# How do I calculate the density of a lined pipe?

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}

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?

• “Rubbed”? Rubber or ribbed? Big difference. – Solar Mike Jun 11 '20 at 14:38
• Sorry about that. I meant rubber - just edited it! – air_nomad Jun 11 '20 at 14:49
• Are you sure the answer neglects the air in the pipe? I do not see how the density could be less than water if both pipe materials have a density greater than water. I ended up with a similar result to yours neglecting air. – mechcad Jun 11 '20 at 14:52
• Ya, I'm sure! I copied the "neglecting the air in the tube" straight from the answer key. Perhaps it's wrong then? :/ In which case, how would I go about this problem including the air in the tube? – air_nomad Jun 11 '20 at 15:11
• Have you checked the detail? Based on you ratio for rubber to steel there needs to be much less steel and the issue of that answer being lower than the sg of water needs looking at. – Solar Mike Jun 11 '20 at 15:16

$$\rho_{pipe} = \frac{m_{rubber}+m_{steel}}{V_{pipe}} = \frac{\rho_{rubber}\cdot V_{rubber} + \rho_{steel}\cdot V_{steel}}{V_{pipe}} \propto \frac{\rho_{rubber}\cdot(R_{inside}^2 - R_{lining}^2) + \rho_{steel}\cdot(R_{outside}^2 - R_{inside}^2)}{R_{outside}^2}$$
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.
• Air has a density of $1.2754\frac{kg}{m^3}$, so it contributes just $0.0012\frac{kg}{m}$ of pipe. Adding this mass gives a density of $990.9\frac{kg}{m^3}$ for the pipe overall. It's a relatively small change. My guess is that whoever wrote the question was probably just trying to save you some work, although if that's the case it seems to have backfired! – realityChemist Jun 12 '20 at 18:16