[Physics] Vertical surface would lose heat faster by convection than a horizontal surface

Question

Is the heat coeff affected by gravity?

Vertical surface would lose heat faster by convection than a horizontal surface?

Answer 1

Short answer to the orientation question: It depends on many factors, but a vertical surface will generally lose heat faster.

Only one type of heat transfer coefficient is strongly affected by gravity: the convection coefficient for natural convection, which occurs because warmer air is less dense than cooler air. Natural convection describes hot air billowing upwards above a hot object and cool air sinking past a cold object, for example.

The other main type of convection, forced convection (e.g., air being blown by a fan past an object), isn't strongly affected by gravity. Nor are conduction or radiation.

Natural convection is extremely complex because it often involves turbulent air moving at various speeds directly past an object of arbitrary roughness and varying temperature. The large number of variables and the resulting uncertainty means that natural convection coefficients often take the form of empirical fits to experiments or simulations rather than theory-derived formulas.

The convection coefficient $h$ modulates the heat flow $Q$ through a cross-sectional area $A$ resulting from a difference in temperature $\Delta T$ (which in this case is the temperature difference between the surface and the ambient environment):

$$Q=hA\Delta T\tag{1}$$

It is common to express the convection coefficient as a function of the so-called Nusselt number $\mathrm{Nu}$, the thermal conductivity of the fluid (here, air) $k$, and a characteristic length $L^*$, which may be the length of the plate in the direction of flow or the area of the surface divided by its perimeter:

$$h=\mathrm{Nu}\times\frac{k}{L^*}\tag{2}$$

In addition, certain dimensionless numbers demarcate ranges over which certain behaviors hold. These include the Prandtl number $\mathrm{Pr}$, Grashof number $\mathrm{Gr}$, Reynolds number $\mathrm{Re}$, and Rayleigh number $\mathrm{Ra}=\mathrm{Pr}\times\mathrm{Gr}$.

At the end of it all, you might have an empirical formula such as

$$\mathrm{Nu}=0.68+\frac{0.67\mathrm{Re}^{1/4}}{[1+(0.492/\mathrm{Pr})^{9/16}]^{4/9}}\tag{3}$$

which is applicable for a vertical plate with height $L=L^*$ when the Rayleigh number satisfies $0.1<\mathrm{Ra}<10^9$ and the material properties are taken at temperature $(T+T_\infty)/2$, except for the thermal expansion coefficient (which appears in the Grashof number), which is taken at the ambient temperature $T_\infty$.

With this information in mind, we can address your question by looking at the Nusselt number $\mathrm{Nu}$ as a function of inclination angle. You can find one summary in Section 7.7 here. See also here. You can also find charts of the Nusselt number for horizontal and vertical flat plates on p. 14 here. Note that the bottom chart has log axes. For a given Rayleigh number, the Nusselt numbers are pretty similar. In fact, the dependence on inclination angle goes as $(cos\,\theta)^{1/4}$ in an additive factor, where $\theta=0^\circ$ describes a vertical surface. The exact details depend on whether the bottom or top of the surface is being heated.

Thus, the heat transfer from the vertical plate is somewhat higher, but generally not more than an order of magnitude larger, for example. (One exception is a very large heated surface directed straight down, which would lose essentially no energy through natural convection. A slight tilt could be expected to increase the heat transfer rate substantially.) However, the actual results can also be expected to vary depending on the roughness of the surface, its exact temperature variation, and the orientation and spacing of the surrounding objects, among other variables.