how Bernollies law applies for fountains calculaion

[247401]

I have a question which occupied my mind for a while:
when we sqeeze a hose for watring, who we can calculate the distance of water drop? how Bernollies law applies in this case? how about fountains? does the total pressure convert to velocity pressure?
appreciated for your answers
thanks
Mehran Nassseri
Vancouver
Canada

 

[kcj]

Think about Bernoulli, at least in theory and neglecting losses. Pressure plus kinetic enrgy plus head pressure, i.e. total energy is constant (without losses or pump input) just changing form.

-after the pump, the pressure is high, kinetic from motion is medium (fairly slow velocity) and head (height of reference) we will asume zero as the reference point.

-coming out of the nozzle, pressure is almost atmospheric, exposed to air at the sides, and head may be only a couple feet higher than the pump so most of the energy is in the kinietic, motion and velocity.

-At highest point of the fountain, velocity is zero as the water just starts to fall back down. Pressure is atmospheric, so all of the energy has been converted to head, height above the pump.

the units are nasty to work with, but the concept is simple.

kcj

 

[BobPE]

kcj:

your statement is neglecting the beauty of nozzle design. Let me clarify for you if I may. Out of the nozzle, pressure is no longer system pressure and takes that of the surrounding atmoshpere and not a bit more. Energy not converted to velocity in the nozzle is lost to the efficiency of the nozzle as the EGL collpases to meet the top surface of the stream as the stream leave the confines of the nozzle. The nozzle comes into its own in converting as much head energy as possible into velocity head.

You are right though, the concept is quite simple, but only for us engineers...trying to explain this concept to laypersons is quite the chalange.

take care...

BobPE