mechanicaldup
Mechanical
what formula shoud i use to determine how far a water jet will carry in terms of velocity , pressure etc?
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water jet
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From Design News Reprint RS 618, V18#12,6/12/63, D.P. Costa gives V=1.158 x/y^0.5, Q=3.61 Ax/y^0.5 where
x=Horizontal distance to vertical flow centerline,in.
y=Vertical distance to start vertical column flow, in.
A=Pipe area, in^2 V=Velocity,ft/sec Q=Flow,GPM = VA/K where K=1/(144*0.002228)
x=Horizontal distance to vertical flow centerline,in.
y=Vertical distance to start vertical column flow, in.
A=Pipe area, in^2 V=Velocity,ft/sec Q=Flow,GPM = VA/K where K=1/(144*0.002228)
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mechanicaldup
Mechanical
what is the formula for X (hor distance) then?
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mechanicaldup
Mechanical
do you know of a formula that account for the inclination with horizontal?
High school physics will give you a close enough answer.
The stream velocity is determinable based on volume/second and area.
Now, from memory,
velocity = u + at, and
distance = ut + 0.5at^2,
where a=acceleration. u = initial velocity. t=time.
I think those formulas are right - highschool was over 20 years ago!
A particle of nominal mass issuing horizontally on earth has no vertical velocity, but will gain it at 9.8m/s/s, i.e., gravitational acceleration.
So if the hose is Xm above grade, then work out the time it takes for a particle to drop that distance. The horizontal distance is simply stream velocity x time.
If you aim up by a certain angle, you now have a stream velocity with both a vertical and horizontal component.
Starting with the initial vertical component, work out the time taken for the particle to decellerate to zero vertical velocity, t1.
Using the distance formula, find out how high the particle goes in this time. Now work out the time for the particle to fall from this new attained height to the ground, t2.
The horizontal distance is then horizontal velocity by (t1+t2).
Mind you, in the real world, wind resistance, wind speed, inherent losses within a stream of water, density of the atmosphere, temperature etc etc etc will all impact on the results.
Cheers
Rob
The stream velocity is determinable based on volume/second and area.
Now, from memory,
velocity = u + at, and
distance = ut + 0.5at^2,
where a=acceleration. u = initial velocity. t=time.
I think those formulas are right - highschool was over 20 years ago!
A particle of nominal mass issuing horizontally on earth has no vertical velocity, but will gain it at 9.8m/s/s, i.e., gravitational acceleration.
So if the hose is Xm above grade, then work out the time it takes for a particle to drop that distance. The horizontal distance is simply stream velocity x time.
If you aim up by a certain angle, you now have a stream velocity with both a vertical and horizontal component.
Starting with the initial vertical component, work out the time taken for the particle to decellerate to zero vertical velocity, t1.
Using the distance formula, find out how high the particle goes in this time. Now work out the time for the particle to fall from this new attained height to the ground, t2.
The horizontal distance is then horizontal velocity by (t1+t2).
Mind you, in the real world, wind resistance, wind speed, inherent losses within a stream of water, density of the atmosphere, temperature etc etc etc will all impact on the results.
Cheers
Rob
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