How do I calculate the density of a lined pipe?

Question

I'm having some trouble figuring out how to calculate the density of a lined pipe given the diameter measurements and the specific gravity of the materials.

The Problem

Calculate the density of an empty rubber lined steel pipe with an inside diameter of 0.075 m, and an outside diameter of 0.079 m. The rubber lining reduces the pipe inner diameter of the pipe to 0.069 m. Assume that the SG of steel is 7.9 and the SG of the rubber is 1.52.

My Attempt

I multiplied the ratio of the cross-sectional area of the steel part to the cross-sectional area of the whole pipe with the density of steel. I did the same for the rubber lining and then added the two values together.

$$\begin{align} \rho_{substance} &= SG * \rho_{water} \\ \rho_{steel} &= 7.9 * 1000 kg/m^3 = 7900\text{ kg/m}^3 \\ \rho_{rubber} &= 1.52 * 1000 kg/m^3 = 1520\text{ kg/m}^3 \\ r &= \dfrac{\phi}{2} \\ r_{outer} &= \dfrac{0.079}{2} = 0.0395\text{ m} \\ r_{inner} &= \dfrac{0.075}{2} = 0.0375\text{ m} \\ r_{lining} &= \dfrac{0.069}{2} = 0.0345\text{ m} \\ A_{pipe} &= \pi(0.03952^2 - 0.0345^2) = 0.00116\text{ m}^2 \\ A_{steel} &= \pi(0.0395^2 - 0.0375^2) = 0.000484\text{ m}^2 \\ A_{lining} &= \pi(0.0375^2 - 0.0345^2) = 0.000679\text{ m}^2 \\ \text{steel:pipe} &= \dfrac{0.000484}{0.0011} = 0.417 \\ \text{lining:pipe} &= \dfrac{0.000679}{0.00116} = 0.585 \\ \rho_{pipe} &= \text{steel:pipe} * \rho_{steel} + \text{rubber:pipe} * \rho_{rubber} \\ &= 0.417 \cdot 7900 + 0.585 \cdot 1520 = 4184\text{ kg/m}^3 \end{align}$$

Correct Answer

According to the key, the answer is supposed to be 990 kg/m³ (neglecting the air in the tube).

Is my entire approach wrong? I know how to find the composite density of, say, a slurry stream (water + solids) but how do I find the composite density of a lined pipe?

Answer 1

What the answer key seems to mean is to neglect the mass of air in the pipe, not its volume. That is, we want:

$\rho_{pipe} = \frac{m_{rubber}+m_{steel}}{V_{pipe}} = \frac{\rho_{rubber}\cdot V_{rubber} + \rho_{steel}\cdot V_{steel}}{V_{pipe}} \propto \frac{\rho_{rubber}\cdot(R_{inside}^2 - R_{lining}^2) + \rho_{steel}\cdot(R_{outside}^2 - R_{inside}^2)}{R_{outside}^2}$

Note that I canceled out $\pi$ and $L$, the length of a section of pipe (which is arbitrary) for compactness above, but you may want to show them in your work. It depends on whether your professor is really keen on showing everything or not.

Doing this with the numbers you give above results in an answer of $990.2 \frac{kg}{m^3}$, so this must be what they want. I think it's poorly phrased though. Simply adding the word "mass" to the answer key would clear up a lot of this confusion.