Fuel Consumption Issue for Gas Turbine 9fa


Thread Starter

Mohamed nawar

This issue about gas turbine frame 9fa. I want to ask why fuel consumption at low loads is higher than base load. For example at base load 245 mw fuel flow rate is 14 kg/s and load 150 mw is 10 kg/s.

This is related to firing temp and it must be constant all the time or different with load?
As this is a 9FA ,it will be running a DLN 2.6 fuel system. This means at different stages, it will be using different fuel nozzles.

I think that you need to read up on DLN before trying to analyze flows.
Mohamed nawar,

Every GE-design heavy duty gas turbine control system I ever worked on came with a document (sometimes three documents) called the 'Control Specification.' That document detailed much of the operating parameters, expected and design, for the turbine and auxiliaries.

Section 05 of the document contained a section entitled 'Expected Fuel Characteristics.' It was a crude table of fuel flow-rates and expected FSR (Fuel Stroke Reference) values and expected loads for the fuel that was specified during the design and requisition phase of the turbine and auxiliaries. The fuel analysis for the fuel expected to be available on site was supplied by the purchaser to GE or one of its packagers and this was used to order fuel control valves and develop the table in the 'Expected Fuel Characteristics' section.

This included rows for firing FSR (the process of establishing flame); warm-up FSR (just after flame was established), FSNL (Full Speed-No Load--100% rated speed) FSR; 1/4 load FSR; 2/4 load FSR; 3;4 load FSR; and 4/4 load (Base Load) FSR. If there were higher loads purchased and available, they were also listed in the table.

Each row had the expected (calculated) FSR value, the fuel flow-rate (sometimes in various Engineering units), etc.

This information is invaluable in understanding the expected fuel flow-rates at various operating conditions.

One thing that is important to understand and remember: It requires a certain amount of fuel just to achieve and maintain rated speed (FSNL). Once the generator breaker closes and the unit is "loaded," the fuel flow-rate increases, usually at a relatively linear rate, to Base Load.

These numbers are just for exemplary purposes; I didn't pull them from any Control Specification, but they are very typical of most GE-design heavy duty gas turbines. It usually requires approximately 25% of rated fuel flow-rate just to maintain rated speed (FSNL). (The axial compressor requires a LOT of energy to keep it spinning at rated speed and moving air through the unit.) And it requires approximately 75% of rated fuel flow to make power from generator breaker closure (0 MW) to Base Load.

So, let's say the FSR at FSNL was 18%, and the FSR at Base Load was 72% (FSR is just as much a measure of fuel flow-rate as it is valve position--in fact, it's really about flow, not so much position; it just so happens that the configuration of valves that GE formerly used (non-DLN combustor-equipped machines) flow-rate was directly proportional to stroke). So that means that if we divide 72% by 4 we end up with 18%, which is one-quarter of rated fuel flow-rate--and which corresponds to FSNL (rated speed, 0 MW). Now, that leaves (72% - 18%) 54% of FSR to produce rated power (from FSNL to Base Load). So, 1/3 of rated load would occur at (18% + 18%) 36% FSR; and 2/3 of rated load would occur at approximately (18% + 18% + 18%) 54% FSR. And full load would occur at approximately (18% + 18% + 18% +18%) 72% FSR, which is 100% of rated fuel flow-rate.

So, the fuel flow-rate <i>at any load</i> includes the amount of fuel required to maintain FSNL PLUS the amount of fuel required to achieve that load. And, as can be seen from the example, the amount of load is proportional to the amount of fuel (over and above the amount of fuel required to maintain FSNL). In other words, load change is roughly linear with the change in fuel flow-rate--over and above the amount of fuel required to maintain FSNL.

Finally, because of the way the IGVs are operated on units with DLN combustors, and if IBH (Inlet Bleed Heat) is used (and it is on most GE-design heavy duty F-class gas turbines) the efficiency of the unit at part load (just the gas turbine) is lower than it would be if the unit didn't have DLN combustors, but just slightly so. As the unit approaches Base Load the efficiency gets better (as IBH is shut off, and the IGVs reach full open operating angle).

The takeaway from all of this is: First, you should find the Control Specification for your unit (they are specific to every turbine), and then go to the 'Expected Fuel Characteristics' section for expected flows and loads. Second, you need to remember that the fuel flow-rate at any load includes the amount of fuel required just to maintain rated speed (FSNL; 0 MW), and so when comparing fuel flow-rates at different loads one must subtract the FSNL fuel flow-rate from both loads to arrive at the true fuel flow difference between the two loads being compared. I think you will find that the fuel flow-rate between FSNL (0 MW) and Base Load is very linear (proportional to) with respect to the change in load, or the change in load is very linear (proportional to) with respect to the change in fuel flow-rate.

I hope this helps!

Mohamed nawar

Dear CSA,

Thank you for your brief description and details. but I ask if I use 14 kg/s at 245 mw, why at 150 mw I didn't consume half value of fuel compared to 245 mw for ex 7 kg /s?

This mean there is other controlled parameter which make more fuel consumed. I think it is firing temp as at low loads cpd decrease so firing temp decrease. so increase fuel fuel to increase firing temp to increase efficiency of combined cycle as it increased by increasing exhaust temp, which it is indication to firing temp and used to control firing temp. so I think we it consume more fuel at low loads to increase exhaust temp and increase efficiency as combined cycle. That's I think.

So I ask does firing temp is constant and must be reached all time? or it different from load to load and different from simple cycle and combined cycle?
I was pretty certain it was going to come to this.

If it takes 25% of rated fuel flow-rate <i>just to maintain FSNL</i>, then this amount of fuel must be subtracted from any calculation of fuel versus electrical power produced to compare apples to apples--that amount of fuel is common to both loads being compared.

As to your question about firing temperature, for simple cycle machines with conventional combustors the firing temperature varies with load at most all loads between 0 MW and Base Load. For units with conventional combustors being used in combined cycle mode (meaning that the IGVs are being used to maximize exhaust temperature), the firing temperature still varies with load. And, for units with DLN combustors using IBH (Inlet Bleed Heat) firing temperature still varies with load.

Since you are discussing working with a GE-design 9FA heavy duty gas turbine, it will most likely have DLN combustors, and so the turbine control system will be calculating firing temperature; the value is either TTRF (for older units) or TTRF1 for newer units. Monitor TTRF (or TTRF1) during loading from 0 MW to Base Load. You will see that it varies mostly with load, so as load increases TTRF (or TTRF1) will increase, fairly linearly.

Now, if you want to look at fuel flow-rate versus load, trend FQG (for gas fuel flow-rate) versus DW during loading (which begins at 0 MW after the unit has reached FSNL and is synchronized to the grid). So, the amount of fuel flowing at FSNL just prior to synchronization must be subtracted from any loads which are compared AFTER the generator breaker closes and the unit is producing electrical power.

In the example above, the maximum fuel flow-rate (at Base Load) was 72%. And the fuel flow-rate for FSNL was 18% (or 25% of the maximum fuel flow-rate, 72%). So, to compare fuel flow-rates at any load one must subtract the FSNL fuel flow-rate from BOTH fuel flow-rates at the two loads being compared to determine if the amount of fuel is proportional to the loads being produced.

To accurately compare how much load is being produced at any two (or more) loads, one MUST SUBTRACT the amount of fuel required to hold FSNL--that fuel isn't producing any load, but it's still required to maintain rated speed, which is critical to any AC power generation system.

So, you said it took 14 kg/s to produce 245 MW (100% of rated power), and you said it took 10 kg/s to produce (150 MW/245 MW) approximately 61% of rated power. Let's use the typical value of 25% of rated fuel flow to maintain FSNL, so 25% of 14 kg/s would be 3.5 kg/s. Now, subtracting 3.5 kg/s from 15 kg/s leaves 10.5 kg/s for 0 MW (0% load) to 245 MW (100% load). So, to determine approximately how much fuel would be required to produce 61% of rated load one would multiply 10.5 kg/s times .61 to arrive at approximately 6.405 kg/s, then add 3.5 kg/s (the approximate amount of fuel required to maintain FSNL) to arrive at 9.905 kg/s. That number is just about equal to the 10 kg/s you reported (and since you didn't report any decimal places, if we round 9.905 to the closest whole number, that makes exactly 10 kg/s!).

You can't compare the fuel flow-rates at two loads without subtracting the amount of fuel required to maintain rated speed (FSNL; 0 MW) from the fuel flow-rate at each load. Using the numbers you provided and from the paragraph above, the fuel flow-rate should increase by approximately (10.5 kg/s /10) 1.05 kg/s for every 10% increase in load (or, 24.5 MW).

At 0 MW, the fuel flow-rate would be (3.5 kg/s + 0 kg/s) 3.5 kg/s.

For 24.5 MW (10% rated load), the fuel flow-rate would be (3.5 kg/s + 1.05 kg/s) 4.55 kg/s.

For 25 % rated load (.25 * 245 MW=61.25 MW), the fuel flow-rate would be (3.5 kg/s + (2.5 * 1.05 kg/s)) 6.125 kg/s.

For 50% rated load, the fuel flow-rate would be (3.5 kg/s + (5 * 1.05 kg/s)) 8.75 kg/s.

For approximately 61% rated load (approximately 150 MW), the fuel flow-rate would be (3.5 kg/s + (6.1 * 1.05 kg/s) 9.905 kg/s (or 10 kg/s, since we're not reporting decimal places and we're rounding to the nearest whole number!).

For 75% of rated load (.75 * 245 MW=184 MW), the fuel flow-rate would be (3.5 kg/s + (7.5 * 1.05 kg/s)) 11.375 kg/s.

For 100% of rated load (245 MW), the fuel flow-rate would be (3.5 kg/s + (10 * 1.05 kg/s) 14 kg/s.

As a final check, let's calculate how much fuel flow would be required to make rated power by adding the fuel 25% fuel flow-rate and the 75% fuel flow-rate. (2.5 * 1.05 kg/s) + (7.5 * 1.05 kg/s) = (2.625 kg/s + 7.875 kg/s) = 10.5 kg/s PLUS 3.5 kg/s (the amount of fuel required to maintain FSNL) = 14 kg/s. And we previously determined that using a typical value of approximately 25% of rated fuel flow is required to maintain rated speed that would mean that (14 kg/s - (.25 * 14 kg/s)) = 10.5 kg/s, which is equal to the amount of fuel required to produce 100% of rated power from 0 MW (rated speed) to Base Load (245 MW). Add 3.5 kg/s (the amount of fuel required to maintain 0 MW (rated speed)) and you get 14 kg/s. And checking it with the one data point you provided, 150 MW (approximately 61% of rated load), we can see that all the numbers just about add up perfectly. (Whew!)

I hope this clears at least the maths part of this up. (Where's Phil when I'm doing maths?!?!?!) Load can only be compared to the amount of fuel being used to produce load; the amount used to maintain FSNL (rated speed; 0 MW) isn't valid in this comparison. Yes; you factor total fuel flow in when determining efficiency or heat rate, but we're just trying to determine if the fuel flow-rate is proportional to load, and we determined--first using an example, then using the whole numbers you provided--that the fuel flow-rates you provided were pretty proportional to the fuel flow-rate that is used to produce load (when the amount of fuel used to maintain rated speed is subtracted from each load--because it's common to each load). If you want to calculate how much fuel is required to produce a certain load, you need to know how much fuel is required just to produce electricity then add the amount of fuel required to maintain rated speed.

And that's the end of my maths for this year!
Thank you very much for your effort for explaining and make more calculations this year as it seem it is boring for you

Thank you to all in this site
Mohamed nawar,

The maths aren't boring. I don't much like maths as an answer to a question because it sometimes muddles the answer and the question.

But mostly, I prefer when people do the maths themselves and ask for clarification if needed. Personally, when I work through an answer (using maths if necessary) and derive the answer myself I retain the information much better and for longer than if someone just gives me the answer. I find many--but not all--people are like this; you seem not to be one of the many, but I didn't know that before.

So, please excuse my weak attempt at humor and frustration with maths. Hopefully you found the information helpful. (The reference to Phil was to a former contributor who chided me for not using maths in my answers to electrical questions.)