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 balance | mair h1 + mw hw = mair h2 (1) |
| Water balance | mair W1 + mw = mair W2 |
| or | mw = mair (W2 - W1) (2) |
| From (1) and (2) | (h2-h1)/(W2-W1) = hw |
| or | h2 = hw (W2-W1) + h1 (2a) |
| | | | | |
| Units | Moist Air | Steam | Outlet Mixture | |
| Pressure | bar | 1.01325 | 1.01325 | 1.01325 | |
| Dry air flowrate | kgdry air/s | 2 | | 2 | |
| Dry bulb temperature | oC | 20 | 110 | 21.14 | |
| Wet bulb temperature | oC | 8 | | 16.08 | |
| Dew Point temperature | oC | -8.72 | | 13 | |
| Humidity ratio | kgw/kgdry air | 0.0018 | | 0.0094 | |
| Enthalpy of moist air | kJ/kgdry air | 24.69 | 2690.95 | 45.06 | solved by equation (2a) |
| Relative humidity | % | 12% | | 60% | |
| Steam flowrate | kg/s | | 0.0151 | | solved by equation (2) |