# Thermodynamics: Calculate ideal enthalpy at turbine exit

1. Feb 11, 2016

### jdawg

1. The problem statement, all variables and given/known data
Ideal Rankine cycle includes irreversibilities in the adiabatic expansion and compression processes. Find the isentropic efficiency of the turbine.
***My book uses a subscript 2s to denote the ideal state

State 1: p1=60 bar h1=2784.3 kJ/kg x1=1 s1=5.8892 kJ/kg*K
State 2: p2=1.5 bar h2=2262.8 kJ/kg x2=0.8065 s2=6.1030 kJ/kg*K

So I know s1=s2s, but I need to find h2s
In my book they interpolate to find the h2s value. I tried to do that and it didn't work. The picture I'm attaching is the solution, but they don't show how the find h2s.

If someone could please show me how to find this value I would really appreciate it!

2. Relevant equations

3. The attempt at a solution

#### Attached Files:

• ###### h2s.png
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2. Feb 11, 2016

### SteamKing

Staff Emeritus
Why don't you show us what you did that didn't work. Linear interpolation is a procedure which should be mastered for looking up data in tables.

3. Feb 11, 2016

### jdawg

I tried interpolating between the entropy and enthalpy values they gave me.
y=y1+(x-x1)[(y2-y1)/(x2-x1)]
h2s=2784.3+(5.8892-5.8892)[(2262.8-2784.3)/(6.1030-5.8892)]
Which is wrong :(

4. Feb 11, 2016

### SteamKing

Staff Emeritus
That's definitely not how to do a linear interpolation.

Your first mistake was not realizing that you can only interpolate against values given directly in the steam tables.

You know that the entropy at P = 1.5 bar is s = 5.8892 kJ/kg-K, which is the isentropic state line end point at the turbine exhaust.
If you check the steam tables for P = 1.5 bar = 0.15 MPa, you won't find a value of s = 5.8892 kJ/kg-K tabulated directly. Why not?
Because the steam at this condition is no longer superheated, it's saturated, and there's a certain percentage of moisture present which you must determine.

If there were 100% vapor at the isentropic exit condition, the steam would be said to have 100% vapor quality. You don't have this, but you must calculate the quality in order to calculate the specific enthalpy.

https://en.wikipedia.org/wiki/Vapor_quality

You need to look up the saturated values of entropy and enthalpy for the vapor and liquid phases at P = 0.15 MPa in order to finish the calculation.

5. Feb 11, 2016

### jdawg

THANK YOU SO MUCH!! That really cleared up my confusion!