Calculate Relative Humidity and Water Vapor Mass in Air - Definition and Problems

Relative humidity is a parameter which express condition of air. If relative humidity is too high, it makes uncomfortable for human body.

Humidity is expressed in two ways, relative humidity and absolute humidity. Relative humidity is expressed as a percentage and absolute humidity is expressed as moisture content per unit volume.


In this tutorial, we first look definitions and equations of relative humidity and then discuss the problems.


What is saturated air?

When air is saturated, air cannot hold more water vapor and start to condense as liquid water.


Relative humidity definition - RH

the amount of water vapour present in air expressed as a percentage of the amount needed for saturation at the same temperature.

Relative humidity does not have units and it depends on temperature. RH is expressed as a percentage.


Relative humidity equations

Relative humidity equation is expressed in two ways, from amount (mass) and vapor pressure.


Relative humidity equation in water vapor amount (mass)


relative humidity equation in mass basis

Relative humidity equation in water vapor pressure


relative humidity equation in vapor pressure

Variation of saturated vapor pressure against temperature

Saturated vapor pressure increases with increase of temperature.

Variation of water vapor pressure against temperature

Download Variation of water vapor pressure against temperature - Excel sheet

Understand why relative humidity increases when temperature is decreased

When temperature increases, saturated water vapor pressure increases. Then look the equation of relative humidity which was expressed in pressure terms. In that equation, saturated vapor pressure term is below the division. So, when that number decreases ( with temperature decrement), related humidity value increases until air gets saturated.


But remember that, absolute humidity remains constant while relative humidity decreases.



Problem 1

A tank has 1m3 volume and contains air. It is initially at 500C air and it is started to cool upto to 200C. Initial relative humidity (RH) is 80% and finally air is saturated due to cooling process. Calculate followings.

  1. Total amount of water vapor in the tank in kg?
  2. After saturating, the total condensed mass of water inside the tank?

You are provided following data.

  • Water vapor constant (R) = 461.5 J/kg K
  • Vapor pressure of water at
  • 500C , P10 = 12.171 kPa
  • 200C , P20 = 2.302 kPa

Humidity changes with cooling - saturated air

Assumptions you have to make in this question

  • Water vapor acts as ideal gas.
  • No volume change.


Answer


Relative humidity increases when temperature decreases.

When air is cooled, relative humidity increases because water vapor amount needed for saturation is also decreased. At one point, air is saturated from water vapor. Then, with furthermore cooling, water vapor starts to condense to liquid water. With that, mass of water vapor in the air decreases.


First, mass of water vapor (m1) in the initial air should be calculated. Then final mass of water vapor (m2) is calculated. Difference between m1 and m2 is the answer. ( m1 > m2 )

Condensed Water Vapor Mass = m1 - m2



Equations of Relative Humidity

Relative humidity (RH) = Water Vapor Pressure (P) / Saturated Water Vapor Pressure (P0)

RH = P/P0

  • Saturated Water Vapor Pressure (P0) depends on the temperature.

Two states


State 1

  • P1 = water vapor pressure at state 1
  • RH = 80%
  • P10 = 12.171 kPa

State 2

  • P2 = water vapor pressure at state 2
  • RH = 100% (saturated)
  • P20 = 2.302 kPa

Determine pressure in state 1

  • P1 = RH * P10
  • P1 = (80 /100)* 12.171
  • P1 = 9.737 kPa

Determine pressure in state 2

  • P2 = RH * P20
  • P2 = (100/100)* 2.302 = 2.302 kPa
  • P2 = 2.302 kPa


Determine moisture (water vapor) mass in the state 1

Apply PV = mRT for water vapor

P1 = 9.737 kPa

  • 9.737 * 103 * 1 = m1 * 461.5 * 323
  • m1 = 0.065 kg
  • m1 = 65 g

Determine moisture (water vapor) mass in the state 2

Apply PV = mRT for water vapor

P2 = 2.737 kPa

  • 2.737 * 103 * 1 = m2 * 461.5 * 293
  • m2 = 0.065 kg
  • m2 = 17 g

Condensed water vapor mass = m1 - m2

Condensed water vapor mass = 48 g



We have solved above problem from equations. Also we can use psychrometric chart to solve this problem very easily. If you did not learn psychrometric chart, learn how to read pstchrometric chart and its axes.