Bernoulli's Equation

[ven2083]

So I was applying the Bernoulli's equation to real life objects, such as a high-pressure hose (used to clean driveways, etc.) and found that something is odd....if the radius of a hose is smaller, than the pressure is decreased, and its the velocity of the fluid that is increased. So why do they call a high-pressure hose "high-pressure" when in fact the hose and nozzle that it comes out is smaller than a garden hose you attach to it. It seems that it should be called a "high-velocity" hose since the fluid that is going from the garden hose (larger radius, larger pressure) to the smaller nozzle (lower radius, lower pressure, high velocity)...I feel like I am missing something here, but not sure. Anyone have an idea?

I am studying TBR Physics Part II Book page 76 Example 7.9b for reference. This problem is what got me thinking in the first place.

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[sdm33]

I like the way you are thinking. May be they should called it a high velocity hose instead, because as pressure and radius decrease the velocity increase to keep the flow constant , but if we increase the pressure while keeping the radius constant then the flow will increase in this case it can be called high pressure hose.

 

[ven2083]

Ah! I knew I was missing something, thanks!!

 

[MedGrl@2022]

So I was applying the Bernoulli's equation to real life objects, such as a high-pressure hose (used to clean driveways, etc.) and found that something is odd....if the radius of a hose is smaller, than the pressure is decreased, and its the velocity of the fluid that is increased. So why do they call a high-pressure hose "high-pressure" when in fact the hose and nozzle that it comes out is smaller than a garden hose you attach to it. It seems that it should be called a "high-velocity" hose since the fluid that is going from the garden hose (larger radius, larger pressure) to the smaller nozzle (lower radius, lower pressure, high velocity)...I feel like I am missing something here, but not sure. Anyone have an idea?

I am studying TBR Physics Part II Book page 76 Example 7.9b for reference. This problem is what got me thinking in the first place.

I think the pressure of the "high pressure hose" refers to the pressure of the water that is actually coming out of the hose... and if you think of pressure as force times area then the pressure of the water that is coming out of a high pressure hose actually may have a higher pressure as it is coming out at a great velocity (I am not sure what equation would relate the two... some other SDNer should feel free to help us with that)... this water also has greater uniform translational kinetic energy and power

The pressure in Bernoulli's equation actually refers to the pressure of the random translational motion of the fluid particles... that is why velocity and pressure are inversely proportional in Bernoulli's equation because if all the energy goes into uniform translational kinetic energy then in order to conserve energy then random translational motion (pressure) would have to decrease...

also in order for the water to be coming out of the tube the pressure of the water must be greater than atmospheric pressure... I am also assuming it might have something to do with (change in)P=QR... change in pressure=(volume flow rate)(resistance to flow)... but I am not sure...

I hope this helps...

 

[Meredith92]

I like the way you are thinking. May be they should called it a high velocity hose instead, because as pressure and radius decrease the velocity increase to keep the flow constant , but if we increase the pressure while keeping the radius constant then the flow will increase in this case it can be called high pressure hose.

Just wanted to clarify one thing- if radius is constant and we increase pressure wouldn't velocity decrease which decreases flow? Since v decreases due to bernoullis and therefore flow deceased due to q=av ( and a is constant)
Thanks!

 

[milski]

Just wanted to clarify one thing- if radius is constant and we increase pressure wouldn't velocity decrease which decreases flow? Since v decreases due to bernoullis and therefore flow deceased due to q=av ( and a is constant)
Thanks!

In general no. When you change the pressure and keep the radius (and everything else) constant, you change the flow. Bernoulli's is applicable only for a given flow. Q=AV will be a constant again, but a different constant.