# Homework Help: Liquid water turbine

1. Apr 10, 2006

### JSBeckton

A liquid water turbine recieves 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:

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

But the anwser should be 1.99 kJ/kg
Can anyone tell me what I did wrong? Are my specific enthalpys right?
Thanks

2. Apr 10, 2006

### Cyrus

Arg, please stop using omega for specific work, its lower case w, not omega!

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

A TURBINE MUST HAVE STEAM AT THE INLET!!!!

What is this guys problem????

Does he also have pumps compressing vapor?....:uhh:
[/rant]

Last edited: Apr 10, 2006
3. Apr 10, 2006

### Pengwuino

Maybe it uses the kinetic theory of gas for a steam input, not a liquid input.

4. Apr 10, 2006

### Cyrus

No, it does not.

5. Apr 10, 2006

### Pengwuino

oh 15m/s, your right, for once.

6. Apr 10, 2006

### Cyrus

7. Apr 10, 2006

### Pengwuino

Wait what are you talking about.

8. Apr 10, 2006

### JSBeckton

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?

9. Apr 10, 2006

### JSBeckton

Is there not such a thing as a liquid turbine?

10. Apr 10, 2006

### Cyrus

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.

11. Apr 10, 2006

### Cyrus

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! ...mumbles lazy...

12. Apr 10, 2006

### JSBeckton

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:

$$\begin{array}{l} \frac{{84.41 - 83.94}}{{500 - 2.339}} = \frac{{h - 83.94}}{{100 - 2.339}} \\ \\ h = 84.03 \\ \end{array}$$

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

13. Apr 10, 2006

### Cyrus

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. Apr 10, 2006

### JSBeckton

at 100 kPa, T=99.62C

It is a compressed liquid right? my compressed liquid tables start out at 500kPa, thats 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. Apr 10, 2006

### Cyrus

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. Apr 10, 2006

### siddharth

Cyrus, It's possible to have liquid turbines with a liquid at the inlet.

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

One more thing,
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: May 2, 2017
17. Apr 10, 2006

### Cyrus

Well I'll be dammed.

I am used to thinking of turbines as Gas turbines.

Damn you and your infinite smartness Siddharth!

You little genius!

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

Edit: Yep, good catch sidd, at 20C.

Last edited: Apr 10, 2006
18. Apr 11, 2006

### JSBeckton

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. Apr 11, 2006