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Differences in the behaviour of pinching a garden hose and closing a tap. Why does flow rate decrease by pinching the hose?

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user131140
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  • For a given pressure difference, the flow rate depends on the flow section area. Less area = less flow. – John Alexiou Sep 25 '16 at 01:19
  • Then what I am thinking about, is that the flow rate depends on the pressure difference between the 2 point which was result of reduction in cross section are? – user131140 Sep 25 '16 at 01:29
  • A good reference for these type of problems: http://web2.clarkson.edu/projects/subramanian/ch330/notes/Pipe%20Flow%20Calculations.pdf – John Alexiou Sep 25 '16 at 01:56
  • but this reference talking about the losses, what I can't understand is why the flow rate decreases by decreasing area of hose? – user131140 Sep 25 '16 at 02:02

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To make the analysis easier assume that the flow is non turbulent.
The rate of flow of water $\dot Q$ through a pipe of radius $r$ and length $l$ is given by Poiseuille's equation

$$\dot Q=\frac {\pi \Delta P r^4}{8\eta l}$$

where $ \eta$ is the viscosity of the water.

The greater the pressure gradient $\dfrac{\Delta P}{l}$ the greater the flow rate.

Let the pressure at the pump end be $P_{\rm pump}$ and the other end of the hose be at atmospheric pressure $P_{\rm atmos}$

With a hose of length $L$ and radius $B$ the flow rate is

$$\dot Q=\dfrac {\pi (P_{\rm pump}- P_{\rm atmos} ) B^4}{8\eta L}$$

Now pinch the end of the hose which reduces the radius of the hose to $b$ and let the length of the constriction be $l$. To have the same flow rate as before the pressure at the entry point of the constriction $P_{\rm entry}$ would have to rise quite a lot because of the ${\rm radius}^4$ dependence with the length of the unconstructed hose changing hardly at all.
So the pressure gradient across the unconstructed pipe would decrease and so the flow rate though the unconstricted hose would decrease.

Perhaps it is easier to understand by using the electrical analogy with the flow rate $\dot Q$ being the current $I$, the pressure difference $\Delta P$ being the potential difference $V$ and $\dfrac {8\eta l}{\pi r^4}$ being the resistance $R$?
The addition of the constriction increase the total resistance of the hose and because the potential difference is constant, the current decreases.

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