The manufacturers of a new oatmeal product want to determine the least expensive
ID: 531090 • Letter: T
Question
The manufacturers of a new oatmeal product want to determine the least expensive way to safeguard and transport their product. They chose lyophilisation (freeze drying), which removes most of the water from the product, making it lighter and therefore less costly to transport over long distances, and substantially increases its shelf life. Lyophilisation also removes some of the impurities introduced by industrial farming methods. A solid feed stream containing 70 wt% oatmeal, 27% water (ice), and the balance organic impurities enters a heated vacuum chamber at a rate of 1 times 10^3 kg/hr at a temperature of -10 degree C. In the vacuum chamber, 97% of the water and 99% of the organic impurities in the feed sublime (vaporise). The dried product in then packaged and shipped. A vapour stream emerges from the chamber at 15 degree C. During the process, 7.95 times 10^5 kJ/hr of heat is transferred to the system. (a) Draw and completely label a flowchart of the process and perform a degree-of-freedom analysis. (b) Calculate the compositions and flow rates of the product and waste streams. (c) Find the temperature of the product stream, using the following values for heat capacity and heat of sublimation and neglecting the contribution of the organic impurities to the energy balance. C_p[H_2O(s)] = 2.11 kJ/(kg. degree C) C_p[H_2O(v)] = 1.86 kJ/(kg. degree C) C_p[oatmeal] = 1.5 kJ/(kg. degree C) (Delta H_subl)_H2O = 28 45 kJ/kg (d) Evaporation is a more conventional way to dry a wet solid - that is, heating the solid at atmospheric pressure to a temperature close to the boiling point of the liquid and holding it there long enough to evaporate all of the liquid. Give at least two reasons why freeze drying is a better alternative for drying the oatmeal.Explanation / Answer
Known :feed compostion and flow rate
Unkown : vapor , flow rate and product flow rate.
there are 3 components and there are 3 unknowns in the product.
overall mass balance - 1 equation
component balance - 2 equations , 1 for water and 1 for orgnic matter.
degree of freedom= no of variables- no of independent equations= 3-3=0
Flow rate of solid feed= 1000 kg/hr, it contains 70 wt% oatmeal= 700 kg/hr oatmeal, 27% water= 270 kg/hr water and organic matter= 1000-700-270= 30 kg/hr
Water to the tune of 97% is vaporized. Hence water vaporized= 0.97*270= 262 kg/hr and water remaining at 15 deg.c in the product= 270-262= 8 kg/hr. Oranic matter in the dried product = 30*1/100= 0.3 kg/hr
Vapor stream from the dryer contains = 262 kg/hr water and 30-0.3= 29.7 kg/hr organic matter.
Compostion of waste : water vapor= 262/(262+29.7)= 89.8% , organic matter= 100-89.8=10.2%
Product contains 700 kg/hr oatmeal, water= 8 kg/hr and organic matter= 0.3 kg/hr
Composition of product : Oat meal= 700/(700+8+0.3)= 98.8%,water= 8/708.3= 1.13 and organic matter= 100-98.8-1.13=0.07%
262 kg/hr of water has to be evaporated from initial temperature of -10 deg.c to 15 deg.c. This is done by
This is heat added to vaporize and this heat= 5275+711250+5625+1114=723264 Kj/hr
Heat supplied= 795000 Kj/hr
Heat added to the product stream = 795000-723264= 71736 Kj/hr
Product contains 700 kg/hr oat meal, 8 kg/hr water and 0.3 kg/hr organic matter. Let their temperature of product be T.
Hence 700*1.5*(T+15) +8* 2.11*(T+15)= 71736
T+15= 67, T= 52 deg.c
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