Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have hot air flowing vertically down a circular duct into the ambient cold air. I'm trying to find the power $P$ that resists the flow.

I know the:

  • Volumetric flow rate $Q$ $[m^3/s]$
  • Flow speed $v$ $[m/s]$ (calculated from flow rate and duct diameter)
  • Density of the ambient air $\rho$ $[kg/m^3]$

So far I have assumed that the air flowing from the duct is an object that displaces the ambient air - much like if you were to submerge a solid object in water. I've used Archimedes' Principle to calculate the buoyant force $F$ $[N]$ and multiplied it by the flow speed $v$ $[m/s]$ to get the power $P$ $[W]$.

Method 1 - Convert mass to force

$m$ $[kg]$ $=$ $V$ $[m^3]$ $*$ $\rho$ $[kg/m^3]$

($V$ is gotten from the volumetric flow rate $Q$, assuming that we're looking at one second.)

$F$ $[N]$ $=$ $m$ $[kg]$ $*$ $g$ $[m/s^2]$

$P$ $[W]$ $=$ $F$ $[N]$ $*$ $v$ $[m/s]$

Method 2 - Impulse

$m$ $[kg]$ $=$ $V$ $[m^3]$ $*$ $\rho$ $[kg/m^3]$

$J$ $[kg*m/s]$ $=$ $m$ $[kg]$ $*$ $v$ $[m/s]$

(Once again $V$ is from the flow rate and $v$ is from the flow speed.)

$F$ $[N]$ $=$ $J$ $[kg*m/s]$ $/$ $t$ $[s]$, ($t$ $=$ $1s$)

$P$ $[W]$ $=$ $F$ $[N]$ $*$ $v$ $[m/s]$

Method 2 produces a significantly larger power $P$ and I feel that it's closer to the truth. (This is an impulse, right?) However, I am no physicist and the whole assumption of the flowing air being comparable to a solid object may be entirely wrong.

Can you approximate the power with the Archimedes' Principle or what should I attempt to get the resisting power $P$ caused by the less dense hot air flowing into the denser ambient air - is that buoyancy even significant?

Thank you for taking the time to read this!

share|cite|improve this question
up vote 0 down vote accepted

the whole assumption of the flowing air being comparable to a solid object may be entirely wrong

If not entirely wrong, it's problematic as it does not cover the compressibility of gases. That issue set aside, our analysis is weak on a few points.

  • First, you used only one density, $\rho$, where there in fact are two densities involved, the one of the cold (ambient) air and of the hot air flowing through the pipe. As I understand your question, that difference in density is the whole point of having gases with different temperatures.
  • What you calculated in Method 1 is the buoyant force of the hot gas expelled in one second. It is highly non-trivial how the expelled gas will flow once it left the pipe, but at some point it will "turn around" and head upwards. This depends on the velocity $v$ but also on things like the viscosity of the air (at different temperatures!). In terms of energy though, no constant power has to be provided to accomplish this once some equilibrium situation is reached. When you turn on the air flow or vary its velocity, the amount of displaced cold air will vary and you'd have to pay energy for that but you do not have to pay any more energy once this is accomplished. (Remember: work is force integrated over distance).
  • In Method 2 you're viewing the situation as some sort of rocket which produces thrust by expelling hot gas. This is an entirely different situation as it makes no reference to buoyancy whatsoever. Indeed, the "rocket" would still work in the absence of gravity (which is why rockets are used in space). That tells you that the two methods are not different ways of calculating the same thing but are just different effects. But still I claim there will be no net force if the system is somehow closed, i.e. if the hot air expelled at the bottom is somehow sucked in at the top.
  • What we haven't dealt with so far is also the equilibration of temperature. This depends on so many things like thermal conductivity, but also on the hydrodynamic motion (how fast will the gases mix), the viscosity, the diffusion constant -- if you know the properties of air very well, you might be able to do a simulation of it, but I don't think you can quantify this effect reliably otherwise.
  • Finally, the issue I've been dancing around so far: energy and power. What do you actually mean by that? As stated above, you'd have to invest energy to get things flowing in the first place but then there is not energy to invest other than what you need to keep the temperature difference up (i.e. to fight the equilibration of temperature) and to account for losses due to internal friction. It's like stiring with a spoon in a cup of coffee. Once you set the circulation in motion, you can take out the spoon and it will continue to circulate and will only stop due to internal friction.

IMO, the upshot is this: the power you have to put in is really difficult to quantify. Simple considerations of buoyancy won't get you anywhere, neither will the rocket analogy. More to the point: the two methods you laid out are different effects and are certainly not expected to give the same result. Sorry to give you such a "destructive" answer.

share|cite|improve this answer
By all means, this is just what I was looking for. I was assuming that this was something much more complicated! To answer the question - The heat is only a part of the entire picture, the "real" issue here is to find out if a pump capable of producing x kW could keep air flowing out of the duct at a specific rate - initially the duct would be filled with the ambient air. Other factors, such as the friction seem to be relatively simple to deal with - but this density difference really confuses me. Thank you for taking the time to answer! – natancodes Jun 19 '14 at 8:34

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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