Improving the Power factor of a steam turbine alternator

[dubstep]

Hi Guys,

A certain sugar factory has a 10MW @0.8PF condensing turbine alternator installed. upon a recent site visit i have concluded the TA is running at an average of 0.62 PF and no one seems to have an issue with this.

Using simple calculations i have determined that if i improve the PF from 0.62 to 0.8 i will reduce the current significantly for a constant MW load.

Now with the lower current to produce the same power, i can state that the I^2R copper losses will be reduced and the excitation will be reduced, resulting in the power in power out conversion effeciency of the alternator been dramatically improved.

For less power in ,i will be getting the same power out. Now power is equal to the product of P(kW)= wT . Therefore i have conclude that there will be a steam saving on the turbine inlet. This can be calcualted as for an average a STC (specific steam combustion) = 5T of steam per 1MW. for a condensing steam turbine.

Increasing the operating power factor of a Turbine alternator not only saves electrical installtion costs due to the lower current, but also saves fuel, that can be calculated back to tonnes of coal needed for the boiler.

The machine is rated at 10MW @0.8PF, my question is, what would the result be if i were to raise the PF to 0.9. would there be any negative effects.

what are your guys thoughts on this issue/ am i missing some technical considerations regarding apparant and reactive power?

Thanks

William

 

[waross]

The usual method to raise power factor is to add capacitors. This may lead to serious over-excitation in the event that the load drops unexpectedly. I would suggest improving the PF of individual large motors.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[dubstep]

Thanks for the reply.

Am i rite in stating, a Generator (turbine alternator in my case) supplies reactive power (VAR's) and real power (MW) into the system. A motor is an inductive load that is why we compensate by using capacitors which is a reactive load and brings the apparant power closer to unity. ( power triangle)

So the main question is, what will the result be if i increase the PF above the rated PF on a Turbine alternator.

your feedback is appreciated.

 

[davidbeach]

Is your generator supplying an islander load by itself, or is it running in parallel with a larger utility system? In the former case the load power factor is what it is and has to supplied by the generator. If the latter case, change your excitation to change the power factor.

 

[rcw retired EE]

Your generator is rated 12.5 MVA @ 0.8 pf. MW=0.8 x 12.5 = 10 MW.

O.8 power factor is the lowest power factor that the generator can operate at when generating 10 MW. Lower power factor = higher MVA. At .65 power factor the MW load needs to be around 8 MW or less to prevent overheating the generator rotor. The actual value depends on the generator's capability curve. See attachment for an example.

As David said, if you are islanded, you can't change the power factor; just check the capability curve to see if your operating point is within the capability lines.

If the plant is tied to a utility, the generator may be supplying MVAR's to the plant loads to improve the power factor at the utility meter to save extra charges for low power factor. Getting MVAr's from a generator is a lot cheaper than installing power factor correction capacitors. Reducing teh generator MVAr output could increase costs.

To raise the power factor closer to 1.0 you need to reduce excitation, but that will also reduce the output voltage which could cause problems with the loads.

As far as saving kW by running at a higher power factor, you will save some I2R losses but they will be in the order of 0.5%- 1.0% of machine rating or about 50-100 kW.

Tell us more about your system and we can provide more suggestions.

 

[Hoxton]

Why not take a step back and ask "where did 0.8pf come from?"

This does not seem to be defined by standards, only by specifiers.

Taking davidbeach's comment: are you operating in island or utilty parallel?

The decision on power factor comes from island mode operation.

Question for when you specify a generator for island mode operation: What power factor should I use to make sure that my generator is never overloaded (in VA not W)? The object is to supply kW for the loads.

Experience shows that most power systems operate at say 0.85pf, so if I size my generator at 0.8pf (ie 20% more kVA than kW) then my generator will not have to generate more kVA (current) than needed to support the load.

The benefit is a generator that never overloads and its temperature rises stay within limits.

The other good news is that because the generator current is less than design, then the generator losses are lower than design, and your primer mover (steam turbine) has to produce less kW to produce the same kW for the load. So fewer buckets of coal per kW!

In island mode, to improve the power factor, you have to use capacitors (or buy new motors with better power factors)....

Improving (raising) the power factor reduces the current and the I2R losses -always a goood idea!

If you are in mains parallel, then changing the generator excitation (field) current will change the generator power factor. Remember that someone else has to supply the total VAr for the system!

 

[dubstep]

Hi Guys,

I would like to thank you for the responses. RCWILSON and HOXTON , the TA runs in a island mode supplying the plant.

The TA generates at 11kV. and the plant has a 11kV cap bank rated at 3MVAR's. The mill complains if they swicth the cap bank on its starts to trip loads on on the LV side.

This is the reason they chose to run at the low power factor. I am busy with moddeling the plant in a power analyses programme. (powermaster)

Thanks alot for your valuable info, i know understand things clearer. One of my suggestions depending on the power model, might be to change the cap bank to 3 smaller units that can be autimatically stepped in and out as needed.

So we agree at the lower power factor you will need higher current to attain the same output power? this higher current
can cause over heating etc inside the alternator. this will force the effeciency of the machine down, which will inturn result in more mechanical power to get the same electrical power. (ie:more fuel ,coal or bagasse,)

i will try and locate the generators capability curve to shed some light on the current operation at 0.62 PF. i am interested to know what the maximun power output the generated is rated at for that specific PF.

Thanks alot for the info.

 

[rasevskii]

As davidbeach and rcwilson have said, in island mode you cannot change the generator PF, it is determined b the plant loads. Adding cap banks will help, theroretically. If you change the excitation , that is, the voltage setting on the AVR, you will be immediately in big trouble with voltage levels on the plant system. In fact, you could easily black out the plant. This would not be popular with the owners...

Adding caps in smaller steps, as you say, would seem a better idea.

rasevskii