The process shown in the figure is a 2-step process for separating a mix of meth
ID: 492266 • Letter: T
Question
The process shown in the figure is a 2-step process for separating a mix of methane (M), ethane (E) and propane (P). It is operating at steady-state and no reactions occur. a. Write an overall methane (M) balance. b. Write a total mass balance for Sep 1 (you need to properly label the output for Sep 1 first). c. Write a propane (P) balance on Sep 2. d. Write the recovery equation for the statement: "75% of the methane fed to the system is recovered in product 4". e. What is the fraction of M in the Feed? What is the fraction of E in P4?Explanation / Answer
Overal methane balance:
From feed:
inlet fraction of methane = (1-0.20-0.15) = 0.65 is methane
so
inlet = 250*0.65 = 162.5 kg of methan inlet
it must go out so:
inlet = outlet
recall that mass of methane is = mass fraction in stream * Mass flow in steam:
F(x-m) = P3*(0.96) + P4*(0.42)
F(0.65) = P3*(0.96) + P4*(0.42)
b)
total mass balance for Sep1:
inlet = outlet, no reactions
therefore
Feed = Product 2 + X
X = outlet of separator 1 (this is the label I choose)
c)
propane balance for Sep 2.
inlet = outlet
X*(x-p) = P3*0 + P4*0.10
note that, we can0t further proceed, since x-p is not known,
X*(x-p) =P4*0.10
d)
recovery equatio nfor --< 75% of methane fed to system is in Product 4...
so
inlet methane = 250*(0.65) = 162.5 kg of methane go inside
so, 75% is recovered --> 0.75*162.5 =121.875 kg of methane are recovered in P4:
so
0.42*P4 = 121.875
P4 = 121.875/0.42 = 290.17 kg
the overall equation is expressed as:
0.75*F*(x-mF) = P4*x-m4
0.75*250*(0.65) = P4*0.42
e)
fraction of M in feed....
this is clearly just a fractional balance:
total sum = 1
1 = M + E + P
1 = M + 0.20 + 0.15
M = 1+(-0.2-0.15) = 0.65
for fraction of E in P4:
once again
1 = sum
1 = 0.42 + E + 0.10
E = 1-0.42-0.10 = 0.48
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