Fluid mechanics

Question

A water tank has a spigot near its bottom. If the top of the tank is open to the
atmosphere,
determine the speed at which the water leaves the spigot when the water level is 0.500 m
above the spigot.

Answer 1

Hi John!

 

I think we can get at this problem with the given information using Bernoulli's principle.  Often, we make one reasonable assumption with this kind of problem, which is that the tank width is so much larger than the spigot opening that the water at the top of the tank is barely moving, even when the spigot is open.  Of course, the water level in the tank will fall slowly, but its speed will be so much less than the speed of water coming out of the spigot that the speed of water at the top of the tank is negligible.

 

With that in mind, let's turn to Bernoulli's equation:

 

P + (1/2)ρv² + ρgh = constant

 

where

 

P = pressure of the fluid

ρ = density of the fluid

v = speed of the fluid

h = height of the fluid (relative to some h = 0 level we can choose)

 

This is essentially a statement of conservation of energy in fluids, and holds for a wide variety of situations.  The pressure represents a kind of "internal energy" for the fluid; the (1/2)ρv² represents kinetic energy (it even looks like (1/2)mv², which is kinetic energy for an object of mass m); and the ρgh term represents the potential energy (compare with mgh for objects).

 

What Bernoulli's equation says is that this quantity of total energy -- internal + kinetic + potential -- is constant at any point within a contained fluid that is not being pumped in any way.  So, if we calculate the left side of the above equation for one point in the fluid, it will be the same for any other point:

 

P₁ + (1/2)ρv₁²+ρgh₁ = P₂+(1/2)ρv₂²+ρgh₂

 

for any points "1" and "2" in the fluid.  We can use this to answer the question posed.

 

Let's look at the top of the water first.  I will set the height of the spigot to be h = 0.

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P_top = P_atm  (because the top of the water is said to be open to the atmosphere, and therefore must be in pressure equilibrium with it)

v_top = 0  (our assumption due to the size of the tank being so much larger than the size of the spigot)

h_top = 0.5 m  (since the top of the water is said to be 0.5 m above the spigot, which is h = 0 for me)

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Then, let's look at the water right at the opening of the spigot.

 

P_spigot = P_atm

This may seem counterintuitive, since the water comes flying out of the spigot when it is opened.  But since the exact position of the spigot is open to the atmosphere, the water right at that point is pressure-equalized with it, just like any fluid open to the atmosphere.  Now, the pressure inside the bottom of the tank, away from the opening, is greater, which is why the water gets moving and comes out, but that is taken care of by another part of Bernoulli's equation.  At the exact point where it is open to the atmosphere, its internal pressure equals atmospheric pressure.

 

v_spigot = what is to be solved for

h_spigot = 0

 

So, with this information, we can set up Bernoulli's equation:

 

P_top+(1/2)ρv_top²+ρgh_top = P_spigot+(1/2)ρv_spigot²+ρgh_spigot

 

which becomes

 

P_atm + (1/2)ρ(0)²+ρg(0.5m) = P_atm + (1/2)ρv_spigot²+ρg(0) --> P_atm+0+ρg(0.5m) = P_atm+(1/2)ρv_spigot² + 0

 

(note that P_atm+ρg(0.5m) is the pressure inside the bottom of the tank due to the weight of the water above.  Putting the "potential energy" term into Bernoulli's equation accounts for the pressure differential in this way.

 

The P_atm terms cancel on each side of the previous equation, leaving:

 

ρg(0.5 m) = (1/2)ρv_spigot²

 

Note that the density of the fluid cancels out, and does not matter to the velocity coming out of the spigot.  This is a lot like how the speed of a falling object does not depend on its mass if it is falling only under the influence of gravity (a force that also conserves mechanical energy).

 

With that in mind, the last equation is one equation with one unknown, v_spigot, that you can solve for.

 

I hope this helps!  If you have any other questions about this, or want to check an answer, just let me know.

 

Answer 2

It is very important to know the weight of the water at the point of the spigot (note: since the tank is open to the atmosphere at the top, there is no less-than-atmospheric pressure due to vacuum or greater-than-atmospheric pressure like in a pump).

 

It is also very important to know the area of the cross-section of the spigot.  This is because the problem asks for "the speed at which the water leaves the spigot."  PLZ note that water leaves the spigot faster when the tank is full (whatever that is) than when the tank is at "water level of 0.500 m."  It leaves at rate that is slower still when the tank is almost empty.  [imagine a hose attached to the spigot]

 

And, "near the bottom" might not be accurate enough because only the water weight between the water level of 0.500 m and the position of the spigot contributes to the pressure that pushes the water out of the spigot. [that is, once the water level reaches the level of the spigot, flow stops, unless there is more-than-atmospheric pressure in the tank].

 

More information is needed !