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Calculation of heat flux on a surface

Physics Asked by Andreu Payne on July 27, 2021

I have a basic question about calculating heat flux applied to a surface.

Suppose you have a solid cylinder that has a height of 1 $text{cm}$ and a radius of 1 $text{cm}$, giving it a lateral surface area of 2π $text{cm}^2$.

You take a heat tape device that has the dimensions 1 $text{cm}$ by 24 $text{cm}$ and outputs 144W. The surface power density is then 6 W/$text{cm}^2$. You then wrap the heat tape around the cylinder. Note that, since the surface area of the heat tape is larger than the lateral surface area of the cylinder, only part of the heat tape will be in contact with the cylinder.

What would be the heat flux acting upon the side of the cylinder? Is it still 6 W/$text{cm}^2$, or does it change since not all of the heat tape is in contact with the cylinder?

One Answer

This constitutes a nontrivial transient heat transfer problem. You cannot assume that the heat flux of 6 W/cm² is somehow always directed inward toward the cylinder. In fact, over time, less and less heat will flow inward, as the cylinder will asymptotically reach an equilibrium temperature such that all 6 W/cm² is directed outward and is dissipated through convection or radiation, for example.

It's essential to estimate (and, if you wish, try to control) heat losses from convection and radiation here, as these will govern the temperature of the cylinder over time.

Ignoring the loose tape and thus assuming axisymmetry, and performing an energy balance, we can write

$$frac{alpha}{r}frac{partial }{partial r}left(rfrac{partial T(r,t)}{partial r}right)=frac{partial T(r,t)}{partial t}$$

within the cylinder (applying the Laplacian in polar coordinates), where $alpha$ is the cylinder thermal diffusivity, and

$$-kfrac{dT}{dr}+q^{primeprime}-h(T-T_infty)-sigmaepsilon(T^4-T_infty^4)=0$$

at the cylinder/tape surface (assuming the tape has negligible thickness), where $k$ is the cylinder thermal conductivity, $q^{primeprime}=6 mathrm{W/cm}^2$, $h$ is the convective coefficient, $sigma$ is the Stefan–Boltzmann constant, $epsilon$ is the tape surface emissivity, and $T_infty$ is the ambient temperature. (Possibly the last term on the left side—the radiative term—can be considered negligible.)

One could derive an analytical equation for the flux and temperature at very short times, when the heat transfer into the cylinder can be idealized as heat transfer into a semi-infinite body. From Incropera & DeWitt's Fundamentals of Heat and Mass Transfer, this is

$$T(x,t)=T_infty+frac{2q^{primeprime}(alpha t/pi)^{1/2}}{k}expleft(frac{-x^2}{4alpha t}right)-frac{q^{primeprime}x}{k}mathrm{erfc}left(frac{x}{2sqrt{alpha t}}right),$$

where $x$ is the depth relative to the surface and $mathrm{erfc}$ is the complementary error function.

Or, one could estimate an equilibrium temperature knowing the dissipated power and the heat-loss mechanisms.

Or, one could do a more detailed spatial and temporal study at intermediate times using finite element analysis, for instance, which would also provide predictions of the internal cylinder temperature and its variation.

All of this is independent of the fact that you have additional loose tape unwinding adjacent to the cylinder that's also emitting 6 W/cm². This aspect just adds an additional complication.

Does this all make sense?

Correct answer by Chemomechanics on July 27, 2021

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