Ch4 Material Balances-part2(1)

Ch4 Material Balances-part2(1) - Problem4.36 m5 m2 m2 3m1...

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Problem 4.36
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2 m 5 m 3 2 1 m 3m 6 m 7 m 8 1 2 3 m m m m 1 2 m m 1 2 m m 1 1 m Beans 0.13m oil 1 Solid 0.87m liquid 0 13m (oil) 1 0.87m solid 1 2 0.13m (oil) m (hex) 3 3 m Cake 0.75m solid 0 25 li id 4 m Oil 3 0.25m liquid hex Oil 2 1 m 3m 1 2 3 4 5 unknowns :m ,m ,m ,m ,m Balances: oil, hexane, solid * Assume Basis: 1 100 kg m
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1 m 2 m 1 2 m m 3 m 2 1 hexane : m oil : 0.13m 3 0.25m liquid 2 2 1 m hexane ratio m 0.13m 3m 3 3 0.13 oil 0.25m 3.13 1 1 3.13m 3.13 0 13m 0 13 3 3 hex 0.25m 3.13 1 2 1 0.13m 0.13 oil ratio m 0.13m 3.13
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3 unknowns: m 3 ,m 4 ,m 5, m 1 = 100 kg, basis 3 Balances: Oil : (Overall) 0.13 m 1 =m 4 +0.25m 3 0.13 3.13 5 3 3 m 0.25m 3.13 Hexane : 1 3 3 (0.87)m 0.75m m 116kg cake Bean Solid : 5 m 27.8kg new hexane (fresh fed) 4 m 11.8kg oil per 100kg bean H 300k R 272k 2 2 5 Hexane:m 300kg, so R m m 272kg 272 Recycle: 9.78 27.8
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