How can I calculate pressure change due to reaction?

I'm studying a reactor, that can be imagined simply as a pipe in which a reaction advances.
In this reactor, the inlet and outlet flow are constant as is the temperature.

If that reaction is equi-molar no pressure change happens, otherwise in a reaction like steam reforming, where product moles are greater than reactant ones, I should expect an increase in pressure.

$$CH_4+H_2 O \rightleftharpoons CO+3H_2$$

How can I calculate the pressure change due to change in the number of moles?

PS: In a closed system (like a vessel) I'd use the ideal gas law: $pV=nRT$ ... but this is not the case!

steady state grants me $\dot m|_{in} = \dot m|_{out}$
I can express mass flow as $\dot m = \dot n \cdot MM$ where MM is molar mass
and as $\dot m = A \cdot u \cdot \rho$ where A is the cross sectional area (constant) u is mean speed and $\rho$ is density
I express density with ideal gas law $\rho = {p \over {R \cdot T}} \cdot MM$
finally I get $\dot n = A \cdot u \cdot {p \over {R \cdot T}}$
now, since A,R,T are constants taking the reaction abovementioned, if $\dot n$ rise up $u \cdot p$ have to follow this trend.

How, who, what, does make pressure constant while speed rise up ?

• Your math is correct; think of it this way: if the reactor pressure increased a little, it would cause the pressure gradient to go backwards, towards the inlet, causing fluid to flow in the wrong direction. The only two options to prevent this are to either compress the inlet stream to a higher pressure (driving the flow towards the outlet) or to keep the reactor pressure low by letting fluid exit at a higher velocity. Sorry, I'm on vacation and only have my phone, when I get home I'll update my answer with pictures to explain better. – Carlton Sep 19 '15 at 17:48
• good thought! I look forward your further stuff – mattia.b89 Sep 19 '15 at 23:01
• I updated my answer, hope it helps. – Carlton Sep 22 '15 at 17:24

The change in pressure depends on the design of the reactor, specifically its boundary conditions like pressure and volumetric flow rate at the inlet and outlet. Without any kind of compressor at the inlet, a reaction like the one you described will generally result in the reactor operating at the same pressure as the feed, but with the exit velocity and volumetric flow rate of the outlet stream being larger/faster than the feed. If the reactants and products are all in the gas phase and are ideal gasses, and the temperature is held constant, you can calculate the change in inlet flow rate to outlet flow rate (volumetric) with this equation:

$$\Delta \dot V = \xi \left( \nu_p - \nu_r \right)$$

Here, $\xi$ is the extent of reaction, and $\nu_p$, $\nu_r$ are the total stoichiometric numbers of the products and reactants. Note that if $\nu_p=\nu_r$, the reaction is equimolar and the volumetric flow rate entering the reactor is the same as the flow rate exiting the reactor (which is intuitive).

To better explain why the velocity increases while the pressure stays more or less the same, consider that you always need a negative pressure gradient through the reactor (assuming steady-state operation) so that the fluid will always flow from inlet to outlet. That is, the outlet pressure, $P_{out}$, must always be less than the inlet pressure $P_{in}$.

Doing a mass balance on the reactor shows that the mass flow rate coming out must be exactly equal to the inlet flow rate (no gains/losses):

$$\dot m = \rho_{in} u_{in} A = \rho_{out} u_{out} A$$

where $\rho$ is density, $u$ is velocity, and $A$ is the cross-sectional area of the reactor (constant). We also have the ideal gas equation of state (assuming the reactants and products are in-fact ideal gasses):

$$P=\rho RT = \rho \frac{\bar R}{MW} T$$

where $P$ is pressure, $R$ is specific gas constant, $\bar R$ is universal gas constant, $MW$ is the average molecular weight, and $T$ is temperature. We can solve the equation of state for $\rho$ and plug it in to the mass balance:

$$P_{in} \, MW_{in} \, u_{in} = P_{out} \, MW_{out} \, u_{out}$$

where $T$, $A$, and $R$ have canceled out. If we assume that the pressure gradient is very small, $P_{in} \approx P_{out}$ (just enough to get a little bit of flow), then we can also cancel the pressure terms and rearrange:

$$\frac{MW_{in}}{MW_{out}} = \frac{u_{out}}{u_{in}}$$

Now it should be more apparant that, if the outlet average molecular weight is less than the inlet (i.e. products have a higher stoichiometric number than the reactants), then the outlet velocity must be higher to satisfy our mass balance and equation of state.

• can you explain better why pressure should be the same? – mattia.b89 Sep 18 '15 at 10:24
• If there isn't a pump or compressor at the reactor inlet, then there's no way for the reaction to build up pressure, assuming everything is running at steady state. – Carlton Sep 18 '15 at 16:51
• that I don't understand! I added some other info in the OP – mattia.b89 Sep 19 '15 at 8:30
• sorry for my pedantry and stubbornness! but in this way momentum balance fails because if $p_{out}-p_{in}<0$ then $u_{out}-u_{in}<0$ or not? – mattia.b89 Sep 23 '15 at 8:30
• No, just the opposite. The momentum equation says that the velocity gradient is proportional to the negative of the pressure gradient, so if the pressure gradient is decreasing ($P_{out}<P_{in}$), then the velocity gradient should be increasing ($u_{out}>u_{in}$). See the Wikipedia article on the Euler equations; you can see that the pressure gradient term in the momentum equation has a negative sign. – Carlton Sep 23 '15 at 13:13

• this statement doesn't convince me... I believe (as you) that temperature is the main reason why pressure in an engine cylinder rise up independently by reactions that occur BUT this statement comes from ideal gas law: $pV=nRT$, so why pressure can't rise up due to a mole change? – mattia.b89 Sep 19 '15 at 23:12