A process for concentrating orange juice takes a feed of 100 kg/h, with a sugar
ID: 934240 • Letter: A
Question
A process for concentrating orange juice takes a feed of 100 kg/h, with a sugar mass percentage of 12%, and concentrates it to 45% using the flow diagram given below.
A proportion of the feed bypasses the evaporation stage and is blended back with the concentrated product. If the sugar mass percentage leaving the evaporator is 65%, calculate all the other flow rates on the flow diagram.
A process for concentrating orange juice takes a feed of 100 kg/h, with a sugar mass percentage of 12%, and concentrates it to 45% using the flow diagram given below. A proportion of the feed bypasses the evaporation stage and is blended back with the concentrated product. If the sugar mass percentage leaving the evaporator is 65%, calculate all the other flow rates on the flow diagram.Explanation / Answer
1.
A. Overall Mass Balance
F = P + W
B. Sugar Component Balance
wfF = wpP + wwW
(0.12)(100 kg/hr) = (0.45)(P) + 0
P = 26.667 kg/hr
W = F - P
W = 100 kg/hr - 26.667 kg/hr
W = 73.333 kg/hr
2. Blender Balance
E + B = P
E + B = 73.333 kg/hr eq 1
wEE + wBB = wPP
(0.65)(E) + (0.12)(B) = (0.45)(26.667 kg/hr) eq 2
Solving 2 equations with 2 unknowns
E = 16.604 kg/hr
B = 10.063 kg/hr
3. Splitting Point Balance
F = S + B
S = 100 kg/hr - 10.063 kg/hr
S = 89.937 kg/hr
Final Answers:
P = 26.667 kg/hr
W = 73.333 kg/hr
E = 16.604 kg/hr
B = 10.063 kg/hr
S = 89.937 kg/hr
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