Which would drain faster, the tank pictured or an upside-down conical tank of th

Question

Which would drain faster, the tank pictured or an upside-down conical tank of the same dimensions draining through a hole of the same size (1-cm diameter)? How long would it take to drain the upside-down tank? Project C: Torricelli's Law of Fluid Flow Project How long does it take for water to drain through a hole in the bottom of a tank? Consider the tank pictured in Figure 2.14, which drains through a small, round hole. Torricelli's law states that when the surface of the water is at a height h, the water drains with the velocity it would have if it fell freely from a height h (ignoring various forms of friction). 30 cm 50 cm A(h) Figure 2.14 Conical tank Procad

Explanation / Answer

dV/dt = -a*sqrt(2gh)

a = area of hole

g = gravity

h = height of water in tank

V = volume of water in tank

For the cone with the point down:

Volume = (1/3) *pi* r^2 h

where:

r = radius at the surface of the water

h = height of water

Let:

r0 = radius of tank

h0 = height of tank

Since the tank and water are the same shape, you can find the relation:

r = h (r0/h0)

So now you have your Volume in terms of h only:

V = (1/3) *pi* (r0/h0)^2 h^3

dV/dt = pi* (r0/h0)^2 h^2 dh/dt = -a*sqrt(2gh)

Bringing all the h's to the left, and all the constants to the right with the dt, gives:

h^2/sqrt(h) dh = [-a(h0)^2*sqrt(2g) / pi*(r0)^2] dt

Integrate to get:

(2/5)h^(5/2) = [-a(h0)^2*sqrt(2g) / pi(r0)^2] t + C

To find C, use the value at t=0, h=h0

C = (2/5)h0^(5/2)

Solving for t when h=0 (when tank will be empty) gives:

t1 = (2/5) [pi(r0)^2*sqrt(h0) / a*sqrt(2g)]

For the cone with the point up:

Volume of the water will be the total volume of the tank (V0) minus the volume of the empty cone above the water.

Using the same relationship between the radius and height from above, and that the height of the cone of empty

space will be

(h0 - h)

where h is the height of the water

We get the equation for Volume of water in terms of height of water as:

V = V0 - (1/3)*pi*(r0/h0)^2(h0-h)^3

dV/dt = pi*(r0/h0)^2(h0-h)^2 dh/dt = -a*sqrt(2gh)

(h0-h)^2/sqrt(h dh) = [-a(h0)^2*sqrt(2g) / pi*(r0)^2] dt

Integrating and solving for the constant of integration with the initial value (t=0, h=h0) gives us:

t = [(2/5)h^2sqrt(h)-(4/3)(h0)hsqrt(h)+2(h0)^2*sqrt(h)] - [(16/15)(h0)^2*sqrt(h0)]

all over [-a(h0)^2*sqrt(2g) / pi*(r0)^2]

I'm not going to bother simplifying, because I just need to find t at h=0, which is:

t = [(16/15)(h0)^2*sqrt(h0)] / [-a(h0)^2*sqrt(2g) / pi*(r0)^2]

which simplifies to:

t2 = (16/15) [pi*(r0)^2*sqrt(h0) / a*sqrt(2g)]

And from above, we remember that the time for the cone pointing down to empty is:

t1= (2/5) [pi(r0)^2*sqrt(h0) / a*sqrt(2g)]

Everything inside the brackets [ ] is the same, so

(2/5) [x] < (16/15) [x]

we can see that the time it takes for the tank with the point down to empty is less than the time it takes the tank with

the point up to empty by the ratio of:

(2/5) / (16/15) = 3/8

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