Thermodynamics - Quasi 1-D Air Mixture

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SUMMARY

The discussion focuses on solving a thermodynamics problem involving a constant area mixing tube with two air streams, referred to as Primary (p) and Secondary (s). The parameters provided include velocities (u_p = 300 m/s, u_s = 30 m/s), temperatures (T_p = 900K, T_s = 300K), and pressures (p_1 = 0.1 MPa). The key equations utilized include the continuity equation, the first law of thermodynamics, and the conservation of momentum. The participant seeks to determine the pressure, velocity, and temperature at station (2) but struggles with the assumptions necessary for applying the perfect gas law and calculating p_2.

PREREQUISITES
  • Understanding of 1-D flow dynamics in fluid mechanics
  • Familiarity with the continuity equation and conservation of momentum
  • Knowledge of the first law of thermodynamics
  • Proficiency in applying the perfect gas law (p = ρRT)
NEXT STEPS
  • Study the principles of compressible flow in thermodynamics
  • Learn about the application of the perfect gas law in varying conditions
  • Explore the concept of stagnation enthalpy and its relevance in flow problems
  • Investigate the characteristics and applications of mixing tubes in fluid dynamics
USEFUL FOR

Students and professionals in mechanical engineering, particularly those specializing in thermodynamics and fluid mechanics, will benefit from this discussion. It is also relevant for anyone involved in the design and analysis of mixing systems in engineering applications.

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

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