In 1825, Michael Faraday began a tradition of annual Christmas lectures at the Royal Institution in London, intending to inspire young audiences with the wonders of science. Despite uninspiring titles such as “Course of Six Elementary Lectures on Chemistry, adapted to a Juvenile Auditory,” his lectures often contained deep amounts of physics. In “The Chemical History of a Candle,” he claimed:
There is not a law under which any part of this universe is governed which does not come into play, and is touched upon in these phenomena.Indeed, his candle lectures included the capillary action required to feed the flame of an oil lamp, the formation of vortices and their appearance as tongues of flame, the process of combustion, the pressure and density of gases, and the chemistry of respiration. He did not, however, discuss the height of a candle flame. It is this basic shortcoming that we amend here.
Consider combustion of the wax vapor: a hydrocarbon molecule burns with oxygen to produce carbon dioxide and water. Each paraffin wax molecule is a long chain, containing some number of carbons — as order-of-magnitude physicists, we’re not sure how many, but it will hardly matter. The molecule must look like C\(_u\)H\(_{2u+2}\), since each carbon (valence 4) is bonded with two hydrogens and two carbons, except at the ends of the chain where it has an additional two hydrogens. Balancing the equation to produce carbon dioxide and water, we need around \(3u/2\) oxygen molecules. Hence, for a mass of \(m \sim 14u\cdot N\) of \(N\) molecules of wax, we need \((3m/28N)(N) \sim 0.1m\) moles of oxygen to fully combust our \(m\) grams of fuel. This defines our combustion ratio $$\begin{aligned} C \sim 0.1~\mathrm{mol/g} \end{aligned}$$ which specifies the number of moles of O\(_2\) required to combust one gram of wax.
Visiting the Yankee Candle website informs me that a 20-ounce candle jar takes 60-90 hours to burn, or about 0.002 grams/second. Of course, a significant fraction of the weight is the jar and not the wax, so perhaps \(b = 0.0015\) grams/second is a better estimate of our burn rate. Taking a molar fraction \(f_{\mathrm{O}_2} \sim 0.2\) of oxygen in air and applying the ideal gas law, we require a volume flow rate \(r\) of air to burn the candle given by $$\begin{aligned} r \sim \left(\frac{Cb}{f_{\mathrm{O}_2}}\right)V_0, \qquad V_0 = \frac{N_Ak_BT}{P}, \end{aligned}$$ where \(N_A\) is Avogadro’s constant, \(k_B\) is Boltzmann’s constant, \(P\) is the pressure, and \(T\) is the temperature. The quantity \(V_0\) is the volume of one mole of air. The rate \(r\) has units of volume per time; to obtain the characteristic length of scale of a candle flame, we might expect to divide by the diffusion coefficient \(\kappa\) (area per time), i.e., \(H=r/\kappa\). Diffusion can be notoriously difficult to get factors of \(\pi\) correct, so let’s be a bit careful. Writing the diffusion coefficient as \(\kappa \sim L^2/\tau\) gives a handy physical interpretation: a particle can diffuse a distance \(L\) in time \(\tau\). For simplicity, let’s rewrite our volumetric flow rate as \(r=V/\tau\), arbitrarily choosing the same time scale as \(\kappa\). Imagine the flame as rising directly up from the wick of the candle: at any given height, air up to a radius \(L\) away can diffuse into the axis of the wick within time \(\tau\). Consequently, the height \(H\) of the cylinder that encloses the full volume \(V\) required for combustion has height \(H = V/\pi L^2 = r/\pi\kappa\). Here, the cylindrical geometry induced by the flame incurred one factor of \(\pi\).
To estimate the diffusion coefficient, we take the product of the mean free path of a traveling molecule and the speed at which it moves. Assuming gas molecules are spheres of radius \(s = 3~\mathrm{Å}\), the mean free path is easily expressed in terms of number density and molecule size: $$\begin{aligned} \lambda \sim \left[\left(\frac{N_A}{V_0}\right)s^2\right]^{-1} \sim \left[\left(\frac{P}{k_BT}\right)s^2\right]^{-1}. \end{aligned}$$ The speed at which gas molecules move is given by the speed of sound, $$\begin{aligned} v \sim \sqrt{\frac{P}{\rho}} \sim \sqrt{\frac{k_B T}{m_\mathrm{air}}}, \end{aligned}$$ where \(m_\mathrm{air}\) is the mass of an air molecule. Evaluating \(\kappa\) gives $$\begin{aligned} \kappa \sim \lambda v \sim \frac{(k_BT)^{3/2}}{Ps^2\sqrt{m_\mathrm{air}}}, \end{aligned}$$ which yields a flame height of $$\begin{aligned} H \sim \frac{r}{\pi \kappa} \sim \frac{1}{\pi}\left(\frac{Cb}{f_{\mathrm{O}_2}}\right)\sqrt{\frac{m_\mathrm{air}}{k_BT}}N_As^2 \sim \frac{1}{\pi}\left(\frac{Cb}{f_{\mathrm{O}_2}}\right)\frac{N_As^2}{v}. \end{aligned}$$ We see that the speed of sound has been restored as a quantity, meaning we don’t have to remember Boltzmann’s constant. We know that at room temperature \(T_0\), the speed of sound is \(v_0 \sim 343\) m/s; hence, at temperature \(T\), the speed of sound is \(v_0\sqrt{T/T_0}\). Although the flame is hot, the diffusion into the flame occurs from the ambient environment, so perhaps we don’t need to correct for the temperature of the flame itself. Plugging in numbers gives a candle flame height of 4 cm for a large, fast-burning household candle and 2 cm for a small candle [1].
For a bonus round, let’s see if we can also say something about the width of the flame. Suppose the flame travels upwards due to the buoyancy of hot air at flame temperature \(T_f\). From the ideal gas law, it will accelerate upwards at about \(a = (T_f/T - 1)g\) given room temperature \(T\). Neglecting the terminal velocity of hot air traveling in cold air, kinematics give a time of \(t = \sqrt{2h/a}\) for the fire to travel up to the flame height \(h\). Given our diffusion constant \(\kappa = w^2/t\) that sets the scale of how far air can diffuse in a given time, we then expect a width of $$\begin{aligned} w \sim \sqrt{\kappa t} \sim \sqrt{\frac{(k_BT)^{3/2}}{Ps^2}\sqrt{\frac{2h}{(T_f/T-1)m_\mathrm{air}g}}}, \end{aligned}$$ which is about 0.3 cm for a wax boiling temperature of 1500 K and a flame height of 3 cm. We successfully recovered that candle flames are taller than they are wide, which is comforting!