# Example: Heat Transfer - conduction

## Understanding the situation

You are building a house and are trying to save energy by insulating it well so that you don't have to burn too much fuel to keep it warm in the winter. To get some sense of what the parameters are, let's create a simple model: The space inside your house is going to be at a higher temperature than the outside. Consider the flow of heat from the volume inside your house (the "hot region" at temperature TH) to the outside (the "cold region" at temperature TC). We'll consider the flow of heat out of an area, A, of your wall. Assume we put a thickness of insulation L between the inside and the outside.

Our equation for the flow of heat through a block is

$$\Delta T = \bigg(\frac{L}{\kappa A}\bigg)\Phi = Z \Phi$$

You buy the batts of insulation by area. But the quality of the batt is determined by its thickness (thicker means more resistance so better insulation) and heat conductivity (lower heat conductivity means slower heat flow so more resistance so better insulation). For this reason, the quality of the insulation batts is given by the reciprocal of $Z$ without the factor of the area:

$$R = \frac{L}{\kappa}$$

so the resistance $Z = R/A$. Bigger $R$ therefore means higher resistance to heat flow.{Regrettably in the US, R-values are quoted in the English units of ft2-oF-hr/BTU instead of the SI units m2-oC-s/J. We'll use the SI units here.}

## Presenting a sample problem

I recently put an addition on my home and, since I want to use it year round, I insulated the exterior walls and windows to an R value of 3.3 (SI units). The room is added on to the house so there are three new exterior walls. Suppose that most of the heat in the winter will be lost through the three new side walls (ignoring heat loss through the roof and floor since they will be much better insulated). Can I compensate for all the extra heat loss in winter with a small electric heater? My floor plan is 4m x 4m and the walls are 3m high. I want to consider a cold winter day when the outside temperature is -5C and my inside temperature is 20C.

## Solving this problem

All we need to calculate is the heat flow out through 3 walls. Each wall is 4m x 3m = 12 m2. I have 3 walls so the total area we are considering is 36 m2. My temperature difference is 25C, so I can calculate the rate of heat flow. I solve the equation for Φ and put in R before putting in numbers. Here goes:

$$\Phi = \frac{\Delta T}{Z} = \frac{\Delta T}{R/A} = \frac{A \Delta T}{R}$$

Putting in the numbers

$$\Phi = \frac{A \Delta T}{R} = \frac{(36 \mathrm{m}^2)(25 \mathrm{^oC})}{3.3 \mathrm{m^2} \mathrm{^oC/W}} = 272 W$$

That's better than I had hoped! It's not very much at all. Of course it's probably a significant underestimate since I have doors and windows and their insulation is not as good at the walls with the $R$ = 3.3 batts, but it suggests that a 1500 W heater would not need to be run all the time even on a very cold day.

Joe Redish 11/24/14 and 3/4/19

Article 437