# Steam injected gas turbine, Calculation Exercise

Steam injected gas turbine, Calculation Exercise

Steam injected gas turbine, Calculation Exercise

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<strong>Steam</strong> injection <strong>gas</strong> <strong>turbine</strong> – <strong>Exercise</strong> 4Created by CE<strong>Steam</strong> injection <strong>gas</strong> <strong>turbine</strong>, <strong>Exercise</strong> 4A small <strong>gas</strong> <strong>turbine</strong> unit working with light oil as fuel has <strong>turbine</strong>, generator and compressor ona single shaft. Following data is given:Air mass flow40 kg/sPressure ratio compressor 10Isentropic efficiency compressor 0.85Ambient temperature 20ºCAmbient pressure1 barIsentropic exponent of air during compression κ = 1.39Turbine inlet temperature 800ºCIsentropic efficiency <strong>turbine</strong> 0.89Pressure loss in combustor 2%Mechanical efficiency 99%Generator efficiency 98%Exhaust temperature in <strong>turbine</strong> outlet 380ºCThe <strong>gas</strong> <strong>turbine</strong> is modified with steam injection in the combustion chamber to recover hotexhaust <strong>gas</strong> from the <strong>turbine</strong> and to increase the power output. The steam is produced in acommon single-pressure heat recovery steam generator (HRSG), consisting of economizer,evaporator and superheater. All steam produced in the HRSG is <strong>injected</strong>, i.e. no process steamis delivered<strong>Steam</strong>-<strong>injected</strong> modification, Data:<strong>Steam</strong> superheat temperature 320ºC<strong>Steam</strong> pressure10 barFeed water temperature25ºC.Gas temperature measured in HRSG outlet (stack) 245ºCTasks:a) Estimate the steam injection mass flow (kg/s) assuming that the <strong>gas</strong>content without steam is x ass = 0.20b) Assuming the steam flow as calculated in a) to be the correct value; howlarge is the difference (in %) between the calculated value of the <strong>gas</strong>content, x calc , (to be calculated here) and the assumed value of the <strong>gas</strong>content above (x ass )?c) The electrical power output (kW) when steam is <strong>injected</strong> (based on x in b))and the electrical efficiency (%)Follow this model for steam-<strong>injected</strong> <strong>gas</strong> <strong>turbine</strong> calculations:• The compressor remains unaffected by the steam injection• The fuel control system maintains the <strong>turbine</strong> inlet temperature at a constant value(which means that the fuel flow must increase a little to heat the <strong>injected</strong> steam up tothe <strong>turbine</strong> inlet temperature)• Assume that the <strong>gas</strong> <strong>turbine</strong> outlet pressure and temperature remain at the samevalues as without the HRSG.1

<strong>Steam</strong> injection <strong>gas</strong> <strong>turbine</strong> – <strong>Exercise</strong> 4SOLUTIONCreated by CEa) <strong>Steam</strong> injection mass flowThe steam injection mass flow is found through a heat balance over the HRSGHeat Recovery <strong>Steam</strong> Generator54<strong>Steam</strong> +Gas(x = 0.20)245 deg C,1 bar<strong>Steam</strong> +Gas(x = 0.20)380 deg C,1 barFeedwater:10 bar,25 deg C<strong>Steam</strong>:10 bar,320 degCHeat balance:m st *(h sh – h fw ) = m tot *(h 4 – h 5 ) [1]The term on the right hand side can be further divided:m tot *(h 4 – h 5 ) = m <strong>gas</strong> *(h 4,<strong>gas</strong> – h 5,<strong>gas</strong> ) + m st *(h 4st – h 5st ) [2]where "<strong>gas</strong>" denotes air and combustion <strong>gas</strong>es without the steam.Combining (2) and (1) gives:m st *(h sh – h fw ) = = m <strong>gas</strong> *(h 4,<strong>gas</strong> – h 5,<strong>gas</strong> ) + m st *(h 4st – h 5st ) [3]Solving for m st :h4,<strong>gas</strong>− h5,<strong>gas</strong>⋅( hsh,320− hfw) − ( h4ST− h5,&ST= m&<strong>gas</strong>[4]ST)mGas flow:x = 0.20 andβ= (1 + 14.52) ⋅1+βx , which gives that β assumed = 0.013The flue <strong>gas</strong> flow (without steam) ism <strong>gas</strong> = m air *(1+ β assumed ) = 40*(1+0.013) kg/s = 40.52 kg/s2

<strong>Steam</strong> injection <strong>gas</strong> <strong>turbine</strong> – <strong>Exercise</strong> 4Enthalpies (h-s graph for steam and table for combustion <strong>gas</strong>es):h 4, <strong>gas</strong> = (380ºC, x = 0.20) = 389.8 + 0.20*29.2 kJ/kg = 395.64 kJ/kgCreated by CEh 5, <strong>gas</strong> = (245ºC, x = 0.20) = 248.45 + 0.20*17.0 kJ/kg = 251.85 kJ/kgh sh,320 = (10 bar, 320ºC) = 3090 kJ/kgh fw = c p *t = 4.18*25 = 104.5 kJ/kgh 4,ST = (1bar, 380ºC) = 3240 kJ/kgh 5,ST = (1bar, 245ºC) = 2965 kJ/kgIntroducing the values in [4], the steam mass flow can be calculated:395.64 − 251.85m&ST= 40 .52 ⋅kg / s = 2.15kg/ s(3090 −104.5)− (3240 − 2965)Answer a) The steam flow is 2.15 kg/sb) <strong>Calculation</strong> of the <strong>gas</strong> content with steam injection flow 2.15 kg/sThe specific fuel consumption can be found through a heat balance over thecombustion chamber:m air *h 2, AIR + m fuel *LHV + m ST *h sh,320 = (m air + m fuel ) *h 3, GAS + m ST * h sh,800 [5]where h 3, GAS = h 3, AIR + x*DH 3 [6]andβ= (1 + 14.52) ⋅1+βx [7]Inserting [7] in [6] and combining with [5] following expression for the specific fuelconsumption is obtained:=h3, AIRm&st− h2,AIR+ ⋅ ( hsh,800− hm&airLHV − h −15.52⋅ DH3, AIRsh,320β[8]3)To find the enthalpy of air after the compressor, h 2,air , the temperature in this point isneeded. The temperature increase of air in the compressor is calculated with:3

<strong>Steam</strong> injection <strong>gas</strong> <strong>turbine</strong> – <strong>Exercise</strong> 4κ −10.39T ⎛1p ⎞2 (20 + 273) ⎛ ⎞1.. 39T2 T ⎜ κ−1= ⋅ ( ) 1⎟⎜101⎟− = ⋅ = 313°C(orK )SKp10.85−η⎝ ⎠⎝ ⎠Created by CEThis gives us that t 2 = 20+313 °C = 333°CThe enthalpy after the compressor is:h 2,air = (333°C, x =0) = 340.2 kJ/kg(interpolated in the table)The fuel is light oil which has a lower heating value of LHV= 42.3 MJ/kg.Point 3, the <strong>turbine</strong> inlet, has a temperature oft 800°C. Looking in the enthalpy table forair and flue <strong>gas</strong> from light oil we find thath 3,AIR = (800°C, x=0) = 856 kJ/kgDH 3 = (800°C) = 76 kJ/kgThe enthalpy for steam at 800°C and 10 bar is found either in a steam table or in anenthalpy-entropy chart:h sh,800 = 4155 kJ/kgThe specific fuel consumption becomes:β =2.15856 − 340.2 + ⋅ (4155 − 3090)4042300 − 856 −15.52⋅ 76= 0.0142The calculated <strong>gas</strong> content is thusxcalcβ 0.0142= ( 1+14.52) ⋅ = 15.52 ⋅ = 0.21781+β 1.0142The difference between the calculated <strong>gas</strong> content and the assumed one is:difference =xcalcx− xassass0.2178 − 0.20⋅100 % =⋅100%= 8.9%0.20Answer b): The calculated <strong>gas</strong> content is about 9% larger than the assumedvalue.c) <strong>Calculation</strong> of the electrical power outputP GT = η G * (P T * η m – P C )The compressor power is4

<strong>Steam</strong> injection <strong>gas</strong> <strong>turbine</strong> – <strong>Exercise</strong> 4P C = m air·(h 2,air – h 1,air )Created by CEThe enthalpy of air in the compressor inlet is found in enthalpy table at 20°Ch 1,air = (20°C, x=0) = 20.1 kJ/kgThe compressor power need thus becomes:P c = 40 · (340.2 – 20.1) kJ/s = 12 804 kWThe <strong>turbine</strong> power output is composed by expansion of the flue <strong>gas</strong> (without steam)and the expansion of the steam:P T = (m air + m fuel ) · (h 3,GAS – h 4,<strong>gas</strong> ) + m ST· (h sh, 800 – h 4,ST )Observe that the steam stays in superheated form throughout the expansion!The enthalpies for <strong>gas</strong> in 3 and 4 have to be determined based on the calculated valueof the <strong>gas</strong> content; x = 0.2178.The enthalpy in point 3 for flue <strong>gas</strong> becomes:h 3, GAS = h 3, AIR + x*DH 3 = 856 + 0.2178*76 = 872.6 kJ/kgThe enthalpy in the <strong>gas</strong> <strong>turbine</strong> outlet becomes (obs! The outlet temperature is given):h 4, <strong>gas</strong> = (380ºC, x = 0.2178) = 389.8 + 0.2178*29.2 kJ/kg = 396.2 kJ/kgThe <strong>turbine</strong> power can now be calculated:P T = [40·(1+0.0142)·(872.6 –396.2)] + [2.15·(4155-3240)] kW = 21 294 kWThe electrical power output:P GT = 0.98· (21 294·0.99 – 12 804) kW = 8112 kWEfficiencyηGTP=Q>Fuel=m&FuelPGT⋅ LHV8112== 0.33760.0142 ⋅ 40 ⋅ 42300Answers c: the power output is 8112 kW and the efficiency 33.8%5