Question

3. Natural gas is a mixture of hydrocarbons and recent developments in directional drilling and hydraulic stimulation has seen that the majority of natural gas is now produced from shale gases. These gases may contain higher proportions of higher alkanes. In order for these gases to be used via the pipeline, they must meet certain pipeline quality standards. These standards are even stricter when the natural gas is to be used by natural gas vehicles. Suppose natural gas from a shale well enters your furnace. The pressure and temperature of the mixture are 17 °C and 100 kPa. A molar analysis shows the gas is 80% methane, 10% ethane, and 10% propane. Determine the following:

a.) Mass fraction of each component (10 points)

b.) The partial pressure of each component (10 points)

c.) The mass flow rate in kg/s for a volumetric flowrate of 1m3 /sec (10 points)

Answer 1

Basis : 1 mol

Moles : CH84=0.8, C2H6= 0.1 and C3H8= 0.1

Molar masses :CH4= 16, C2H6= 30 and C3H8= 42

mass : moles * molar mass , CH4= 16*0.8= 12.8, C2H6= 0.13*30 =3 and C3H8 =0.1*42 =4.2

total mass =12.8+3+4.2= 20

mass fractions : CH4 = 12.8/20 =0.64, C2H6= 3/20 =0.15 and C3H8= 4.2/20=0.21

b) partial pressure = Moel fraction* total pressure

since one mole is the basis, moles of the componebnt = mole fraction ofthe component

partial pressures : CH4= 0.8*100 =80 Kpa, C2H6= 0.1*100 Kpa =10 Kpa nad C3H8= 0.1*100 =10 Kpa

c) flow rate is given as 1 m3/sec and desity need to be caclulated.

1 moles of any gas occupies 22.4 Liters at STP

density = 20 * mass/ 22.4 L =-0.892 g/L

this is the density at 273.15K and 101. 3 Kpa P1= 101.3 Kpa and T1=273.15K

at P2= 100 Kps and T2= 17+273.16=290.15K

since P* mass= densitty* RT

since P2/P1= d2T2/d1T1

d1 and d2 are densities at condition 1 and 2

d2= (P2T1/ P1T2)*d1 = 0.892*(100/101.3)* 273.15/297.15 =0.83 g/L=0.83 kg/m³

Mass flow rate = volumetric flow rate* density =1*0.83 kg/s =0.83 kg/s