LTR: Combined Convection and Radiation

As is probably perfectly clear to you, it is of only limited value to understand these modes as purely separate entities. So we have explored how these things may act together.

EXAMPLE:

A room heater is in the form of a thin vertical panel of length, $L = 20 cm$ and area, $A = 0.5 m^2$. We will use an averaged temperature for the surface of the heater at 30C (303K). The air temperature far from the heater is constant at 20C (293K), and the emissivity of the heater is $\epsilon = 0.9$. What is the rate of heat flow away from the heater?

We know that the heat leaving the surface of the "radiator" may leave as either radiation or convection.

So

EQUATION:

$q = q_{rad} + q_{conv}$

where we know that the two q's are given by:

$q_{rad} = 4\epsilon\sigma T_m^3A\Delta T = h_r\Delta T$

and

$q_{conv} = hA\Delta T$

Which gives us that:

$q = hA\Delta T + h_rA\Delta T = (h+h_r)A\Delta T$

Given that the convective heat transfer coefficent is $3.79W/m^2K$, we can then calculate our radiative heat transfer coefficient (with $T_m = 298 K$):

$h_r = 4\epsilon\sigma T_m^3 = 4(0.9)(5.676x10^{-8}W/m^2K^4)(298K)^3 = 5.41W/m^2K$

$q = (3.79W/m^2K + 5.41W/m^2K)(0.5m^2)(303K - 293K) = 46W$