Bernoulli equation (watering can)

In summary, the gardener waters a bush with the help of a watering can. The water level in the can lays at height H = 1m, and the water flows horizontally out of the can at height h. a) at which distance x from the bush must the water flow when the mouth of the can is located at h = 0.8m, in order for the water to reach the bush? b) for which height h is x maximum? (we consider H is constant)
  • #1
JulienB
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Homework Statement



Hi everybody! Here is a classical Bernoulli problem, which I'd like you to review to check if I (finally!) make a proper use of Bernoulli's equation!

A gardener waters a bush with the help of a watering can (see attached pic). The water level in the can lays at height H = 1m, and the water flows horizontally out of the can at height h.
a) at which distance x from the bush must the water flow when the mouth of the can is located at h = 0.8m, in order for the water to reach the bush?
b) for which height h is x maximum? (we consider H is constant)

Homework Equations



Bernoulli equation, free fall and maybe a little bit of derivative?

The Attempt at a Solution



I think I get the point of the problem, but I often get confused :biggrin: Hopefully this time I get it right:

a) I first set up a Bernoulli equation between points (0) (behavior at height H) and (1) (mouth of the can at height h):

ρg(H - h) = ½⋅ρ⋅v12
(Here I considered P0 = P1 = 0 at gauge pressure, v0 = 0 and h the reference height)
⇔ v1 = √(2⋅g⋅(H - h)) = √(0.4⋅g)

Now from basic equations of motion I know:

horizontal displacement: x = v1⋅t
vertical displacement: -h = -½⋅g⋅t2

I solve for t using the second equation and get:

t = √(2h/g)

...and substitute it in the first one:

x = v1⋅t = √(0.8⋅h) = 0.8m

Is that correct?

b) Here I used the equation I found for velocity in a):

x = v⋅t = √(2g⋅(H - h))⋅√(2h/g) = 2√(hH - h2)

Here I wasn't sure what I should do so I simply took the derivative of (hH - h2) with respect to h and solved for which h it is equal to 0 (that is, when the function reaches a local critical point):

f'(h) = (hH - h2)' = H - 2h
f'(h) = 0 ⇔ h = ½⋅H = 0.5 m

Does that make sense? I get x = 1m, which is at least indeed bigger than in a). I'm sure there is also an easier method to get that result, but I couldn't figure it out yet. Any idea?Thank you very much in advance for your answers, I appreciate it.Julien.
 

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  • #2
All looks good, well done.
 
  • Like
Likes JulienB
  • #3
@haruspex thank you, that is definitely thanks to your huge help! :) I hope this post can help other people too!

Julien.
 

1. What is the Bernoulli equation and how does it relate to a watering can?

The Bernoulli equation is a fundamental equation in fluid mechanics that describes the relationship between fluid pressure, velocity, and elevation. In a watering can, the Bernoulli equation can be used to explain how the water flows out of the spout at a constant rate due to the changing pressure within the can.

2. Can you use the Bernoulli equation to calculate the velocity of water coming out of a watering can?

Yes, the Bernoulli equation can be used to calculate the velocity of water coming out of a watering can by considering the changes in pressure and elevation between the water inside the can and the water exiting the spout.

3. What is the significance of the shape of a watering can on the flow of water?

The shape of a watering can plays a significant role in the flow of water due to the Bernoulli equation. The shape of the can affects the pressure distribution inside the can, which in turn affects the velocity and flow rate of water coming out of the spout.

4. How does the water level in a watering can affect the flow rate of water?

The water level in a watering can is directly related to the pressure inside the can. According to the Bernoulli equation, as the water level decreases, the pressure decreases, which results in a higher velocity and flow rate of water coming out of the spout.

5. Can the Bernoulli equation be applied to other fluid systems besides a watering can?

Yes, the Bernoulli equation can be applied to a wide range of fluid systems, including pipes, pumps, and airfoils. It is a fundamental equation in fluid mechanics that is used to analyze and understand the behavior of fluids in various applications.

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