# Calculate Relative Humidity and Water - Questions, Answers

Example 1

A tank has 1m^{3} volume and contains air. It is initially at 50^{0}C air and it is started to cool
upto to 20^{0}C. Initial relative humidity (RH) is 80%
and finally air is saturated due to cooling process. Calculate followings.

- Total amount of water vapor in the tank in kg?
- 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
- 50
^{0}C , P_{1}^{0} = 12.171 kPa
- 20
^{0}C , P_{2}^{0} = 2.302 kPa

### 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 (m_{1}) in the initial air should be calculated. Then final mass of water
vapor (m_{2}) is calculated. Difference between m_{1} and m_{2} is the answer.
( m_{1} > m_{2} )

Condensed Water Vapor Mass = m_{1} - m_{2}

## Equations of Relative Humidity

Relative humidity (RH) = Water Vapor Pressure (P) / Saturated Water Vapor Pressure (P^{0})

RH = P/P0

- Saturated Water Vapor Pressure (P
^{0}) depends on the temperature.

### Two states

State 1

- P
_{1} = water vapor pressure at state 1
- RH = 80%
- P
_{1}^{0} = 12.171 kPa

State 2

- P
_{2} = water vapor pressure at state 2
- RH = 100% (saturated)
- P
_{2}^{0} = 2.302 kPa

### Determine pressure in state 1

- P
_{1} = RH * P_{1}^{0}
- P
_{1} = (80 /100)* 12.171
- P
_{1} = 9.737 kPa

### Determine pressure in state 2

- P
_{2} = RH * P_{2}^{0}
- P
_{2} = (100/100)* 2.302 = 2.302 kPa
- P
_{2} = 2.302 kPa

### Determine moisture (water vapor) mass in the state 1

Apply PV = mRT for water vapor

P_{1} = 9.737 kPa

- 9.737 * 10
^{3} * 1 = m_{1} * 461.5 * 323
- m
_{1} = 0.065 kg
- m
_{1} = 65 g

### Determine moisture (water vapor) mass in the state 2

Apply PV = mRT for water vapor

P_{2} = 2.737 kPa

- 2.737 * 10
^{3} * 1 = m_{2} * 461.5 * 293
- m
_{2} = 0.065 kg
- m
_{2} = 17 g

Condensed water vapor mass = m_{1} - m_{2}

Condensed water vapor mass = 48 g