Alex614:
The following is from your original posting:
I just need to know how to find gas consumption in m3/hr of a gas engine. Basically, I have the energy input data of the engine which is 3959kW (which of course comes from the LHV of gas). Now, I'll be using this engine with gas supply LHV of 48284 KJ/kg and SG of gas at 0.687 (computed from the properties of gas).
In order to find the gas consumption, is it right to divide the supply gas LHV by engine energy input, which you'll get kg/s and convert it to m3/s by using the specific gravity?
The above quoted material raises a number of questions:
(1) You first state that you want to find the
m3/hr of fuel gas consumption ... but you later talk about determining the
m3/s of gas consumption. Which one is correct?
(2) Are you sure that the energy input of 3959 kW was determined from the gas LHV and the gas consumption? Or is the 3959 kW perhaps the engine's nameplate rated output of the engine ... which would significantly affect the fuel gas consumption calculated below.
(3) I assume that the gas specific gravity (which you state to be 0.687) is the specific gravity relative to air. Is that correct?
(4) As pointed out by 25362, you fail to state at what temperature and pressure you wish to determine the fuel gas volumetric flow rate.
(5) As also pointed out by 25362, you would
NOT divide the gas LHV by the engine's energy input ... instead, you would divide the engine's energy input by the gas LHV.
Assuming the 0.687 specific gravity is relative to air, then:
0.687 = gas MW/air MW
gas MW = 0.687(28.95) = 19.89
gas density = (19.89 kg/kgmol)(1 kgmol/22.414 Nm3) = 0.887 kg/Nm3
gas LHV= (48.28 MJ/kg)(0.887 kg/Nm3)(1 kWh/3.6 kJ) = 11.9 kWh/Nm3
Assuming that the 3959 kW is in fact the engine input energy:
gas consumption = (3959 kW)/(11.9 kWh/Nm3) = 333 Nm3/h
where:
MW = molecular weight
LHV = lower heating value = net heating value
Nm3 = cubic meters of gas at 0 [°]C and 1 atmosphere
Milton Beychok
(Contact me at
www.air-dispersion.com)
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