Dynamic modeling and simulation of a water supply system with applications for improving energy efficiency

Abstract

This paper presents the dynamic simulation of an urban water supply system based on a phenomenological model of distributed parameters capable of predicting the hydraulic behavior and the energy consumption. This case study involves a segment of the supply and distribution water system in the city of Salvador (Brazil), and the analysis focuses on the pumping station, water mainline, and distribution tanks. The current operation of the system is performed through heuristic rules based on expert knowledge, and the control of discharge flow pumps is essentially on/off. The comparison between simulation results and measured data shows that the model describes the hydraulic network satisfactorily as it is able to predict the transient behavior of the flow along the water main line. Considering the high capacity and dimensions together with the intrinsic dynamic features (disturbances caused by the maneuvers in the pump operation and variable demand throughout the day), the use of dynamic modeling for the case studied is really needed for the simulation results to be consistent. Furthermore, the dynamic model and strategy used enables a careful analysis of the water supply system and represents a potential tool for the evaluation of control strategies and energy efficiency. The results show the potential gain in energy efficiency with the inclusion of frequency converters in the pumping system without considerable changes in the level of reservoirs and ensuring the total service demand.

Appendix—Equation of motion from equation of thermal energy

From the complete equation of energy that considers the rate of accumulation of internal and kinetic energy, it is possible to derive the equation of thermal energy (by analogy with the equation of mechanical energy) that only considers the rate of gain of internal energy (Bird et al. 1960). Then, for an infinitesimal element of fluid in one-dimensional flow, the following equation is valid:

$$ \rho \cdot \frac{Du}{Dt}=-\left(\nabla \cdot q\right)- p\cdot \left(\nabla \cdot v\right)-\left(\tau :\nabla v\right) $$

(A.1)

u is the internal energy per unit mass of fluid in the element (control volume), v is the local fluid velocity, p is the static pressure, τ is the stress tensor and q is the heat flux vector. Du/Dt is the substantial derivative (derivative following the fluid motion) of u. − (∇ ⋅ q) represents the rate of internal energy input by conduction (per unit volume), − p ⋅ (∇ ⋅ v) and − (τ : ∇v) represents reversible rate of internal energy increase by compression and irreversible rate of internal energy increase by viscous dissipation (friction loss), respectively, both per unit volume. Assuming incompressible flow, (∇ ⋅ v) = 0, and integrating Eq. A.1 over the control volume:

$$ {\displaystyle \underset{V}{\int}\rho \cdot \frac{Du}{Dt} dV}={\displaystyle \underset{V}{\int }-\left(\nabla \cdot q\right) dV}-{\displaystyle \underset{V}{\int}\left(\tau :\nabla v\right) dV} $$

(A.2)

Applying the Leibnitz formula for differentiating integral and equation of continuity,

$$ {\displaystyle \underset{V}{\int}\rho \cdot \frac{Du}{Dt} dV}={\displaystyle \underset{V}{\int }-\left(\nabla \cdot q\right) dV}-{\displaystyle \underset{V}{\int}\left(\tau :\nabla v\right) dV} $$

(A.3)

where dU/dt is the total time derivative, and U is the total internal energy in the system. An unsteady-state macroscopic energy balance, assuming no shaft work and that the time derivative of total potential energy is negligible, leads to Eq. A.4:

$$ \frac{dU}{dt}=-\varDelta \left[\left( u+\frac{p}{\rho}+\frac{1}{2}\cdot {\left\langle v\right\rangle}^2+ g\cdot z\right)\cdot w\right]+ Q-\frac{dK}{dt} $$

(A.4)

z is the quota, w is the mass rate of flow, 〈v〉 is the average velocity over conduit cross-section, Q is the net rate of heat added to the system, and K is the total kinetic energy in the system.

Using the Gauss divergence theorem:

$$ {\displaystyle \underset{V}{\int}\left(\nabla \cdot q\right) dV}={\displaystyle \underset{S}{\int}\left( n\cdot q\right) dS}= Q $$

(A.5)

Using Eqs. A.3, A.4, and A.5 in Eq. A.2:

$$ \varDelta \left[\left( u+\frac{p}{\rho}+\frac{1}{2}\cdot {\left\langle v\right\rangle}^2+ g\cdot z\right)\cdot w\right]+{\displaystyle \underset{V}{\int}\left(\tau :\nabla v\right) dV}+\frac{dK}{dt}=0 $$

(A.6)

Using Eq. 2 and expressing $K={\displaystyle \underset{V}{\int}\frac{1}{2}\rho {v}^2 dV}\kern1em \mathrm{and}\kern0.75em {\displaystyle \underset{V}{\int}\left(\tau :\nabla v\right) dV}={\displaystyle \underset{V}{\int}\rho {\hat{E}}_v dV}$, where${\widehat{E}}_v$ denotes the friction loss per unit of mass rate flow:

$$ \varDelta \left[\left( u+ g\cdot H+\frac{1}{2}\cdot {\left\langle v\right\rangle}^2\right)\cdot w\right]+{\displaystyle \underset{V}{\int}\rho {\hat{E}}_v dV}+{\displaystyle \underset{V}{\int}\frac{\partial }{\partial t}\left(\frac{1}{2}\rho {v}^2\right) dV}=0 $$

(A.7)

For incompressible and isothermal flow, Δu = 0. The other net rate added can be expressed as volume integrals:

$$ \begin{array}{l}\varDelta \left[\left( g\cdot H+\frac{1}{2}\cdot {\left\langle v\right\rangle}^2\right)\cdot w\right]={\displaystyle \underset{S}{\int}\left( n\cdot gH\rho v\right) dS}+\frac{1}{2}{\displaystyle \underset{S}{\int}\left( n\cdot {\left\langle v\right\rangle}^2\rho v\right) dS}=\\ {}{\displaystyle \underset{V}{\int}\left(\nabla \cdot gH\rho v\right) dV}+\frac{1}{2}{\displaystyle \underset{V}{\int}\left(\nabla \cdot {\left\langle v\right\rangle}^2\rho v\right) dV}\end{array} $$

(A.8)

Replacing in Eq. A.7:

$$ {\displaystyle \underset{V}{\int}\left\{\left(\nabla \cdot gH\rho v\right)+\frac{1}{2}\left(\nabla \cdot {\left\langle v\right\rangle}^2\rho v\right)+\rho {\hat{E}}_v+\frac{\partial }{\partial t}\left(\frac{1}{2}\rho {v}^2\right)\right\} dV=0} $$

(A.9)

For an incompressible and one-dimensional flow, Eq. A.9 provides the following equation:

$$ g\frac{\partial H}{\partial x}+{v}_x\cdot \frac{\partial {v}_x}{\partial x}+\frac{\partial {v}_x}{\partial t}+\frac{{\widehat{E}}_v}{v_x}=0 $$

(A.10)

v _x is the component of velocity in the x-direction.

The Eq. 6 is derived directly from Eq. A.10 considering that the term v _x  ⋅ ∂v _x /∂x is significantly smaller than the term ∂v _x /∂t (Chaudhry 1979), using Darcy-Weisbach formula to compute the friction loss and writing Eq. A.10 in terms of volumetric flow (Q).