[bushindo]

Suppose that we have two rectangular boxes that are exactly the same. Both are placed upright and filled to the brim with identical amount of water. Suppose that we drill a circular hole with radius r at the bottom of box 1, and we drill 2 separate holes, each also with radius r, in the bottom of box 2. Would box 1 take twice as long to completely drain all the water compared to box 2? Would box 1 take longer to drain? Shorter?

Edited May 9, 2010 by bushindo

[Guest]

two holes will drain faster than one hole, but not twice as fast. After a small amount of time, dt, the level in box2 will have dropped twice as much as the level in box1. Thus the pressure at the outlet holes in box2 will be a proportionately small amount less than the pressure at the outlet hole in box1. Thus the rate at which water flows through each of the holes in box2 will be proportionately slower than the flow rate through the hole in box1. This flow rate discrepancy will increase progressively as the levels drop. However, the level in box1 cannot catch up with the level in box2, otherwise the pressures would then have become equal again. So box2 will empty faster than box1, but not twice as fast.

[bushindo]

two holes will drain faster than one hole, but not twice as fast. After a small amount of time, dt, the level in box2 will have dropped twice as much as the level in box1. Thus the pressure at the outlet holes in box2 will be a proportionately small amount less than the pressure at the outlet hole in box1. Thus the rate at which water flows through each of the holes in box2 will be proportionately slower than the flow rate through the hole in box1. This flow rate discrepancy will increase progressively as the levels drop. However, the level in box1 cannot catch up with the level in box2, otherwise the pressures would then have become equal again. So box2 will empty faster than box1, but not twice as fast.

This is a good start. I should have clarified my question better in the OP. My main question is would box 1 take twice as long to completely drain all the water compared to box 2? Would box 1's drainage time be longer or shorter than twice the time of box 2?

Edited May 9, 2010 by bushindo

[Guest]

Imagine a videotape of the box with with one hole, emptying out completely. Say it takes ten seconds. Run the tape at double speed. It takes five seconds and at all times the rate of flow will be twice the rate of the original tape. But two holes is twice the rate of the original tape. It's twice the area. So it should take half the time for the 2-hole box to empty, even though the flow will decrease during the entire time.

Edited May 9, 2010 by mpmp

[Guest]

Interesting question, it's mostly about physics and since I'm not into physics as other science I won't try to answer...

But I have a spin-off question, if in Box1 there was a hole with radius r, and in Box2 there were two holes with radius r/sqrt(2) (in other words two holes with size sum same as the hole in Box1) would they drain at same speed?

[Guest]

That is how I understood the question, and that is the question I think I have answered. Box1's drainage time is shorter than twice the time of box2, as pressure at hole in box1 is (inreasingly) greater than the pressure at the holes in box2, whilst draining.

This is a good start. I should have clarified my question better in the OP. My main question is would box 1 take twice as long to completely drain all the water compared to box 2? Would box 1's drainage time be longer or shorter than twice the time of box 2?

This is a good start. I should have clarified my question better in the OP. My main question is would box 1 take twice as long to completely drain all the water compared to box 2? Would box 1's drainage time be longer or shorter than twice the time of box 2?

[Guest]

Suppose that we have two rectangular boxes that are exactly the same. Both are placed upright and filled to the brim with identical amount of water. Suppose that we drill a circular hole with radius r at the bottom of box 1, and we drill 2 separate holes, each also with radius r, in the bottom of box 2. Would box 1 take twice as long to completely drain all the water compared to box 2? Would box 1 take longer to drain? Shorter?

Fluid dynamics is not my specialty, but I will give it a shot.

I think Bernoulli's Principle is probably an acceptable model to use here.

I might be doing this wrong, because I haven't used this in several years.

From wikipedia:

v^2/2 +gz + p/d = const

where d is density

Let's consider 2 points, one at the surface of the water and one at the hole in the bottom of the box.

The water here is exposed to the atmosphere, so pressure is the same. We can sort of say

v^2 + 2*gz = const

v2^2 - v1^2 + 2g(z2-z1) = 0

Assume the box has Width*Length*Height dimensions.

Let A = area = Width*Length

The volume of water is A*h.

If the radius of a single hole at the bottom is r, then the area is a = pi*r^2

If the speed of water leaving the hole is v2 = dx/dt, then in time dt, a volume of water equal to dx*a leaves the container.

This corresponds to a change in water level dh.

The change in volumes should be the same by conservation of mass (since density is constant).

So A*dh = a*dx, and v1 = -dh/dt = a/A * v2

Letting also, z2 = 0, and z1 = h, we have

(A^2/a^2 - 1)*v1^2 = 2gh

or

v1 = a*sqrt(2*g*h)/sqrt(A^2-a^2) = -dh/dt

the solution of the differential equation

dh/dt = -c*sqrt(h) is

h = -c^2/4*t^2 + C

So we have

h(t) = h0 - (a^2/2)*g/(A^2-a^2)*t^2

The tank will empty when h(t) = 0

or when

t = sqrt(2*h0*(A^2-a^2))/a

If we double the area a, to b=2a

the if t1 = sqrt(2*h0*(A^2-a^2))/a

t2 = sqrt(2*h0*(A^2-4a^2))/2a

The fraction is

t1/t2 = 2*sqrt((A^2-a^2)/(A^2-4a^2))

If I did everything correctly, which I am not sure is the case,

then it should take MORE than twice as much time for box 1 to empty than box 2.

This is because A^2-a^2 is greater than A^2-4a^2

If a is much less than A, then this will barely be noticeable.

if a is 1 in some unit, and A is 100, then

t1/t2 = 2*sqrt(9999/9996) = 2.000300098

[Guest]

Fluid dynamics is not my specialty, but I will give it a shot.

I think Bernoulli's Principle is probably an acceptable model to use here.

I might be doing this wrong, because I haven't used this in several years.

From wikipedia:

v^2/2 +gz + p/d = const

where d is density

Let's consider 2 points, one at the surface of the water and one at the hole in the bottom of the box.

The water here is exposed to the atmosphere, so pressure is the same. We can sort of say

v^2 + 2*gz = const

v2^2 - v1^2 + 2g(z2-z1) = 0

Assume the box has Width*Length*Height dimensions.

Let A = area = Width*Length

The volume of water is A*h.

If the radius of a single hole at the bottom is r, then the area is a = pi*r^2

If the speed of water leaving the hole is v2 = dx/dt, then in time dt, a volume of water equal to dx*a leaves the container.

This corresponds to a change in water level dh.

The change in volumes should be the same by conservation of mass (since density is constant).

So A*dh = a*dx, and v1 = -dh/dt = a/A * v2

Letting also, z2 = 0, and z1 = h, we have

(A^2/a^2 - 1)*v1^2 = 2gh

or

v1 = a*sqrt(2*g*h)/sqrt(A^2-a^2) = -dh/dt

the solution of the differential equation

dh/dt = -c*sqrt(h) is

h = -c^2/4*t^2 + C

So we have

h(t) = h0 - (a^2/2)*g/(A^2-a^2)*t^2

The tank will empty when h(t) = 0

or when

t = sqrt(2*h0*(A^2-a^2))/a

If we double the area a, to b=2a

the if t1 = sqrt(2*h0*(A^2-a^2))/a

t2 = sqrt(2*h0*(A^2-4a^2))/2a

The fraction is

t1/t2 = 2*sqrt((A^2-a^2)/(A^2-4a^2))

If I did everything correctly, which I am not sure is the case,

then it should take MORE than twice as much time for box 1 to empty than box 2.

This is because A^2-a^2 is greater than A^2-4a^2

If a is much less than A, then this will barely be noticeable.

if a is 1 in some unit, and A is 100, then

t1/t2 = 2*sqrt(9999/9996) = 2.000300098

I just realized that +C is not allowed in that particular differential equation.

Perhaps someone more experienced with fluid mechanics can lend a hand.

[bushindo]

Fluid dynamics is not my specialty, but I will give it a shot.

I think Bernoulli's Principle is probably an acceptable model to use here.

I might be doing this wrong, because I haven't used this in several years.

From wikipedia:

v^2/2 +gz + p/d = const

where d is density

Let's consider 2 points, one at the surface of the water and one at the hole in the bottom of the box.

The water here is exposed to the atmosphere, so pressure is the same. We can sort of say

v^2 + 2*gz = const

v2^2 - v1^2 + 2g(z2-z1) = 0

Assume the box has Width*Length*Height dimensions.

Let A = area = Width*Length

The volume of water is A*h.

If the radius of a single hole at the bottom is r, then the area is a = pi*r^2

If the speed of water leaving the hole is v2 = dx/dt, then in time dt, a volume of water equal to dx*a leaves the container.

This corresponds to a change in water level dh.

The change in volumes should be the same by conservation of mass (since density is constant).

So A*dh = a*dx, and v1 = -dh/dt = a/A * v2

Letting also, z2 = 0, and z1 = h, we have

(A^2/a^2 - 1)*v1^2 = 2gh

or

v1 = a*sqrt(2*g*h)/sqrt(A^2-a^2) = -dh/dt

the solution of the differential equation

dh/dt = -c*sqrt(h) is

h = -c^2/4*t^2 + C

So we have

h(t) = h0 - (a^2/2)*g/(A^2-a^2)*t^2

The tank will empty when h(t) = 0

or when

t = sqrt(2*h0*(A^2-a^2))/a

If we double the area a, to b=2a

the if t1 = sqrt(2*h0*(A^2-a^2))/a

t2 = sqrt(2*h0*(A^2-4a^2))/2a

The fraction is

t1/t2 = 2*sqrt((A^2-a^2)/(A^2-4a^2))

If I did everything correctly, which I am not sure is the case,

then it should take MORE than twice as much time for box 1 to empty than box 2.

This is because A^2-a^2 is greater than A^2-4a^2

If a is much less than A, then this will barely be noticeable.

if a is 1 in some unit, and A is 100, then

t1/t2 = 2*sqrt(9999/9996) = 2.000300098

I agree with everything here, but my version has a small change with respect to the integration.

From the above post, we need to solve a differential equation

v1 = a*sqrt(2*g*h)/sqrt(A^2-a^2) = -dh/dt

Let c = a/sqrt(A^2-a^2), we have

-dh/dt = c*sqrt(2*g)* sqrt(h)

-dh/sqrt(h) = c*sqrt(2*g)* dt

Integrate both sides from h1 to h2 and from the corresponding t1 to t2,

h^1/2₁ - h^1/2₂ = (1/2) * c * sqrt(2*g) * ( t₂ - t₁ ).

The time it takes to completely drain a tank, t_d, is

h^1/2/ [ (1/2) * c * sqrt(2*g) ] = t_d.

The conclusion comes out to be the same though. It should take MORE than twice as much time for box 1 to empty than box 2 because sqrt(A^2-a^2) is greater than sqrt(A^2-4a^2).

Edited May 9, 2010 by bushindo

[Guest]

I agree with everything here, but my version has a small change with respect to the integration.

From the above post, we need to solve a differential equation

v1 = a*sqrt(2*g*h)/sqrt(A^2-a^2) = -dh/dt

Let c = a/sqrt(A^2-a^2), we have

-dh/dt = c*sqrt(2*g)* sqrt(h)

-dh/sqrt(h) = c*sqrt(2*g)* dt

Integrate both sides from h1 to h2 and from the corresponding t1 to t2,

h^1/2₁ - h^1/2₂ = (1/2) * c * sqrt(2*g) * ( t₂ - t₁ ).

The time it takes to completely drain a tank, t_d, is

h^1/2/ [ (1/2) * c * sqrt(2*g) ] = t_d.

The conclusion comes out to be the same though. It should take MORE than twice as much time for box 1 to empty than box 2 because sqrt(A^2-a^2) is greater than sqrt(A^2-4a^2).

Great! Glad I was on the right track, except for the integral mistake.

[Guest]

The rate of drainage is proportional to the area of the hole. You can write the differential equations using mass and energy balance to prove it. The answer is exactly twice as long. Great question!

Bonus question: If the volume of liquid in box 1 is only half that of box 2 to start with, will they now empty simulataneously? HINT: Write out the diff. equations before answering!