Calculate Relative Humidity and Water - Questions, Answers

Example 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

Cooling the air

When air is cooled, relative humidity increases. At one point air become saturated from water vapor. Then, with furthermore cooling, water vapor starts to condense. 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