Why water can fly faster (then slower) when you press the water hose with your thumb?

If you press your thumb on a water hose, you can make the water flow faster, but if you try to make the hole even smaller, you will find that the water velocity starts to decrease. Why?

Also, the velocity of the water should be

$v = \sqrt{\frac{2P}{ρ}}$
where P is water pressure and $ρ$ is water density.

This means that the water pressure must increased when you decreased the hole area, but shouldn't the water pressure be independent to the flow rate?

The pressure at the pinched end of the hose asymptotically approaches the source pressure in the water main for decreasing values of the flow rate through the hose. Since the flow rate is set by the source pressure and the orifice area of the pinched end of the hose, as you pinch the end of the hose off, the pressure right inside the pinched end stops increasing as its area gets small. This sets a limit on how fast the water can get shot out the end of the hose.

For very small values of the size of the opening at the end of the hose, viscous losses in that "orifice" get large and the velocity of the exit jet starts decreasing. This effect is made worse by surface tension effects, which oppose the creation of the jet for very small values of the opening size. At that point, the water exiting the hose begins to dribble slowly out the hose end- that is, its exit velocity approaches zero.

Continuity equation:

$$A_{1}V_{1}=A_{2}V_{2};$$

where $$V_{2}$$ is the hose velocity. If you reduce the area $$A_{2}$$, $$V_{2}$$ increases.

• V1 and V2 in this case means the velocity coming into the pipe and velocity coming out of the pipe, which is not useful in this scenario. This equation only tells you the conservation of volume flow rate. (For example, I can cover the end of pipe entirely to make both A and V zero; and then if I release my thumb both A, V, and their product will be positive number, so AV is not conserved in this case) Feb 9, 2018 at 0:28

This is explained by continuity equation. Since water isn't that compressable, then it density stays constant throughout the flow.

(Mass/t)in = (mass/t)out

(Density×Volume/t)in = (density×Volume/t)out

Density is same on both sides so it drops out.

Volume=A×l, where l is length of the mass element moving from inside hose to outside. Then equation becomes: (A×l/t)in=(A×l/t)out, let velocity of water be v=l/t, then equation becomes: (Av)in=(Av)out Blocking part of hose opening with thumb, leads to Ain>Aout which means that vout>vin. So the water has to exit hose faster so that the rate water mass enters the hose, is same as exit rate.