# A room with a heater in contact with walls as a heat sink

I have to calculate the increase of my room temperature as a function of the heater's power. The easiest way of doing it is:

$$T_1=\frac{P\Delta t}{C_v V n }+T_0$$

$$T_1$$ is the final room temperature, $$T_0$$ the initial room temperature, $$P$$ the machine power, $$\Delta t$$ the machine running time, $$C_v$$ the specific heat of the air, $$V$$ the volume of the room, $$n$$ the air density.

It is clear that the room's temperature increases with $$\Delta t$$ without a bound. In reality, some heat of the room will be transfered to the outside through walls. I would like to include this effect.

What differential equation does the system satisfy? Perhaps, the simplest 'second estimate' would be to treat the problem in 1D with the heat source at $$0$$ and the wall at $$L$$. For $$r>L$$, $$T(r)=const.$$ The room mean temperature would then be a satisfactory estimate.

• Are you considering the thermal inertia of the walls too, and, what about convective heat transfer coefficients between the room and inside of the wall and between the outside of the wall and the outdoors? Nov 15, 2021 at 14:02
• I suppose it would be appropriate to include these terms. Nevertheless, maybe it is sufficient to omit them for the time being. Any answer of the form $T(r)$, with a temperature bound would be a better approximation to what I already have, Nov 15, 2021 at 14:58
• Excellent point. Nov 15, 2021 at 15:41

I'm not going to provide a full solution here, only point out an important facet of any solution you may contemplate.

Consider the following composite wall:

It has $$3$$ conduction zones, as well as $$1$$ inside convection zone and $$1$$ outside convection zone.

We assume steady state, so all temperatures are invariant to time. The red curve is the temperature profile throughout the wall.

It can then be shown that the heat flux $$\dot{Q}$$ flowing through the wall is given by:

$$\dot{Q}=UA\left(T_e-T_o\right)$$

($$^{\dagger}$$ see note)

where $$U$$ the overall heat transfer coefficient is given by:

$$\frac1U=\frac{1}{h_e}+\sum_i \frac{t_i}{\lambda_i}+\frac{1}{h_o}$$

Here, the $$h$$ are convection coefficients (see Newton's law of cooling), $$t_i$$ the thicknesses of the wall components and their respective heat conductivities $$\lambda_i$$.

Note that this doesn't take into account any thermal inertias but those don't feature in a steady state regime.

Without knowing $$U$$ no solution, transient or steady state, can be developed.

($$^{\dagger}$$ and not $$T_{10}$$ as written in the figure)

What differential equation does the system satisfy?

One possible model is as follows.

Assume an object with internal heat generation $$P$$, inside uniform temperature $$T_e$$, mass $$M$$ and heat capacity $$M$$, then for an infinitesimal time interval the heat balance is:

$$Mc\mathrm{d}T_e(t)=P\mathrm{d}t-UA\left(T_e(t)-T_o\right)\mathrm{d}t$$ $$Mc\frac{\mathrm{d}T_e(t)}{\mathrm{d}t}=P-UA\left(T_e(t)-T_o\right)$$

Substitute: $$\left(T_e(t)-T_o\right)=\Theta$$, then: $$\mathrm{d}T_e(t)=\mathrm{d}\Theta$$

So:

$$Mc\frac{\mathrm{d}\Theta}{\mathrm{d}t}+UA\Theta=P$$

$$\Theta'+\alpha \Theta=\frac{P}{Mc}$$

where $$\alpha=\frac{UA}{Mc}$$

The ODE solves to:

$$\Theta=c_1\exp(-\alpha t)+\frac{P}{UA}$$

Initial condition:

$$\Theta(0)=T(0)-T_o$$

$$T(0)-T_o=c_1+\frac{P}{UA}$$

$$c_1=T(0)-T_o-\frac{P}{UA}$$

$$\boxed{T(t)-T_o=\left[T(0)-T_o-\frac{P}{UA}\right]\exp(-\alpha t)+\frac{P}{UA}}$$

Note that for $$t \to +\infty$$, $$T(\infty)-T_o=\frac{P}{UA}$$, the steady state temperature.

Another approach is the Fourier heat equation with load:

$$\frac{\partial T}{\partial t}=\alpha \nabla^2 +\dot{Q}$$

which for the steady state reduces to:

$$\alpha \nabla^2 +\dot{Q}=0$$

Assuming a square/rectangular geometry:

$$\alpha \left(T_{xx}+T_{yy}+T_{zz} \right)+\dot{Q}=0$$

which requires three boundary conditions.

Simplifying further, by assuming the object is a perfect cube, then: $$T_{xx}=T_{yy}=T_{zz}$$

So:

$$3\alpha T_{xx}+\dot{Q}=0$$

No partials are needed here, so:

$$3\alpha \frac{\mathrm{d}^2T}{\mathrm{d}x^2}+\dot{Q}=0$$

Set a corner of the cube at the origin of the Cartesian coordinate system and call the side of the cube $$L$$.

Assume the heat loss from the cube's surfaces is by convection only, then we can write:

$$k\left(\frac{\mathrm{d}T}{\mathrm{d}x}\right)_{x=0}=h(T(0)-T_o)$$

and: $$k\left(\frac{\mathrm{d}T}{\mathrm{d}x}\right)_{x=L}=h(T(L)-T_o)$$

Make a substitution:

$$u=T-T_o$$

Then our ODE and BCs become:

$$3\alpha u''+\dot{Q}=0$$ $$u'(0)=\frac{h}{k}u(0)$$ $$u'(L)=\frac{h}{k}u(L)$$

The ODE solves easily:

$$u'=-\frac{\dot{Q}}{3\alpha}x+c_1$$ $$u=-\frac{\dot{Q}}{6\alpha}x^2+c_1 x+c_2$$

Now use the BCs. $$c_1=\frac{h}{k}c_2$$ $$-\frac{\dot{Q}}{3\alpha}L+c_1=-\frac{h}{k}\left[\frac{\dot{Q}}{6\alpha}L^2+c_1 L+c_2\right]$$

$$-\frac{\dot{Q}}{3\alpha}L+c_1=-\frac{h}{k}\left[\frac{\dot{Q}}{6\alpha}L^2+c_1 L+\frac{k}{h}c_1\right]$$

$$-\frac{\dot{Q}}{3\alpha}L+c_1=-\frac{h}{k}\frac{\dot{Q}}{6\alpha}L^2-\frac{h}{k}c_1 L-c_1$$

$$\left(2+\frac{h}{k} L\right)c_1=\frac{\dot{Q}}{3\alpha}L\left(1-\frac{h}{k}\frac{L}{2}\right)$$

With $$c_1$$ and $$c_2$$ known and $$u=T-T_O$$ the core and boundary temperatutes $$T(0)$$ and $$T(L)$$ can be calculated if so wished.

Note that the $$\alpha$$ and $$\dot{Q}$$ are not the same as in the derivation above this one.

• Thank you for your insight. Then assuming I have $\dot{Q}$, I would proceed by applying Fouriers Law to find $T$. Or is there maybe a simpler way that I'm missing? Nov 16, 2021 at 12:10
• The simpler way is what I presented here. Fourier is generally far more demanding mathematically, because it also gives you the spatial temperature gradients. However, a Fourier steady state solution might be quite doable.
– Gert
Nov 16, 2021 at 13:52
• I'm a little busy right now but later I'll summarise the Fourier solution for steady state.
– Gert
Nov 16, 2021 at 14:08