1
$\begingroup$

I was wondering about this problem (which is NOT homework, it's one of my mental problems due to what surrounds me): supposing I'm in a room, and it's winter. Heating is on, the boiler works and the temperature in the room is about $290$ K.

Now let's say I want to heat the room until it reaches a temperature of $305$ K. The formula for estimating the amount of energy (heat) required is

$$Q = m\cdot \mathcal{C} \Delta T = \rho_{\text{air}}\cdot V_{\text{room}}\mathcal{C}\Delta T$$

Where $\rho$ is the air density, about $1.23\ kg/m^3$, and $\mathcal{C}$ is the air specific heat, about $1.00\ J/(kg\cdot K)$ Considering $\Delta T = 15$ k one would find

$$Q = 18.45\cdot V_{\text{room}}$$

where clearly $V_{\text{room}}$ is the volume of the room and it's easily estimable. For the sake of amusement, let's say I'm in a perfect cubic room, whose volume is $64 m^3$, so I get

$$Q = 1180.8\ J$$

Here come the two main questions

1) How can I calculate the amount of time necessary for the room to be heated?

2) Assuming I turn off the boiler when the room reached the desired temperature, how much time it will take before the temperature goes down to $x$ degrees?

There maybe are some assumption that I didn't write, like the boiler's power (if it makes sense) and whatever.

I just need some clarification about the procedure / the formulas involved!

Thank you so much!

$\endgroup$
3
$\begingroup$

Alright - I promised an in-depth answer, so here it is!

First: we outline our assumptions.

  1. The walls are perfectly well insulated.
  2. There is a small heater in the center of the room.
  3. This is a one-dimensional room. This greatly simplifies the problem, but the result should be correct within an order of magnitude.

Now, we need a way to implement our heater here. You already did the calculation (although there is a units mistake - the specific heat of air is about 1 kJ per mass temperature. Whoops! Regardless, let's continue.

Next, we try to look at the equation we are trying to solve. We can't use the normal heat equation - the initial distribution is a flat initial temperature $T_0$, and the normal heat equation does not include a heat source term! So, we have to alter it by adding a forcing term. How does the temperature of the room change in time? Well, in a given region $\mathrm{d}x$, we have heat flowing in from all sides (which gives rise to the term $\alpha \frac{d^2T}{dx^2}$). But we another source - at the origin of the problem, we assume a furnace. Let's say the source gives off heat with a power $P(x,t)$ (given in watts). Clearly, this adds heat to the system (at least at that point). So how do we account for that? We add an additional term $+\frac{P(x,t)}{c\rho}$, where $c$ is the specific heat of the air and $\rho$ is the density. Note that here, $\alpha$ is the thermal diffusivity of air (a number we can look up, but it is composed of the thermal conductivity, the air density, and the heat capacity). Thus, the equation we are looking to solve is

$$\frac{\partial T}{\partial t} = \alpha \frac{d^2T}{dx^2} + \frac{P(x)}{c \rho}.$$

So, for this particular problem, we are assuming a point heat source - so, let's define $$P(x,t) = P_0\delta\left(x-\frac{L}{2}\right)\theta(t_0-t).$$ Here, the $\delta$ is just the Dirac delta, which I hope you're familiar with. The $\theta(t_0-t)$ is a Heaviside function which controls how long the heater is on - for $t>t_0$, $\theta(t_0-t) = 0$, so the heater "shuts off." The number $P_0$ controls the power output of the heater. Note that this power source function is centered in the room - the values that $x$ can take on are $0\le x \le L$, so the power source is located at $\frac{L}{2}$.

Next, because our walls are well insulated, that means that heat cannot flow out of the system. How do we consider that, mathematically? We simply say that $\frac{dT}{dx} = 0$ on the boundaries!

The last consideration is the initial temperature distribution. But this is easy - we just say that it is a constant temperature, $T_0$.

So now we have completely finished the setup of the problem. In summary, our differential equation and conditions looks like

$$\frac{\partial T}{\partial t} = \alpha \frac{d^2T}{dx^2} + \frac{P_0}{\rho c}\delta\left(x-\frac{L}{2}\right)\theta(t_0-t),$$

$$T'(0,t) = T'(L,t) = 0,$$

$$T(x,0) = T_0.$$

So now comes the tricky part - solving this equation. Fortunately, it's not too hard, even though it would take a long time to write out how we solve this, so see this site for an in-depth explanation of the calculations I did. Note that the most important in solving an physical system is determining, mathematically, what the equations look like and what conditions you must impose, which we have just finished! So now we simply solve the equation as described in the link above, and we produce a final series solution to the differential equation, satisfying the bounds.

Fortunately, I've done that all for you! The final result is all sorts of nasty, but I'll post it anyway (after my computer algebra finishes simplifying it - it's really a nasty expression.

$$\sum _{n=0}^{\infty }{-\frac {4}{{\pi }^{2}{n}^{2}c\rho\,\alpha} \left( -\frac{LP_{0}}{2}\, \cos \left( \frac{n\pi}{2} \right) \theta \left( t-t_{0} \right) {{\rm e}^{{\frac {\alpha\,{n}^{ 2}{\pi }^{2}t_{0}}{{L}^{2}}}}}+L \left( \theta \left( t-t_{0} \right) -1 \right) \frac{P_{0}}{2} \cos \left(\frac{n\pi}{2} \right) {{\rm e}^{{\frac {\alpha\,{n}^{2}{\pi }^{2} t}{{L}^{2}}}}}+\frac{LP_{0}}{2} \cos \left( \frac{n\pi}{2} \right) \right) \cos \left( {\frac {n\pi \,x}{L}} \right) {{\rm e}^{ -{\frac {\alpha\,{n}^{2}{\pi }^{2}t}{{L}^{2}}}}}} $$

Now, I told you it was nasty! But that's okay - all that we care about is that this is some function, which takes in a large set of parameters, a position, and a time, and spits out a temperature. And that's what it does.

The list of parameters are the following:

  • $L$ - length of the system. Chosen to be 6.4m.
  • $\alpha$ - thermal diffusivity of air. Can be looked up, I used $1.9 \times 10^{-5} \text{ m$^2$/s}$
  • $c$ - mass specific heat of air. I used 718 J/kgK
  • $\rho$ - air density. I used 1.225 kg/m$^3$.
  • $t_0$ - the length of time the heater is on. I chose 100 seconds.
  • $P_0$ - power output of the heater in watts. I chose 1000 W.
  • $T_0$ - initial temperature of the "room." I chose 290 K.

So with those parameters in mind (note that I made your room 10 times smaller - otherwise, you'll have to wait a long time to heat the room!), and the total heat that I chose to put into the system is on the order of 100kW. If you had used the correct heat capacity of air, your room would have required 1MW - I reduced the room size by a factor of 10, so we only need 100kW to bring this room up to temperature.

So, all that said, I plotted some data points of the temperature as a function of position and time, below:

Plot of $T(x,t)$ where $0\le x \le 6.4$ m and $0 \le t \le 2\times 10^5$ s

Note that, in this plot, it doesn't look the temperature is initially 290K uniformly - that's just an artifact of using uniform sampling for data points. If you evaluated the function at $t=0$, you would find that it is 290K everywhere. Further, the temperature isn't fully equalized until $t$ is maximum, which is around 55 hours!

So, in final answer to your question - it takes a long time to heat a room if you just use thermal diffusion! That's why heaters blow air around the room - that will make the temperature more uniform extremely quickly. Further, there is a reason air is used as an insulator - it's not a very good thermal conductor. This result would be starkly different if we were in a metal, for example.

I hope this answer explained everything in depth! I can edit this question later to clarify more if more is needed.

$\endgroup$
  • $\begingroup$ @user66309 - That's nothing compared to what it was. One of my edits was changing from an approximate delta function to an actual delta function. The delta function cleaned it up significantly. $\endgroup$ – Sam Blitz Jan 22 '16 at 19:02
  • $\begingroup$ That was AMAZING. I will read it with lots of calm and I'll tell you if I need something to be clarified/added. Thank you a lot, forsooth! $\endgroup$ – Les Adieux Jan 23 '16 at 0:24
  • $\begingroup$ No worries! But if you really like it, make sure to choose it as the answer! But no worries - even if you do, I'll continue to update it. $\endgroup$ – Sam Blitz Jan 23 '16 at 0:50
1
$\begingroup$

The time required for the wall to attain its steady state temperature profile is going to be on the order of $L^2/\kappa$, where $\kappa$ is the thermal diffusivity of the wall material (averaged over the materials that comprise the wall), and L is the wall thickness. This characteristic time arises as a result of reducing your transient heat conduction differential equation to dimensionless form.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.