Adiabatic mixing of airstreams

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The discussion focuses on the adiabatic mixing of two airstreams with distinct temperatures and relative humidities. The first airstream has a temperature of T1=5°C and relative humidity of φ1=90%, while the second airstream has T2=25°C and φ2=50%, both at a flow rate of 20 m³/min. The calculated specific humidity of the mixed airstream is ω3=7.34×10-3 kg H2O/kg air, which aligns with the expected result. Additionally, the discussion includes calculations for molar flow rates and partial pressures of water vapor in both streams, crucial for determining the final temperature and relative humidity after mixing.

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I'm having a lot of trouble with an adiabatic mixing process.
This is what I know:
There's an airstream coming from outside and its temperature is T1=5 'C, its relative humidity being \varphi1= 90% at a rate of 20 m3/min.
Another airstream is coming from inside and its temperature is T2= 25 ´C, relative humidity \varphi2= 50%, at the rate of 20 m3/min.
The airstreams are mixed in a chamber with the pressure p = 1 bar.
I should find the densities of the first and second mixtures (meaning the airstreams including the water). This I've accomplished, the densities being \rho1=1.249 kg/m3 and \rho2= 1.162 kg/m3. I've also found the mass rates for the water in both streams; ^{.}mw1= 2.0 g/s and m^{.}w2= 3.8 g/s.

What I haven't been able to find is this: the specific humidity \omega3 of the mixed airstreams. I simply don't know what to do: I've tried finding first the specific humidities of the airstreams before mixing and stuff like that, but I just can't get the right result! This is what it should be: \omega3= 7.36*10-3 kg H2O/kg air.

Can someone please help me?

I should also find the temperature after mixing, as well as the new relative humidity, but I don't know how to do that either...
 
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The molar density of a gas is given by the equation $$\rho=\frac{P}{RT}$$At 25 C and 1 bar, this is equal to 40.36 moles/m^3 and at 5 C and 1 bar, it is equal to 43.27 moles/m^3.

At 5 C, the equilibrium vapor pressure of water is 0.0087 bars, and at 25 C, it is 0.0317 bars. So, for the 5 C stream, at 90% humidity, the partial pressure of water vapor is (0.0087)(0.9)=0.0078 bars, and, for the 25 C stream, at 50% humidity, the partial pressure of water vapor is 0.0159 bars. These are also the mole fractions of the water vapor in the streams (since the total pressure is 1 bar).

The molar flow rate of the 5C air stream is (20)(42.27)=845.4 moles/minute, and the molar flow rate of the 25 C stream is (20)(40.36) = 807.2 moles/min. The water vapor flow rate in the 5 C stream is (845.4)(0.0078)=6.59 moles/min, and the water vapor flow rate in the 25C steam is (807.2)(0.0159)=12.83 moles/min. So that total molar flow rate of water vapor is 6.59 + 12.83 = 19.43 moles/min and the total flow rate of combined air is 845.4 + 807.2 = 1661.4 moles/min. So the mole fraction water vapor in the combined stream is 19.43/1661.4 = 0.0117 .

Taking as a basis 1 mole of air, the mass of water vapor is (0.0117)(18) = 0.211 gm and the mass of bone dry air is (1-0.0117)(29)=28.66 gm. So the specific humidity is 0.211/28.66 = 0.00734 kg/kg.

This agrees with the answer provided.
 
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