I am working on a problem at the moment and would appreciate some of your expertise regarding this issue. If you see the image below, I am trying to calculate heat transfer from a fluid entering a vessel/tank on the grey cuboid within. To begin with, I am focussing on side C.

I would like to understand how we would work out the Velocity of the fluid at point C when water is pumped in at A (intake), and, when discharging from point (B), i.e A flow, B return.

The overall problem I am trying to solve is how long it would take to heat the cuboid from convection heat transfer (forced + also natural). At the moment I am calculating (Re) and need to determine velocity.

It is not a home work question, it is for general research. There is no actual question I’m working from.

I would like to understand how to work out velocity at C when there is a flow into A and out of B. This will allow me to work out Re at C.


  • $\begingroup$ Welcome to Engineering! This looks like a homework question. In order for such questions to be answered in this site, we need you to add details describing the precise problem you're having. What have you tried to solve this yourself? Please edit your question to include this information. $\endgroup$
    – Wasabi
    Jan 21, 2018 at 12:18
  • $\begingroup$ You need to add size of aperture A, B and length between a and b and width just to begin, then intake v, losses and outlet v etc etc $\endgroup$
    – Solar Mike
    Jan 21, 2018 at 12:40
  • $\begingroup$ hi @85l00k85 thank you for providing this sketch. Could you maybe also add some equations you are using right now? And please add some numbers to is as well. $\endgroup$
    – rul30
    Jan 21, 2018 at 13:13

1 Answer 1


Assuming incompressibility you can work with volume-flow to calculate (a first guess) the velocity, $u$ with the following equations.

Probably you have some volume or mass $\dot{m}$ flow entring at (A). This mass-flow is entering the box through a opeing with a certain cross-section $A_\mathrm{A}$. The relation is:

$u_\mathrm{A} = \frac{\dot{m}}{\rho\, A_\mathrm{A}}$

The same relation should also be valid for the exit B.

$u_\mathrm{B} = \frac{\dot{m}}{\rho\, A_\mathrm{B}}$

For the velocity around the cube we will now assume a constant velocity $u_\mathrm{C}$ which is of course not really the case but it is a start.

Here the equation has the same structure you just need to put in the correct area ($A_\mathrm{C}$) which is the area of the box minus the area of the cube.

enter image description here

$u_\mathrm{C} = \frac{\dot{m}}{\rho\, (A_\mathrm{box} - A_\mathrm{cube})}$

  • 1
    $\begingroup$ Thankyou Rul, much appreciated. Yes, I will assume constant velocity. Is this based on the continuity equation a1v2 = a2v2? $\endgroup$
    – 85l00k85
    Jan 21, 2018 at 14:02
  • $\begingroup$ Right, continuity but especially we assume that the flow is so slow, that it nicely flows around the edges, without flow separation, which is not true, but this depends on your Reynolds number. $\endgroup$
    – rul30
    Jan 21, 2018 at 14:23

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