Solving word problem in hydrodynamics

[HP8]

1. A circular hole 2 cm in diameter is cut in the side of a large stand pipe 10 m below the water level in the standpipe. Find:
a. the velocity of the efflux
b. the volume discharged per unit time[/b]

2. P + 1/2 pv^2 + pgh
Q = A1 V1 = A2 V2
where =
a = area
v = volume
p = pressure[b/]3. Can't really plug any values here since my professor didn't really discussed about it, more like self-study so I thought of seeing solution to understand concept. [/b]

 

[Solibelus]

Look into Torricelli's law for the velocity.
The volume discharged is just dV/dt. What exactly confuses you here?
If it's a really big stand pipe, you can maybe ignore the rate at which the water decreases when it flows out of the tiny hole :)

The first formula you posted there is Bernoulli's principle for incompressible fluid.
To understand it, try deriving it. Remember that it's based on energy conservation. You can easily spot kinetic energy and potential energy there :)
Try thinking about the definition of work and how it may play a part.

The second formula is true when the flow is uniform. It's basically a continuity equation. Think about it, it makes sense: say your fluid is moving 1m/s. During that second, it moved 1m across a pipe with the cross section of say A1. Now it flows into a pipe with c.s. of A2, which is, let's say, much smaller. Since the flow is uniform and incompressible, the same amount of fluid must come out, so the velocity will have to increase. I hope that makes sense.

 

[Nathanael]

The water level and the hole should both have the same pressure (atmospheric pressure) because they're both exposed to air (I'm assuming the water level is exposed to air and not a vacuum)

Bernoulli's equation (the one you first wrote) states that the density of energy in a liquid is conserved.
So, this means, if one part of the liquid has a greater amount of one type of energy*, it must have that amount less of other types of energy*. (PER VOLUME)**

Since the pressures are the same, the only differences in energy are that the top has more gravitational potential energy and the bottom (the hole) has more kinetic energy

(Since it didn't give you the size of the pipe you can just assume that the kinetic energy of the water level is negligable, so call it zero)

 

[HP8]

Look into Torricelli's law for the velocity.
The volume discharged is just dV/dt. What exactly confuses you here?
If it's a really big stand pipe, you can maybe ignore the rate at which the water decreases when it flows out of the tiny hole :)

The first formula you posted there is Bernoulli's principle for incompressible fluid.
To understand it, try deriving it. Remember that it's based on energy conservation. You can easily spot kinetic energy and potential energy there :)
Try thinking about the definition of work and how it may play a part.

The second formula is true when the flow is uniform. It's basically a continuity equation. Think about it, it makes sense: say your fluid is moving 1m/s. During that second, it moved 1m across a pipe with the cross section of say A1. Now it flows into a pipe with c.s. of A2, which is, let's say, much smaller. Since the flow is uniform and incompressible, the same amount of fluid must come out, so the velocity will have to increase. I hope that makes sense.

With your reply, I tried to do the problem and this is what happened;
-at a.
I applied torricelli theorem, plugging the value 10 m to h and I got 14 m/s

-at b.
I used dV/dt = which was A * v and I got 0.0043 (m^3/s)?

Am I doing it right?

 

[HP8]

Off-Topic: Can I still post one more question related to this type of question?

 

[Nathanael]

Off-Topic: Can I still post one more question related to this type of question?

I'm not an expert on the forum rules, but I'm sure it would be fine.


About the original question, sorry for my last post. Reflecting on it, I see that it's an awful explanation.

A better way to explain it requires one perspective, that pressure energy is essentially internal kinetic energy (by "internal" I basically mean that the net velocity is zero) (It helps to visualize a bunch of atoms squeezed together and constantly bouncing around each other but never moving anywhere on the whole.)
This is my natural way of viewing pressure, I find it intuitively useful.

Now you can probably imagine that the only way to be in this state of nonzero pressure, is to have something "holding it together" (otherwise it would be like little balls that bounce apart)

So, in your problem, we have a little circle of that "something that holds it together" that is removed.

What will happen?

All of the pressure will "become" velocity. (Really, nothing is becoming anything new, it's just that the net result is no longer zero.)

Except, it won't be ALL of the pressure that becomes velocity, because there is still a little bit of "something that holds it together," namely, the air pressure is trying to keep the water in.

So you will have the (energy of the) velocity will be equal to the (energy of the) pressure minus the (energy of the) air pressure.

But we know, (I'm assuming you know) that the (energy of the) pressure is equal to the air pressure plus density*height*gravity so the "air pressure minus air pressure" part becomes zero, and you're left with the equation:
[itex]ρgh=\frac{ρv^2}{2}[/itex]




(Which could be verbally interpreted as "gravitational potential energy = kinetic energy" but I think the pressure perspective is superior)

 

[Nathanael]

With your reply, I tried to do the problem and this is what happened;
-at a.
I applied torricelli theorem, plugging the value 10 m to h and I got 14 m/s

-at b.
I used dV/dt = which was A * v and I got 0.0043 (m^3/s)?

Am I doing it right?

I'm not the most experienced, (I don't know what torricelli theorem is) but for what it's worth, my answers were the same
EDIT:
These answers are only true for the first instant, though.

Because in the equation [itex]ρgh=\frac{ρv^2}{2}[/itex] from my last post, h is not a constant. h (the water level) decreases with time.

However, if the pipe has a (relatively) large area, then the decrease in the height of the water level with time could be considered negligable.
(A certain amount of water loss per second will have less of an effect on the change in height of the water level if the cross sectional area of the water is larger)