Question

please solve this questuon

Fresh orange juice contains 13% solids and rest water. It is concentrated to contain 45% solids. In the present process, the evaporator is bypassed with a fraction of fresh juice. Juice that enters the evaporator is concentrated to 60% solids and is mixed with bypass juice to achieve the desired final concentration. Calculate the amount of final juice per kg of fresh juice fed to the process and the fraction of the feed in percentage that bypasses the evaporator. TRICO

Answer 1

r evaporated No solid Content se Product 11 the 13% Solid Forces on Feed (1 ks Fush Juice DC1-B)13 | Evaperater Heate a mixny 27 J Evaporator 60% Solid {(kg) @ mixing point us % Solid B kg By Pass (13 % solid Take Basis of Fresh Juice 1kg as fresh feed "solid concentration in fresh feed and By pass Stream care same because Just spliting the Stream No Physical / chemical change"

Suppose B is fraction d fresh staken as By pass with same ax fresh Juica overall material Balance around stream that concentration evaporator (l-B) = a + y Solid content mass Balance around evaporator (1-3)( 3 ) = 0 + + (1-B)(0.13) = q (0.60) %= (-B) C13) = 0.2166 (1-B) Гоо

Now Material Balance an und mixing point - 2 + B = P P2 kg a produal Solid content Balance around mixing point Shor) + B(13) = (45) (0.60)(2) + 10.13) B = 10.45) p = 0.45 (Y+ B) 0.6(%) + 0113(B) - Oius (y +13) Coot - 0.45) - BC0145 -0.13) 012166 (1-B) 10.6-0.45) = B10145-013) put value from previous equation

or 03249 (1-B) = 0.328 01032u9 = 10:32 + 0103249) B B = 0.03249 = 0.09217 kg 0.3524g B = 0.09217 f By Pass = 10.09217 kg % I= 9.217 % Put valul of B and find y Yo 0.2166 C1-B) = 0.2166 C1- 0.09217) 1 = o.196635 kg Product P = y +B=10.196635 + 0.09217) = 0.2888 final juice = 0.2888 kg / kg of fresh juice