6 Engineering Example 2

6.1 Estimating the mass of a pipe

Introduction

Sometimes engineers have to estimate component weights from dimensions and material properties. On some occasions, engineers prefer use of approximate formulae to exact ones as long as they are sufficiently accurate for the purpose. This Example introduces both of these aspects.

Problem in words

1. Find the mass of a given length of pipe in terms of its inner and outer diameters and the density of the pipe material.
2. Find the wall thickness of the pipe if the inner diameter is 0.15 m, the density is $7900\text{kg}{\text{m}}^{-3}$ and the mass per unit length of pipe is $40\text{kg}{\text{m}}^{-1}$ .
3. Find an approximate method for calculating the mass of a given length of a thin-walled pipe and calculate the maximum ratio of inner and outer diameters that give an error of less than 10% when using the approximate method.

Mathematical statement of problem

1. Denote the length of the pipe by $L$ m and inside and outside diameters by $d_i$ m and $d_o$ m, respectively and the density by $\rho \text{kg}{\text{m}}^{-3}$ . Assume that the pipe is cylindrical so its cross section corresponds to the gap between concentric circles (this is called an annulus or annular region - see HELM booklet  2.6). Calculate the difference in cross sectional areas by using the formula for the area of a circle ( $A=\pi r^2$ where $r$ is the radius) and multiply by the density and length to obtain mass ( $m$ ).
2. Rearrange the equation in terms of wall thickness ( $d$ m) and inner diameter. Substitute the given values to determine the wall thickness.
3. Approximate the resulting expression for small values of ( $d_o-d_i$ ). Calculate the percentage difference in predictions between the original and approximate formulae for various numerical values of $d_i∕d_o$ .

Mathematical analysis

1. The cross section of a cylindrical pipe is a circular annulus. The area of a circle is given by $\pi r^2=\frac{\pi }{4}{d}^2$ , since $r=d∕2$ if $d$ is the diameter. So the area of the outer circle is $\frac{\pi }{4}{d}_o^2$ and that of the inner circle is $\frac{\pi }{4}{d}_i^2$ . This means that the mass $m$ kg of length $L$ m of the pipe is given by

   $m=\frac{\pi }{4}\left(d_0^2-d_i^2\right)L\rho$

2. Denote the pipe wall thickness by $\delta$ so $d_o=d_i+25$ .

   Use $\left(d_o^2-d_i^2\right)=\left(d_o-d_i\right)\left(d_o+d_i\right)=2\delta \left(2d_i+2\delta \right)$ . So $m=\pi \delta \left(d_i+\delta \right)L\rho$

   Given that $m∕L=40,d_i=0.15$  and  $r=7900$ ,

   $40=pd\left(0.15+d\right)7900.$

   Rearrange this equation as a quadratic in $\delta$ ,

   ${\delta }^2+0.15\delta -4\pi ∕790=0$

   Solve this quadratic using the standard formula with $a=1,b=0.15$ and $c=4\pi ∕790$ . Retain only the positive solution to give $\delta =0.072$ , i.e. the pipe wall thickness is 72 mm.

3. If $\delta$ is small then $\left(d_o-d_i\right)$ is small and $d_i+\delta \approx d_i$ . So the expression for $m$ in terms of $\delta$ may be written

   $m\approx \pi \delta d_{i}L\rho$

   The graph in Figure 3 shows that the percentage error from using the approximate formula for the mass of the pipe exceeds 10% only if the inner diameter is less than 82% of the outer diameter.

The percentage error from using the approximate formula can be calculated from

(exact result $-$ approximate result)/(exact result) $\times$ 100% for various values of the ratio of inner to outer diameters. In the graph the error is plotted for diameter ratios between 0.75 and 1.

Figure 3

Comment

The graph shows also that the error is 1% or less for diameter ratios $>0.98$ .

Exercises

Solve the following quadratic equations by using the formula. Give answers exactly (where possible) or to 4 d.p.:

Answer