The air flows in against the heat

Incompressible is not a property of the fluid separated the two assumptions an incompressible calculation makes without saying so. One is thermodynamic: that pressure changes are too small to move the density. That is the statement that the Mach number is small, since the density change a flow’s pressure makes is about half the square of its Mach number. The other is caloric: that nothing else changes a parcel’s density — no heating, no mixing, no change of phase. That one has nothing to do with the Mach number, and the essay named the flame as the flow that breaks it.

It named it and did not compute it. This essay does. The result is sharper than a warning that heat can matter. In a gas at constant pressure, heat and volume are the same quantity in different units, and the velocity a gas acquires is set by where its heat is going. The clearest case has no flame in it at all: a spot of hot air cooling in still air, at a Mach number so small that no textbook would give it a second thought.

A joule is a volume #

At a low Mach number the pressure in a gas is uniform to within a fraction of order $M^2$, so for an ideal gas the product ρT is fixed: a parcel’s volume is proportional to its temperature. Continuity says the divergence of the velocity is the rate at which a parcel’s volume grows, relative to its size, so $\nabla\cdot\mathbf u = (1/T)\,DT/Dt$. The energy equation says what changes the temperature. At constant pressure a parcel’s temperature rises with the heat released inside it, $\dot q$, and falls with the heat it conducts away, $\nabla\cdot\mathbf q$, where q is the heat flux:

$$\rho c_p \frac{DT}{Dt} = \dot q - \nabla\cdot \mathbf q .$$

For an ideal gas $\rho c_p T = \gamma p/(\gamma - 1)$, and substituting gives the whole of the result:

$$\nabla\cdot\mathbf u = \frac{\gamma - 1}{\gamma p}\left(\dot q - \nabla\cdot\mathbf q\right).$$

Every joule added to a region at constant pressure adds (γ − 1)/(γp) to its volume, and every joule conducted out takes the same away. For air at one atmosphere that is 2.8 cubic centimetres per joule. The Mach number appears nowhere. A gas at rest whose temperature is changing has a divergence, and it is as large as the heat makes it.

The velocity is the heat flux turned round #

For a sphere cooling in still gas the divergence can be integrated at once. By symmetry the velocity is radial, and the volume flux out through a sphere of radius r is the divergence integrated over the ball inside it. The part coming from $\nabla\cdot\mathbf q$ integrates to the heat flowing out through the same sphere:

$$u(r) = -\frac{\gamma - 1}{\gamma p}\, q(r).$$

The gas moves at the heat flux, reversed. Heat flows outward from the hot spot, so the gas flows inward, everywhere, at every moment, towards the spot that is losing heat. There is no outflow anywhere. The spot’s hot core is contracting as it cools, and the gas round it moves in to fill the space.

The computation follows the gas rather than a fixed grid. The sphere is divided into 240 shells of fixed mass. Each shell’s volume is its mass times its temperature, since the density is the inverse of the temperature in these units. Heat conducts between neighbouring shells with a conductivity rising as $T^{0.8}$, as air’s does. The shells’ radii are recomputed from their volumes after every step, so the velocity is the shells’ own motion and owes nothing to the formula above. The heat flux is computed separately, from the temperature gradient and the conductivity. The two agree to 7 × 10⁻⁵ of the peak flux, which is the size of one time step’s error.

In units of the spot’s radius a and the time $a^2/\alpha$, the inflow peaks at 2.7 α/a just inside the spot’s edge at the first moment drawn, and falls as the spot cools. For a spot of a millimetre radius at 900 kelvin in air at 300 kelvin, α/a is 0.022 metres a second and $a^2/\alpha$ is 45 milliseconds. A heat flux of order kΔT/a, 16 kilowatts a square metre, drives the gas at 4.4 centimetres a second. That is Mach 1.3 × 10⁻⁴, and the gas it moves changes its volume by a factor of three.

The hot gas gives its volume away #

Follow the gas that started inside the hot sphere. As it cools its volume falls: from 2.89 radius-cubes at the first moment drawn to 1.52 by t = 1. The gas that started cold round it warms and grows by exactly the same amount. The total volume does not change at all — computed at two times, it holds to 7 × 10⁻¹⁶, which is rounding error.

That is not a coincidence of the numbers, and it is not the outer boundary holding anything in. At constant pressure an ideal gas’s volume is (γ − 1)/(γp) times its enthalpy. Conduction moves enthalpy from place to place and creates none, so it moves volume from place to place and creates none. The hot spot does not push any air away as it cools; it hands its volume to its neighbours. Far from the spot, where the heat has not yet arrived, nothing moves. Mass has nowhere to go states the kinematic rule that mass is conserved. Here it is volume that is conserved, for a different reason: at constant pressure volume is a way of counting energy.

A flame accelerates its gas by its density ratio #

A flame is the same identity with the heat-release term switched on. In a steady planar flame the mass flux ρu is the same at every point, since nothing accumulates, and ρT is constant, since the pressure is. So the velocity is proportional to the temperature: $u/S_L = T/T_u$, where $S_L$ is the speed at which the flame consumes fresh gas. A flame whose burnt gas is seven times hotter than its fresh gas sends its products away seven times faster than the fresh gas arrives. For a hydrocarbon flame burning at 40 centimetres a second that is 2.8 metres a second out of a zone half a millimetre thick, still at a Mach number below 10⁻².

The divergence is confined to the flame, peaks at its middle, and integrates across it to the velocity jump, six times $S_L$. The integral matches to 4 × 10⁻⁹. By the identity it must be (γ − 1)/(γp) times the heat released per unit area per second, and it is: the jump $S_L(T_b/T_u - 1)$ is that heat release written in velocity. This is also why a flame cannot be computed with an incompressible solver even though sound plays no part in it. The solver would impose $\nabla\cdot\mathbf u = 0$ and delete the expansion that is the flame’s main mechanical effect.

A heated pipe speeds its own gas up #

The flame’s rule — the velocity follows the temperature — holds for any steady gas flow along a channel of fixed cross-section, heated or cooled through its walls. The mass flux is the same at every section, so as the gas warms and thins it must move faster. Air entering a tube at 300 kelvin and 10 metres a second and leaving at 600 kelvin leaves at 20. Nothing pushed it harder: it expanded, and the only direction it could expand in was along the tube.

That acceleration has to be paid for, and the payment is a pressure drop. The momentum flux $\rho u^2$ rises from entry to exit as ρu × u, the mass flux times a velocity that has doubled, and the pressure must fall by the same amount: $\Delta p = \rho_1 u_1^2(T_2/T_1 - 1)$, 118 pascals in this example. That is about the friction of half a metre of a centimetre-bore tube at that speed, and an incompressible pressure-drop calculation leaves it out entirely, because it takes the velocity to be constant along a duct of constant area. Heat-exchanger designers call it the acceleration pressure drop and book it separately.

Carried to high speed, the same effect is the Rayleigh line of compressible flow: heating a subsonic gas in a duct drives it towards Mach 1, and two ways to choke shows that heat and friction both reach the same sonic limit. At low speed nothing chokes, but the mechanism is already there, and it is this essay’s identity. The heat adds volume, and a duct turns added volume into speed. How far down the tube the heating takes effect — how far before the heat arrives — then sets where along it the gas speeds up.

What an incompressible model keeps, and what it drops #

The model that most often handles heated low-speed flow is the Boussinesq approximation. It keeps the density difference where gravity multiplies it, in the buoyancy force, and sets the divergence to zero everywhere else. The identity here says exactly what that costs. The divergence it drops is (γ − 1)/(γp) times the net heating, which relative to the flow’s own strain rate is of order the fractional temperature difference, ΔT/T. For a layer of water a few degrees warmer at the bottom that is a few thousandths and the approximation is excellent. For air over a radiator, or in a fire, it is of order one and the approximation is simply wrong.

The departures show up first as asymmetries. A Boussinesq layer heated from below is statistically the same upside down, and its convection cells are rolls. Let its properties vary with temperature and the symmetry goes, and hexagonal cells appear, as hexagons remember how the heat was turned up computes. The low-Mach equations used for fires and flames go the other way: they keep the whole of the heat-driven divergence and still filter out sound, so their pressure obeys a Poisson equation as an incompressible solver’s does — pressure has no speed in either. The difference is a source term, the rate of change of the divergence that heat is making.

Where the Mach number is the wrong question #

The standard test for incompressibility is the square of the Mach number: air at Mach 0.1 changes its density by about one per cent, at Mach 0.3 by about nine, and a flow below Mach 0.3 is conventionally treated as incompressible. The test is right about what it measures. It measures what pressure does to density. The two heated flows sit a thousand times lower in Mach number and change their volume by factors of three and seven. The ordering of these four flows by Mach number is exactly the reverse of their ordering by compressibility.

So the question to ask of a low-speed flow is not how fast it is going but whether any parcel is gaining or losing heat. A gas flow with walls at the same temperature as the gas, and no reaction, is incompressible at low Mach number. A gas flow over a hot plate, through a heat exchanger, in a room with a radiator, or past a flame is not, however slowly it moves. The equations built for it — low-Mach-number or variable-density equations — keep the pressure uniform in the equation of state and keep the divergence in the continuity equation, because the first is small and the second is not.

The same fact heard as sound #

The identity has a consequence that is heard rather than seen. If the heat released in a region fluctuates, the volume it adds fluctuates with it, and a region whose volume pulses is a monopole: the most efficient source of sound there is. The sound that only leaves explains why a flow on its own is so poor a radiator — it has no monopole available and must make its sound from quadrupoles that nearly cancel. A flickering flame has a monopole, of volume strength (γ − 1)/(γp) times its fluctuating heat release. That is why combustion is loud out of all proportion to the speed of the gas.

It is also the mechanism of thermoacoustic instability. Rayleigh stated the criterion in 1878: heat added to a gas where the pressure of a sound wave is high, and removed where it is low, feeds the wave. In the terms used here, the heat release pumps volume into the gas in phase with the compression, which is work done on the sound. A gauze heated in the lower half of a vertical tube — Rijke’s tube of 1859 — sings for that reason, and the same coupling shakes gas-turbine combustors and rocket engines hard enough to destroy them.

What the picture cannot show #

The hot spot has no buoyancy. The calculation keeps the spot spherical and still, which is true only for a time shorter than the time for it to rise. A millimetre spot in air begins to move upward within tens of milliseconds, the same order as its cooling time, and a larger one rises before it has cooled at all. The inward flow is still there, carried inside a rising plume.

The pressure is uniform, not constant. The identity uses the uniform thermodynamic pressure, which is what the low-Mach limit keeps. The small dynamic pressure that drives any real flow is still there. It does not change the density to first order, but it is what the momentum equation needs to decide the part of the velocity field that has no divergence — every flow is two flows, and the heat decides only one of them.

The flame’s profile is a model. A real flame’s temperature rises through a preheat zone and a reaction zone of different thicknesses. The drawing uses one smooth step. The velocity-follows-temperature result does not depend on the shape; the divergence’s shape does.

The gas is ideal and does not radiate. At flame temperatures radiation carries some of the heat, which the identity would count as a heat flux like any other, and real burnt gas has a changing heat capacity, which shifts the numbers by some per cent without changing the argument.

The convention the numbers depend on #

The hot spot is measured in units of its starting radius a, time in $a^2/\alpha_\infty$ where $\alpha_\infty$ is the surrounding gas’s thermal diffusivity, temperature in the surrounding temperature, and density in the surrounding density, with $c_p = 1$. In those units ρT = 1 and (γ − 1)/(γp) = 1, so the velocity equals the heat flux reversed with no factor. The conductivity rises as $T^{0.8}$. The flame is measured in its own thickness δ and its burning speed $S_L$, with a temperature ratio of seven. The dimensional scales use air at 300 kelvin and one atmosphere, with k = 0.0262 W/m/K and $\alpha = 2.2\times10^{-5}\ \mathrm{m^2/s}$.

How each number was checked #

The calculation is built so that its checks are not the calculation restated. The volume is computed from the shells’ temperatures and the enthalpy from the same temperatures with a different weight. That they stay equal is the constant-pressure identity, and the scheme could break it if its fluxes did not balance between shells. The velocity comes from the shells’ motion and the heat flux from the temperature gradient, and they could disagree if the time stepping were inconsistent. The gas that started hot must shrink monotonically and the total must not move. The flame’s divergence must integrate to its jump. The calculation refuses a negative temperature ratio, fewer than twenty shells, a flame whose burnt gas is colder than its fresh gas, and a volume tolerance below zero.

Who found it, and when #

That an ideal gas’s volume at fixed pressure is proportional to its temperature is Charles’s and Gay-Lussac’s, from around 1800. The low-Mach-number equations that keep it while filtering out sound were set out by Rehm and Baum in 1978 for fire plumes, and by Majda and Sethian in 1985 for combustion. The flame’s velocity jump is as old as the study of flames: Darrieus and Landau used it in the 1930s and 1940s to show that a flat flame is unstable because its expansion bends the flow in front of it. Rayleigh’s criterion for thermoacoustic instability is from 1878, and Rijke’s singing tube from 1859.

Still open: the flame front the expansion bends #

The flame here is flat and steady. The Darrieus–Landau instability says it cannot stay that way. The expansion across a curved flame deflects the fresh gas ahead of it, and the deflection makes the curvature grow, at a rate set by the density ratio alone. The calculation that follows would compute that growth rate from the density ratio of seven and show why every large laboratory flame is wrinkled. Beside it is the hot spot with gravity switched on: the inward flow of this essay against the upward flow of buoyancy, and the time at which the second takes over from the first.