# Gas mixture flow rate equation [closed]

2 gas flows connect and mix in a pipe. The following is known:

$\dot{m_1}$; $\dot{m_2}$ - mass flow rate of the inlet gases $(\frac{kg}{s})$

$A_1$; $A_2$; $A_{mix}$ - the cross section of the pipes $(m^2)$

$\rho_1$; $\rho_2$ - the densities of the 2 gases

$T_1 = T_2 = T_{mix}$

We are looking for the following:

$Q_{mix}$ - flow rate of the mixture $(\frac{m^3}{s})$

So far I have been able to get only this:

$\dot{m_1} + \dot{m_2} = \dot{m}_{mix}$ (1)

$\dot{m}=\rho*A*V$ (2)

$\dot{m} = \rho*Q$ (3)

$\dot{m_1} + \dot{m_2} = \rho_{mix}*Q_{mix}$ (4)

$Q_{mix} = \frac{\dot{m_1} + \dot{m_2}}{\rho_{mix}}$ (5)

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## closed as off-topic by Nathaniel, Dan, Dilaton, Qmechanic♦Jul 25 '13 at 13:16

This question appears to be off-topic. The users who voted to close gave this specific reason:

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It should be

$$\rho_{mix}= \frac{\dot m_1 + \dot m_2}{\dot Q_1 + \dot Q_2}$$

and not $$\rho_{mix}= \frac{\dot Q_1 + \dot Q_2}{\dot m_1 + \dot m_2}$$

You can easily verify this by considering $\dot m$ and $\dot Q$ as $\frac{dm}{dt}$ and $\frac{dV}{dt}$ respectively, where $V$ is the volume. So the equation becomes $$\rho_{mix}= \frac{\frac{dm_1}{dt} + \frac{dm_2}{dt}}{\frac{dV_1}{dt} + \frac{dV_2}{dt}}$$ $$\rho_{mix}= \frac{dm_1 + dm_2}{dV_1 + dV_2}$$ $$\therefore \rho_{mix}= \frac{dm_{total}}{dV_{total}}$$

and that is the standard definition of density($\frac mV$).

P.S. you should use $v$ instead of $V$ in equation (2). I was confusing it with volume instead of velocity till now!

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