Skip to main content

Home  Services  Products  Downloads  Support  Order  About  Contact  Site Map   
 

MoistAirTab™ Example 5 - Humidification

Example 5

Steam (or liquid water) injection is frequently used in air conditioning for humidification. Moist air at 20°C db and 8°C wb is being humidified by injecting saturated steam at 110°C to a final stream with a dew point of 13°C. The rate of dry air flow is 2 kg/s. What is the final db of the moist air and what is the steam injection rate?

Solution

The solution is presented below as a table generated from Excel. The gray shaded areas represent the input data.

Key concepts:.
1. Adiabatic mixing assumes that no net gain or loss of total energy in the mixing process
2. The dew point temperature specifies the moisture content in the final mixture.
Overall Enthalpy balancemair h1 + mw hw = mair h2  (1)
Water balancemair W1 + mw = mair W2
ormw = mair (W2 - W1) (2)
From (1) and (2)(h2-h1)/(W2-W1) = hw
orh2 = hw (W2-W1) + h1 (2a)
UnitsMoist AirSteamOutlet Mixture
Pressurebar1.013251.013251.01325
Dry air flowratekgdry air/s22
Dry bulb temperatureoC2011021.14
Wet bulb temperatureoC816.08
Dew Point temperatureoC-8.7213
Humidity ratiokgw/kgdry air0.00180.0094
Enthalpy of moist airkJ/kgdry air24.692690.9545.06solved by equation (2a)
Relative humidity%12%60%
Steam flowratekg/s0.0151solved by equation (2)