Thermodynamics - Quasi 1-D Air Mixture

[Kolodny]

Homework Statement

Constant area mixing tube with stations (1) and (2) (i.e., (1)-->-->--(2))
At station (1):
- Mixture of two air streams, defined as Primary (p) and Secondary (s)
u_p = 300 m/s
T_p = 900K
u_s = 30 m/s
T_s = 300K
p_1 = 0.1 MPa
(A_p)/(A_s) = 1/3

1-D flow assumed

Find: Pressure, Velocity, and Temperature at station (2)

Homework Equations

$\dot{m}_{1}=\dot{m}_{2}$ (Continuity)
$\dot{m}_{1}h_{01}=\dot{m}_{2}h_{02}$ (1st Law)
$\sum \vec{F}=\frac{d}{dt}\int{\rho\vec{u}(\vec{u}\hat{n})dV}+\int{\rho\vec{u}(\vec{u}\hat{n})dA}$(Conservation of Momentum)
$p=\rho R T$ (perfect-gas law)

The Attempt at a Solution

From Continuity:
$\dot{m}_{p}+\dot{m}_{s}=\dot{m}_{2}$
$\rho_{p}u_{p}A_{p}+\rho_{s}u_{s}A_{s}=\rho_{2}u_{2}A_{2}$
$\frac{1}{4}\rho_{p}u_{p}+\frac{3}{4}\rho_{s}u_{s}=\rho_{2}u_{2}$

From 1st law:
$\dot{m}_{p}(h_{p}+\frac{1}{2}u_{p}^{2})+\dot{m}_{s}(h_{s}+\frac{1}{2}u_{s}^{2})=\dot{m}_{2}(h_{2}+\frac{1}{2}u_{2}^{2})$ (incidentally, anyone know why this LaTeX isn't working properly?)
$\frac{1}{4}\rho_{p}u_{p}(c_{p_{p}}T_{p}+\frac{1}{2}u_{p}^{2})+\frac{3}{4}\rho_{s}u_{s}(c_{p_{s}}T_{s}+\frac{1}{2}u_{s}^{2})=\rho_{2}u_{2}(c_{p_{2}}T_{2}+\frac{1}{2}u_{2}^{2})$

From Conservation of Momentum:
$\int{\rho\vec{u}(\vec{u}\hat{n})dA}=p_{1}-p_{2}$
$\frac{1}{4}\rho_{p}u_{p}^{2}+\frac{3}{4}\rho_{s}u_{s}^{2}-\rho_{2}u_{2}=p_{1}-p_{2}$

I can get $\rho_{2}u_{2}$ and the stagnation enthalpy at station 2, but beyond that I'm not sure how to proceed. I feel like there's an assumption that I'm missing which would allow me to calculate $p_{2}$ or some other condition at station 2, and then use the perfect-gas law to fill things out, but I'm not sure what that is. I freely admit that there's probably something about how the conditions necessary to apply these equations interact which I'm not understanding.

 

[rude man]

Sorry you have no replies yet. I personally don't know what a mixing tube is. A diagram would help.