Lesson 6 - Radiation and Combined

Lastly, let’s look at the total heat loss from a hot body to its surroundings, which includes losses by conduction-convection and radiation.

As an example, a hot steam line in a room loses heat nearly equally by each of the two mechanisms: conduction-convection and radiation. These two mechanisms of heat loss occur in parallel (at the same time). Assuming black surroundings, we can write the total heat transfer rate as:

Where

q_T is the total heat transfer rate

q_c is the conduction-convection heat transfer rate per unit area

q_r is the radiative heat transfer rate per unit area

h_c is the convective heat-transfer coefficient

σ is the Stefan-Boltzmann constant = 5.672 × 10^-8 Wm^-2 K^-4

ε is the emissivity of the surface

T_w is the temperature of the surface

T is the temperature of the surroundings

The previous equation is often written as:

Where h_r is a radiation heat transfer coefficient defined as:

The radiation heat transfer coefficient varies with application and should be specific for each scenario.

The resistance to heat transfer via convection/conduction is independent of heat transfer via radiation. In other words, they are happening simultaneously in parallel.  Therefore the resistances can be added instead of adding the reciprocal as we do in convection/conduction only cases.

Let’s do an example.

Calculate the heat loss per unit area from the surface of an un-lagged steam pipe to its surroundings. The surface of the steam pipe is 430 K, while the ambient temperature is 310 K. The convective heat transfer coefficient is 10 Wm^-2K^-1 and the radiative heat transfer coefficient is 5 Wm^-2K^-1.

To calculate this, we will use the equation for total heat loss from a surface to its surroundings.

We can simply enter the values given to calculate the heat loss per unit area as:

We can then conclude that the heat loss from the steam pipe is substantial and thus the pipe work would benefit from insulation.