0
$\begingroup$

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.

$\endgroup$
2
  • 1
    $\begingroup$ You forgot to consider the heat of vaporization. $\endgroup$ Oct 10, 2018 at 12:05
  • 1
    $\begingroup$ To make @ChesterMiller's point more explicitly: the Rankine cycle involves change of phase. $\endgroup$ Oct 10, 2018 at 14:12

1 Answer 1

0
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ 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, 2018 at 16:42
  • $\begingroup$ @user209194 I believe you can still select the "accept" button (one with checkmark) $\endgroup$
    – Bob D
    Oct 11, 2018 at 19:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.