Wet sugar that contains one-fifth water by mass is conveyed through an evaporato
ID: 501753 • Letter: W
Question
Wet sugar that contains one-fifth water by mass is conveyed through an evaporator in which 85.0% of the entering water is vaporized. (a) Taking a basis of 100 kg feed, calculate (i) x_W, the mass fraction of water in the wet sugar leaving the evaporator, and (ii) the ratio (kg H_2O vaporized/kg wet sugar leaving the evaporator). (b) If 1000 tons/day of wet sugar is fed to the evaporator, how much additional water must be removed from the outlet sugar to dry it completely, and what annual revenue can be expected if dry sugar sells for $0.15/lb_m? (c) The evaporator is built to achieve the production rate of part (b), installed, and started up, and the water content of the partially dried sugar is measured on successive days of operation. The results follow. In subsequent runs, the evaporator is to be shut down for maintenance if x_W falls more than three standard deviations from the mean of this series of runs. Calculate the endpoints of this range. (d) Considering the results of parts (a) and (c) together, what can you conclude about the recently installed evaporator?Explanation / Answer
. Wet sugar that contains one-fifth water by mass is conveyed through an evaporator in which85.0% of the entering water is vaporized.
W = Water
S = Sugar
(a) Feed stream:m= 100 kg
xW= 0.20à20 kg W
xS= 0.80à80 kg S
Products:
20 kg W * 0.15 = 3 kg W
80 kg S
xW= 3 / (80 + 3) = 0.036
17 kg W evaporated / 83 kg WS =.205 kg W/kg WS
3 remaining kg W/100 kg WS
================================================================================================================================================================
b) 1000 tons WS * 0.03 extra W =30 tons of water per day
1000 tons WS/day * 0.80 tons S/ton WS * 2000 lbm/ ton * 0.15 $/lbm* 365 days/year =8.8 *10^7$/day
=================================================================
(c) Average xW= 1/10 * (0.0513 + 0.0486…+0.0511+0.0494) = .0504 kg W/kg
SD =sqrt 1/9 (0.504 0.0513)´^2… + (0.504 0.494)´= 0.00181 kg W/kg
Lower limit = 0.0504 – 3 * 0.00181 = 0.0450
Upper limit = 0.0504 + 3 * 0.00181 = 0.0558
=====================
(d) The evaporator is not working properly because
xW= 0.036 < 0.450
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.