Restoration Cert Lab · Free practical tool

Drying condition comparison

Compare affected, unaffected and outside air before explaining a structural-drying strategy. A cooler air sample with high relative humidity may still carry less water per mass of dry air.

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Air locations
Entry 1
Entry 2
Entry 3
Your result
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The method

Calculate humidity ratio and dew point independently for each air state at the common absolute pressure.

Before using the result

  • Compare moisture per mass of dry air rather than RH alone. Outdoor-air suitability also depends on temperature, contaminants and the work plan.

Worked examples

These original training examples illustrate the method for IICRC Water Restoration Technician (WRT). Use the matching “Load example” button above to explore a variation; example values are not equipment settings or prescribed criteria.

1. Worked example

Compare affected, unaffected and outside air before explaining a structural-drying strategy. A cooler air sample with high relative humidity may still carry less water per mass of dry air.

Absolute atmospheric pressure
101.325 kPa
Supplied air locations
LocationDry-bulb temperature (°C)Relative humidity (%)
Affected room2570
Unaffected room2250
Outside air1580
  1. Affected room: W = 13.921934095722953 g/kg dry air; dew point = 19.149933798669693 °C.
  2. Unaffected room: W = 8.224239345969671 g/kg dry air; dew point = 11.110092965013603 °C.
  3. Outside air: W = 8.488898908939891 g/kg dry air; dew point = 11.581588542887152 °C.
  • Locations compared: 3

2. Comparison example

Compare a different supplied condition using the same method. For Restoration Cert Lab, the important check is that the input meaning, units and method assumptions still match the problem.

Absolute atmospheric pressure
101.325 kPa
Supplied air locations
LocationDry-bulb temperature (°C)Relative humidity (%)
Affected room2860
Unaffected room2245
Outside air1080
  1. Affected room: W = 14.248500309832545 g/kg dry air; dew point = 19.514064572804493 °C.
  2. Unaffected room: W = 7.392040606333415 g/kg dry air; dew point = 9.533503738507774 °C.
  3. Outside air: W = 6.089102676754953 g/kg dry air; dew point = 6.713005486940737 °C.
  • Locations compared: 3

Check your understanding

1. Which relationship is used?

Calculate humidity ratio and dew point independently for each air state at the common absolute pressure.

2. Which assumption should you check first?

Compare moisture per mass of dry air rather than RH alone. Outdoor-air suitability also depends on temperature, contaminants and the work plan.

3. How should you verify a different unit system?

Convert every input to the displayed calculation units first. Equivalent physical inputs must give equivalent results; a unit label alone does not convert a number.

Continue with the matching exam topic

This tool connects to IICRC Water Restoration Technician (WRT) free practice. Work through a question independently, then use the calculation steps to identify where your method differs. The tool stays free whether or not you choose paid preparation.

Tool terms and notices

Drying condition comparison — free with working | Restoration Cert Lab