= Differential thermal model of a sun valve
{title2=$\delta=\alpha_LL(T_b-T_r)$}
Take a stipulated pair of effective rods with common <length> $L$, <linear thermal expansion coefficient> $\alpha_L$, <heat capacity> $C$, projected illuminated <area> $A$, and <thermal conductance> $G$ to an environment at constant <temperature> $T_a$. One reference represents the symmetric polished-rod assembly. Write $\theta_b=T_b-T_a$ and $\theta_r=T_r-T_a$. Let their effective <radiometric absorptances> be $a_b>a_r$ for a fixed incident <optical spectrum>, and let $E(t)$ be the common <irradiance>. A spatially uniform, linear heat-balance model gives
$$
C\dot\theta_b=a_bAE(t)-G\theta_b,\qquad
C\dot\theta_r=a_rAE(t)-G\theta_r.
$$
Here $G$ includes a linear approximation to the relevant losses; changing the surface treatment can also change those losses in a real device. We stipulate equal $C$ and $G$ to isolate the response to absorbed illumination, with thermal coupling from the burner neglected. Small, mechanically unconstrained extensions then give a differential displacement
$$
\delta=\alpha_LL(\theta_b-\theta_r),\qquad
\dot\delta+\frac{\delta}{\tau}
=\frac{\alpha_LLA(a_b-a_r)}{C}E(t),\qquad
\tau=\frac CG.
$$
The <time constant> $\tau$ describes thermal memory. A common imposed <temperature> shift cancels from the extension difference under the matched-coefficient assumptions; mismatched materials, geometry or thermal coupling can spoil that cancellation.
For a step from darkness to constant <irradiance> $E_0$, with initially equal rod <temperatures>, the <first-order linear differential equation> gives
$$
\boxed{\delta(t)=\delta_\infty(1-e^{-t/\tau}),\qquad
\delta_\infty=\frac{\alpha_LLA(a_b-a_r)E_0}{G}.}
$$
Choose $L=0.20\,\mathrm m$, $\alpha_L=20\times10^{-6}\,\mathrm{K^{-1}}$, $A=0.0020\,\mathrm{m^2}$, $a_b=0.9$, $a_r=0.1$, $C=5.0\,\mathrm{J\,K^{-1}}$, $G=0.20\,\mathrm{W\,K^{-1}}$ and $E_0=600\,\mathrm{W\,m^{-2}}$. The calculated difference of absorbed <powers> is $0.96\,\mathrm W$, giving
$$
\boxed{\tau=25\,\mathrm s,\qquad
T_b-T_r\longrightarrow4.8\,\mathrm K,\qquad
\delta_\infty=19.2\,\mathrm{\mu m}.}
$$
For an illustrative ideal valve that closes at $\delta=g=10\,\mathrm{\mu m}$, the closing time is
$$
t_{\rm close}=-\tau\log\left(1-\frac{g}{\delta_\infty}\right)
\simeq18.4\,\mathrm s.
$$
This threshold is crossed only when $0<g<\delta_\infty$. These are chosen model parameters, not a fit to a historical valve.
\Image[/sun-valve-thermal-lag-and-switching.png]
{title=Thermal memory delays both closing and reopening in a chosen sun-valve model}
{description=Calculated response to $600\,\mathrm{W\,m^{-2}}$ illumination for $180\,\mathrm s$, followed by darkness. Equal thermal <time constants> isolate differential absorption. The $10\,\mathrm{\mu m}$ switching threshold and all material parameters are stipulated; the curves are not historical measurements. The gas trace represents permission to supply the main burner, rather than a measured flow rate.}
After illumination stops at $t=t_1$, the displacement decays as $\delta(t)=\delta(t_1)e^{-(t-t_1)/\tau}$. With $t_1=180\,\mathrm s$, the model reopens about $16.3\,\mathrm s$ later. Thus a <sun valve> responds to the thermal history of illumination; it does not read a clock or directly measure the beacon's brightness.
Back to article page
