Biological & Biosystems Engineering DISCUSSION

Grain drying: wet-basis vs dry-basis moisture content, and how much water must be removed?

Started by akira grain moisture contentwet basisdry basisdrying calculationlatent heat of evaporation
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Latest activity · 30 Sep 2026

Grain drying: wet-basis vs dry-basis moisture content, and how much water must be removed?

#1

I have to size a dryer for maize harvested at 24 percent moisture that must be brought down to 14 percent for storage. My first estimate was that drying 1000 kg removes 100 kg of water, since the moisture drops by ten percentage points, but my instructor says that is wrong.

What is the correct mass balance, what is the difference between wet-basis and dry-basis moisture content, and how do I estimate the energy needed?

Community replies 4

Re: Grain drying: wet-basis vs dry-basis moisture content, and how much water must be removed?

#2

Moisture content on a wet basis is water mass divided by total mass; grain trading and moisture meters use it. The total mass changes as the grain dries, so percentage points on a wet basis cannot simply be subtracted. The quantity that stays constant is the dry matter.

1000 kg at 24 percent contains 240 kg of water and 760 kg of dry matter. At 14 percent wet basis the dry matter is 86 percent of the total, so the final mass is 760 / 0.86 = 883.7 kg. Water removed = 1000 - 883.7 = 116.3 kg, not 100 kg. A compact formula: water removed = m_initial × (M1 - M2) / (1 - M2), with M as wet-basis fractions.

Re: Grain drying: wet-basis vs dry-basis moisture content, and how much water must be removed?

#3

Dry basis is water mass divided by dry matter mass, and it is the form used in drying theory because the denominator does not change. Conversion: M_db = M_wb / (1 - M_wb). So 24 percent wet basis is 0.24 / 0.76 = 31.6 percent dry basis, and 14 percent is 0.14 / 0.86 = 16.3 percent.

On a dry basis the subtraction does work: 760 kg × (0.316 - 0.163) = 116.3 kg of water, the same answer. Always state which basis a figure is on; at high moisture contents the two differ a lot.

Re: Grain drying: wet-basis vs dry-basis moisture content, and how much water must be removed?

#4

For energy, the floor is the latent heat of evaporation, about 2.4 MJ per kg of water at typical drying temperatures. 116.3 kg × 2.4 MJ/kg = 279 MJ, about 78 kWh per tonne of wet grain. Real heated-air dryers use noticeably more than this, because the exhaust air leaves warm and not fully saturated, the grain itself is heated, and there are casing losses. The specific energy consumption per kg of water evaporated is the figure to ask a dryer manufacturer for.

Multiply by the throughput to get burner power: 10 tonnes per hour means 1163 kg of water per hour, which is 775 kW at the theoretical minimum.

Re: Grain drying: wet-basis vs dry-basis moisture content, and how much water must be removed?

#5

The drying air does the transport, so check the air side as well. Heating ambient air lowers its relative humidity and raises the amount of water each kilogram of air can pick up; a psychrometric chart gives the figure as the difference in humidity ratio between inlet and exhaust. If each kg of dry air leaves with 10 g more water than it came in with, removing 116.3 kg of water needs 11,630 kg of air.

Air temperature is limited by the product: seed grain tolerates much lower temperatures than feed grain, because overheating damages germination and causes stress cracks. Grain also settles towards an equilibrium moisture content set by the air's temperature and humidity, so it can over-dry near the air inlet of a deep bed.

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