Mixing Material Balance on The
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Mixing material balance on the The basis methane of 100kmol/hr The ratio of methane to air = 0.3 (from the reference ) The ow rate of air =100/0.3 =317.5 kmol/hr A: methane : air= !"# $: %ro&'ct Fa + Fb = Fab 100 +317.5 = 417.5 kmole hr FA = 100 K mole / hr FB = 317.5 K mole / hr FO = 92.2 K mole hr FN = 125.3 K mole / hr The mole fraction input of mixing ole fraction of methane=1 ole fraction of o *+en=,,., kmol/hr ole fraction of nitro+en = 0.7- kmol/hr The mole fraction o't%'t of mi in+ ole fraction of methane = 100 ∗ 1 = x ∗ 417.5 x = 0.24 ole fraction of o *+en = 317.5 ∗0.22 = 417.5 ∗ x
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Transcript of Mixing Material Balance on The
Mixing material balance on theThe basis methane of 100kmol/hrThe ratio of methane to air = 0.3 (from the reference )The flow rate of air =100/0.3 =317.5 kmol/hrA: methaneB: air= O+NC: product
The mole fraction input of mixingMole fraction of methane=1Mole fraction of oxygen=22.2 kmol/hrMole fraction of nitrogen = 0.79 kmol/hrThe mole fraction output of mixingMole fraction of methane=
Mole fraction of oxygen =
Mole fraction of nitrogen =
Kg A=Kg O=Kg N=The total mass input = 6583.6KG/hr
Mitral balance on the reactor
Convertion of methan = 100% (Frome the refefrence ) Input to Reactor
Output From The Reactor
Total Mass 0f the output from reactor =
Flash Separator Antonio equation table
C B A component
266.55255.686.49457 nitrogen
222.309 1533.3137.38782 CH3CooH
2301473.117.87863 CH3OH
Tav=Tb*Xi Tav=(118.2*0.29)+(64.7*0.181)+(-195.8*0.53)=-57PT=0.754 mmHgBy trial and error of Antonio equation C0Assume T=-50
By trial and error
