I do not frequently post on this community, so hopefully my question fits within the community guidelines. Below is a picture of two pipe systems. All values that are color coded in green are identical between the two pipe systems (e.g. $v_o$, the inflow velocity, in the upper system is equivalent to $v_o$ in the bottom system).

The primary parameters that I am interested in is how a change in the middle chamber's diameter ultimately affects the final velocity $v_f$ exiting the smallest chamber of the pipe system. For clarification, $D_1 \lt D_2$. I wanted to confirm that, correspondingly, $v_{f-1} \lt v_{f-2}$.

If this is correct, I would greatly appreciate any intuitive idea as to why this is the case. I assume it has something to do with the kinetic energy of fluid molecules in the top system being lower than the kinetic energy of the fluid molecules in the bottom system after transiting through the middle chamber (perhaps as a consequence of experiencing more total resistance during the passage through the pipe system).

In the event that any further assumptions are required to best answer this question, assume the following:

  1. the length of each chamber is the same between the two systems
  2. the length of the connecting pieces (i.e. the sloped transition sections) are the same between the two systems. However, these transition sections are obviously increasing their diameter at different rates (but I don't think this really affects the answer)
  3. gravity is either acting downwards OR right-to-left (but I don't think this really affects the answer)
  4. the fluid is water (so it can be assumed to be incompressible)

Thank you! Cheers~

Upstream Dilation

The following are orders of magnitude that the parameters referenced in the above picture will tend to float around:

  1. diameter $\lt 4$mm
  2. pressure $\lt 80$ mmHg
  3. velocity $\lt 20$ cm/sec
  4. Re $\lt 20$
  • $\begingroup$ Is vo the same in both cases? Is the fluid incompressible? $\endgroup$ Feb 18, 2020 at 21:43
  • $\begingroup$ @ChetMiller Yes, $v_o$ is the same in both case. The fluid is water...I will put this in the assumptions. So, yes, it is incompressible. $\endgroup$
    – S.C.
    Feb 18, 2020 at 22:51
  • $\begingroup$ A simple momentum balance on the control volume of your pipe should tell you the answer. $\endgroup$ Feb 18, 2020 at 23:01
  • $\begingroup$ @Drew I have no clue how to perform that calculation...so it is not "simple" with respect to my skill set. $\endgroup$
    – S.C.
    Feb 18, 2020 at 23:19
  • $\begingroup$ @Drew after looking up the strategy that you refer to, either I am grossly misapplying the technique (which is certainly possible) or the assumptions imposed in this technique are wildly violated in the above depicted situation. Using a more "real world example", if I stepped on a hose in the middle of its length, the velocity that exits the tip of the hose would certainly be less than if I had not stepped on the middle of hose. Your strategy does not predict this. Presumably because the "viscous force" assumptions are violated. $\endgroup$
    – S.C.
    Feb 19, 2020 at 0:22

3 Answers 3


The Change $D_1<D_2$ causes the exit velocity to be lowered $v_{f-1} \lt v_{f-2}$ through higher velocity stack which simple means smaller effective $D_c$ for the flow.

This higher velocity stack is caused by the velocity difference $V_{D1}>V_{D2}$ It should be noticed that there is no possibility for velocity stack after $D_A$ because there is no "defining nozzle".

The maximum flow velocity (pressure is zero) is thus created on the entrance of $D_C$. This is the defining nozzle for the whole system, and all other velocities can be calculated through it. More information is avaible in this answer; Air core Vortex; Physical explanation of the "air Entrainment Hook" at $F_{co}=0.7$

  • $\begingroup$ Thanks for the input. I just wanted to confirm that this is applicable for water. The examples I’ve seen online seem to only reference air flow. $\endgroup$
    – S.C.
    Feb 24, 2020 at 20:43
  • 1
    $\begingroup$ It's applicable for water and any fluid. My link to the Q/A Air entrainment hook" is excatly about water. The incompressibilty is exactly the factor which limits the amount of flow to V-max @ P-zero. Btw, it sounds like you have done experiment about this, is it so? $\endgroup$
    – Jokela
    Feb 24, 2020 at 23:19
  • $\begingroup$ Ahhh, great. Thank you for the clarification. And, sort of. It's moreso a field that I am in (functional magnetic resonance imaging). Just trying to understand some of the blood flow dynamics within living organisms. The numbers and geometry that I provided above are simplified idealizations of the blood flow within the vasculature. $\endgroup$
    – S.C.
    Feb 25, 2020 at 0:06
  • $\begingroup$ @S.Cramer Oh, thats interesting. As a natural system this flow must then follow the Froude-law. If this is some medical thing you are trying to optimize, I can offer my hand on these flow aspect. As I have made quite some research on this field. $\endgroup$
    – Jokela
    Feb 25, 2020 at 9:13
  • $\begingroup$ i just wanted to confirm that your answer is saying $v_{f-1} \lt v_{f-2}$ $\endgroup$
    – S.C.
    Feb 27, 2020 at 18:54

Applying a mass balance to the entire system gives:

$$ D_0V_0=D_cV_c$$

where $V_c$ is the final velocity. So, the final velocity won't change if:

  1. You keep $D_0$, the middle diameter, and $D_c$ constant (i.e., the tube is rigid and can't deform).
  2. The fluid is incompressible.

Consider your real world example of the hose; if you step on the hose, the water is actually going to speed up to satisfy the mass balance. If you step on it enough, however, the water will slow down due to an appreciable change in viscous forces, affecting the momentum balance, requiring a larger pressure to pump. Your pump doesn't work harder to maintain new pressure drop that you just required, so the flow slows down.

Check out the momentum balance:

$$ \sum Forces= \rho A_cV_c^2-\rho A_0V_0^2$$ $$ P_0A_0-P_cA_c - F_{viscous}= \rho A_cV_c^2-\rho A_0V_0^2$$

and $F_{viscous}=A_s\tau_w$ is the pipe surface area times the shear stress at the wall (given from Hagen-Poiseuille flow in a circular pipe). If your pipe does not deform, however, I think that energy would then be dissipated and result in your fluid gaining thermal energy at the expense of kinetic energy. So yes, the final velocity should be lower in that case, but I don't know how to reconcile this with the mass balance. Maybe others can weigh in?

  • $\begingroup$ Recalling from my undergrad...isn’t there areas of recirculation at the transition zones between the different sizes chambers? As such, the mass input actually is NOT the same as the mass output. $\endgroup$
    – S.C.
    Feb 20, 2020 at 4:56
  • $\begingroup$ @S.Cramer If mass input was not the same as mass output, then your system would be gaining or losing mass. How would that make sense? $\endgroup$ Feb 20, 2020 at 16:19
  • $\begingroup$ Ahhh, I guess you’re right. Once the system reaches a steady state, I suppose the mass input will adjust accordingly in response to the regions of recirculation. $\endgroup$
    – S.C.
    Feb 20, 2020 at 16:46

I think this is most helpfully framed as optimising the middle diameter.

The keys are to 1.Minimise the momentum change at the joins. 2.Realise that a horizontal pose will enhance vorticity and minimise momentum change angles and consequently energy loss.

As an optimisation it appears to be something like minimising the sum of the v.v.Sin(theta) of the two joins. As a rule of thumb I would guess that the optimal diameter squared is about midway, ie


Thereby weighting the minimisation to the primary loss at the small join. Vorticity is also likely to scale as V2, noting that gh is same order as V2 and the static pressure is significantly larger than the mass terms.


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