Partial Vol of horizonal cylinder with 2 frustums

[firmulator]

Homework Statement

I am trying to solve for the partial volume of a horizonal cylinder that has two frustums attached to the sides. The cylinder radius is equal to the large radius of each frustum. The frustum bases and caps are open. The container is full when liquid begins to run out of the sides of the container at the bottom of the caps.

Homework Equations

Constants -
L=cylinder length
R1=cylinder radius = frustum base radius
R2- frustum cap radius
d=frustum distance from base to cap
h=height of liquid
Equations -
1) Cylinder volume is: V = Pi * R1^2 * L
2) Frustum full volume is: V=Pi * (d/3) * (R1^2 +R2^2 + R1*R2)
3) Partial volume of Cylinder is: V= L*(R1^2 * acos((R1-h)/R1) - (R1-h)*(2*R1*h-h^2)^.5

The Attempt at a Solution

I attempted to subsitute (R1^2 +R2^2 + R1*R2) for the cylinder's R1 into#3 for one frustum's partial volume but it doesn't seem to work out. Can you help? I'm interested in the liquid volume that would begin to overflow out of the small frustum cap and smaller h's also.

 

[firmulator]

equation 3 change:
3) Partial volume of Cylinder is:
V= L*[(R1^2 * acos((R1-h)/R1) - (R1-h)*(2*R1*h-h^2)^.5]

 

[dynamicsolo]

I've been working on this question when I've had some free time and notice that no one else had jumped on it...

While it is true that you can get the transverse cross-section of the filled portion of the cylinder without calculus, such is not the case (unfortunately) for the situation you've asked about. For the cylinder, the filled cross-section is the section of a circle below a horizontal chord a distance h from the bottom (or R-h below the center). So geometry and a little trig will give you

$$A = [ R^2 \cdot arccos(\frac{R-h}{R}) ] \, - \, [ (R-h) \cdot \surd( h \cdot [2R-h] ) ]$$

The volume is then found by multiplying this area by the length L of the cylinder. (Here is a graph of the cross-section function, with the height of the water level h and the filled area A scaled to the radius: http://usera.imagecave.com/dynamicsolo/PFpix/frustum0808a.jpg . As one would expect, the ratio A/R² runs from zero at h/R = 0 to pi/2 at h/R = 1 , a filled semicircle.)

I am reprising the formula you provided, in order to point out what must be altered to produce the partial-frustum volume formula you seek. As I am understanding this, you want the frustum to have its symmetry axis horizontal (in fact, coaxial with the cylinder) with the base radius matching the cylinder's radius.

We will first look at the case where the water level just reaching the "lip" of the open frustum end. We take the radius of the frustum base to be equal to the cylinder's radius and call it R₁; the other end of the frustum has radius R₂; and we call the length of the frustum d.

Since the longitudinal cross-section of the frustum is a trapezoid, it has straight sides, so variations in relevant quantities follow linear functions. Over the length of the frustum, the radius changes linearly from R₁ to R₂ , so we have a radius function

$$R(x) = R_1 + (R_2 - R_1) \cdot (\frac{x}{d}) = R_1 - (\frac{\rho }{d}) \cdot x$$

where we take x = 0 at the end attached to the cylinder and x = d at the open end of the frustum, and $$\rho = R_1 - R_2$$, a difference which will be convenient to use here and there.

By the same token, the transverse cross-section of the frustum is filled to a depth which also varies linearly, from h = R₁ - R₂ at the cylinder end to
h = 0 at the open end. So we will also construct a height function

$$h(x) = ( R_1 - R_2 ) + (\frac{R_2 - R_1}{d}) \cdot x = \rho - (\frac{\rho}{d}) \cdot x$$

using the same conventions as for R(x). It will also be worthwhile to work out

$$R(x) - h(x) = [ R_1 - (\frac{\rho }{d}) \cdot x ] - [ \rho - (\frac{\rho}{d}) \cdot x ] = R_2$$

Now, we are ready to make the necessary replacements in the formula for the transverse cross-section of the partially-filled cylinder. The area formula becomes

$$A(x) = [ R^2(x) \cdot arccos(\frac{R(x)-h(x)}{R(x)}) ] \, - \, [ (R(x)-h(x)) \cdot \surd( h(x) \cdot [2R(x)-h(x)] ) ]$$

(in words, everything that was a constant in the cylinder formula has become a linear function)

$$\Rightarrow A(x) = [ [R_1 - (\frac{\rho }{d}) \cdot x]^2 \cdot arccos(\frac{R_2}{R_1 - (\frac{\rho }{d}) \cdot x}) ] \, - \, [ R_2 \cdot \surd( [\rho - (\frac{\rho}{d}) \cdot x ] \cdot [2R_1 - \rho - (\frac{\rho}{d}) \cdot x] ) ]$$

Here is a graph of the transverse cross-sectional area A(x) as a function of x, for a frustum with R₁ = 1 , R₂ = 0.5 , rho = 0.5 , and d = 2 :
http://usera.imagecave.com/dynamicsolo/PFpix/frustum0808b.jpg , the function being marked in black. We will refer to it further below.

This would mean that the volume of the partially-filled frustum with water just reaching the mouth of the open end is given by the integral

$$V = \int_0^{d} A(x) \, dx$$

$$= \int_0^{d} \, [ [R_1 - (\frac{\rho }{d}) \cdot x]^2 \cdot arccos(\frac{R_2}{R_1 - (\frac{\rho }{d}) \cdot x}) ] \, - \, [ R_2 \cdot \surd( [\rho - (\frac{\rho}{d}) \cdot x ] \cdot [2R_1 - \rho - (\frac{\rho}{d}) \cdot x] ) ] \, dx$$

As you might imagine, this one doesn't appear in tables of integrals, largely, I suspect, because of that argument of the arccosine function. I would guess that either the integral is done numerically or the matter of finding the volume is simply done empirically (say, by making a full-sized figure or a model, putting water in it, and measuring the volume involved).

However, it is still possible to say something about the volume and maybe come up with a simple approximation for it. The area under the black curve in that second attached graph would give us the volume, so we can take a crack at estimating it.

Let's call the area at the cylinder end (x = 0), A_o, which is

$$A_0 = [R^2_1 \cdot arccos(\frac{R_2}{R_1})] \, - \, [R_2 \cdot \surd(2R_1 \rho - \rho^2)]$$

If we just draw a straight line from end to end (the red line), the area of the triangle produced would give a volume estimate of V_hi = (1/2)·A_o·d , which plainly overestimates the water volume in the frustum.

If we instead use a parabolic curve connecting the ends of the area function curve, we have a curve of the form

$$x = \frac{d}{A^2_0} \cdot (y-A_0)^2 \Rightarrow y = A_0 \cdot [1 - \surd(\frac{x}{d})]$$

which is marked in green on the graph. When we integrate this curve to find the area under it (the estimate for the volume in the frustum), we find

$$V_{lo} \, = \int_0^{d} A_0 \cdot [1 - \surd(\frac{x}{d})] \, dx = \, \frac{1}{3} \cdot A_0 \cdot d$$

which somewhat underestimates the volume in the frustum. So we can bracket the result for the volume by $$\frac{1}{3} \cdot A_0 \cdot d < V < \frac{1}{2} \cdot A_0 \cdot d$$. We might get even closer by averaging these limits to give the result as

$$V \approx \frac{5}{12} \cdot A_0 \cdot d$$

or, to one significant figure, even just

$$V \approx 0.4 \cdot A_0 \cdot d$$

with A₀ as given above.

The last part you asked about was how to adapt this to a frustum filled to lower levels. Since the variations are linear, the rescaling isn't too difficult (just a bit of a nuisance). The depth of water we've looked at is what we called rho = R₁ - R₂ . If the depth were instead a fraction of that, which we could call f · rho (f between 0 and 1), then only a couple of our quantities would need to be rescaled: R₁ would be unaffected and we would now replace R₂ with f · R₂ and d with f · d . You would then substitute these appropriately into our expression for A(x) , or into A₀ to use our volume estimate.

(Wow -- longest post I've ever made...)