Suppose a closed sphere with a hull and an $R$-value of $10$ m2 K/W and a radius of $1$ m. Suppose the temperature inside of the sphere is Ti, the temperature outside is To. The inside temperature is kept at Ti with a heat pump.

A common way to measure R-value is to assume $T_i \gg T_0$. The thin layer of air causes a slightly different temperature at the inside surface of the sphere, $T_f$. The $R$-value of such air layer is known and hence: $R_{air} \ / \ (T_i-T_f) = R \ / \ (T_f-T_0)$ and can be solved for $R$. But this requires $T_i >> T_0$.

Now, suppose in $I$ plot $T$ versus $t$ (time). When the heat pump turns off, the sphere will be slowly getting colder as heat moves out. If I wait long enough $T_i = T_0$. The speed at which this occurs should depend on the $R$ value (and surface area).

So, I am wondering, given $T(t)$ and/or $∆T /∆ t$, can I estimate the sphere's $R$-value? If yes, how? If not, why not?

  • 1
    $\begingroup$ sure. you'd need to make some assumptions about the inside of the sphere, esp that it's much more thermally conductive than the insulation. you'd still need to know the R for convection. you'd need to know the thermal mass inside the sphere. then you'd have exponential decay of the T inside, whose time constant you could observe with a couple measurements, from which you'd compute the sum of R_convection (known) and R_insulation $\endgroup$
    – Pete W
    Commented Jan 26 at 2:00
  • 2
    $\begingroup$ I see. I’m a circuits person, so there Id measure tau=RC. You say I need to know C and then I can calculate R=Rconv+Rinsulation? Why do I need assumptions that the inside (=air) is more conductive? Why can’t I ignore R for convection? Does C correspond to thermal mass? I’d really appreciate an answer with all of these elaborated. Sounds like a promising approach! $\endgroup$
    – divB
    Commented Jan 26 at 6:29
  • 1
    $\begingroup$ As the heat "moves out" well heat energy moves from hotter to colder, the driving force ie Temperature difference reduces so as the temperatures get closer to equilibrium the rate of heat movement slows which tends to show R and surface area will be effectively constant - although the sphere should change in size as it changes temperature - is that change relevant? Can you measure it? $\endgroup$
    – Solar Mike
    Commented Jan 26 at 6:34

1 Answer 1


Here's the idea, drawn in the analogy of an electrical network

enter image description here


  • solid sphere (or spherical shell with negligible thermal mass inside it), with very high thermal conductivity compared to everything else
  • surrounded by insulation with negligible thermal mass
  • surrounded by air


  • sphere has specific heat $c$ , and mass $m$
  • conduction thru the insulation characterized by $R_1$, and effective area $A_1$
  • convection between insulation and outside air characterized by $R_2$ and effective area $A_2$ (i.e. is linear vs ${\Delta}T$, perhaps questionable)
  • no other significant thermal resistances


  • $T_i$ = inside temp
  • $T_o$ = outside air temp
  • $T_x$ = temp at representative point (taking into account bouyancy effects etc) of boundary layer


  • If you know $T_i$ , $T_o$ , and $T_x$ , then you can compute the ratio of the R's, with a single measurement.
  • If you know $T_i$ , $T_o$ , and $(c.m)$ , then you get the heat flow in the loop, and you can compute the sum of the two R's. Still with a single measurement. Good if $T_x$ were troublesome to measure.
  • If you know $T_o$ , $T_x$ , and $(c.m)$ , then you can observe the exponential decay's time constant with multiple measurements, and from that compute the sum of the two R's. Good if $T_i$ cannot be measured for some reason. Also in theory it should be less sensitive to systematic error measuring $T_x$, as may be the case if we don't know exactly where to find a "representative point" on the sphere.
  • 1
    $\begingroup$ ps - drawing the circuit like that is abusing the metaphor a little bit, just to be clear. heat flux doesn't really return around the loop, I don't think. $\endgroup$
    – Pete W
    Commented Jan 26 at 18:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.