Calculate Work & Power of Liquid Water Turbine

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SUMMARY

The discussion focuses on calculating the specific work and power of a liquid water turbine receiving water at 2 kg/s, 2000 kPa, and 20°C, with an exit pressure of 100 kPa. The initial specific enthalpy (hi) is 85.82 kJ/kg, and the exit specific enthalpy (he) is 84.03 kJ/kg. The calculated specific work (ω) is 1.9025 kJ/kg, but the expected answer is 1.99 kJ/kg. The discrepancy arises from the use of different enthalpy tables and the misunderstanding of the turbine's operation with liquid versus gas inputs.

PREREQUISITES
  • Understanding of thermodynamics principles, specifically energy equations.
  • Familiarity with specific enthalpy values for liquids and gases.
  • Knowledge of interpolation techniques for thermodynamic tables.
  • Experience with liquid water turbine operations and their characteristics.
NEXT STEPS
  • Research the properties of saturated liquids at various pressures, specifically at 100 kPa and 20°C.
  • Learn about the differences between gas turbines and liquid turbines, including their operational principles.
  • Study the use of thermodynamic tables and the importance of accurate interpolation methods.
  • Explore software tools for calculating thermodynamic properties and their reliability compared to traditional tables.
USEFUL FOR

Mechanical engineers, thermodynamics students, and professionals involved in turbine design and performance analysis will benefit from this discussion.

JSBeckton
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A liquid water turbine receives 2kg/s water at 2000kPa, 20C, and velocity of 15m/s. The exit is at 100kPa, 20C and very low velocity. Find specific work and power produced.

From my tables I get

hi=85.82 kJ/kg
he=84.03 kJ/kg

Energy equation reduces to:

<br /> \begin{array}{l}<br /> h_i + \frac{{V_i^2 }}{2} = \omega + h_e \\ <br /> \omega = h_i + \frac{{V_i^2 }}{2} - h_e \\ <br /> \omega = 85.82\frac{{kJ}}{{kg}} + \left( {\frac{{15m/s^2 }}{2} \times \frac{{1km}}{{1000m}}} \right) - 84.03\frac{{kJ}}{{kg}} \\ <br /> \omega = 1.9025\frac{{kJ}}{{kg}} \\ <br /> \end{array}<br />

But the anwser should be 1.99 kJ/kg
Can anyone tell me what I did wrong? Are my specific enthalpys right?
Thanks
 
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Arg, please stop using omega for specific work, its lower case w, not omega! :cry:

[rant]
Man, another poorly written problem. Throw your textbook in trash!

A TURBINE MUST HAVE STEAM AT THE INLET!

What is this guys problem?

Does he also have pumps compressing vapor?...:rolleyes:
[/rant]
How did you find your enthalpy values? Your method looks good.
 
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Maybe it uses the kinetic theory of gas for a steam input, not a liquid input.
 
No, it does not.
 
oh 15m/s, your right, for once.
 
Pengwuino said:
oh 15m/s, your right, for once.

What are you talking about? :confused:
 
Wait what are you talking about.
 
Sorry about the omega, I used the computer tables that came with my crappy book to find the enthalpys because its not inthe table and I'm too lazy to interpolate. I've used the program many times and it hasn't failed me yet. Do you get different values?
 
Is there not such a thing as a liquid turbine?
 
  • #10
A turbine runs on GAS/STEAM

A COMPRESSOR RUNS ON LIQUIDS

I don't know. Look it up yourself. Look it up for the saturated liquid at that TEMPERATURE.

Pengwuino, stop playing around in here. :mad:
 
  • #11
The difference in your anwser is because your using a program and not the enthalpy in your table. Its an error of the second decimal place. Use your TABLE! and see what happens. There is a reason why they publish those tables for you! :wink: ...mumbles lazy...
 
  • #12
The program is basically expanded tables. I do not have tables for 100kPa and 20C I have:

P=500 Kpa
h=84.41

and

P=2.339
h=83.94

I interpolate to find:

<br /> \begin{array}{l}<br /> \frac{{84.41 - 83.94}}{{500 - 2.339}} = \frac{{h - 83.94}}{{100 - 2.339}} \\ <br /> \\ <br /> h = 84.03 \\ <br /> \end{array}<br />

Exactly what my computer tables told me, that's why i asked if you had tables that said differently. What is the problem?

Thanks for your help.
 
  • #13
100kPa and 20C

You sure do. Look at the table under pressure for 100kpa. What is the saturation temperature? Is it below 20C? If it is, your exit is a superheated steam. If it is above, then you have a saturated stream at the exit (or possibly only liquid at the exit).

P.S., you don't interpolate in 500kpa intervals. (That would be a bad idea)
 
  • #14
at 100 kPa, T=99.62C

It is a compressed liquid right? my compressed liquid tables start out at 500kPa, that's why I used that high value. I only have the saturated value and the lowest compressed value to work with, what else can i do?
 
  • #15
Ok, so you are below there, and therefore yes, you should have a compressed liquid. That means that the enthalpy is just the value of the enthalpy for a SATURATED LIQUID at 100KPa. Hf
 
  • #16
Cyrus, It's possible to have liquid turbines with a liquid at the inlet.

For example, http://en.wikipedia.org/wiki/Water_turbine" have water flowing in the inlet.

One more thing,
That means that the enthalpy is just the value of the enthalpy for a SATURATED LIQUID at 100KPa. Hf

Should it not be, The enthalpy is just the value of the enthalpy for a saturated liquid at 20 C, and not 100Kpa?

And JSBeckton, there isn't much of a difference between 1.90 and 1.99. Maybe, the difference is due to the tables.
 
Last edited by a moderator:
  • #17
Well I'll be dammed. :smile:

I am used to thinking of turbines as Gas turbines. :frown:

Damn you and your infinite smartness Siddharth! :wink:

You little genius! :devil:

I still think the wording in many of his problems are careless though...

Edit: Yep, good catch sidd, at 20C.
 
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  • #18
Thanks guys, I used the value given at saturated 20C and got 1.99! My question still remains, in my compression tables the specific enthalpy increaces with pressure albeit not by a lot So was my anwser not more accurate than theirs, and hence would i have not lost points? I guess nearly 5% error seems somewhat significant to me.
 
  • #19
You should ask that to your prof or TA.
 

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