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Okay, I am trying to work backwards from an exam which I messed up and I am trying to figure out what the lecturer did based on his answers posted online. Looking back, I realise one mistake I made was to use the specific heat of water given to calculate the enthalpy change in a transition from one pressure to the next.

In a transition within a Rankine cycle: I have P1 = P2 = 200 bar. Moving from 1 to 2, the water vapour turns to two-phase water and the enthalpy at point 2 is given as 2,353 kJ/kg. We also know that the enthalpy at 1 is 145 (worked out from a constant volume work problem and it checks out with the lecturer's answer).

Long story short: if I had tried to use the specific heat of water to determine the change in enthalpy and work from there to the enthalpy, I would have been way off. Instead, students were meant to use the given enthalpy value and then find the correct temperature from a table of saturated water provided.

I understand that the specific heat of water will change with temperature, but looking up the temperatures I am working at (changes from 30 C to 388 C) I can see that even an averaged out specific heat does not allow for a computation of a change in enthalpy from 145 to 2,353 (the second value was given). So I'm trying to figure out what I don't understand; why is a specific heat/temperature change calculation insufficient to calculate the enthalpy change here.

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    $\begingroup$ You forgot to consider the heat of vaporization. $\endgroup$ – Chet Miller Oct 10 '18 at 12:05
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    $\begingroup$ To make @ChesterMiller's point more explicitly: the Rankine cycle involves change of phase. $\endgroup$ – dmckee --- ex-moderator kitten Oct 10 '18 at 14:12
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I think you mean the water turns to two-phase water, not the water vapor turns to two-phase water.

See figure below.

200 Bar and h = 2353 puts point 2 in the two-phase region.

From point 1 to point 1A you can use specific heat to calculate enthalpy change. From point 1A to 2 you have liquid and vapor at constant temperature. The change in enthalpy (527) is entirely due to the heat of vaporization.

Hope this helps. enter image description here

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  • $\begingroup$ Okay, that makes sense. I'm sorry but my upvotes won't be worth much until I increase my credibility score or whatever. $\endgroup$ – Abed Oct 11 '18 at 16:42
  • $\begingroup$ @user209194 I believe you can still select the "accept" button (one with checkmark) $\endgroup$ – Bob D Oct 11 '18 at 19:31

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