Chemical Engineering - Mass Transfer - Discussion
Discussion Forum : Mass Transfer - Section 14 (Q.No. 18)
18.
In a single stage extraction process, 10 kg of pure solvent S (containing no solute A) is mixed with 30 kg of feed F containing A at a mass fraction xf = 0.2. The mixture splits into an extract phase E and a raf-finate phase R containing A at xB = 0.5 and xR = 0.05 respectively. The total mass of the extract phase is (in Kg)
Discussion:
4 comments Page 1 of 1.
Om Patel said:
3 years ago
Why xS is zero? Please explain me.
K Mahek said:
5 years ago
@Snehal.
From overall balance;
40 = E + R.
From overall balance;
40 = E + R.
Snehal chandanshive said:
5 years ago
How come (40-E) ? Please explain in detail.
Zeeshan Ulhaq said:
5 years ago
F+S = E + R.
10+30 = E+R.
xaF + xaS = xaR + xaE.
0.2 * 30 + 0 = 0.05(40-E) + 0.5E.
So, E = 8.89.
10+30 = E+R.
xaF + xaS = xaR + xaE.
0.2 * 30 + 0 = 0.05(40-E) + 0.5E.
So, E = 8.89.
(2)
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