Adiabatic mixing of airstreams

[Appa]

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 $$\varphi$$₁= 90% at a rate of 20 m³/min.
Another airstream is coming from inside and its temperature is T₂= 25 ´C, relative humidity $$\varphi$$₂= 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 $$\rho$$₁=1.249 kg/m³ and $$\rho$$₂= 1.162 kg/m³. I've also found the mass rates for the water in both streams; $$^{.}$$m_w1= 2.0 g/s and m$$^{.}$$_w2= 3.8 g/s.

What I haven't been able to find is this: the specific humidity $$\omega$$₃ 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: $$\omega$$₃= 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...

 

[Chestermiller]

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.