Thermodynamics: Calculate ideal enthelpy at turbine exit

In summary: So I used the x1 and x2 to find the saturated values at that pressure, and then used linear interpolation to find the quality at the isentropic exit condition, and from there I was able to find the enthalpy of the 100% vapor state. In summary, to find the isentropic efficiency of the turbine in an Ideal Rankine cycle, you must first determine the vapor quality at the isentropic exit condition by interpolating between the entropy values given in the steam tables. Then, use the vapor quality to find the specific enthalpy of the 100% vapor state, which can then be used to calculate the isentropic efficiency of the turbine.
  • #1
jdawg
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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

 

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  • #2
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.
 
  • #3
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
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.
 
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  • #5
THANK YOU SO MUCH! That really cleared up my confusion!
 

FAQ: Thermodynamics: Calculate ideal enthelpy at turbine exit

1. What is thermodynamics and why is it important?

Thermodynamics is the study of energy and its transformations between different forms. It is important because it helps us understand how energy is used and transferred in various systems, such as power plants or engines.

2. What is ideal enthalpy at turbine exit and why is it calculated?

Ideal enthalpy at turbine exit is the theoretical amount of energy that would be available at the exit of a turbine, assuming all processes are reversible and there are no losses. It is calculated to determine the maximum possible energy output of a turbine and to optimize its performance.

3. What factors affect the ideal enthalpy at turbine exit?

The ideal enthalpy at turbine exit is affected by the temperature and pressure of the fluid entering the turbine, as well as the efficiency of the turbine itself. Other factors such as the type of fluid and the geometry of the turbine can also play a role.

4. How is ideal enthalpy at turbine exit calculated?

Ideal enthalpy at turbine exit can be calculated using the equation: Hexit = Hentry + (W/Q) + (Vexit^2 - Ventry^2)/2, where Hexit is the ideal enthalpy at turbine exit, Hentry is the enthalpy at the turbine inlet, W is the work done by the turbine, Q is the heat input to the turbine, and Vexit and Ventry are the velocities at the turbine exit and inlet, respectively.

5. What are some real-world applications of calculating ideal enthalpy at turbine exit?

Calculating ideal enthalpy at turbine exit is important in the design and operation of power plants, gas turbines, and steam turbines. It is also used in other industries such as aerospace, where it can help optimize the performance of jet engines. Additionally, it is relevant in renewable energy systems, such as geothermal power plants, where the efficiency of turbines is crucial for maximizing energy output.

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