Assignment 1Assignment 1submitted via moodle within 31 Aug 11submitted via moodle within 31 Aug 11

Problem 1.Problem 1. 50 kg of food at a moisture content of 400 % dry basis is dried to 25 % wet basis. Calculate the amount of water removed

F = 50 kg of raw product

(400 % d.b.)

Product

(25 % w.b.)

Water removed

Drying process

Solution P = 13.33 kg , W = 36.67 kg

Problem 2.Problem 2. A membrane separation system is used to concentrate the liquid food from 12 % to 45 % total solid (TS). The process is accomplished in two stages, in the first stage a low total solid liquid stream is released. For the second stage, there are two streams released. The first stream is final product with 45% TS and the others is coming out with 1.5% TS and recycled to the first stage. Determine the magnitude of the recycle stream, the waste stream from first stage contains 0.25 % TS and the stream between stages 1 and 2 contains 30 % TS . The flow rate of 45 % TS final product stream is at 120 kg/min.

Feed

12 % TS B

30 % TS

R

1.5 % TS

120 kg/ min of

product

45 % TS

W , 0.25 % TS

first stage Second stage

Solution F = 457.02 kg/min , W = 337.02 kg/min and R = 63.16 kg / min

Problem 3. Steam is used for peeling of potatoes in a semi-continuous operation . Steam is supplied at the rate of 5 kg per 80 kg of unpeeled potatoes. The unpeeled potatoes enter the system with a temperature of 22 C and the peeled potatoes leave at 35 C . A waste stream from the system leaves at 65 C . The specific heats of unpeeled potato, waste stream and peeled potatoes are 3.7 , 4.2 and 3.5 kJ/ (kg K ) , respectively. If the heat content of steam is 2800 kJ /kg , determine the quantities of the waste stream and the peeled potatoes from the process

Solution P = 17.89 kg and W = 67.11 kg

Problem 4. Potato flakes ( 62 % w.b.) are being dried in a concurrent flow dries. The moisture content of the air entering the drier is 0.05 kg of water per 1 kg of dry air. The moisture content of air leaving the drier is 0.125 kg water per 1 kg of dry air. The air flow rate in the drier is 150 kg dried air per hour. As show in a picture , 80 kg of wet potato flakes enter the drier per hour. At steady state , Calculate the following

A. What is the mass flow rate of dried patatoes.B. What is the moisture content , dry basis , of “ dried potatoes ” exiting the drier

Solution mass flow rate = 68.75 kg/h and MC = 126 %db

Air in Exit Air

Feed Product