# Volumetric flow rate ratio calculation

I'm using two tanks of nitrogen and oxygen and mixing them in different ratios to change the oxygen partial pressure in the system. System is opened to the atmosphere through the venting pipe which therefore sets the total pressure of the system to 1 bar. System is opened and gases flow through it. Schematic is provided to give better insight:

To create atmospheric composition of the mixture, mole fraction of oxygen needs to be 0.21 while nitrogen needs to be 0.79. Therefore, molar flow rate ratio, $$A$$ is: $$A = \frac {\dot n(O_2)}{\dot n(N_2)} = \frac {0.21}{0.79} = 0.266$$

Gas tanks are equipped with volumetric flowmeters. Relating volumetric and molar flow rates can easily be done if the assumption of ideal gas is valid. Since I'm using air at atmospheric pressure and high temperatures (500 – 800 $$^\circ$$C), the assumption should hold.

Volumetric flow ratio is: $$\frac {\dot V(0_2)}{\dot V(N_2)} = \frac {\frac {\dot n(O_2)RT}{P(O_2)}} {\frac {\dot n(N_2)RT}{P(N_2)}} = A \frac {P(N_2)}{P(O_2)}$$

Since we're talking about the ideal gas, Dalton's law holds. Partial pressure ratio is equal to the mole fraction ratio of the components in the mixture. Therefore: $$\frac {P(N_2)}{P(O_2)} = \frac {1}{A}$$

Leading to the volumetric flow rate ratio being always equal to 1. This doesn't seem correct, what am I doing wrong?

I think you are wrongly relating the variables to the flow streams. Maybe the following image can help orienting yourself:

• Nitrogen and oxygen amount balances \begin{align} y_{\ce{N2},3}\dot{n}_3 = (1)\dot{n}_1 \rightarrow \dot{n}_1 = 0.79\dot{n}_3 \tag{1} \\ y_{\ce{O2},3}\dot{n}_3 = (1)\dot{n}_2 \rightarrow \dot{n}_2 = 0.21\dot{n}_3 \tag{2} \\ \end{align} and taking the quotient we obtain $$\dot{n}_2/\dot{n}_1 = 0.21/0.79$$ which is what you obtained.

• Those volumetric flow rates from your second equation refer to streams $$1$$ and $$2$$ $$$$\frac{\dot{V}_2}{\dot{V}_1} = \dfrac{\dfrac{\dot{n}_2RT_2}{P_2}}{\dfrac{\dot{n}_1RT_1}{P_1}} \rightarrow \frac{\dot{V}_2}{\dot{V}_1} = \left(\frac{\dot{n}_2}{\dot{n}_1}\right) \left(\frac{P_1}{P_2}\right) = \left(\frac{0.21}{0.79}\right) \color{blue}{\left(\frac{P_1}{P_2}\right)} \tag{3}$$$$ since $$T_1 = T_2$$ as I see in your math. This is again what you obtained. However, the streams refer to $$1$$ and $$2$$. Thus, $$P_1$$ is the pressure of the nitrogen tank, and $$P_2$$ the pressure of the oxygen tank.

• The mistake comes when you make the quotient in the last equation. We have that $$$$\frac{P_{\ce{N2},3}}{P_{\ce{O2},3}} = \frac{y_{\ce{N2},3}P_3}{y_{\ce{O2},3}P_3} \rightarrow \color{blue}{\frac{P_{\ce{N2},3}}{P_{\ce{O2},3}}} = \frac{0.79}{0.21} \tag{4}$$$$ and the question is, are the blue parts in Eqs. (3) and (4) the same? The answer is no. One thing are the pressures of the pure compounds in their respective tanks, and another one the partial pressures of the constituents in stream $$3$$. A clear example would be an initial situation where the tanks are at the same pressure, hence $$P_1/P_2 = 1$$. However, the relation $$P_{\ce{N2},3}/P_{\ce{O2},3}$$ cannot be equal to one, and is in fact $$0.79/0.21$$. The pressure in stream $$3$$, upon mixing, is almost $$P_3 = \pu{5 atm}$$.

If you want to ensure, you have to do more calculations where the unknowns are $$\dot{V}_3$$ and $$P_3$$.

Firstly, we have the global balance of amount of substance \begin{align} \dot{n}_3 &= \dot{n}_1 + \dot{n}_2 \\ \rho_3\dot{V}_3 &= \rho_1\dot{V}_1 + \rho_2\dot{V}_2 \\ \left(\frac{\color{red}{P_3}}{RT_3}\right)\color{red}{\dot{V}_3} &= \left(\frac{P_1}{RT_1}\right)\dot{V}_1 + \left(\frac{P_2}{RT_2}\right) \dot{V}_2 \tag{5} \\ \end{align} Secondly, we have the global conservation of mechanical energy, under the assumptions that there is no work transfer and the friction is absent \begin{align} \left(\frac{P_3}{\rho_3} + \frac{\color{red}{v_3}^2}{2} + gz_3\right)\dot{m}_3 = \left(\frac{P_1}{\rho_1} + \frac{v_1^2}{2} + gz_1\right)\dot{m}_1 + \left(\frac{P_1}{\rho_2} + \frac{v_2^2}{2} + gz_2\right)\dot{m}_2 \tag{6} \end{align} where the $$\dot{m}$$'s are the mass flow rates, and $$z$$'s are the heights. This is Bernoulli's equation but with three streams. Note that in Eq. (6), the only unknown is $$v_3$$ since $$P_3/\rho_3 = RT_3/M_3$$ which is a value that you can calculate. After this, you can calculate $$\dot{V}_3 = v_3S$$, where $$S$$ is the area perpendicular to the flow. Afterwards, go back to Eq. (5) to know the final pressure $$P_3$$.

• Yes, I thought that the pressure on which the gas is used, $P_3$ is what relates mass and volumetric flow rate. I'm not sure why this isn't the case on intuitive level, though. Why is it that pressure in the tank relates the two? Jan 24 at 12:06
• If that's the case, that means that the volumetric flowrate increases as the tank is emptied (pressure in the tanks drops) to keep the mass flow rate constant. Mass flow rate should remain constant in time due to mass conservation. Jan 24 at 12:08
• Oh well, then the situation is mathematically more involved. In here, all the balances are stationary (don't depend on time). I imagined that your tanks were an enormous source of nitrogen an oxygen, and they will not get depleted as to decrease its pressure (or $\dot{m}$ or $\dot{V}$). You should see that mass and volumetric flow rates upstream have constant values throughout the experiment. Is that the case? Jan 24 at 12:13
• I think $P_3$ is the total pressure of the mixture which is set to the atmospheric value as the system is opened to the influence of the atmosphere through the venting pipe. Jan 24 at 12:15
• For the first seconds of the run, all the variables will change a little and then they will stabilise (they acquire a steady state value). For example, you go from 0 $\pu{m^3/h}$ in the volumetric flow rates to some constant value. These balances focus on what happens after that transient time. Jan 24 at 12:24