Potato that has reached equilibrium with drying air at 60 degree C and a humidit
ID: 108423 • Letter: P
Question
Potato that has reached equilibrium with drying air at 60 degree C and a humidity of 0.04 kg of moisture per kg of dry air is packaged in a bag that is impermeable to moisture. The bag initially had air at 60 degree C and 0.008 kg of moisture per kg of dry air. We would like to know what will be the final moisture content of the potato in the sealed bag. Consider the package to have 95% by weight of potato and 5% by weight of air. 1) Formulate the problem, i.e., write the equation from which the final moisture content of the potato in the package can be calculated. 2) Solve for the final moisture content in step 1. The equilibrium moisture content of potato at 60 degree C is shown in Figure 9.15.Explanation / Answer
As potato dries, it releases its moisture into the drying air and consequently loses weight.
The weight of potato after drying may be found using the following equation:
W2 = W1 – W1(M1-M2) / 100 – M2
where:
W1 = Weight of undried potato (kg)
W2 = Weight of dried potato (kg)
M1 = Moisture content of undried potato (percent)
M2 = Moisture content of dried potato (percent).
Now, if 0.038 kg of potatoes at 32 percent moisture content are dried to 19 percent moisture content, what is the weight of the dried potato?
W2 = 0.038 – 0.038(32-19) / 100 – 19 = 0.038 – 32.1 = 1679 kg.
When the moisture content of the potato to be dried has been determined, it is possible to check the progress of the drying process by using the following procedure. Then, using the initial weight, the initial moisture content and the newly observed weight in the following equation, the current moisture content at that specific level may be calculated:
M2 = 100 – W1 (100-M1) / W2
Therefore the final moisture content is = 93.3 %.
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