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Steam Turbine performance with variation in inlet steam pressure 2

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icecool

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Sep 6, 2010
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Dear All,

I am estimating a future problem while still in erection phase for 60 MW steam turbine. I may need your help to estimate a solution.

Due to some reason my process heater is not able to generate steam at the desired pressure (other parameters are remain same). Therefore I want to know the effect of reduce inlet pressure in the HP Steam turbine output. Th steam parameters are mention below

Steam Turbine is Superheated, Condensing type with Reheater (HP-LP combination)

1. Normal condition
Manufacturer : GE
HP Steam inlet pressure : 98 bar
HP Steam inlet temperature : 365 Degree C
HP Steam Turbine outlet pressure 23 bar
HP Steam Turbine outlet Temperature 220 Degree C

There is 1 extraction of steam from the turbine to HP Heater

2 Revised condition

HP Steam inlet pressure : 75 bar
HP Steam inlet temperature : 365 Degree C

Can somebody suggest the role of pressure in steam turbine. Will steam turbine able to work properly even at low inlet pressure?
 
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No. Performance and efficiency and reliability (against transients or load changes) will decrease significantly. can you get some useable energy out? Maybe.
 
For sure your turbine will produce less power. The best way to look to TS diagram. Less pressure cause less enthalpy of inlet flow. Also low inlet pressure leads to lower mass flow rate (if you can't preserve it with some control means). As power is in general the production of heat drop (difference of inlet and outlet enthalpies) an flow rate your power will be deteriorated .
To make fast estimations you may need some simple steam cycle (heat balance) calculation tool.Check Google.


Good luck
 
hermit47,Turbine will produce less power. are you sure?
because my temperature remains same.Therefore actually the inlet enthalpy will increase little bit. Theoretically the power output might be same?
But what about the kinetic energy turbine needed for driving force. As I understand the kinetic energy will increase with decrease in inlet pressure.
my question is remain same..what will be the effect on turbine if there is reduction in inlet pressure if rest of the parameters including mass flow rate and temperature will same.than what is the role of inlet pressure? will it effect the HP turbine exhaust pressure. as HP Turbine exhaust pressure is not driven by any force (like in LP turbine).
Plz suggest.
 
Sorry, I was a bit wrong about enthalpy. Really it slightly increased when pressure drops to 75 bar.
The real problem is Mass Flow Rate (MFR).
I am not clear are you about new turbine design or drop pressure on existing turbine.
For 1st case you can design new turbine with 75 bar inlet pressure that provides same 60 MW output power for sure.
If you are about existing turbine you can estimate an effect of pressure drop using “turbine constant”
Ct= M*sqrt(Tk/(Pin^2-Pout^2)).
M- mass flow rate in kg/sec
Pin – inlet pressure of component in bar
Pout – outlet pressure of component in bar
Tk temperature on component inlet in K

Very rough estimation for Pin= 98 bar. Pout=23 bar, Tin= 365C, eff=0.9 gives Ct=58.647 and MFR= =221.18 kg/s
When we’d use this Ct (i.e same turbine ) wit Pin=75bar we’ll get MFR = 165.7 kg/s and N= 39.5 MW
This figures were obtained in heat balance calculation tool so the effects on kinematic, losses, etc. were not taken into account. To do this it is necessary to know flow path and cascades geometry of your turbine
 
At the reduced pressure, you are much closer to the Tsat at the final condenser, therefore you will get much higher water percents in the final stages -> Higher water erosion over time.
 
hermit47,

Could you please tell me more about the Turbine constant. I am not aware of it. Also how did you come to a number 221.8 kg/s and 39.5 MW? as I mention it is a HP-LP combination turbine. The total mass flow rate at the inlet of the turbine is 70 kg/s.

racookpe1978
Could you please elaborate how the last stage of LP turbine is effected by moisture?
 
2 racookpe1978

It is funny but process simulation shows that 75-bar-cycle (green line) lays completely outside wet area so it looks more safe from wet erosion point.

Process.JPG


2 icecool

Turbine constant is derivative from "Stodola Cone Law". It is simple correlation between inlet temperature, pressure drop and mass flow rate.It well suited for 1-stage but can be also applied for raw estimation of multistage turbine.

MFR of 221 kg/s I get in assumption that the section with pressure drop 98-23 bar produced 60 MW of power.

Taking in account MFR of 70 kg/s the power of section at 98 bar will be ~18.99 MW and turbine constant 18.56 . Using this constant value for 75 bar we'll get MFR 52.44 kg/s and power 12.49 MW. See cycles pictures below.

98-75.JPG
 
Thanks hermit47

Few more questions:

1. I know the turbine upstream pressure. But will the HP Turbine downstream pressure will remain same ( I mean the HPT exhaust pressure is 23 bar when the HPT inlet pressure was 98 bar. Now what happens when the inlet pressure goes down to 75 bar? I know the LPT exhaust pressure is defined by the condenser but what about the HPT)

2. Is the 52.45 kg/s is the max flow at 75 bar condition? if not then how to calculate the choked flow?

Probably I am going more theoretical. But, I am thinking from every possible way.



 
To define new pressure distribution it is nacessary to run meanline analysis of whole flow path at least. As I wrote early to do it it is necessary to know flow path and cascades geometry.
Other way is to try to remodel your turbine with preliminary design tool.
What you know about it? Is HP section governing (Curtis) stage? Are there reheater after HP section? Are there any other reheaters? How much and at what pressure steam is extracted to deaerator and feedwater heaters?

Do you know model number of this GE turbine?
 
The largest effect of reduced pressure at the turbine inlet will be a reduction in the flow capacity of the turbine. This reduction is mainly due to the change in specific volume of the inlet steam. The specific volume of steam at 98 bar is 0.0244 m^3/kg; at 75 bar, it is 0.0339 m^3/kg. (I presume that all pressures are absolute pressures.) Flow is proportional to the square root of (p/v), where 'p' is the pressure and 'v' is specific volume at the turbine inlet conditions. The flow capacity of the turbine for the reduced inlet pressure may be estimated using the ratio of the square roots for each inlet condition. (The ratio of the square roots is 0.742.) If the maximum capacity at 98 bar is 70.0 kg/s, then the flow capacity at 75 bar will be approximately 51.9 kg/s.

If the guaranteed capacity at 98 bar is 70.0 kg/s, the turbine manufacturer has probably included some flow margin in the design (typically, 2% to 3%). In any event, the maximum capacity at the 75 bar will be the maximum capacity at 98 bar multiplied by 0.742.

I suspect that the generator electrical output at the lower inlet pressure will also be lower. As a first approximation, I would assume that the efficiency of the HP turbine remains constant. If the enthalpy drop through the HP turbine remains constant, the electrical output will be less due to the reduction in flow capacity.

It is difficult to predict the effects on all the equipment in the cycle (boiler including reheater, feedwater heater, condenser, generator, etc.) without the aid of a performance program like GateCycle or PEPSE.

Best of luck!
 
as stgme indicates, a constant speed turbine is a volumetric machine, and the volumetric flowrate downstream of the governing stage valves into the first row of turbine blades is a fixed value at all turbine loads. If you drop the full load inlet pressure from 98 bar to 75 bar then the mass flow of steam thru the first row of blades will drop by almost 25% , hence the power output of the STG will drop by about 25%. So there seems to be several alternate conclusions to this exercise:
-accept the 25% reduction in STG load and mass flowrate, or
- buy a larger steam turbine designed for the new design conditions, or
-revise the upstream process to deliver the originally specified 98 bar steam pressure (and find out how this mistake was not found earlier), or
-add a gearbox and run the STG at another RPM- not a likely scenario


"Whom the gods would destroy, they first make mad "
 
Hermit47,

The HP turbine exhaust is connected to moisture separator which is connected to reheater. After reheating the steam enters inside the LP Turbine. HP Heater 1 extraction is at 36 bar, 256 degree C, and flow is 5.2 kg/s. The Deaerator and other HP heater take extraction from the HPT exhaust.
Choking depends on upstream pressure only if the ratio with downstream is approx more than 2. Thats why I was thinking of any possibility of increasing mass flow rate. But choking might not be practical in HP turbine?

Stgrme, davefitz ,

Thanks for your input. Your calculation of power output and calculation with equation suggested by Hermit47 matches very closely.

I understand there is no possibility other than increasing the pressure for getting more power out.

Could you suggest what dictates the Exhaust pressure of a turbine. Also is there any way of calculating it theoretically.

Thanks again for all your suggestions.


 
2 Icecool

I U'd give me temp after reheater and data about other extractio I can remodel thi turbine. Are theer any references to its original specification?
 
The exhaust pressure of a condensing turbine is determined by the ACC surface area A, convective heat transfer coefficient ( fins to air U) , and ambient air temperature Ta. The known heat loss Q to be convected to the air will depend on the temperature difference between the exhaust steam's saturation temperature and the ambient air temperature. If one has under-sized the ACC then the saturation temperature must increase in order to transfer the same amount of heat to ambient, and the saturation pressure ( eg backpressure) is a known function of the saturation temperature.

If we consider the ACC as a "compact heat exchanger" then a simplified view would be that e= effectiveness = 1-exp(-NTU), which works for the special case of one fluid ( condensing steam) apparently having an infinite Cp. NTU= U*A/(W*Cp), air,, where W= mass flow of air over the fins and Cp= air heat capacity.

e= effectiveness = (Tao-Tai)/(Tss-Tai)
Tai= ambient air temp
Tao= air outlet temp, determined via Q= W*Cp,a*(Tao-Tai)
Tss= steam saturation temperature.

These equations are sufficient to determine the back pressure ( ie saturation pressure) if one knows Wa from the fan curves. If the curves are no available, one can estimate the value of e from the vendor supplied performance predictions.

"Whom the gods would destroy, they first make mad "
 
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