2025-01-10
Density is a straightforward concept, defined mathematically by the formula in equation 1. It represents the average mass per unit volume. When measuring density, it is typically assumed that the material is homogeneous and uniformly distributed within the volume. What if it wasn’t? I have some questions I want to examine:
Density is defined as:
\[ \rho = \frac{m}{v} \qquad{(1)}\]
Where:
If we add two different materials to a cylinder, we have the scenario is illustrated in fig. 1. In our case, let’s assume:
NOTE: The math can easily be extended to multiple materials. This article does not explore that route.
The variables we’ll need to define are:
The volume of the cylinder is:
\[ V = \pi r^2 \cdot h \qquad{(2)}\]
Where:
The average density is defined in equation 1. The average density of the two material system is defined in equation 6. We need to solve equation 6 and simplify the results.
\[ \rho_t = \frac{m_t}{V_t} \qquad{(3)}\]
\[ m_t = m_a + m_b \qquad{(4)}\]
\[ V_t = V_a + V_b \qquad{(5)}\]
Using equations eq. 4 and eq. 5 we can expand eq. 3 as follows:
\[ \rho_t = \frac{m_a + m_b}{V_a + V_b} \qquad{(6)}\]
Where:
\[ m_a = \rho_a \cdot V_a \qquad{(7)}\]
\[ m_b = \rho_b \cdot V_b \qquad{(8)}\]
\[ V_a = \pi r^2 \cdot h_a \qquad{(9)}\]
\[ V_b = \pi r^2 \cdot h_b \qquad{(10)}\]
Using eq. 7, eq. 8, eq. 9, and eq. 10:
\[ \begin{aligned} \rho_t &= \frac{\rho_a \cdot V_a + \rho_b \cdot V_b}{\pi r^2 \cdot h_a + \pi r^2 \cdot h_b} \\ &= \frac{\pi r^2 \cdot \left( \rho_a \cdot h_a + \rho_b \cdot h_b \right)}{\pi r^2 \cdot \left(h_a + h_b \right)} \end{aligned} \]
\[ \rho_t = \frac{\rho_a \cdot h_a + \rho_b \cdot h_b}{h_a + h_b} \qquad{(11)}\]
The final average density can be calculated from eq. 11 in terms of the height and density of each material.
What if we had a target density for the whole cylinder, \(\rho_t\), and we knew the material densities, \(\rho_a\) and \(\rho_b\)? Can we determine \(h_a\) and \(h_b\)? Let’s rearrange eq. 11 and see if we can figure it out.
\[ \begin{aligned} \rho_t \cdot h_a + \rho_t \cdot h_b &= \rho_a \cdot h_a + \rho_b \cdot h_b \\ \rho_t \cdot h_a - \rho_a \cdot h_a &= \rho_b \cdot h_b - \rho_t \cdot h_b \\ h_a \cdot \left(\rho_t - \rho_a \right) &= h_b \cdot \left( \rho_b - \rho_t \right) \end{aligned} \]
\[ \frac{h_a}{h_b} = \frac{\rho_b - \rho_t}{\rho_t - \rho_a} \qquad{(12)}\]
Equation 12, shows the ratio of \(h_a\) and \(h_b\).
\[ h_a = \frac{\rho_b - \rho_t}{\rho_t - \rho_a} \cdot h_b \qquad{(13)}\]
We know:
\[ h_t = h_a + h_b \qquad{(14)}\]
Therefore, if we substitute eq. 13 into eq. 14 and simplify:
\[ h_t = \frac{\rho_b - \rho_t}{\rho_t - \rho_a} \cdot h_b + h_b \]
\[ h_t = h_b \cdot \left(1 + \frac{\rho_b - \rho_t}{\rho_t - \rho_a}\right) \qquad{(15)}\]
Rearranging and solving for \(h_b\):
\[ h_b = \frac{h_t}{\left( 1 + \frac{\rho_b - \rho_t}{\rho_t - \rho_a} \right)} \qquad{(16)}\]
If we know the shell dimensions, the densities of the materials and the average target density, we can solve for the height of the cylinder for material b, \(h_b\) (equation 16). Once you know \(h_b\) you can solve for \(h_a\) using equation 14. If you are interested in the volume of either material cylinder use equation 9 and 10. If you want the mass of the material cylinders using equations 7 and 8.
To solidify the concept, let’s use a practical example:
NOTE: Normally, you would convert them so they are all using the same unit system. In this case, as long as the density units are the same and the length units are the same, there will not be a problem. The units cancel.
\[ \begin{aligned} h_b &= \frac{6.25}{\left( 1 + \frac{0.93 - 1.00}{1.00 - 1.30} \right)} \\ h_b &= \frac{6.25}{\left(1 + 0.23 \right)} \\ h_b &= 5.08 \end{aligned} \]
From equation 14, we have:
\[ \begin{aligned} h_a &= h_t - h_b \\ h_a &= 6.25 - 5.07 \\ h_a &= 1.18 \end{aligned} \]
Therefore, to achieve the target density of \(\rho_t = 1.00 \, \frac{g}{\text{cc}}\), \(h_b\) is 5.06 inches and \(h_a\) is 1.19 inches.
It occurred to me, since I wrote the article yesterday, that equation 12 is not the most convenient form to work with. A ratio of both materials is not ideal. What would be convenient is a ratio of one of the material heights to the height of the cylinder. Something like this:
\[ \frac{h_b}{h_t} \qquad{(17)}\]
From equation 17, we can see that:
\[ \frac{h_a}{h_t} + \frac{h_b}{h_t} = \frac{h_a + h_b}{h_t} = 1 \qquad{(18)}\]
Why is the relationship in equation 18 valid? We have a cylinder of height, \(h_t\). We fill it with material a to a height of \(h_a\). We fill it with material b to a height of \(h_b\). From the diagram in fig. 1, it follows that \(h_a + h_b = h = h_t\).
We can establish the material height ratios to the cylinder heights. We derive relationships from equation 18 in the following:
\[ \begin{aligned} \frac{h_a}{h_t} &= 1 - \frac{h_b}{h_t} \\ \frac{h_b}{h_t} &= 1 - \frac{h_a}{h_t} \end{aligned} \qquad{(19)}\]
From equation 12, let:
\[ \frac{h_a}{h_b} = \frac{\rho_b - \rho_t}{\rho_t - \rho_a} = r \qquad{(20)}\]
This leads to:
\[ h_a = r \cdot h_b \qquad{(21)}\]
Working from equation 19:
\[ \begin{aligned} \frac{h_b}{h_t} &= 1 - \frac{r \cdot h_b}{h_t} \\ \frac{h_b}{h_t} + \frac{r \cdot h_b}{h_t} &= 1 \\ \frac{h_b \left(1 + r \right)}{h_t} &= 1 \end{aligned} \]
\[ \frac{h_b}{h_t} = \frac{1}{1 + r} \qquad{(22)}\]
With equation 22 we have a solution in terms of the ratio of the material b height to the cylinder height. Let’s simplify and derive the final equation. Let:
\[ \frac{h_b}{h_t} = \frac{1}{1 + r} = \frac{1}{m} \]
Let:
\[ \begin{aligned} m &= 1 + r = 1 + \frac{x}{y} \\ x &= \rho_b - \rho_t \\ y &= \rho_t - \rho_a \end{aligned} \]
\[ m = 1+ \frac{x}{y} = \frac{y}{y} + \frac{x}{y} = \frac{x + y}{y} \]
\[ \begin{aligned} m &= \frac{(\rho_b - \rho_t) + (\rho_t - \rho_a)}{\rho_t - \rho_a} \\ m &= \frac{\rho_b - \rho_a}{\rho_t - \rho_a} \end{aligned} \]
\[ \frac{h_b}{h_t} = \frac{1}{m} = \frac{1}{\frac{\rho_b - \rho_a}{\rho_t - \rho_a}} \]
All of this simplifies to the final form:
\[ \frac{h_b}{h_t} = \frac{\rho_t - \rho_a}{\rho_b - \rho_a} \qquad{(23)}\]
Equation 23 is the final form. This form places material height of b in a ratio with the height of the cylinder. It is more convenient to work with. We can use the equations listed in eq. 19 to solve for the other material height ratio.
To confirm, let’s use the numbers from the previous section:
\[ \frac{h_b}{h_t} = \frac{1.00 - 1.30}{0.93 - 1.30} = 0.8108 \]
\[ h_b = 0.8108 \cdot 6.25 = 5.07 \]
The numbers match our previous calculations!