Download presentation
Presentation is loading. Please wait.
Published byDouglas Hawkins Modified over 8 years ago
1
Assignment 1 submitted via moodle within 31 Aug 11
2
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
3
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 stageSecond stage Solution F = 457.02 kg/min, W = 337.02 kg/min and R = 63.16 kg / min
4
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
5
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 FeedProduct
Similar presentations
© 2024 SlidePlayer.com Inc.
All rights reserved.