Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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)

share|improve this question

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:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Nathaniel, Dilaton, Qmechanic
If this question can be reworded to fit the rules in the help center, please edit the question.

1 Answer 1

up vote 0 down vote accepted

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!

share|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.