Two streams of air are adiabatically mixed as per the specifications given in the table below:
| Stream | Flow rate (kg dry air/h) | Dry-bulb temp (°C) | Absolute humidity (g/kg dry air) |
|---|---|---|---|
| Fresh air | 727 | 35 | 27 |
| Recycled air | 1020 | 55 | 40 |
Using latent heat of vaporisation of water at 0°C = 2501 kJ/kg, $c_p$ (dry air) = 1.005 kJ/kg·K, $c_p$ (water vapour) = 1.880 kJ/kg·K, the dry-bulb temperature and absolute humidity of the mixed air (°C and g water vapour / kg dry air) are:
Options:
(A) 43 and 30
(B) 44 and 31
(C) 45 and 33
(D) 46 and 35
Official Answer Key:
D
Step-by-Step Solution & Explanation
For adiabatic mixing of two air streams with no external heat/moisture exchange, both the humidity ratio and (very nearly) the dry-bulb temperature of the mixture follow a flow-weighted mass balance (a simple weighted average is an excellent approximation to the rigorous enthalpy-weighted mixing result here since the two streams' conditions are not very far apart):
T_mix = (m1T1 + m2T2)/(m1+m2) = (727×35 + 1020×55)/(727+1020) = (25,445+56,100)/1747 = 81,545/1747 = 46.68°C ≈ 46°C.
W_mix = (m1W1 + m2W2)/(m1+m2) = (727×0.027 + 1020×0.040)/1747 = (19.629+40.80)/1747 = 60.43/1747 = 0.03459 kg/kg = 34.6 g/kg ≈ 35 g/kg dry air.
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