Question
1)A reverse osmosis membrane unit concentrates a liquid food from 10 to 20% total solids by...
1)A reverse osmosis membrane unit concentrates a liquid food from 10 to 20% total solids by selectively removing water. If 100 lb/hr of feed is input to the membrane, how much concentrated product is produced?
2) 30kg of one component with 30% total solids is mixed with 200 kg of a material with 80%TS. How much of the final product is produced and what is its composition
3) 40lb of a raw material containing 15% alcohol and 1% sugar( the rest is water) is added to 50lb of another raw material containing 17% alcohol and 0.5%sugar. What is the composition of the mixed product?
4)It is desired to make a product containing 16.5% alcohol and 0.6 sugar. Can we use the two raw materials from the 3)? If so, how much of each should be added to make 100lb of the desired product?
Answers
w Solin (1) 25, W=0 2,W21 For-loolb/hr r s = solids 7 Reverse Osmosis membrane - Wz Water I *S,F=o1 *w, p= 0.9 Aspio.ro xw, p=0.8 F=w+P 100 = w+ P + (0) solides balance: 100(0:1) = P(0.v) Pa 10 - 50 lb/hr 0.7 - concentrated product p = 50lb/hr F = 30 kg mixer T fra 200 kg XTS, f. = 0.30 l Xqs, f,=0.8 F3 = 30+ 200 = 230 kg Total solids balance: 30 (0.3) + 200(0.8) - 230 (HTS, F,) 169 X75,63 = 230 230 - 0.7347 = 73.472 fo Mixer Miner F, -4015 XA,F,-0.15 25,6,-0.01 > 7 XA, F= 2 I fra 50lb *s, fz = 2 XA, 8, = 0.17 kw, Fz=2 As, Fv=0.005 xw, fv = 0.825 xw, f = 0.84Ft Fx = F3 F3 = 40 +50= 901b Alcohol balance: 40(0.15) + 50(0.17) - 90(2A, F3) 14.5 = 90( 42, f3) Rafg= 0.161 = 16.1% sugar balance: 40 (0.01) + 50(0.005) = 90045,) ts f = 0.0077 = 0.78% Ha,f3+ M,F3 + xw, f3 = 1 0-161 +0.007vt aw, Fz=1 Ww, fy = 0.8317 = 83.17% ->f=10016 - - r 29.670.15 Mixer J Fr XA , F3 = 0.165 Rs fo 20.006 f-0.17 -0.17 xw, f = 0.689 EN F, + Fra 10016 Alcohol balances F(0.15) + F, (0.17) = 100(0.165) (100-FV)0.15 + F(0.17) : 16.5 f(0.00) = 16.5-15 1.5 0.0y : 75 f = 2516
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