I am trying to calculate the amount of energy required to heat 12,300 kg of water from 70 °C at 25 MPa to 120 °C at 40 MPa. I have thought of using Q = mcT but I know that the specific heat capacity of water varies slightly with pressure. How can I calculate the energy required since c varies?
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2$\begingroup$ Please enlighten us as to how it requires less energy to cause a change in temperature just because the pressure is higher... So, if I run a bath and when the tap is closed the water is at 9 Bar, once it is in the bath at atmospheric it must be colder.... $\endgroup$– Solar MikeNov 29, 2017 at 21:25
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$\begingroup$ AFAIK the specific heat of liquid water is independent of the pressure. Take a look at CRC ("rubber book") tables or equivalent. $\endgroup$– Carl WitthoftNov 30, 2017 at 15:42
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1$\begingroup$ Specific heat of liquid water is usually treated as constant, since it only varies by a few percent, but it does vary. In the example above, it's 4074 J/(kgK) at 25MPa, and 4029 J/(kgK) at 40MPa, or a reduction of 1.1%. $\endgroup$– Jonathan R SwiftDec 1, 2017 at 13:36
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1$\begingroup$ I can't provide an official answer to the question since it's on hold, but, the energy required to make this change is the difference in internal energy between the two states, so using the numbers shown in my comment above, subtract the initial energy Q1=12300*4070*343.2 from the final energy Q2=12300*4029*393.2, to give you Q2-Q1 = 2305x10^6 Joules $\endgroup$– Jonathan R SwiftDec 1, 2017 at 13:43
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1$\begingroup$ @SolarMike - Yes, what you described is exactly the case. The vapour pressure is lower at atmospheric pressure than at 9barg, meaning some of the liquid water that was in the pipes will turn to vapour once outside the pipe. This loss of sensible heat (now stored as latent heat in the vapour) causes the remaining liquid (in the bath) to cool down. The total internal energy of the liquid+vapour hasn't changed, but the temperature of the liquid certainly has. $\endgroup$– Jonathan R SwiftDec 4, 2017 at 16:17
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1 Answer
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The energy required to make this change is the difference in internal energy between the two states
$ ∆Q=m((c_{v_2}×T_2 )-(c_{v_1}×T_1 )) $
where
$ m=12300kg $
$ c_{v_1}=4074\;Jkg^{-1}K^{-1} $
$ c_{v_2}=4029\;Jkg^{-1}K^{-1} $
$ T_1=343.2\;K $
$ T_2=393.2\;K $
so
$ ∆Q=12300((4029×393.2)-(4070×343.2))=2305MJ $
answered Dec 4, 2017 at 17:33