Which container will drain first: one big hole or several small holes?

[Carolyn]

if we have two same containers with the same amount of water

and we drill one big hole in one container and several small holes in the other container so that the total area of the small holes are equal to the area of the big hole.

The water in which container will be drained out completely first?

I think it really depends on if the condition is ideal or not. If we don't consider friction or water surface tension, then I would think they require the same amount of time. What do you think?

Thanks in advance.

 

[Astronuc]

if we have two same containers with the same amount of water, and we drill one big hole in one container and several small holes in the other container so that the total area of the small holes are equal to the area of the big hole.

The water in which container will be drained out completely first?

I think it really depends on if the condition is ideal or not. If we don't consider friction or water surface tension, then I would think they require the same amount of time.

OK, let us assume that your conclusion is reasonable. Why would one conclude that?

What can one say about mass flow rate, density, water velocity, and flow area?

 

[Carolyn]

OK, let us assume that your conclusion is reasonable. Why would one conclude that?

What can one say about mass flow rate, density, water velocity, and flow area?

Hi, the way I think about this is as the following:

the reason that water can be drained is because gravity is pulling on them and there is no base to support them. If we assume the density of water is the same throughout. Then the unit mass of water would be the same throughout. F=mg, so the force on each unit mass would be the same, thus acceleration is the same. So the flow rate is the same?

I also assumed that if the water is flowing in cylinder-like units. V=pi*r^2h.
since the area of the circle add together to be the area of the large circle,
in a certain fixed amount of time, the volume of water flowing out is the same?

So it would take the same amount of time to drain both?

 

[Astronuc]

Assuming that the containers are the same size, and the depth of the water in both containers is the same, then yes, the pressure at the same depth is the same. In the same gavitational field and same atmosphere, the pressure is dependent on height (depth) of water, not on cross-sectional area. Cross-sectional area of the vessel would however determine the total amount of water transport.

I am trying to get you to write the equation which relates mass flow rate $\dot{m}$ to fluid (water) density ($\rho$), water velocity (v), and flow area (A).

 

[Carolyn]

Assuming that the containers are the same size, and the depth of the water in both containers is the same, then yes, the pressure at the same depth is the same. In the same gavitational field and same atmosphere, the pressure is dependent on height (depth) of water, not on cross-sectional area. Cross-sectional area of the vessel would however determine the total amount of water transport.

I am trying to get you to write the equation which relates mass flow rate $\dot{m}$ to fluid (water) density ($\rho$), water velocity (v), and flow area (A).

hi, I'll thank for your reply in the first place.

In fact I have never learned about mass flow rate.

But I searched on google:

If the fluid initially passes through an area A at velocity V, we can define a volume of mass to be swept out in some amount of time t. The volume v is:

v = A * V * t

A units check gives area x length/time x time = area x length = volume. The mass m contained in this volume is simply density r times the volume.

m = r * A * V * t

To determine the mass flow rate mdot, we divide the mass by the time. The resulting definition of mass flow rate is shown on the slide in red.

mdot = r * A * V


******************************************************

so based on this equation:

if r and A and V are the same, then the flow rate is the same?

Is it why the time required is the same?

 

[Astronuc]

hi, I'll thank for your reply in the first place.

In fact I have never learned about mass flow rate.

But I searched on google:

If the fluid initially passes through an area A at velocity V, we can define a volume of mass to be swept out in some amount of time t. The volume v is: v = A * V * t

A units check gives area x length/time x time = area x length = volume. The mass m contained in this volume is simply density r times the volume.

m = r * A * V * t

To determine the mass flow rate mdot, we divide the mass by the time. The resulting definition of mass flow rate is shown on the slide in red.

mdot = r * A * V

******************************************************
so based on this equation:

if r and A and V are the same, then the flow rate is the same?

Is it why the time required is the same?

Yes - mass flow rate $\dot{m}$ = $\rho$ * V * A.

As long as the flow areas are the same, and excluding other effects like friction and surface tension, then as long as the other properties are equal, the containers empty at the same time, which also assumes that it is the same amount of water.

 

[civil_dude]

If you do consider friction, those losses will be greater with the small holes than with the large hole because they are a proportionl function of the velocity head and the one with the large hole would drain first.

 

[fig Newton]

i know this is really late, but I was going to add that the two equidimensional containers would only drain completely in the same amount of time if the smaller holes summing up to equal the area of the larger hole, were aligned so that they could all have steady flow over time. a better assumption would be that the rates of flow/time, Qdot would be equal if the fluid provided was steady (infinite) and inviscid (frictionless). understand what I'm saying?

if the topmost smaller holes were done draining water due to the diminishing depth of fluid remaining, then the new surface area of the smaller holes that are fully immersed in water is a lot less than the surface area of the larger hole if it were placed towards the bottom of the cylinder.

 

[DaveC426913]

i know this is really late, but I was going to add that the two equidimensional containers would only drain completely in the same amount of time if the smaller holes summing up to equal the area of the larger hole, were aligned so that they could all have steady flow over time. a better assumption would be that the rates of flow/time, Qdot would be equal if the fluid provided was steady (infinite) and inviscid (frictionless). understand what I'm saying?

if the topmost smaller holes were done draining water due to the diminishing depth of fluid remaining, then the new surface area of the smaller holes that are fully immersed in water is a lot less than the surface area of the larger hole if it were placed towards the bottom of the cylinder.

No one said where the holes are drilled. Thery could be drilled in the bottom, in which case, it's moot.

But point taken, the different pattern of drainage that the smaller holes can be configured in can potentially lead to a difference in flow rates once the water level is below the top of the first holes. But it would be a small difference.