Food Engineering Lecture 2

Lecture 8 - 10 am (Mon.) September 2, 2024 Lerjun M. Peñaflor PhD Agricultural and Biosystem Engineering Department BUCAF Moisture Content Food Sample = Food Solid + Food Liquid Mass of Product = Mass of dry solid + Mass of water in food Mass of water in fo...

Lecture 8 - 10 am (Mon.) September 2, 2024 Lerjun M. Peñaflor PhD Agricultural and Biosystem Engineering Department BUCAF Moisture Content Food Sample = Food Solid + Food Liquid Mass of Product = Mass of dry solid + Mass of water in food Mass of water in food Mass of dry solid Food Sample Moisture Content, dry basis Moisture Content, wet basis Example 1. Convert moisture content of 85% wet basis to moisture content dry basis. Example 2 A food is initially at a moisture content of 90% dry basis. Calculate the moisture content in wet basis Example 3 A 10 kg of food sample has a moisture content of 75% wet basis. Determine the mass of dry solid. Solution: Amount of water = 10 kg (0.75) = 7.5 kg water Mass of Solid = 10 kg - 7.5 kg = 2.5 kg Material Balance The Principle of conservation of mass states that: Mass can be neither created nor destroyed. However, its composition can altered from one to another. Example 4 Example 5 20 kg of food at a moisture content of 80% w.b. is dried to 50% w.b. Calculate the amount of water removed. Solution: OMB: F=W+P 20 = W + P...... eqn. 1 CMB: 20 kg (1- 0.80) = P (1 - 0.50) P = 8 kg threfore , Mass of water removed = 20 kg - 8 kg W = 12 kg water Example 6 A membrane separation system is used to concentrate the liquid food from 10% to 30% total solid (TS). the product is accomplished in two stages, in the first stage, a low total solid -liquid stream is obtained. In the second stage there are two streams , the first one is final product stream with 30% TS and the second is recycled to the first stage. Determine the magnitude of the cycle stream when the cycle contains 2% TS. The waste streams from the first stage contains 0.5% TS and the stream between stage 1 and 2 contains 25% TS. The final Product is 100 kg/min. with 30% TS. = 334.944 Example 7 Steam is used for peeling of potatoes in a semi-continuus operation. It is supplied at the rate of 4 kg per 100 k g of unpeeled potatoes. The unpeeled potatoes enter the system with a temperature of 17 o C and the peeled potatoes leaved at 35 oC. A waste stream from the system leaves at 60 oC. The specific heat of unpeeled potatoes waste stream and peeled potatoes are 3.7, 4.2, and 3.5 kJ/kg oK, respectively. If the heat content of stream is 2750 kJ/kg, determine the quantities of the waste stream and the peeled potatoes from the process. Solution Laboratory exercise no. 2 / Board work Compute the amount of cooling water required to chill 330 kg of pineapple from 24 o C to 9 o C. The cooling water temperature increases from 1 oC to 6.5 oC.  In the processing of soymilk , 100 kg of soybeans (mc = 11%) is soaked overnight increasing the moisture content to 59%. It is then comminuted in a rietz disintegrator while water is added at a rate of 1:3.4 (soybean:water) by weight to facilitate grinding The resulting slurry is then passed through a filter press giving out a stream of pulp(31% solid) and a stream of milk (3.9 % solid). Calculate the amount of soymilk (sp.gr. = 1.08) produced in liters. It is desired that the temperature of one liter of soymilk (5% soild) be reduced from 85 oC to 10 oC. If ice at -1.4 oC is added in order to do this, what will be the percent solid of the dilute soymilk.?