# Moisture Content

## Content of moisture in products like wood - wet and dry basis

** Moisture content** is the amount of water present in a moist sample of a product like wood, soil or similar. Moisture content can be be expressed on wet or dry basis.

### Moisture Content on Dry Basis

Moisture content on dry basis is the *mass of water* to the *mass of dry solid*:

*MC _{d} = m_{h2o} / m_{d }(1)*

*where *

*MC _{d} = moisture content on dry basis*

*m _{h2o} = mass of water (kg, lb) *

*m _{d} = mass of dry solid (kg, lb) *

Moisture content on dry basis is commonly used in the timber industry. Note that it is common to multiply values with 100%.

### Water Content on Wet Basis

Water content on wet basis is the *mass of water* to the *mass of water and mass of solid*:

*MC _{w} = m_{h2o} / m_{w }*

_{ }= * m _{h2o} / m_{h2o} + m_{d }*(2)

*where *

*MC _{w} = moisture content on wet basis*

*m _{h2o} = mass of water (kg, lb) *

*m _{w} = total mass of moist - or wet - sample - mass of solid and mass of water (kg,lb) *

### Dry vs. Wet Moisture Calculator

*m _{H2o} - mass of water (kg, lb)*

* m _{d} - mass of dry solid (kg,lb)*

*m _{H2o} - mass of water (kg, lb)*

* m _{w} - total mass (kg,lb)*

### Example - Water Content in Birch on Wet Basis

The density of air-dried seasoned dry Birch is *705 kg/m ^{3}* with

*20% (0.2)*water content. The amount of water per unit volume can be calculated by transforming (2) to

*m _{h2o} = m_{w }MC_{w} *

* = (705 kg/m ^{3}) (0.2)*

* = 141 kg/m ^{3}*

The amount of solids can be calculated as

* m _{d }*=

*(705 kg/m*

^{3}) (1 - 0.2)* = 564 kg/m ^{3}*

### Example - Moisture Content in Birch on Dry Basis

The density of air-dried seasoned dry Birch is *705 kg/m ^{3}* with

*20% (0.2)*moisture content. Equation (1) can be modified to

*MC _{d} = m_{h2o} / (m - m_{h2o}) _{ }(1b)*

*where *

*m = mass water and solid (kg, lb)*

This equation can be transformed to

* m _{h2o} = MC_{d} m / (1 + MC_{d} ) (1c)*

_{ }

The moisture content per unit volume can be calculated as

*m _{h2o} = (0.2) (705 kg/m^{3}) / (1 + (0.2)) *

* = 117.5 kg/m ^{3}*

The amount of solids can be calculated as

* m _{d }*= m -

*m*

_{h2o}_{ }

* = (705 kg/m^{3}) - (117.5 kg/m^{3})*

* = 587.5 kg/m ^{3}*

### Moisture and Water Content Calculator

*mass (kg, lb)*

* moisture/water content (%)*