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Isentropic efficiency vs Real Manufacturer efficiency

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mfqd13

Mechanical
Sep 27, 2007
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Hi all!

Digging on the subject i found in several situations with real manufacturers data, that the mechanical (shaft) power output is much more higher than the the one that we calculate using the isentropic efficiency.

For instance, i can get the following values, as an example:
- Isentropic efficiency: 0,75
- Heat Power available: 5376 kW (difference between enthalpies at the inlet and outlet conditions)
- Mechanical Power delivered by a real steam turbine: 5000 kW
- Real mechanical efficiency: 0,93

So if we use the isentropic efficiency to calculate the output mechanical power, is, in fact, an error.

Does anyone have any explanation?

Thanks!
 
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Show how you calculated the heat power available. The theoretical turbine exit enthalpy in this case should be based on 100% isentropic expansion.
What you are implying is that isen eff is 93%, which cannot be true, so I suspect there is some error in calculating exit enthalpy at 100% isen eff.
 
Hi dear georgeverghese,

Here are the values:

Turbine inlet:
[ul]
[li]Pressure: 60 bar(a)[/li]
[li]Temperature: 450 ºC[/li]
[li]Enthaly: 3304,2 kJ/kg[/li]
[/ul]
Turbine outlet:
[ul]
[li]Pressure: 3 bar(a)[/li]
[li]Temperature: 133,2 ºC[/li]
[li]Enthaly: 2724,9 kJ/kg[/li]
[/ul]
Steam flowrate: 33,41 ton/h
 
Isentropic efficiency is not a measure of power in / power out.

Isentropic efficiency is a comparison of the actual performance / idealized isotropic machine performance.

So, 75% isentropic efficiency means that the real machine operates at 75% of the performance of an ideal machine.

 
That turbine exit condition must be the actual exit condition. To get the isentropic exit parameters, use the steam tables. I've used linear interpolation in a region where curvature appears to be considerable, so values are approximate. I dont have access to a simulator.

Inlet condition
P=60bar, t=450degC=723degK, h=3301kJ/kg, s=6.718kJ/kg/degK

To get isentropic exit condition, s = 6.718kJ/kg/degK ; is constant at exit condition.

Isentropic exit condition
With some interpolation, I get, at 3bar, t =439degK = 166degC, h=2556kJ/kg, s=6.718kJ/kg/degK
So dH at isentropic condition = 3301-2556=745kJ/kg
At M=33410kg/hr = 9.28kg/sec, power extraction at driver shaft end at isentropic condition = 9.28x745=6914kJ/sec=6914kW

So isentropic eff = 5376/6914= 0.78
Isentropic eff for a given turbine does not always remain constant; it is a function of speed, exit pressure and mass flowrate. Fouling of turbine blades also affects isentropic eff. Pls refer to your Uni text on thermodynamics and steam turbine performance. Steam turbine performance is also well covered in Perry Chem Engg Handbook - see also Fig 29-23 in the 7th edn. Using h and s values from this very useful graph instead of tabular data, isen eff could be 80-82%.


 
pierreick,
Correct. Is that.
Thanks with the links.

MintJulep,
So, in your opinion, in practic life, the isentropic value, doesn't gives much of use, correct? We must consider the efficiency given by the manufacturers, is that?

georgeverghese,
It's exactly that. My issue is not with the calculation of the isentropic eff, witch i can do, but what's it's use un practical situation, because the values differ from the real turbine efficiencies given by the manufacturers.

Thanks!!!

 
In many companies, Operations engineers draw up their own curves for machine performance based on real plant data ( turbines, compressors, pumps, strippers, absorbers etc), so they can predict performance for revamps, debottlenecking projects, keep track of fouling etc.
In this case, you can also make necessary corrections for possible flow readout errors caused by off design operating conditions at the flow meter.
 
Suspect this driver is a 9stage multivalve steam turbine. So, also check that the number of steam admission valves in your opersting case is the same as that for which you are quoting what appears to be a typical 75% isentropic eff, in addition to all the other variables that affect isen eff.
Just because there is an apparent 10% difference in isen eff is no reason to throw the baby out along with the bathwater - check you are comparing apple with apple. Why not show us the turbine performance datasheet that shows this operating case with 75% isen eff ?
 
Hi all! Sorry for the late reply. Vacations.

I've searched an alternativa use case for this analysis. Please see below the turbine data and my calculation, also with my questions:

Turbine data:
turbine_oofogp.jpg


My calculations:
calc_bwf0s4.jpg


Questions:
1) My total real power (underlined in red) is below the power indicated by the manufacturer. Any explanation?
2) I've searched in other turbine data and this occurs in everyone: the enthalpy indicated in the condensation outlet conditions is allways bellow than the one indicated in the steam tables and this leads to higher powers, accordingly with the indicated by the manufacturers. But how is defined that outlet enthalpy, that doesn't match the steam tables?

Thanks again!
 
The manufacturer estimates the final exhaust stream has some moisture in it for the expected isentropic eff at this speed for the LP turbine, so the exhaust enthalpy is expected to be lower than 2625kJ/kg/degK. So total power extracted will be somewhat more. There is no harm in having some moisture at LP turbine exhaust, as long as BFW is on spec re silica content. Up to about 10% moisture content in the exhaust stream appears to be the permissible upper limit for condensing turbines. Vendor exit enthalphy of 2493kJ/kg corresponds to 5.7% wt moisture content in exit stream - okay.
 
Dear georgeverghese,

Thank you so much for the feedback, again!
I understood perfectly what you've said!

So, in practical ways, to get the exact power we must get the manufacturer enthalpy value at the outlet, correct? If we are making a study (and we don't have yet this value), the only way to estimate this enthalpy (and therefore the power), is to assume an isentropic efficiency and then calculate it. Do you agree?

Onde other question is regarding the power, with the corrected values (below), i found that my shaft power is a bit higher, despite all the enthalpies are equal. Any idea?
calc2_pji4wu.jpg


For backpressure turbines, should we expect to have steam without moisture or slightly overheated, or is it possible to have moisture content? (Probably is dependable on the use of this steam in the process after).

Thanks!!!
 
I have calculated Steam Turbine-Generator output and turbine overall (HP/LP) efficiency. Please see the attached PDF of my spreadsheet.

You will note that I have included estimated percentages for mechanical and electrical (generator) losses based on my experience. Although the value of the output at the generator terminals, which I calculated, is close to the manufacturer’s value, it does not match exactly.

It is not possible to match the manufacturer’s value without a detailed heat balance diagram. A detailed heat balance must include all shaft leakages with pressures and enthalpies (for the leakage and mix enthalpies) at the return to the turbine blade path. In addition, the heat balance must include the mechanical and electrical losses.

For backpressure turbines, I would not expect any moisture in the LP turbine exhaust, but that depends on the exhaust pressure and isentropic efficiency of the turbine. As the exhaust pressure rises, it is less likely to be wet.

Best of luck!
 
 https://files.engineering.com/getfile.aspx?folder=e95cbe92-dbb4-41cc-b46e-8f1798f7aaf8&file=STG_Output-Efficiency_Calc.pdf
That 3% difference between 2339kW and 2273kW total output may be due to some other losses within or outside the turbine (internal bearings, internal speed reduction gear between HP and LP turbine, gearbox from LP turbine to generator shaft). Presume "power at coupling" means shaft power at generator input after gearbox and flex coupling ?
 
In actual operations, there is no way of telling what the % moisture is at LP turbine exhaust once you are in the 2 phase region, since exit temp will remain constant at any given P for all moisture content values. It would be better to operate with some superheat (say some 5-10degF) at the exit so you know the last row of blades wont suffer droplet impingement - erosion. So in this case, you would raise the exit pressure a little and see if the exit temp corresponds to 5-10degF superheat. There is nothing worse than an unplanned shutdown stemming from high vibration caused by blade failure.
 
Hi all!

Thank you again for all the replies! Very appreciated!

stgrme,
thank you soo much for the file! It was great to have this for double check! Awsome!

georgeverghese,
Thanks again for your valuable insights! Good to have this practical and exerienced feedback!

Thank you all for the feedbacks! I was preaty cleared up about this subject!! It helped me a lot!

[bowleft]
 
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