## Steam state (p,T) between nozzle and impulse turbine wheel

## Steam state (p,T) between nozzle and impulse turbine wheel

(OP)

Hello,

I am trying to calculate the state of steam between the nozzle and impulse turbine wheel for tip and labyrinth leakage purposes for a specific steam loss model, and having some troubles with it.

Using _0 before nozzle, _1 after nozzle but before turbine blades, and _2 after turbine blades. I have the following information from process (simplified):

p

p

T

T

Calculated/figured the following values:

h

h

Δh = 186 kJ/kg

v

v

Assumed nozzle speed ratio v

In isentropic case (ratio is 1):

Δh

h

But my major question is, how do I combine all this information? I don't know the isentropic efficiency for turbine wheel (without nozzle) either. I was going to solve it after knowing (p

I tried to calculate h

How can I account velocity in (p,T,h,S)-table? I have a software to calculate Curtis wheel's isentropic efficiency using (p

And is the energy transfer in wheel 186 kJ/kg * n

If p

I am trying to calculate the state of steam between the nozzle and impulse turbine wheel for tip and labyrinth leakage purposes for a specific steam loss model, and having some troubles with it.

Using _0 before nozzle, _1 after nozzle but before turbine blades, and _2 after turbine blades. I have the following information from process (simplified):

p

_{0}= 1600 kPap

_{2}= 500 kPaT

_{0}= 350 CT

_{2}= 250 CCalculated/figured the following values:

h

_{0}= 3147 kJ/kgh

_{2}= 2961 kJ/kgΔh = 186 kJ/kg

v

_{0}= 60 m/sv

_{1}= SQRT(2000*Δh+v_{0}^{2}) = 612 m/sAssumed nozzle speed ratio v

_{1}/v_{1,is}= 0.95In isentropic case (ratio is 1):

Δh

_{is}= 186/0.95^{2}= 206 kJ/kg, so nozzle loss therefore is:h

_{n}= Δh_{is}-Δh = 20 kJ/kg (and v_{1,is}= 644 m/s)But my major question is, how do I combine all this information? I don't know the isentropic efficiency for turbine wheel (without nozzle) either. I was going to solve it after knowing (p

_{1},T_{1}), since I can calculate the isentropic efficiency for nozzle+wheel combination using p_{0},T_{0}to p_{2},T_{2}(which is 0.65, fairly common for Curtis wheel). Seems I am missing some theory behind this. Can someone shed some light on how to proceed with this? Nozzle loss should be loss in flow energy (644 -> 612), increase in temperature and loss in entropy as far as I know.I tried to calculate h

_{1}=h_{0}-h_{n}and combine that with p_{1}=p_{2}. What actually happens with the temperature T_{1}in relation to T_{0}or T_{2}?How can I account velocity in (p,T,h,S)-table? I have a software to calculate Curtis wheel's isentropic efficiency using (p

_{in},T_{in},is%,p_{out}) so I can't account velocity in as input variable? If I use p_{2}as p, I get 0 W power obviously.And is the energy transfer in wheel 186 kJ/kg * n

_{wheel,is%}or (186-20)=166 kJ/kg * n_{wheel,is%}? is% = isentropic efficiency in wheelIf p

_{0}> p_{1}> p_{2}, how can I calculate it? Can I assume incompressible flow and use Bernoulli equation p_{1}= p_{0}+ 1/2 ρv_{0}^{2}- 1/2 ρv_{1}^{2}(assume ρ=ρ_{0}=ρ_{1}with enough accuracy) and look for T_{1}through h_{1}=h_{0}-h_{n}?
## RE: Steam state (p,T) between nozzle and impulse turbine wheel

Try an iterative approach, assuming no condensation (Ideal gas) for a start and an assumed efficiency for that stage. Get the numbers close for the ideal gas. You will be wrong, but be closer than nothing. (Why impulse blade? ) LP turbine outlet are very wet, but IP or HP turbines are non-condensing (typically).

Then, but only after you have reasonable numbers for an ideal gas, change to a compressible gas. Then change to a condensing gas.

YOu need actual numbers for each stage for temperature, pressure, percent of water.

## RE: Steam state (p,T) between nozzle and impulse turbine wheel

Forgot to mention, this turbine is one-wheel HP turbine (impulse, Curtis wheel) so condensing / wet-% shouldn't be a problem as shouldn't exist in this only stage.

So if I assume nozzle speed ratio to be 0.95, the efficiency for the nozzle is around 0.9 (0.95

^{2}if you don't account v_{0}).Since I know the whole stage's isentropic efficiency (p

_{0}, T_{0}to p_{2}with entropy constant is is-100%, and I know the real enthalpy drop because of process data sensors), can't I assume that nozzle efficiency * wheel efficiency = ~stage efficiency?So after these stages I should be able to make a raw approximation about p

_{1}and T_{1}, but still can't figure out how.## RE: Steam state (p,T) between nozzle and impulse turbine wheel

A curtis wheel is basically a single stage turbine except that the power needed to be split over two buckets because there wasn't a single bucket choice that was suitable. In this regards the second set of nozzles can be treated as just turning the direction of the flow and takes no drop across it.

## RE: Steam state (p,T) between nozzle and impulse turbine wheel

Yes, I have assumed all enthalpy drop (186 kJ/kg, isentropic case 206 kJ/kg -> loss 20 kJ/kg) to be in the nozzle set, but I am a little lost how to calculate the state after I have lost all kinematic energy generated with nozzle.

I highly doubt the state which counts as leakage loss in labyrinth seals before the blade but after the nozzle is same as before the nozzle, so what would be a raw approximation for it? Does it have:

1) around the same pressure 1600 kPa and enthalpy 3147 kJ/kg as before the nozzle

2) around the same pressure 1600 kPa and enthalpy 3147-20 kJ/kg (nozzle loss) as before the nozzle

3) significantly lower pressure than 1600 kPa but bigger than 500 kPa and enthalpy 3147 kJ/kg same as before the nozzle

4) significantly lower pressure than 1600 kPa but bigger than 500 kPa and enthalpy 3147-20 kJ/kg (nozzle loss) as before the nozzle

5) pressure does not change significantly after the nozzle back to previous state and real enthalpy drop is converted to friction

6) pressure does not change significantly after the nozzle back to previous state and real enthalpy drop-nozzle loss (186-20 kJ/kg) is converted to friction?

So which of these scenarios would be the closest approximation in this case, or is there another alternative?

Basically I have to calculate both tip loss for the blade (where kinematic energy remains for most part) and more importantly the steam leakage ratio over labyrinth seals (where all kinematic energy is lost).

## RE: Steam state (p,T) between nozzle and impulse turbine wheel

Assume its an ideal nozzle (100% throttling device). The all of the energy for that particular stage will be converted to kinetic through the pressure drop accross the nozzle only. This assumes 100% impulse bucket with no reaction along the length (small bucket heights). So in this case, the pressure on the entrance side of the nozzle will be higher than the exit edge of the nozzle. Assume no energy is lost and enthalpy is maintained (you havnt extracted any useful work yet). So a decent approximation is that the pressure between the bucket and nozzle is the same as the exhaust pressure just after the blade, meaning the pressure is the same before and after the bucket. . You can use like a .97 factor to adjust pressure to get a better approximation, but i doubt your in the realm of this accuracy anyways.

## RE: Steam state (p,T) between nozzle and impulse turbine wheel

Yes, I have figured this part out that the pressure is the same after the nozzle and after the bucket (afterall, this is the theory behind impulse turbine) but currently I am looking for the pressure before the labyrinth. I used my magical MS Paint skills to draw a picture.

Yellow is the Curtis wheel, green is the nozzle, red is the spot I am interested in. If I assume v = 0 in that spot, should I have similar pressure as p

_{2}(which is same as p_{1})?So question in more simple form is, should I approximate the pressure in red spot to be same as it, or p

_{0}- pressure loss from p_{1}to red due to high velocity? Or just assume that about all velocity is converted into heat so it would be isenthalpic process between p_{0}and red point, but pressure is same as p_{1}?## RE: Steam state (p,T) between nozzle and impulse turbine wheel

The flow will be determined by P1 / Pleakoff (assuming it goes to some other pressure controlled lined) and the flow coefficient of your packing.

## RE: Steam state (p,T) between nozzle and impulse turbine wheel

Thanks for the advise. What do you mean by Pleakoff?

## RE: Steam state (p,T) between nozzle and impulse turbine wheel

## RE: Steam state (p,T) between nozzle and impulse turbine wheel

typical arrangement

Atmos 14.5 psia ~18psia Pleakoff (150 psia??) P1 (Nozzle exit press)

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