How to calculate heat loss of a building using thermal inertia

The rate of heat energy loss, i.e. the heat flux, of an object is described by Newton's Law of Cooling:

$$frac{text{d}Q}{text{d}t}=-hA[T(t)-T_{infty}]$$

where:

• $frac{text{d}Q}{text{d}t}$ is the heat flux
• $T(t)$ is the temperature of the object in time $t$ and $T_{infty}$ the temperature of the surroundings
• $h$ is the heat transfer coefficient and $A$ the surface area exposed to the surroundings.

Note that this law relies on lumped thermal analysis, which means the temperature of the object $T(t)$ is uniform in space, with no spatial temperature gradients. For large objects like buildings that is of course not very realistic.

Does it make it easier if you assume that there is no heating? Or that the temperature is constant?

These are two distinct cases:

• The latter case allows direct calculation of the heat flux, using the formula above.
• For the former case, $T(t)$ will evolve in time, which can be seen by re-writing the heat flux:

$$mc_pfrac{text{d}T(t)}{text{d}t}=-hA[T(t)-T_{infty}]$$

where $m$ is the object's mass and $c_p$ its specific heat capacity.

This is an easy differential equation which solves to:

$$T(t)=T_{infty}-(T_{infty}-T_0)expBig(-frac{t}{tau}Big)$$

where $T_0$ is the initial temperature ($t=0$) of the object and $tau$ is the characteristic time:

$$frac{1}{tau}=frac{h A}{m c_p}$$

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A temperature v. time plot of:

$$lnBig(frac{T(t)-T_{infty}}{T_0-T_{infty}}Big)=-frac{t}{tau}$$

should yield a straight line with $frac{1}{tau}$ as the gradient. If $m$ and $c_p$ are known then $hA$ can be calculated from that.

In the figure above,

$$ln Theta=lnBig(frac{T(t)-T_{infty}}{T_0-T_{infty}}Big)$$

Carry out linear regression on the data points (little circles) to find $1/tau$.