Finding ideal enthalpy at turbine exit using isentropic process

[jdawg]

Homework Statement

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=s_2s, but I need to find h_2s
In my book they interpolate to find the h_2s 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 h_2s.

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

Homework Equations

The Attempt at a Solution

 

[SteamKing]

Homework Statement

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=s_2s, but I need to find h_2s
In my book they interpolate to find the h_2s 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 h_2s.

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

Homework Equations

The Attempt at a Solution

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.

 

[jdawg]

I tried interpolating between the entropy and enthalpy values they gave me.
y=y1+(x-x1)[(y2-y1)/(x2-x1)]
h_2s=2784.3+(5.8892-5.8892)[(2262.8-2784.3)/(6.1030-5.8892)]
Which is wrong :(

 

[SteamKing]

I tried interpolating between the entropy and enthalpy values they gave me.
y=y1+(x-x1)[(y2-y1)/(x2-x1)]
h_2s=2784.3+(5.8892-5.8892)[(2262.8-2784.3)/(6.1030-5.8892)]
Which is wrong :(

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.

 

[jdawg]

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