Convergence of a Hydraulic Solver with Pressure-Dependent Demands

Abstract

This paper analyzes the convergence of a pressure-driven analysis (PDA) model of a water distribution network solver based on Todini’s global gradient algorithm. The PDA model is constructed by embedding a pressure−demand relationship in the EPANET simulator code. To avoid spurious convergence, a residual-based convergence error was used. The introduction of pressure-dependent demands is shown to result in a far poorer convergence. The study of solver convergence as a function of the smoothness of the pressure−demand curve has demonstrated that, statistically, a smooth pressure−demand relationship gives a somewhat better convergence. To improve convergence, use was made of a quadratic approximation of the Hazen–Williams head loss−flow relationship in the vicinity of zero and the correct implementation of the Darcy−Weisbach formula in the solver. To further improve convergence, an iteration step control technique called the line search was used. The analysis of solver convergence for different line search variants has shown that the line search in its usual form is not efficient enough and may result in poorer convergence. A necessary error decrease algorithm, whose use in the line search improves solver convergence, is proposed. It is shown that due to the convergence improvement methods the convergence of the PDA solver is somewhat better than that of the demand-driven analysis solver and sufficient for direct problems such as design, for example.

Appendix

1.1 Implementation of the Darcy−Weisbach Formula in the EPANET Solver

Using the Darcy−Weisbach formula, the frictional head loss ΔH for incompressible flow Q in a pipe can be written as (Bhave 1991)

$$ \varDelta H=\frac{8L}{\pi g{D}^5}f{Q}^2, $$

where L is the pipe length, g is the gravitational acceleration, D is the pipe diameter, and f = f(Q) is the friction factor. As in EPANET, we introduce the resistance coefficient

$$ r=\frac{8L}{\pi g{D}^5}. $$

Then the frictional head loss with the inclusion of the minor loss m in the pipe is

$$ \varDelta H=\left( rf+m\right){Q}^2. $$

The calculation of the matrix D ₁₁ in (2) calls for the calculation of the flow derivative of the head loss. With the inclusion of the flow dependence of the friction factor, the derivative of the head loss will be

$$ \frac{d\kern0.1em \varDelta H}{ dQ}=r{Q}^2\frac{ df}{ dQ}+2Q\left( rf+m\right). $$

According to the EPANET manual, the friction factor f is computed depending on the flow’s Reynolds Number (Re):

For laminar flow (Re < 2, 000) as

$$ {f}_1=\frac{64}{ Re}=\frac{16\pi D\nu}{Q}, $$

(A1)

where $Re=\frac{4Q}{\pi D\nu}$ is the Reynolds Number, and ν is the kinematic viscosity.

For fully turbulent flow (Re > 4, 000) as

$$ {f}_3=\frac{0.25}{{\left[{ \log}_{10}\left(\frac{\varepsilon }{3.7D}+\frac{5.74}{R{e}^{0.9}}\right)\right]}^2}=\frac{{ \ln}^2(10)}{4{ \ln}^2\left[\frac{\varepsilon }{3.7D}+\frac{5.74}{{\left(\frac{4Q}{\pi D\nu}\right)}^{0.9}}\right]} $$

(A2)

For transitional flow (2, 000 < Re < 4, 000) the friction factor f ₂ is calculated using a cubic interpolation from the Moody Diagram.

Let

$$ {f}_2={a}_0+{a}_1x+{a}_2{x}^2+{a}_3{x}^3 $$

(A3)

where

$$ x=\frac{ Re}{2000}-1=\frac{Q}{500\pi D\nu}-1 $$

(A4)

The coefficients of the polynomial are determined from the continuity of the functions f ₁(Q _L ) = f ₂(Q _L ), f ₂(Q _R ) = f ₃(Q _R ) and their derivatives ${\left.\frac{d{f}_1}{ dQ}\right|}_{Q_L}={\left.\frac{d{f}_2}{ dQ}\right|}_{Q_L}$, ${\left.\frac{d{f}_2}{ dQ}\right|}_{Q_R}={\left.\frac{d{f}_3}{ dQ}\right|}_{Q_R}$ at the left (Q _L  = 500πνD) and right (Q _R  = 1000πνD) boundaries. The derivatives are

$$ \frac{d{f}_1}{ dQ}=\frac{-16\pi D\nu}{Q^2} $$

(A5)

$$ \frac{d{f}_2}{ dQ}=\frac{a_1+2{a}_2x+3{a}_3{x}^2}{500\pi \nu D} $$

(A6)

$$ \frac{d{f}_3}{ dQ}=\frac{2\cdot 5.74\cdot 0.9\cdot { \ln}^2(10)}{4Q{\left(\frac{4Q}{\pi \nu D}\right)}^{0.9}\left[\frac{\varepsilon }{3.7D}+\frac{5.74}{{\left(\frac{4Q}{\pi D\nu}\right)}^{0.9}}\right]{ \ln}^3\left[\frac{\varepsilon }{3.7D}+\frac{5.74}{{\left(\frac{4Q}{\pi D\nu}\right)}^{0.9}}\right]} $$

(A7)

Then the coefficients will be

$$ {a}_0=\frac{64}{2000},\kern1em {a}_1=\frac{-64}{2000} $$

$$ {a}_2=-\frac{5.74\cdot 0.9\;{ \ln}^2(10)}{4\cdot {4000}^{0.9}\ c\kern0.24em { \ln}^3c}+\frac{3\;{ \ln}^2(10)}{4\ { \ln}^2c}-{a}_0\kern1em {a}_3=\frac{5.74\cdot 0.9\;{ \ln}^2(10)}{4\cdot {4000}^{0.9}\ c\ { \ln}^3c}-\frac{{ \ln}^2(10)}{2\;{ \ln}^2c}+{a}_0 $$

where $c=\frac{\varepsilon }{3.7D}+\frac{5.74}{4000^{0.9}}.$

However, the resulting expressions for the friction factor and its derivative as they are cannot be used in the EPANET solver. As is evident from Eqs. (A1) and (A5), zero flows will result in division by zero. To avoid this, the source code of the solver should be modified so that the functions return the products Q f(Q) and ${Q}^2\frac{ df(Q)}{ dQ}$ instead of f(Q) and $\frac{ df(Q)}{ dQ}$.