How to calculate overall thermal efficiency of combined cycle power plants – a sample CCGT presented
How to calculate overall thermal efficiency of combined cycle power plants – a sample CCGT presented
Calculating or predicting the overall performance of a combined cycle power plant, specifically a combined cycle gas turbine (CCGT) power plant is sometimes difficult for most design engineers. Your favorite energy technology expert again comes to the rescue – Engineer Marcial T. Ocampo – has derived the following equation to guide the design engineer and project finance modeler or business development engineer in predicting the overall thermal efficiency of the combined cycle.
Here’s the step by step derivation:
Let Total Energy Input = Total Energy Output = 100%.
Then let’s assume that the proportion of the energy input going to the gas turbine (GT) and to the waste heat recovery steam generator (SG) is around:
GT = 1/3 = 33.33%
SG = 1 – 1/3 = 2/3 = 66.67%
The energy input to the steam generator is then split into 85% (up to 90%) being recovered as steam energy in the boiler while the balance of 15% (or 10%) is lost to the atmosphere (smoke stack, radiation and convection loss).
The steam energy is then captured in the steam turbine with an efficiency of 37% and the balance (63%) is lost to the cooling water in the condenser.
The mechanical energy captured in the GT and ST shafts are then coupled in a clutch (100% efficiency) to drive finally the electric generator (EG) with mechanical to electrical conversion efficiency of 98% (balance of 2% lost to heating in the generator windings and generator cooling system).
The final formula is thus: (with correction for GT efficiency per suggestion of George – see his comments)
OE = (GT x GTE + SG x BE x STE) x ME x GE
where OE = overall energy efficiency, % of fuel energy input (as GHV or LHV)
GT = 1/3 = 33.33% (assumption)
GTE = gas turbine efficiency = say 90% – 95% (the reader please comment or advice)
= function of heat loss from GT due to conduction, convection and radiation from GT casing, any heat loss due to friction resulting in higher GT exhaust is captured, however, by the heat recovery steam generator
SG = 1 – GT = 1 – 1/3 = 2/3 = 66.67% (balance)
BE = waste heat recovery boiler efficiency = 85% (up to 90%)
= function of exit flue gas temperature (energy lost), gas turbine exhaust temperature (energy input), boiler design, flue gas composition (fuel, excess air)
STE = steam turbine efficiency = 37% (up to 40%)
= function of steam inlet temperature, steam exhaust pressure, condenser vacuum pressure, cooling system, steam turbine design, steam quality
ME = mechanical drive shaft and clutch efficiency = 100%
= function of drive shaft design, clutch system, bearing lubrication, mechanical design, windage losses (air drag)
GE = electric generator efficiency = 98%
= function of generator design, voltage, windage losses (air drag), type of gas cooling (hydrogen, air), bearing lubrication
Putting all together now, the predicted overall thermal efficiency of the combined cycle power plant is:
OE = (33.33 x 95% + 66.67% x 85% x 37%) x 100% x 98% = 51.58%
For instance, if we raise the gas turbine efficiency from 95% to 98%, the resulting overall efficiency would be raised to
OE = (33.33 x 98% + 66.67% x 85% x 37%) x 100% x 98% = 52.56%
For instance, if we raise the boiler efficiency from 85% to 90%, the resulting overall efficiency of the combined cycle power plant is raised to:
OE = (33.33 x 98% + 66.67% x 90% x 37%) x 100% x 98% = 53.77%
Further raising the steam turbine efficiency to 40% will result in a much higher efficiency of:
OE = (33.33 x 98% + 66.67% X 90% X 40%) X 100% X 98% = 55.53%
I guess this provides an upper limit of what a CCGT could deliver, around 56%.
The only way to go higher than this is to improve further the gas turbine efficiency beyond 98%, boiler efficiency beyond 90%, raising the steam turbine efficiency beyond 40% and optimizing the proportion of energy output thru the gas turbine (currently 1/3) and the steam generator (balance of 2/3).
The author has developed an state of the art project finance model for a 25-year economic life CCGT and may be requested thru this link:
http://energytechnologyexpert.com/technology-data-resource/large-scale-project-finance-models/
The model has the following capability:
1) Given the all-in capital cost (EPC, installation and erection, taxes and duties, project development, regulatory costs, working capital, interest during construction), O&M costs (variable and fixed O&M, recurring regulatory costs, property taxes, ROW and land lease, property insurance, business interruption insurance, etc) and electricity tariff (industries, distribution utilities, national grid, wholesale electricity spot market) ==> it determines the maximum price of the natural gas fuel needed to meet the 15% p.a. DCF IRR for equity investment.
2) Alternatively, if the capital cost, O&M costs, electricity tariff and fuel cost (natural gas, gas oil, bunker oil) are fixed, the model will calculate the net present value and project IRR (return of investment or ROI from the project cash flow), NPV and equity IRR (return on equity or ROE from the equity cash flow) or the more stringent financial analysis tool of bankers today –> the NPV and dividends IRR (return on dividends from dividends cash flow).
Other variations of the model objective include determining the maximum capital cost given the O&M cost, tariff and fuel cost to meet the 15% equity IRR hurdle rate for making equity investments.
The model is currently capable of analyzing several CCGT machines / models from various manufacturers for a number of plant locations with specific type of cooling system (sea water once thru, lake water once thru, river water or deep well cooling tower, and dry cooling with radiator).
By using case switches, the model will go over each pair of sensitivity (machine and plant location/type of cooling) and calculate the fuel cost needed to meet 15% equity IRR. It uses goal seek to set the equity cash flow NPV to zero for each of sensitivity pair.
I would like to invite you and your company to continue supporting this blog thru the DONATE button. You may order my power generation technology articles and project finance models thru the ENERGY DATA page. Thanks!
Marcial T. Ocampo
(Friendly note: All content written by Engr. Marcial T. Ocampo are copyrighted and may not be redistributed in any way or form.)
28 Responses to “How to calculate overall thermal efficiency of combined cycle power plants – a sample CCGT presented”
Leave a Reply
September 23rd, 2009 at 5:57 am
It seems as though you also need to account for the efficiency of the gas turbine. The calculation above does not discount the 1/3 of the energy that goes to the gas turbine…it appears to assume that all of that energy is transferred to the gas turbine shaft. The same was not done for the boiler or steam turbine…the efficiency of each of these two items is included in the calculation.
September 23rd, 2009 at 8:33 pm
Hi George,
You’re absolutely right. I made allowance for inefficiency in the heat recovery steam boiler (85-90%) of the 2/3 heat input while the 1/3 to the gas turbine was only corrected together with the the shaft work for the clutch efficiency (assumed 100%) and generator efficiency (98%). Perhaps you may suggest a gas turbine efficiency level in the order of over 95-98%? I guess our guys from Siemens, Westinghouse, GE, Alstom, etc. could comment on this. Any heat loss at the GT due to friction resulting in higher GT exhaust temperature will, however, be captured by the heat recovery steam boiler. Thus the only significant heat loss at the GT would be due to conduction to surrounding GT supports, convection to surrounding cold air and radiation to the cold sky which could be minimized by installing sufficient insulation and light colored aluminum covering to minimize all these three losses. Thank you so much for your clarification. I stand corrected!
Cheers,
Marcial
September 26th, 2009 at 2:10 pm
Attn Admin,
the efficiency calculatoions based on assemptions cannot lead us to reality.It should be measurable and should calkculated on the basis of input energy and recovered energy in the process.
Efficiency———————=input- losses
— do—in percentage———= Output(KWX860Kcals) X 100
——————
(input(fuel Wt X kcals)
The input should be considered as weight x cal val of fuel fired & out put wil be kwx860(net power out put)
September 27th, 2009 at 12:41 pm
Thanks Don,
My simplified method for calculating the overall efficiency of a combined cycle gas turbine (CCGT) is indeed an approximation:
OE = (GT x GTE + ST x BLE x STE) x ME x GE
where OE = overall efficiency
GT = fraction of energy output to GT
GTE = gas turbine efficiency (to account for heat losses from GT engine)
ST= 1 – GT = fraction of energy output to ST
BLE = heat recovery steam generator (boiler) efficiency (to account for heat losses from boiler)
STE = steam turbine efficiency (to account for heat loss from steam turbine)
ME = mechanical efficiency (to account for shaft work losses)
GE = generator efficiency (to account for generator losses)
Alternatively, a more detailed and rigorous determination of overall efficiency will partake of the form:
Total Input = Total Output
where total energy input = Mf x GHVf + enthalpy of all incoming streams relative to 25 deg Celsius (air, fuel, water) + Pinput
Mf = mass flow rate of fuel, kg/s
GHVf = gross heating value of fuel, kJ/kg
enthalpy of any stream = Ms x CPs x (Ts – 25)
Ms = mass flow rate of any stream (air, fuel, water)
CPs = specific heat of any stream, kJ/(kg – deg Celsius)
Ts = inlet temperature of any stream (fuel, air, cooling water), deg Celsius
Pinput = power input (parasitic load), kW
total energy output = Pgross + Total Losses
Pgross = gross power output of the generator, kW
Total Losses = Surface Losses + Friction Losses + Sensible Heat Losses + Latent Heat Losses + Cooling Water Losses
Surface Losses = conduction + convection + radiation
Friction Losses = bearings + shafts + clutch losses + windage losses
Sensible Heat Losses = flue gas flowrate x specific heat of flue gas x (Tfg – 25)
flue gas flow rate = CO2 + H2O + SO2 + NO2 + CO + O2 + N2 + moisture + ash, kg/s
specific heat of flue gas component, kJ / (kg – deg Celsius)
Tfg = flue gas exit temperature, deg Celsius
Latent Heat Losses = heat of evaporation of moisture and water formed from combustion of hydrogen, kJ/s
= mass (moisture in fuel + moisture in combustion air + water from combustion of hydrogen) x latent heat of evaporation of water
mass = kg/s
Latent heat = kJ/kg
Cooling Water Losses = Mw x CPw x (Tw – 25)
Mw = cooling water flow rate, kg/s
CPw = specific heat of water, kJ/(kg – deg Celsius)
Tw = exit temperature of cooling water
Simplifying, we get the losses by difference (heat loss method):
Total Losses = Total Input – Pgross = total energy input – gross electrical output at generator terminals
The percentage total losses is thus
% Total Losses = (Total Losses / Total Input) x 100%
The overall gross energy efficiency is therefore
% overall gross energy efficiency = 100% – % Total Losses
The above derivation is based on the gross heating value (gross calorific value) of the fuel with water in the flue gas in liquid form, hence, the need for the latent heat of evaporation correction. The overall efficiency is called overall gross efficiency (based on GHV or GCV of fuel).
Alternatively, if the lower heating value (net calorific value) of the fuel is used, then water would be in vapor form in the flue gas.
There would be no need, however, for the latent heat of evaporation correction. In this case, the overall efficiency is called overall net efficiency (based on LHV or NCV of fuel).
Lastly, the power is expressed as net exportable power (Pgross – Pinput). Hence, the losses are adjusted as follows:
Total Losses (net) = Total Input – (Pgross – Pinput)
The percentage total losses (net) is thus
% Total Losses (net) = (Total Losses (net) / Total Input) x 100%
The overall net energy efficiency is therefore
% overall net energy efficiency = 100% – % Total Losses (net)
I hope Don this clarification will also help our readers.
Cheers and Thanks for bringing this out.
Marcial
October 8th, 2009 at 11:02 am
Considering that much of the energy input into a gas turbine must be consumed by the compressor portion of the gas turbine, overall gas turbine efficiencies are much lower than 90%…please see info below regards expected gas turbine efficiency:
The Frame 5-2E is the newest member of GE’s Frame 5 gas turbine fleet, which has compiled more than 16 million hours of service worldwide. The two-shaft machine combines the latest heavy duty gas turbine technology with proven design concepts to achieve 36% efficiency in mechanical drive operation and 35% in generator drive operation.
The waste heat from the gas turbine exhaust is a souce of energy for the overall plant process, but it does not conteibute to the shaft output power of the gas turbine per se.
Best regards,
George
December 27th, 2009 at 6:58 am
[...] http://energytechnologyexpert.com/power-generation/fossil-fuels-and-energy/gas-turbines/combined-cyc... [...]
June 4th, 2010 at 9:02 am
Hi George,
Wow, could not disagree more. I have been analyzing our GT setup here. Having difficulties also w/ the efficiencies,losses, etc. But in my calculation done, I have gone the long method done, but still verifies that your assumptions do not deviate much on actual values.
Cheers!
June 4th, 2010 at 9:17 pm
Hi George!
Thanks again for your nice comment. My method will actually allow you to estimate the ultimate efficiency of each technology given the limitations of each component efficiencies.
Cheers,
Marcial
June 7th, 2010 at 11:25 am
Hi Marcial/George,
Your comments please. Ours is a MF111B GT. Confirming around 36% for GT and 35% @ the generator. I am a bit unsure if 6% losses (not converted to heat energy to flue gas) due to mechnical inefficiencies of the GT is ok to assume. And how about convetive/ radiation losses to the atmosphere, how much do those factors contribute?
Thanks,
Mon
July 25th, 2010 at 8:49 am
Hi. What is the efficiency cost that results im increasing marginal cost of running the each system. I guess there is a % parameter that suggests this ratio. When the generators become older and older, marginal runninng cost of that generator goes up that is given by %. Do you have any idea about it or any literature u have got? Thanx for the information you are sharing with us. I find this site amazing.
October 6th, 2011 at 11:23 pm
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October 28th, 2011 at 12:36 pm
Hi George,
Your articles were really helpful. I want to know how to calculate water and fuel requirement for a Combine cycle power plant. If possible water requirement for both air cooled and water cooled condensers. I hope you revert back soon. Will be waiting for your reply.
Thanks and regards
Anil Kumar
India
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