John Rennie's answer to this question got me thinking about the water supply in a house. I know that water supplies are normally operated at an over-pressure with respect to atmospheric pressure to make sure that if you open the tap the water will flow out due to the lower pressure at the exit of the tap (including a correction for pressure losses due to the plumbing and potential height differences).

What I am wondering is how adjusting the tap results in an adjustment of the flow rate from that tap if the water pressure in the main supply is a constant (probably by approximation)? Because in that case the pressure upstream is fixed, the pressure downstream (atmosphere) is fixed so pinching or opening a valve should not change the flow rate.

Obviously it does change the flow rate so there must be something else going on. The only thing I can think of is that the water supply somehow acts as if there are parallel connected (hydraulic) resistances as illustrated in the schematic below which makes sure that the distribution at the flow split can be adjusted. Is that the right thinking or is there a different argument why the flow is adjustable?

enter image description here

(The dashed line indicates some way of rerouting the water back to the mains, although that cannot be a direct connection because that would allow a pressure drop over this section)

  • $\begingroup$ I'm sure you're learned in your Ph.D. studies in fluid dynamics that pipe networks and wire networks are analogous, and an adjustable tap is just like a variable resistor. $\endgroup$ – Mike Dunlavey Jan 11 '14 at 23:28
  • $\begingroup$ Ok that's what I thought, but I wasn't sure whether that was truly the way it works. So the diagram I'm making is the correct way of interpreting this. Thanks! $\endgroup$ – Michiel Jan 12 '14 at 15:31
  • $\begingroup$ No need to invoke "resistance". If you say the flow velocity is given by the pressure difference, then multiplying that velocity by the area (of the constriction) gives the flow rate - smaller hole = less mass flow. And a half open tap is a "smaller hole". $\endgroup$ – Floris Oct 30 '14 at 15:35

The Bernoulli equation, with friction, has the following form: $$ p_1+\frac{1}{2}{\rho}v_1^2+{\rho}gz_1=p_2+\frac{1}{2}{\rho}v_2^2+{\rho}gz_2+\left(f\frac{L}{D}+{\sum}K\right)\frac{1}{2}{\rho}v_2^2 $$ where $p$ is pressure, $\rho$ desity, $v$ velocity, $z$ height, $g$ the acceleration due to gravity, $f$ the friction factor, $L$ the length of the tube, $D$ the hydraulic diameter and $K$ additional sources of local friction.

A tap falls under the category of additional sources of local friction. Adjusting the tap (its $K$ value) will change the flow-rate, because this factor gets multiplied by the velocity squared.


Consider two holes on the bottom face of a bucket, one hole is larger than the other. Your intuition should tell you that more water will flow from the larger hole. This is analogous to two taps, one fully open and one half open.

The fluid velocity will be identical (ignoring friction) through each tap/hole but the volumetric flow rates will differ as the cross sectional area of the tap/hole is different.

To calculate the fluid velocity from pressure and height differences Bernoulli's equation can be used:



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