Subtopic: Concentration and drying of liquid foods – evaporators, tray, drum and spray dryers
Question Statement:
A single effect evaporator concentrates 9090 kg·h⁻¹ herbal extract of 12 % solids entering at 300 K to a final solid concentration of 20 %. Assuming negligible product loss through vapor, the amount of concentrated product, in kg, is __________. (Answer in integer)
Official Answer Key:
5452 to 5456
Step-by-Step Solution & Explanation
This is a pure solids balance — the solids entering with the feed all leave with the product, since only water is evaporated.
Step 1 — mass of solids in the feed:
S = 9090 × 0.12 = 1090.8 kg·h⁻¹
Step 2 — mass of product carrying the same solids at 20 % concentration:
P = S/0.20 = 1090.8/0.20 = 5454 kg·h⁻¹
Step 3 — check:
Water evaporated = 9090 − 5454 = 3636 kg·h⁻¹
Product solids fraction = 1090.8/5454 = 0.20 ✔
Concentrated product = 5454 kg
(The feed temperature of 300 K is contextual — it affects the heat load, not the mass of product.)
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