 - 2255000 + 3000)/46200
m1/m2 = 48.66
Therefore the ratio of water to steam is 48.66, the book says 50, if I followed the significant figures than I would've acheived 50 I'm pretty sure so, I'm confident I did that right does it look ok? But now here's what I want to know:
I want to know exactly how the pressure changes in the steam would affect the answer(considering it told me to ignore them).
Thanks in advance, someone please respond soon