# Isentropic Efficiency - Nozzle

1. May 17, 2008

### VooDoo

Hey guys,

I have been given that the isentropic efficiency of a nozzle = 90%. Inlet condictions are: 5MPa and 550°C and exit: 100kPa.

Now $$\eta_{n}(h_{i}$$ - $$h_{0}$$) = $$V^{2}_{i}$$/2

h(o) - h(i) is the enthalpy drop across the turbine.

Now next they say that s(o) = s(i). Now how can the entropy at the entrance of the nozzle be equal to the entropy at the exit if the isentropic efficiency of the nozzle is not equal to one!

Here is a picture of my notes:
http://img259.imageshack.us/img259/6140/111nm6.jpg

Last edited: May 17, 2008
2. May 19, 2008

### jaap de vries

Maybe there is heat transfer out of your nozzle?

3. May 20, 2008

### FredGarvin

The isentropic efficiency is a type of efficiency that is defined by dividing the actual work output or KE output to the isentropic output. Why do you assume for your nozzle that it is isentropic when you are given the efficiency?

4. May 23, 2008

### Jupiter6

5. May 23, 2008

### Mechaniac

let me get this straight ,the question is wrong isn't it ?..cos his notes clearly mention that inlet entropy is equal to outlet entropy=7.131 kj/kgK and yet the question says that isentropic efficiency is 90 % how can both of this be true ?

90 % would mean a curved expansion graph, in which the initial entropy would be different from final entropy which means S(I) and S(O) cant be same ,right ?

6. May 23, 2008

### Jupiter6

Yes, the notes are unclear and seem to have no relevance to the problem stated by the original poster. Turbine....nozzle...which is it???