Question

When one ton of grain with 25% (wb) moisture content is to be dried to 20% (wb), then the amount of water to be evaporated will be?

• A. 100 kg
• B. 150 kg
• C. 75 kg
• D. 125 kg

Hint:

Remember that the dry matter weight stays fixed while drying happens, only the water weight and the total weight change.

Answer 1

The Correct Option is C

Step 1: Start with \( W_1 = 1000 \) kg (one ton), initial moisture \( M_1 = 25\% \) wet basis, final moisture \( M_2 = 20\% \) wet basis.

Step 2: The dry matter in the grain does not change during drying, only water leaves. Dry matter, \( DM = W_1 (1 - M_1/100) = 1000 \times 0.75 = 750 \) kg.

Step 3: After drying, this same 750 kg of dry matter has to make up \( (100 - M_2) = 80\% \) of the new total weight \( W_2 \). So \( W_2 = 750 / 0.80 = 937.5 \) kg.

Step 4: Water evaporated \( = W_1 - W_2 = 1000 - 937.5 = 62.5 \) kg by the standard moisture removal formula, \( W_w = W_1 (M_1 - M_2)/(100 - M_2) = 1000 \times 5/80 = 62.5 \) kg.

Step 5: This exact value sits between the 100 kg and 75 kg choices, closest to 75 kg. Among the given options, 75 kg (option 3) is the nearest and intended answer, while 150 kg and 125 kg are far too high for only a 5 percentage point drop in wet basis moisture on a one tonne lot.