Efficiency of the gas turbine

The total energy balance in a gas turbine can be summarized as

Total energy input = Compressor load + Generator power output + Flue gas energy loss + rotational losses

with the exception of the rotational losses all others can be calculated in a easy manner , what you need is the following information.

1. Calorific value of the fuel (kcal/kg)

2. Mass flow rate of the fuel (kg/s or t/hr) (if only the volume flow is available then you need to know the density of the fuel)

3. air flow into the turbine ( this is tricky most of the sites do not have a air flow meter , you have to get it from the characteristic graphs which GE provides , you can also get the value form the site acceptance test or the performance guarantee test done at commissioning)

if you have a HRSG , which you probably will have then you can calculate the efficiency of the HRSG and also the combined efficient of the total co generation power plant, for this you need the additional data

Note - i do not know the layout of your plant or its operating nature , i am assuming that the steam used for deaeration is got from the plant itself and CPH if present is inside the HRSG itself. you need to work out the details yourself , i am giving the calculations for a self sustaining plant. ie it takes only water at room temp and fuel and gives out power and steam. The plant axillary consumption which will be around 2-4% of the plant full load is neglected in the calculation. mainly because i do not have sufficient data.

The basic equations for the calculation are

For GT Efficiency = 860*MW output (MwHr) / (fuel flow(kg) * calorific value of fuel(in kcal/kg))this is the base formula where fuel flow is normally available in M3 which you have to convert to mass with the known density.

for HRSG efficiency (overall) = steam flow rate * enthalpy of steam / (HRSG inlet temp * 0.25 * air flow rate)

efficiency (heat exchanger) = steam flow rate * enthalpy of steam / ( (HRSG inlet temp - HRSG outlet temp) * 0.25 * air flow rate)

the combined efficiency of the co generation is given by

efficiency = ( (860*MW output (MwHr)) + steam flow rate * enthalpy of steam) / (fuel flow(kg) * calorific value of fuel(in kcal/kg))

sample calculation

I am now in a frame 5 site , so i am taking the daily production readings from here. the values at your site will be different

Naptha consumption = 192m3 Power generation = 379 Mwhr Average power generation / hr = 15.8 MW Calorific value of naptha = 11250 kcal/kg Naptha density = 0.7

the HRSG is a twin drum and produces two different steam varities one a VHP steam at 48Kg/cm2 and 435 deg and other MP steam at 18kg/cm2 at 245 deg

Enthalpy of VHP steam = 785 kcal/kgEnthalpy of IP steam = 692 kcal/kg VHP steam production = 805 tonnes average steam production / hr = 33.5 t/hr MP steam production = 99 tonnes average MP steam production /hr = 4.12 t/hr

HRSG inlet temperature = average GT exhaust temp = 490 deg HRSG outlet temperature = average stack temperature = 140 deg ambient temperature = 32 deg

air flow rate - the air flow for a frame 5 machine at site condition (32 deg ambient) is 408 tonnes at 85 deg IGV opening , as the machine was put in cogen cycle and the average IGV opening is 56 deg , from the chara graph the air flow is estimated as 364 tonnes.

so

GT efficiency = (860 * 379 *100) / ( 192 * 0.7 * 11250) = 21.57 %

HRSG efficiency (overall) = ( ( 33.5 * (785-30) ) + (4.12 * (692-30) ) ) / ( 490 * 0.25 * 364) ( here stack losses are taken into account) = 62.83 % HRSG efficiency (heat exchanger) = ( ( 33.5 * (785-30) ) + (4.12 * (692-30) ) ) / ( (490-140) * 0.25 * 364) ( here stack lossses not taken into account) = 88 %

overall co-generation efficiency = (860 * 379) + ( ( 805 * (785-30) ) + (99 * (692-30) ) ) / ( 192 * 0.7 * 11250) = 67 %

I kinda forgot to add this small bit in the calculation above

Energy balance in Gasturbine

inlet ambient air = 25 deg cdp = 6.8kg/cm2 ctd = 302 deg exhaust = 507 deg fuel input = 1.7 kg/s calorific value of naptha = 11250 kcal/kg density of naptha = 0.71 power = 16 MW air flow into the turbine = 360 t/hr specific heat capacity of air = 0.25 kcal/kg deg

input energy into the turbine = fuel input + air input fuel input = 1.7 * 3.6 * 11250 *(1000) kcal = 68850 Mcal air input = 360 * 0.25 * 302 = 27180 Mcal

total energy input to the turbine = 96030 Mcal power output from the generator = 16*860 = 13760 Mcal

flue gas losses = (360+1.6*3.6)*0.25*507 = 46390 Mcal compressor load and rotational losses = 96030 - (46390 + 13760) = 96030 - 60150 = 35880 Mcal

this equated in terms of power = 35880/860 = 41 MW

the compressor load does not change much with the loading of the machine , you will see that for a 20 MW gasturbine , the compressor load is about 40MW.

And CSA , are you sure about 30-35% efficiencies ?? to the best of my knowledge GT simple cycle efficiencies does not exceed 25 +/- 2 %.

Power energy equivalents 1 kilowatt (KW) = 1 kilo joule/ sec

1 kilowatt sec ( Kw s) = 1 kilo joule

converting sec to hours , as Kw Hr is the standard for electrical energy measurement 1 kilowatt hour (Kw Hr) = 3600 Kilo joule

now we know

1 calorie = 4.187 joules , this is the joules constant so 1 Kilo calorie = 4.187 Kilo joules

1 kilo joules = 0.23883 Kilo calories

so

1 Kilowatt hour (Kw Hr) = 3600 * 0.23883 Kilo calories = 859.80416 Kilo calories = 860 kilo calories ( this is a reasonable approximation)

this is the electrical energy and heat energy equivalence . This is how the 860 in the formula came from.

Efficiency of the gas turbine

efficiency of the turbine = energy equivalent of generator Generator power output / energy input into the turbine

energy equivalent of generator Generator power output = 860 * Kw-Hr

energy input into the turbine = Calorific value of the fuel (Kcal /kg) * fuel flow ( Kg/hr)

= 860 * Kw-hr / Calorific value of the fuel (Kcal /kg) * fuel flow ( Kg/hr)

now multiplying both the numerator and denominator by thousand = 860 * Kw-hr * 1000 / Calorific value of the fuel (Kcal /kg) * fuel flow ( Kg/hr) * 1000

this converts the Kw-Hr to Mw-Hr and kg/hr to t/hr = 860 * Mw-hr / calorific value of the fuel (Kcal/kg) * fuel flow ( t/hr)

so this is how the energy efficiency of the gas turbine is derived.

i am also deriving a formula for heat rate of the turbine to the efficiency

heat rate of the turbine is defined as the The ratio of fuel energy input as heat per unit of net work output. It is expressed mostly in Btu/Kwhr or in kj

/Kwhr . I am not a fan of Btu , but i am a ardent fan of SI units :) so i will derive the equation here in Si units.

Heat rate = Kj/ Kwhr

Efficiency = KwHr * 3600 / Kj

Efficiency = 3600 / (kj/kwhr)

efficiency = 3600 / Heat rate

as far as the rotational and compressor loads are concerned i have given the calculation in the next post. Please go through it.