Jetting water for maximum distance
This barrel filled with water is located on the ground and has 7 equally spaced, vertically aligned holes. Hole number four is exactly at the middle of the barrel.
Assuming that there is no air resistance, viscosity, or other hindrance to the water flow, which hole will jet the water farther than any other holes do?
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6 solutions
There was no mention of ground height in the problem. Assuming great ground height gives another answer does it not?
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The problem states that the barrel is located on the ground. I assumed the ground is horizontal. Otherwise you are right, the answer can be anything.
Could you explain it in a simpler manner
That means if y0 approximately equals 0,water will not move out??
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It will move out and it will hit the ground right there.
I thought when we use KE=PE, h should be the height of the hole to the ground? Why is it H-y0, which is the height of the top of the drum to the hole instead?
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The pressure at the hole is created by the water that is higher than the hole. The water that is below the hole does not matter.
Seriously, you should carry out the experiment. The hydrostatic head is greatest at the bottom hole. Do the experiment and you will see how ridiculous your analysis is.
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Did you do the experiment? I did.
the hydrostatic pressure is greatest at the bottom but that does not mean the jet will shoot out the furthest in this case. 7 will fall to the ground before it has time to reach its otherwise maximum distance. Number 4 is an optimization of the distance traveled in a manner analogous to the maximum range for firing a projectile is at 45 degrees. Think of the jet as resolved into x and y vectors and you are trying to maximize the distance traveled by the horizontal x velocity vector, which is constrained not just by pressure but time in flight.
Yes I agree with the comment saying one should conduct an experiment. As we studied in lower classes and experimented with 3 holes in the school lab the hydraulic pressure , due to volume, weight and gravity is more at the bottom hole compared to water pressure on upper holes Hence , the bottom most hole will always jet water farther distance than other holes do
Imagine a situation when water level comes down slowly by jetting out water from all holes. Hole 4 will be blank after some time and still the bottom most hole will jet water farther distance than other holes do
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Your logic is wrong. Larger pressure does not mean longer distance.
I've done this as a demonstration for Dynamic Equilibrium. The bottom hole has much greater flow than any other. . .
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That is true, but here we are not talking about the amount of flow. The question is, how far will it go.
The velocity of efflux,
V
e
=
(
1
−
A
2
a
2
)
2
g
h
where,
h
=
height of hole from the top of Barrel ,
A
is the area of the top surface of Barrel and
a
is the area of the hole.
Since, A > > > a , 1 − A 2 a 2 = 1 .Hence, V e = 2 g h
H = Height of the Barrel.
Let. X be the distance at which the water hits the ground from the bottom of the Barrel, X = 2 g h g 2 ( H − h ) = 4 h ( H − h )
X is maximum at. h = 2 H
Hence, range X will be maximum at hole 4, which is at a distance of h = 2 H .
A N S W E R : 4
I believe the last statement in your solution is wrong. The velocity is maximum at hole 7, not hole 4. However, that is irrelevant.
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Thanks for pointing that out. I by mistake typed velocity instead of range. My fault.
Where does the denominator come from for your initial expression of Ve? Your limiting expression is known as Torricelli's law.
Also, if you've done the experiment, how closely does an actual (low-viscosity) fluid conform to this theoretical result?
EDIT:
Okay I believe I understand where this expression for Ve comes from.
If a isn't negligible compared to A, there will be a non-negligible vertical velocity at the opening. From from the continuity equation, the fluid velocity inside of the barrel will be given by:
Vi A = Ve a ⇒ Vi = (a/A) Ve.
Since the fluid is inviscid, we can apply Bernoulli's equation (which is how Torricelli's law is derived) to write:
p₀ + ρg h + ½ ρ Vo² = p₀ + ½ ρ Ve² ⇒ ρg h + ½ ρ (a/A)² Ve² = ½ ρ Ve² ⇒ ρg h = ½ ρ [1- (a/A)²] Ve².
and your result follows immediately from this.
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Let, v 1 be the velocity of liquid in between surface and hole and v 2 the velocity at the hole.
By equation of continuity,
A v 1 = a v 2 ⇒ v 1 = A a v 2
Bernoulli Principle,
P a t m + 2 d v 1 2 + d g H = P a t m + 2 d v 2 2 + d g ( H − h )
We get, v 2 = v 1 2 + 2 g h
We get, v 2 = 1 − A 2 a 2 2 g h
Actually, this indeed works for low viscosity liquids. As you can see from the formula, the range has got nothing to do with the viscosity of the liquid.
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Thanks Niraj... we cross posted there, but I appreciate your response.
I've solved this problem here using dimensional analysis.
I think it is worth a look.
We are not told the distance from the top of the drum to the top spout, or the bottom of the drum to the bottom spout. The only spout who's position is not affected by this lack of information is the middle one, so that's the answer.
4 is he middle number so 4 is the correct number
You can solve it using Torcelli's Theorem.
Let as assume that the hole is at elevation y 0 from the ground, and the height of the drum is H . The exit velocity of the water is v 0 = 2 g h , where h = H − y 0 . The time of flight of the water is t = 2 y 0 / g , and in that time the water reaches the distance of x = v 0 t = 2 ( H − y 0 ) y 0 . Let us look at x 2 = 4 ( H − y 0 ) y 0 as a function of y 0 and find the maximum. The answer is y 0 = H / 2 . Therefore the correct option is #4.