# Thermodynamics ( steam tables ) isentropic

1. May 6, 2013

### steve2510

1. The problem statement, all variables and given/known data
The power output of an adiabatic steam turbine is 5 MW. If the device can be assumed
to operate as a steady flow device with isentropic expansion, determine the
following:
i) The dryness fraction, x, at the exit from the turbine;
ii) The work output per unit mass of steam flowing through the device;

Intital state : 20 BAR, 400°C C=50M/S Z=10M
Final State : 16 KPA, c = 180m/s z = 6m

2. Relevant equations
Q+W+h1+c12/2 + z1g = H2+c22/2 + z2g

3. The attempt at a solution
I looked up the following in steam tables
@20 Bar
Ts= 212.4
vg=0.1511
hg=3248

@16kpa
Ts = 55.3
Hf = 232
hfg= 2369

So if its dry i know that h=hf + xhfg
Im confused is it hasnt given a value of γ so i'm not sure if i am supposed to use the fact that:
(P2/P1)γ-1/γ = T2/T1
If i do use that and take a value of gamma to be 1.4 i get :
(16x10^3/2x10^6)^2/7 = T2/T1 Thefore T2= 169.4 ° K but thats actually below 0°C, so surely this isn't correct. Could anyone point me int he right direction please

2. May 6, 2013

### SteamKing

Staff Emeritus
I'm puzzled about the values of c and z given in the OP for the initial and final states. Where did these come from? It is highly unusual for the steam entering or leaving a turbine to have such high velocities due to erosion of the internal surfaces of the piping.

3. May 6, 2013

### steve2510

Hm didn't notice the large numbers, its just a past exam question so i guess its Made up. I just dont believes theres enough Information to find X, im puzzled.

4. May 6, 2013

### SteamKing

Staff Emeritus
For the final state, you are going to need hg in order to work out the quality. (I don't know which steam tables you are using; there are several online apps which will give quality along with the other thermo properties.)

I don't know why you are using gamma: you are supposed to use the steam tables to find the point where p = 16 kpa and the entropy is the same as the initial state. Gamma for steam, BTW, is taken to be 1.33, but in your case, the expansion of steam thru the turbine takes you across the saturation line, and you have a significant quantity of moisture mixed in with the vapor, so I don't think the ideal gas law is a valid approach to calculating final temp.

5. May 6, 2013

### SteamKing

Staff Emeritus
6. May 6, 2013

### steve2510

Thanks, well i'm using "Thermodynamic and Transport Properties of Fluids"
So i can use the entropy of the system i guess.
Am i right in saying for a isentropic process dS=0
Therefore;
@ 400c, 20 bar S=7.126
@ 16KPA Sf= 0.772 Sfg= 7.213
So 7.126 = 0.772 + x7.213
X= 0.88
I think i'm actually okay on the rest of the question just that bit was really confusing me, thanks very much.

Lastly Does X relate in any way to temperature, say if you no how dry/wet the vapour is can you find temperature?

7. May 6, 2013

### SteamKing

Staff Emeritus
I'm not aware of any relationship. In Rankine cycle calculations, the temperature at the turbine exit is not as important as the quality. Generally, the quality of the steam at the turbine exit should not be lower than about 90%, as the moisture will cause the turbine blades to erode

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