Thermodynamics: Calculate ideal enthelpy at turbine exit

Click For Summary
SUMMARY

The discussion focuses on calculating the ideal enthalpy at the turbine exit in an Ideal Rankine cycle, specifically addressing the isentropic efficiency of the turbine. The user provided specific states with pressures, enthalpies, and entropies, but struggled with interpolating to find the ideal enthalpy (h2s) at the turbine exit. The correct approach involves using steam tables to find the saturated values of entropy and enthalpy at the given pressure of 1.5 bar (0.15 MPa) and calculating the quality of the steam to determine the specific enthalpy accurately.

PREREQUISITES
  • Understanding of Ideal Rankine cycle principles
  • Familiarity with steam tables for water and steam properties
  • Knowledge of linear interpolation techniques
  • Concept of vapor quality in thermodynamics
NEXT STEPS
  • Study how to use steam tables for calculating enthalpy and entropy
  • Learn about the concept of vapor quality and its significance in thermodynamic cycles
  • Practice linear interpolation with real thermodynamic data
  • Explore the effects of irreversibilities in the Ideal Rankine cycle
USEFUL FOR

Students and professionals in mechanical engineering, particularly those focusing on thermodynamics and power generation systems, will benefit from this discussion.

jdawg
Messages
366
Reaction score
2

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=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!

Homework Equations

The Attempt at a Solution

 

Attachments

  • h2s.png
    h2s.png
    12.3 KB · Views: 7,381
Physics news on Phys.org
jdawg said:

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=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!

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.
 
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 :(
 
jdawg said:
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 :(
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.
 
  • Like
Likes jdawg
THANK YOU SO MUCH! That really cleared up my confusion!
 

Similar threads

Replies
2
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 8 ·
Replies
8
Views
5K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
5
Views
5K
  • · Replies 4 ·
Replies
4
Views
11K
Replies
1
Views
3K
Replies
8
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 3 ·
Replies
3
Views
5K