I am interested in deriving an expression for how an instantaneous exothermic chemical reaction (where the molecules comprising the fluid are converted to a mix of smaller molecules) process influences the velocity of a compressible fluid flow. The fluid flows across a circular boundary and the surface area of the flow is constant before and after the reaction.

I will have to assume the mass stays constant throughout. There will be some increase in pressure due to the exothermic nature of the reaction, which in turn affects the compressibility of the fluid, causing the velocity to increase. I obtained an expression showing this, however I am not convinced I have the full story.


Using the relation $V_m=\frac{M}{\rho}$, we obtain a relationship for the density of the fluid relative to the molar mass of the fluid. Which, after the chemical reaction will have decreased. Now, using the continuity of the fluid we have $$\frac{M_1}{V_{m_1}}(\pi r_1^2)v_1=\frac{M_2}{V_{m_2}}(\pi r_2^2)v_2$$ From here the two areas must be equal in order for the surface area to stay constant, we have: $$\frac{M_1}{M_2}\frac{V_{m_2}}{V_{m_1}}v_1=v_2$$ I am sure that there must be some formula relating the temperature change to the increase in molecules or something along those lines, however I have absolutely no experience with chemistry at all. I could assume the fluid to follow the ideal gas law and get an expression with temperature but this of course will affect the fidelity of the derivation, I am not sure whether it is a valid assumption.

If someone would be able to point me in the right direction or advise that would be great!


3 Answers 3


The problem can be addressed using macroscopic thermodynamics without referring to the molecular picture of the fluid. We however need to know something about the chemical composition of the fluid after the reaction. I suppose that not all of your fluid undergoes the chemical reaction, so the resulting fluid is going to be multi-component. Below, I sketch a solution of the problem under the assumption that all the components move with the same macroscopic velocity, so one can treat the fluid after the reaction as a single-component matter with some average thermodynamic properties. A more detailed solution would depend on the exact nature of your fluid, e.g. on whether the ideal-gas approximation is appropriate.

The state of the fluid is completely described by its temperature $T$, pressure $P$, density $\rho$ and velocity $v$. I assume, as you indicated, that the geometry of the problem is one-dimensional so that the velocity can be treated as a scalar. I will use indices 1 and 2 to refer to the fluid before and after the reaction. There is one relation between $T$, $P$ and $\rho$ supplied by the equation of state, so we need three additional relations to fully determine the unknown values of $T_2$, $P_2$, $\rho_2$ and $v_2$.

Mass conservation. Given that the cross-section area of the flow does not change, this amounts to the simple condition $$ \rho_1v_1=\rho_2v_2, $$ equivalent to your equation.

Momentum balance. The flow of momentum per unit area and per second in a fluid of velocity $v$ is $\rho v^2$, which is just the mass flow times velocity. Then the second law of Newton requires that $$ P_1-P_2=\rho_2v_2^2-\rho_1v_1^2. $$

Energy balance. The above two constraints are purely mechanical. Here is the only place where we need some input about the chemical reaction in the system. I will model the reaction as an instantaneous supply of heat per kilogram equal to $q$. Then the energy balance condition can be expressed in terms of increase of specific enthalpy $h$ (enthalpy per kilogram) plus specific kinetic energy (which is just $\frac12v^2$), $$ h_2+\frac12v_2^2=h_1+\frac12v_1^2+q. $$ The enthalpy itself is in turn determined by the temperature and pressure. For an ideal gas, it can be extracted for instance from the knowledge of specific heat at constant pressure.

The above set of equations together with the equation of state is sufficient to determine the final state of your fluid. As said, a more concrete solution can only be given in case you know the equation of state of your fluid. Hope this helps anyway.

EDIT. In classical thermodynamics, thermodynamic potentials such as enthalpy are only defined up to an arbitrary additive constant, which can be chosen independently for different materials. In order that the comparison of $h_1$ and $h_2$ makes sense, we therefore have to fix the reference values of the enthalpy of the fluid before and after the reaction appropriately. In my answer, I assume that the specific enthalpies of these two different fluids are equal at temperature $T_1$ and pressure $P_1$.

  • $\begingroup$ In the absence of the knowledge of the equation of state, is simply the continuity equation the best we can do? Also, I am aware that the mass will not stay constant and there are many assumptions made in my post which do not reflect accurately the real world scenario, however I wonder how poor of an assumption the ideal gas law is? $\endgroup$
    – benmcgloin
    Commented Dec 26, 2021 at 13:31
  • $\begingroup$ @benmcgloin You cannot solve this problem without knowing anything about the equation of state. We need sufficient amount of information to pin down the thermodynamic state of the fluid uniquely. You say that your fluid is compressible, but all real fluids are. Is it a liquid or a gas then? For gases, the ideal gas equation of state is usually a good approximation, unless you are too close to the critical point. I am also wondering what you mean by saying that mass will not stay constant. Can you elaborate on this point? $\endgroup$ Commented Dec 26, 2021 at 16:35
  • $\begingroup$ @benmcgloin The ideal gas equation of state alone is still not enough to solve the problem. In order to be able to calculate the enthalpy, we need to know something about the specific heat at constant pressure of the gas. Is the specific heat (at least approximately) independent of temperature? If yes, then the sufficient amount of input (apart from initial velocity, temperature and pressure or equivalent data) would be the gas constant of the fluid before and after the reaction, and the constant-pressure specific heat before and after the reaction. $\endgroup$ Commented Dec 26, 2021 at 16:38
  • $\begingroup$ Well I have almost no experience with chemistry but I was under the impression if there is heat flow, the mass of the fluid must change. I was simply interested in the physics behind this it is not a problem I encountered. I am clearly being far too general, I appreciate the insight. $\endgroup$
    – benmcgloin
    Commented Dec 26, 2021 at 16:46
  • $\begingroup$ @benmcgloin The mass of the fluid is made up by the masses of the protons, neutrons and electrons inside. Chemical reactions change bonds between atoms, but cannot affect the total number of these atomic constituents. So the total mass must very accurately stay constant in any chemical reaction. The "very accurately" is just for nitpickers. Due to the relativistic equivalence between mass and energy, the change in chemical binding energy will result in a tiny change of mass. The energy of chemical bonds is of the order of eV, which is nine orders of magnitude smaller than nucleon rest mass. $\endgroup$ Commented Dec 26, 2021 at 19:18

Consider a differential control volume element $dV$ (m$^3$) with a constant reaction rate $\hat{r}$ (mols/m$^3$ s). The materials balance equation inside the volume for a species when a chemical reaction occurs is

$$ d\dot{n} = \hat{r}\ dV $$

where $\dot{n}$ is the molar flow rate in/out for the species (mols/s) and $r$ is the reaction rate (mols/s). Consumption is negative and production is positive.

We know that $\dot{n} = v\ A/ \bar{V}$, where $v$ is the net velocity for the species in/out of the control volume (m/s), $A$ is the area for flow in/out (m$^2$), and $\bar{V}$ is the molar volume (m$^3$/mol). Molar volume is inversely proportional to (mass) density $\rho$ as $\bar{V} = M/\rho$, where $M$ is the molar mass for the species. Therefore

$$ d\dot{n} = \left(\frac{A}{\bar{V}}\right)dv + \left(\frac{v}{\bar{V}}\right)dA - \left(\frac{v\ A}{\bar{V}^2}\right)d\bar{V} $$

This expression states how molar flow rate changes in the control volume as a function of the factors that define ti. The first term accounts for changes in expansion or contraction, the second for changes in area, and the third for changes in molar volume (inverse mass density).

When we have no change in flow area $dA = 0$. Recognize that $dV/V = d\hat{V}/\hat{V}$. The final expression becomes

$$ \hat{r}\ dV = \left(\frac{A}{\bar{V}}\right)\left[dv - \left(\frac{v}{V}\right)dV\right] $$

Consider that the control volume is at specific temperature and pressure $T, p$ with a steady-state reaction throughout. The homogeneous reaction and steady-state assumptions allow us to recognize that all concentrations for all species remain constant and that molar volume remains constant. The change in net velocity $\Delta v$ for a species in a fluid that passes through such a control volume $V$ across surface area $A$ with a reaction rate $\hat{r}$ and an molar volume is

$$ \Delta v = \frac{\hat{r}\ V\ \bar{V}}{A} $$

The above makes physical sense. Consider a reaction A -> B that creates 1 mol B per m$^3$ per second. Run this reaction in a 1 m$^3$ control volume where the fluid molar density is constant at 1 mol per m$^3$. Flow A across an input area of 1 m$^2$, with B exiting across an area of 1 m$^2$. The net change in velocity of B from input (zero) to output is 1 m/s.

All that remains is to apply the relevant expressions for $\bar{V}(p, T)$ inside the control volume. By example, an ideal gas yields

$$ \Delta v = \frac{\hat{r}\ V\ R\ T}{A\ p} $$

Recognize that $p$ is the partial pressure for the species. A reaction for a component in a liquid might be written using the component mass density $\rho$ and molar mass $M$.

$$ \Delta v = \frac{\hat{r}\ V\ M}{\rho(p,T)\ A} $$

Analysis for the behavior of species in reaction volumes that are non-homogeneous could be done by returning to the differential form and integrating the expanding expression appropriately over the entire volume. In such cases, $\bar{V}$ may not be constant, so that $dv$ and $v$ can change in each increment along the integration path.

  • $\begingroup$ The mass flow of a single species is $\dot m=A\rho v$, and it changes because of the chemical reaction. What you are saying is basically that since $A$ is constant (by the geometry of the problem) and $\rho$ is constant (by your assumption), then the velocity must change to account for the change of the mass flow. Why should the assumption that density stays constant hold? For reactions that are not associated with any heat production, I would expect the exact opposite: the velocity to stay constant and the density (of each species involved) to change according to the chemical reaction. $\endgroup$ Commented Dec 26, 2021 at 21:03
  • $\begingroup$ @TomášBrauner I've added a statement that my analysis is for a homogeneous control volume under steady state conditions. How do you propose to keep the velocity of product B in the reaction A -> B at zero (its starting velocity) and increase the density of B without exploding the control volume? $\endgroup$ Commented Dec 26, 2021 at 21:33
  • $\begingroup$ I don't think it's correct to assume that the initial velocity of $B$ is zero. It is its density that is initially zero. Think of an initial state that contains mostly $A$ and a small (even infinitesimal) admixture of $B$, both with the same velocity. Then the chemical reaction will convert part of $A$ into $B$. But if the reaction does not produce any heat, then it perfectly consistent with all conservation laws to conclude that the final velocity of both components in the mixture will remain the same as in the initial state, only the densities of $A$ and $B$ will change. $\endgroup$ Commented Dec 26, 2021 at 21:55
  • $\begingroup$ Isn't it physically strange that your formula for $\Delta v$ does not depend at all on the initial motion of the fluid? Consider the limit in which the initial velocity $v$ of the species $A$ goes to zero. Would your formula still predict the same value for $\Delta v$? In other words, would it predict that the chemical reaction necessarily produces motion in the fluid? $\endgroup$ Commented Dec 26, 2021 at 22:00
  • 1
    $\begingroup$ @TomášBrauner Things that do not exist have no initial velocity. You have none, then you have some, then you make more. This is unsteady state at best and confusion at worst. A control volume (CV) with $v_A = 0$ at its input will support no chemical reaction A -> B inside. The $\Delta v$ is a change for a component after going through a CV (not while in a CV). A CV can move with the bulk fluid. A chemical reaction changes velocities for each species. With only A entering and only B leaving a CV, the change in molar flow change for A is the negative of the change in molar flow for B. $\endgroup$ Commented Dec 26, 2021 at 23:08

I have made success on this kind of issues by calculating the Mean Free Path of the molecules. This is very easy if you have the density and molar weight. Obviously this is not quite right, as the particles of condensed matter are not free, but you can just think that they move their thermal movements on a fixed path of this path.

After that you can basically calculate (rms) particle velocity from the known gas laws; $$v_{rms}=\sqrt{\frac{3k_BT}{m}}$$

Then you can consider molecules to be point particles and think that they travel with this velocity against each other. This is -again- not quite correct and you could use also some particle diameter to reduce the mistake, but as you can't get fully rid of this mistake you can as well start with the point particle-concept due to it's simplicity.

Now with this kind of thought set up you can easily ad the exothermic heat to your single molecules and turn it to kinetic energy and again to velocity, with the help of molecule masses, and you can also then calculate these back to temperature.

You can also calculate the pressure change from these; For that you need

  • the change of molecular bounces on time, which can be calculated from the speed and free path.
  • and the change on the kinetic bounce of the molecule; calculated from the molecular mass and velocity.
  • Think (simplify the system) that the acceleration part of Pressure force remains the same, causing the only variables being 1. the mass of the molecule and 2. the time interval of the elastic collisions causing the pressure.
  • $\begingroup$ As T. Brauner stated in his anser the generality of the question doesnt allow for a concrete answer fully. By using the mean path are we not inherently assuming ideal flow of the fluid? Do you know if this is a valid assumption for this kind of setup? As I mentioned in my comment above I wonder how far this deviates from reality. $\endgroup$
    – benmcgloin
    Commented Dec 26, 2021 at 13:37
  • $\begingroup$ @benmcgloin Yes, as I write in the answer this is not quite right. but then "Newtons Gravitation" or "Ideal gas law" aren't either. I expect this is a valid way to estimate the Change on some range, but if this range is too big, you might need to add correcting factors. The mean free path argument stays valid even if there is flow in the fluid. The Answer of Brauner is basically just about basic fluid dynamics without adding anything to you problem setting about "how an instantaneous exothermic chemical reaction". Enthalpy is the sum of internal energy ="mean particle velocity and free path" $\endgroup$
    – Jokela
    Commented Dec 26, 2021 at 21:23
  • $\begingroup$ @Jokela To the extent that the chemical reaction can be treated as instantaneous, the problem is about basic fluid dynamics. You can find many similar problems in textbooks on (engineering) thermodynamics. $\endgroup$ Commented Dec 27, 2021 at 8:26
  • $\begingroup$ @TomášBrauner Yes. To exactly that extent! I assumed them to be continuous, and happening during the flow. Anyway; as you seem to be quite expert on fluid dynamics too, so pls. do check these out; (Lab result supported solution to NS-Existency & Smoothness) researchgate.net/publication/… and researchgate.net/publication/… $\endgroup$
    – Jokela
    Commented Dec 28, 2021 at 18:15

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