**Appendix A**

The total derivative of strain represents a derivative of pipe inner diameter changes [38]:

$$\frac{d\varepsilon}{dt} = \frac{d}{dt} \left( \frac{D - D\_0}{D\_0} \right) = \frac{1}{D} \frac{dD}{dt}.\tag{A1}$$

These changes in plastic pipes differ from those reported in elastic pipes. The relation between the cross-section derivative and strain derivative is:

$$\frac{dA}{dt} = \frac{d}{dt} \left(\frac{\pi D^2}{4}\right) = \frac{\pi D}{2} \frac{dD}{dt} = \frac{\pi D^2}{2} \frac{1}{D} \frac{dD}{dt} = \frac{\pi D^2}{2} \frac{d\varepsilon}{dt} = 2A \frac{d\varepsilon}{dt}.\tag{A2}$$

The circumferential strain can be decomposed into an instantaneous elastic strain *ε<sup>e</sup>* and a retarded strain *εr*:

$$
\varepsilon = \varepsilon\_{\ell} + \varepsilon\_{r}. \tag{A3}
$$

Using the above decomposition (Equation (A3)) in Equation (A2) gives:

$$\frac{dA}{dt} = 2A\left(\frac{d\varepsilon\_{\varepsilon}}{dt} + \frac{d\varepsilon\_{r}}{dt}\right). \tag{A4}$$

Typically, the instantaneous strain *εe*, which is assumed to be linear–elastic, can be related to the hoop stress as follows:

$$
\varepsilon\_{\ell} = \frac{\not\mathbb{1}\sigma\_{\ell}}{E} \tag{A5}
$$

where *σe*—elastic component of the hoop stress.

The hoop stress is also related to the fluid pressure and the ratio between the pipe inner diameter and wall thickness:

$$
\sigma\_{\ell} = \frac{pD}{2e}.\tag{A6}
$$

Let us take the derivative of the above Equation (A6):

$$\frac{d\sigma\_{\varepsilon}}{dt} = \frac{p}{2e}\frac{dD}{dt} + \frac{D}{2e}\frac{dp}{dt}.\tag{A7}$$

The derivative of the elastic strain component (Equation (A5)) can be written with the help of Equation (A7):

$$\frac{d\varepsilon\_{\varepsilon}}{dt} = \frac{d}{dt} \left( \frac{\xi \sigma\_{\varepsilon}}{E} \right) = \frac{\tilde{\xi}}{2eE} \left( p \frac{dD}{dt} + D \frac{dp}{dt} \right) = \frac{\tilde{\xi}}{2eE} \left( pD \frac{1}{D} \frac{dD}{dt} + D \frac{dp}{dt} \right). \tag{A8}$$

Finally, with help of Equation (A1), one gets:

$$\frac{d\varepsilon\_{\varepsilon}}{dt} = \frac{\Xi}{2E} \left( p \frac{d\varepsilon\_{\varepsilon}}{dt} + \frac{dp}{dt} \right). \tag{A9}$$

After rearrangement:

$$\frac{d\varepsilon\_{\varepsilon}}{dt} = \frac{\frac{\Xi}{2\Xi} \frac{dp}{dt}}{1 - \frac{\Xi}{2\Xi}p} . \tag{A10}$$

Introducing Equation (A10) into Equation (A4) results in:

$$\frac{dA}{dt} = \frac{A\frac{dp}{dt}}{\frac{E}{\Xi} - \frac{P}{\Xi}} + 2A\frac{d\varepsilon\_r}{dt}.\tag{A11}$$

Since in most practical applications *p*/2 *E*/Ξ, then:

$$\frac{1}{A}\frac{dA}{dt} = \frac{\Xi}{E}\frac{dp}{dt} + 2\frac{d\varepsilon\_r}{dt}.\tag{A12}$$

This equation has been used in the manuscript during the derivation of the continuity equation—see Equation (11).
