0
$\begingroup$

Is there a way to calculate how warm air will get when it passes a source of heat knowing how fast the air is moving?

For example, let's say there is an object being heated by 40W of electricity and there is a fan blowing 20C air past the object at 30 cubit feet per minute. Assuming a perfect setup where there is no air leakage and it's purely air in and air out is there a way to calculate this? I'm thinking we might also need to know the surface area of the heated object and if that's the case is it possible to assume it's large enough to perfectly transfer all heat into the air, or if that's not possible I can estimate the surface area to be 0.2 meters squared. Also all surfaces are metal in case that matters.

$\endgroup$
1
  • $\begingroup$ you'll have better posting this on the engineering stack exchange. -NN $\endgroup$ Sep 27 '21 at 2:24
1
$\begingroup$

Yes, we can calculate it. Although the numbers would only be an estimate and deriving a very accurate relation is very difficult as it requires advanced knowledge of heat and mass transfer and fluid mechanics. But we have tables of various properties for various geometries and dimensionless numbers that enable us to estimate the heat transfer reasonably. Moreover, the convective heat transfer coefficient differs for different geometries, air velocities and fluids and estimating it in the first place is a very tedious task requiring extensive experimentation.

After determining the convective heat transfer coefficient, you can apply the following law known as Newton's law of cooling.

$$q' = h(T_w - T_{\infty})$$

Where $q'$ is the heat flux, $h$ is the convective heat transfer coefficient, $T_w$ is the temperature of the surface, and $T_{\infty}$ is the temperature of the surroundings.

$\endgroup$
1
$\begingroup$

The situation may be simpler than the solution of @mechanic.

We can assume that the room heater/hair dryer/paint stripper has reached a steady state.

So, in one second, the external electrical circuit delivers $40$ J of electrical energy to the heating coils. Since we are assuming a steady state, the heating coils must distribute the energy to the passing medium. Otherwise, the coils would heat up/cool off on their way to new steady state.

In that same second, there will be $30/60$ or $0.5$ cubic feet of $20$ C air flowing past the hot coils. That's the only place the heat can go.

So, change cubic feet to cubic metres, ask Google the density of air and the specific heat of air, and you have the answer: $$\Delta T=\frac{\Delta Q}{m \times c}$$

Of course, the design of the heated object will change the efficiency of the heat transfer. A tiny heating element will be white hot, and the small amount of air passing close to it will get very hot, requiring some mixing downstream with the rest of the air. A large porous object, with a lot of surface area, like the fins on a baseboard heater, will not get as hot, more of the air will come in contact with the fins, and the air leaving will be more evenly heated.

But in any case, you're putting a known amount of energy into a known amount of a specific substance, and the end result will be the same.

$\endgroup$

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.