Chemical Engineering - Heat Transfer - Discussion
Discussion Forum : Heat Transfer - Section 9 (Q.No. 40)
40.
Fresh orange juice contains 12% (by weight) solids and the rest water 90% of the fresh juice is sent to an evaporator to remove water and subsequently mixed with the remaining 10% of fresh juice. The resultant product contains 40% solids. The kg of water removed from 1 kg fresh juice is
Discussion:
3 comments Page 1 of 1.
Philip said:
8 years ago
@ALL.
1. Solid will remain unchanged even after evaporation.
2. The 40% solids is the mixture of 10% fresh juice (initially) + 90% evaporated juice.
1. Solid will remain unchanged even after evaporation.
2. The 40% solids is the mixture of 10% fresh juice (initially) + 90% evaporated juice.
Sarthak said:
8 years ago
1*0.12 = 0.4*m.
m=0.3 = kg of thick liquor.
Feed= water evaporated + thick liquor,
1 kg = x + 0.3,
x= 0.7= water removed.
m=0.3 = kg of thick liquor.
Feed= water evaporated + thick liquor,
1 kg = x + 0.3,
x= 0.7= water removed.
(4)
Sangram said:
9 years ago
I am not sure on this, but according to me,
F' = 1 + .1 = 1.1 because 10% of the fresh juice is added.
F' * Xf = P * Xp.
1.1 * .12 = P * .4
Xf will be same as the same 10 percent of fresh juice is added so fraction will not change.
P = .33;
F' = E + P;
E = 1.1 - .33 = .77 Answer.
F' = 1 + .1 = 1.1 because 10% of the fresh juice is added.
F' * Xf = P * Xp.
1.1 * .12 = P * .4
Xf will be same as the same 10 percent of fresh juice is added so fraction will not change.
P = .33;
F' = E + P;
E = 1.1 - .33 = .77 Answer.
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