Moisture Content, Water Activity & Equilibrium Moisture Content Notes

Moisture Content, Water Activity & Equilibrium Moisture Content

Forms of water in grain, moisture content on wet and dry basis with conversions, shrinkage and weight calculations, water activity and its link to microbial growth, equilibrium moisture content (EMC) and equilibrium relative humidity, sorption isotherms and hysteresis, common EMC models (Henderson, Chung–Pfost, Halsey, Oswin, GAB), safe storage moisture levels, and methods of measuring moisture, with worked examples.

Last reviewed 1 Oct 2026 · 6 min read

Moisture content

Agricultural materials hold water as: free (surface) water, capillary water in pores, adsorbed/bound water on surfaces, and chemically bound water (hydrates). Free water dries first; bound water needs more energy to remove.

Wet basis and dry basis

FormulaDefinitions

Let $W_{w}$ = mass of the wet sample, $W_{d}$ = mass of dry matter (bone dry), $W_{w} - W_{d}$ = mass of water.

Wet basis (w.b.): $M_{w b} = \frac{W _{w} - W _{d}}{W _{w}} \times 100$ — used in trade, standards and marketing (cannot exceed 100 %).

Dry basis (d.b.): $M_{d b} = \frac{W _{w} - W _{d}}{W _{d}} \times 100$ — used in engineering calculations of drying (can exceed 100 %).

Conversions: $M_{d b} = \frac{M _{w b}}{100 - M _{w b}} \times 100$ and $M_{w b} = \frac{M _{d b}}{100 + M _{d b}} \times 100$ .

Worked ExampleExample — conversion

A grain at 20 % w.b. has $M_{d b} = \frac{20}{80} \times 100 = 25 %$ . A grain at 14 % w.b. has $M_{d b} = \frac{14}{86} \times 100 = 16.3 %$ . Note that the numerical difference is larger on d.b. as the moisture rises.

Mass change on drying

The dry matter stays constant, so for a lot of initial mass $m_{0}$ at $M_{0}$ (w.b., %) dried to $M_{f}$ (w.b., %):

m_{f} = m_{0} \frac{100 - M _{0}}{100 - M _{f}} water removed = m_{0} - m_{f} = m_{0} \frac{M _{0} - M _{f}}{100 - M _{f}}

Worked ExampleExample — water removed

1000 kg of paddy at 22 % w.b. is dried to 13 % w.b.

$m_{f} = 1000 \times \frac{78}{87} = 896.6 kg$ ; water removed $= 1000 - 896.6 = 103.4 kg$ . (If a seller is paid for 1000 kg of wet grain, 10 % of the weight is just water that has to be dried off.)

Mixing lots of different moisture

For lots $m_{1}$ at $M_{1}$ and $m_{2}$ at $M_{2}$ (both w.b.), the mixture moisture is the mass-weighted average of water: $M = \frac{m _{1} M _{1} + m _{2} M _{2}}{m _{1} + m _{2}}$ . (Dry-matter-weighted averaging is needed on a d.b.)

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