natural gas consumption

[aleks614]

Hi everyone, 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? any comments and correction from all of you will be much appreciated. Thanks

 

[25362]

As I see it, the enthalpy of the exhaust gases should be deducted from the LHV of the fuel gas, to obtain a "Net" HV.

Assuming the various mechanical efficiencies have been incorporated in the estimate of the energy input data, EID, divide EID by Net HV, and proceed as you indicated.

m[sup]3[/sup]/s will be measured at the same P,T conditions used to compute the density of the gas.

 

[mbeychok]

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)
.

 

[mbeychok]

Oops...made a typo!! There are 3.6 MJ per 1 kWH, so my gas LHV equation should be:

gas LHV = (48.28 MJ/kg)(0.887 kg/Nm3)(1 kWh/3.6 MJ) = 11.9 kWh/Nm3

Milton Beychok
(Contact me at www.air-dispersion.com)
.