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If you have a change in the inerts (CO2, N2, etc) there will be a noticeable change in LHV, performance and NOx emissions.

Higher the methane (CH4) content, which also increases the lower heating value (LHV) of the fuel, increases the GT power output and reduces heat rate. Also, NOx emissions decrease. The higher the CH4 content, greater is the water vapour in the products of combustion. This improves the engine...

You can try Gas Turbine Combustion by Lefebvre or Gas Turbine Theory BY Saravanamutoo, Rogers and Cohen. They cover it quite well.

Also have a look at this thread thread1036-173075

If you only want the enthalpy of moist air you can use the formula: h = (1.007t – 0.026) + w (2501 + 1.84t) Where t is ambient temperature in C and w is specific or absolute humidity. h is specific enthalpy kJ/kg. This equation is valid between 0 and 60C.

The increase in flow through the compressor will result in the expander/turbine imposing an increased back pressure on the compressor. This results in an increase in compressor pressure ratio and speed. Imagine this running line on the compressor map, which is approximately parallel to the surge...

As you are aware, compressor maps are normally plotted with pressure ratio VS non-dimensional air or mass flow (W(RT/Gamma)^0.5))/AP) for a series of non-dimensional speeds ND/((GammaRT)^0.5). W = mass flow rate R = gas constant Gamma = cp/cv (ratio of specific heats) T = absolute temperature...

Strictly speaking yes. If you want a ball park figure then constant gamma gives an answer to withing 3% of the actual efficiency.

Yep, Thats right. gtsim

p1 = 14.7 (atmospheric pressure) p2 = 230 psi (compressor discharge pressure) Pressure Ratio = 230/14.7 = 15.646 t1 = 75 deg F = 23.89 deg C = 23.89 + 273 =296.89 = 535 R t2 = 650 deg F = 343.33 = 243.33 + 273 = 616.33K = 1110R Gamma = cp/cv for air (approximately 1.4) C-1=...

What are your temperature units - degrees F I think? Whatever units you use it should be in absolute (R or K). The efficiency should be bewteen 0.85(85%) to 0.9(90%). There is something wrong in your data. It gives 100% efficiency (assuming F for temperature)??? Where did you get these...

To determine the (isentropic) efficiency of a gas turbine compressor, such as the Frame 7 the airflow is not needed. The compressor efficiency can be calculated via: Eff= (C-1)/(T2/T1-1) Where C= Pr^((Gamma-1)/Gamma) Eff is the isentropic efficiency, which is the same as adiabatic Pr is the...

Relative Humidity (RH)=p/ps, where p is the vapour pressure at 37 deg C and ps is the saturated vapour pressure at 37 deg C. From steam tables we can calculate ps at 37 deg C. Therefore vapour pressure, p = RH x ps. Specific Humidity (SH) = 18/MW x p/(P-p), where P is the pressure of the gas...

Briefly, in conventional (diffusion flame) combustors, as saludos says, the fuel/air mixture in the primary zone, where the fuel is burnt, is heterogeneous or stratified. Therefore the fuel/air ratio varies a great deal from lean to near stoichiometric. Thus high flame temperatures result in the...

There isn’t one. The adiabatic efficiency depends on where on the map you are operating. For example, if the operating point moves from the current position to the left on the characteristic at a constant pressure ratio the adiabatic efficiency will increase (i.e. a decrease in corrected flow)...

I am sorry. I should have read your question more carefully. You need a hygrometric chart for a given ambient pressure. Alternatively, you can use the questions given in the site: http://www.srh.noaa.gov/elp/wxcalc/formulas/rhTdFromWetBulb.html to generate the necessary chart. I think your...

Ambient conditions Pressure = 14.696 PSI Temperature = 100F Relative humidity = 10% Wet bulb temperature = 63.753F Airflow rate = 1000 cfm = 70.867 lbs/s After cooling Water flow rate (evaporated) = 0.573 lbs/s Temperature = 65.709F Relative humidity = 90% Cooling = 34.291F Cooling...

Hi toothless, I respectfully disagree that using the isentropic efficiency of the compressor alone would be sufficient to optimise the compressor wash. Knowing only the isentropic efficiency is not sufficient as this efficiency can decrease if the load conditions are such that the compressor...

Norzul, Yes, that’s correct. In your case you should consider reducing your wash frequency. The heat rate would increase by about 1% to 1.5% due to fouling after 6 weeks. It does not increase very much after that. In fact we are monitoring a 40MW gas turbine offshore (exploration and...

toothless, If you wish, go can go to this website www.gpal.co.uk/tryit.htm and download the simulators and subject them ambient temperature and pressure transients and see how the operating point on the compressor map changes in the steady state. Also you can subject these simulators to...