Question

Twenty kg of a fuel consisting of 50% methane, 25% ethane and 25% propane (by mass) is compressed to 180 atm and 100 degrees C. What is the compressed volume using Kay’s rule?

Answer 1

Mass of methane= 20*0.5= 10 kg mass of ethane= 20*0.25= 5 kg and propane= 20*0.25= 5kg

Moles = mass/ Molecular weights, Molecular weights : CH4=16 ethane C2H6= 30 and propane C3H8=44

Moles of methane = 10/16=0.625Kg moles of ethane = 5 /30=0.17 kg moles and moles of propane =5/44=0.113 kg moles

Total mol.es= 0.625+0.17+0.113 =0.908 kg moles=908gmoles

Mole fractions CH4 (Y1)= 0.625/0.908=0.688    C2H6 = 0.17/0.908=0.187 and C3H8= 0.125

For methane TC=-82.3 deg.c = -82.3+273.15 K=190.085 K Pc=673 Psia=673/14.7 atm=45.78 atm

For ethane ,Critical temperature , TC : 32.17 °C =32.17+273.15K   =305.32 K , Critical pressure : 48.72 bar

For propane Critical temperature . TC= 96 deg.c= 96+273.15= 369.15K and critical pressure = 42.04 atm

Mixture critical temperature TC= Y1*TC1+ Y2*TC2+ Y3*TC3   = 0.688*190.085+0.187*305.32+0.113*369.15=229.58 K

Similarly,

Mixture P_C= 0.608*45.78+0.187*48.72*0.9869+0.113*42.04=41.57 atm

Tr= T/TC and Pr= P/PC

Tr= (100+273.15)/229.58=1.63   and Pr= 180/41.57=4.33

From compressibility chart, Z= 0.825

V= nZRT/P= 908*0.825*0.08206L.atm/mole.K * 373.15/180 =127.43 L