Question 8, Case Analysis, Afternoon Session, Registered Chemical Engineering Exam 2011
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8. An existing 2% (by mass) salt solution is continuously fed into a single-effect evaporator at 28°C and concentrated to 3% (by mass). The heat transfer area of the evaporator is 69.7 m2, and the heating steam is saturated water vapor at 110°C. The feeding rate is 4950 kg/h, and the specific heat capacity of the material fluid C0 = 4.1 kJ/(kg·℃). The operation is carried out at 101.3 kPa, with an evaporation latent heat of 2258 kJ/kg; the boiling point elevation is negligible. When all other conditions (heating steam and feed temperature, feed concentration, operating pressure) remain unchanged, if the feeding rate is increased to 7480 kg/h, to which of the following values can the concentration of the solution be reduced (in mass percentage)? (A) 4.2% (B) 2.4% (C) 1.8% (D) 3.4% Solution: W = F(1 – X0/X1) = 4950(1 – 0.02/0.03) = 1650. Q = 1650 × 2258 + (4950 – 1650) × 4.1 × (100 – 0) – 4950 × 4.1 × (28 – 0) = W’ × 2258 + (7480 – W’) × 4.1 × 100 – 7480 × 4.1 × 28. The value of W’ is 1246, and X1’ = 7840 × 0.02 / (7840 – 1246) = 0.0238. Questions: 1. I think there is an issue with this solution; since it refers to the latent heat of evaporation, the formula that takes into account the negligible effect of heat of dilution should be used. Moreover, the boiling point of salt solutions is not necessarily 110 degrees℃
This post was last edited by zwp997 on 2015-8-1 22:42; there is indeed a slight issue. .
Where does the 100 in the answer come from?
Operation is at atmospheric pressure, and the boiling point elevation is ignored; therefore, the boiling point of the solution is taken as 100°C.
The solution to this problem provided by the original poster overturned my views. Why is it necessary to deduct the heating heat for evaporating water? ?
My answer: W = F(1 – X0/X1) = 4950(1 – 0.02/0.03) = 1650. Q = 1650 × 2258 + 4950 × 4.1 × (100 – 28) = W’ × 2258 + 7480 × 4.1 × (100 – 28). Following your guidance, I obtained W’ = 1319 and X1’ = 7840 × 0.02/(7840 – 1319) = 0.024