I am doing some simple flow experiments and my instruments are giving me strange results - I think they are broken, I need a sanity check for my understanding of the flow in my system to confirm it.

Here is a schematic. An inlet has a gauge pressure of 19 inches of water gauge (19"wg). The stream gets split up into two, pipes A and B. Both pipes lead to the same atmospheric pressure (Pe = 0).

I have denoted the net resistance in the two lines with $R_A$ and $R_B$.

Here is what I am wondering about: Do $P_A$ and $P_B$ necessarily have to be equal to 19"wg? If I double the resistance RB, how will that affect the pressure PA? If RB doubles, then the resistance in line B will be more, diverting more flow into line A. That means the velocity in line A has suddenly increased (since RA did not change). According to Bernoulli then, the pressure PA has to drop. Is this correct?

The principle of continuity applies where $Q_{in} = Q_A + Q_B$ and the energy per unit mass according Bernoulli, i.e. $\frac{P_{in}}{\rho} + \frac{V_{in}^2}{2} = \frac{P_A}{\rho} + \frac{V_{A}^2}{2} = \frac{P_B}{\rho} + \frac{V_{B}^2}{2}$

I am trying to think about this intuitively with a sort of hydraulic analogy - in terms of resistors, pressures and flows ($\frac{P}{R} = Q$).

Split Pipe Flow


1 Answer 1


I'm assuming incompressible fluid here and rigid pipes (no compliance).

Before the actual physical split is there any appreciable resistance compared to RB and RA? If not then PA = PB = 19" wg.

If there is a significant resistance you need to include it in the model as Rin, and then

PA = PB = 19" *(RA || RB)/(Rin + RA || RB)

In any event PA = PB

  • $\begingroup$ Why is $P_A$ = $P_B$. Doesn't the pressure depend on downstream conditions. If one of the resistances increase, more flow is going to be diverted into the other pipe (velocity increases) and according to Bernoulli, the pressure is going to drop? $\endgroup$
    – l3win
    Commented Oct 11, 2014 at 23:56
  • $\begingroup$ No. Not the pressure, the pressure drop or rather delta-pressure if you prefer. Your model is a 'lumped' component model, and that's how I treated it. With that assumption the resistances essentially collapse at single points in space. And anything outside of the resistance is considered zero resistance, so no pressure drops. The pressure is the same at each resistance inlet (upstream in the split). If one leg of the split has more resistance than the other there will be more pressure drop relative to the other resistance and the flow rate will be less. $\endgroup$
    – docscience
    Commented Oct 12, 2014 at 0:17
  • $\begingroup$ It's the flow rate that changes not the upstream pressure. $\endgroup$
    – docscience
    Commented Oct 12, 2014 at 0:17
  • $\begingroup$ Bernoulli's equation comes from the conservation of energy and energy is conserved. You can express your model in terms of mass & energy conservation, but it's easier (for me at least) to use the analogy of electric circuits. That's how I derived the equations above - it's just a voltage divider. $\endgroup$
    – docscience
    Commented Oct 12, 2014 at 0:22
  • $\begingroup$ So....if one resistance increases, the flow in that circuit, and therefore the total flow is going to drop? $\endgroup$
    – l3win
    Commented Oct 12, 2014 at 0:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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