Optimal control of pump rotational speed in filling and emptying a reservoir: minimum energy consumption with fixed time

Abstract

An effective way to save energy in pumping systems with low static head is to control the pump’s rotational speed with a variable-speed drive (VSD), which allows changing of the rotational speed when necessary. VSDs can be utilized to control batch transfer systems, for example, in filling or emptying a tank or a reservoir. In the literature, such processes have been optimized only with respect to energy consumption, but the time limit has been ignored. This means that pumping time can be very long. Our paper deals with this optimization problem and considers both pumping time and energy demand, which are often conflicting criteria. We derived a general optimal control law for rotational speed, which can easily be implemented in existing VSDs in the market. Minimum energy and minimum time schemes are special cases of this general new scheme. A constant flow rate scheme, suggested in the literature, is verified to give an optimum solution if the efficiency of the pump remains constant during operation. In addition to energy consumption, rotational speed control can have a favorable effect on the pump’s lifetime, as pointed out in the paper.

Appendix

Calculus of variations

The following theory can be found, e.g., in chapters 1–3 in (Smith 1998).

Consider a functional $F(y)$ defined as a definite integral:

$$ F(y)={\displaystyle {\int}_a^bf\left(y(x)\right)dx} $$

(25)

where f is some known function. We seek to find the unknown function y(x) in the interval a ≤ x ≤ b to minimize F(y). Henceforth, we denote y(x) by y.

The variation of the functional $F(y)$ is defined as

$$ \updelta F\left(y;\Delta y\right)=\underset{\epsilon \to 0}{ \lim}\frac{F\left(y+\epsilon \Delta y\right)-F(y)}{\epsilon } $$

(26)

where Δy is an arbitrary function in the interval a ≤ x ≤ b. The minimum of F(y) occurs when the variation is δF(y; Δy) = 0 for every function Δy. By inserting Eq. (25) into Eq. (26) and changing the order of limit and integral, we get

$$ \updelta F\left(y;\Delta y\right)={\displaystyle {\int}_a^b\underset{\epsilon \to 0}{ \lim}\frac{f\left(y+\epsilon \Delta y\right)-f(y)}{\epsilon}\mathrm{d}x={\displaystyle {\int}_a^b\frac{\mathrm{d}f}{\mathrm{d}y}\Delta y\mathrm{d}x}} $$

(27)

Since Eq. (27) must be zero for every Δy, we specifically choose Δy = df/dy, and the integrand becomes (df/dy)². The integral of this non-negative function is zero only if df/dy = 0, which is the condition for a minimum of F(y).

Now consider the minimization of F(y) subject to the constraint

$$ G(y)={\displaystyle {\int}_a^bg\left(y(x)\right)\mathrm{d}x-{G}_0=0} $$

(28)

where ${G}_0$ is some constant. The necessary condition for y to be a minimum is that there is a Euler-Lagrange multiplier C such that

$$ \updelta F\left(y;\Delta y\right)+C\updelta G\left(y;\Delta y\right)=0 $$

(29)

for every $\Delta y$. We insert Eqs. (25) and (28) into Eq. (29) and continue as with Eq. (27) above, which finally gives

$$ \frac{\mathrm{d}f}{\mathrm{d}y}+C\frac{\mathrm{d}g}{\mathrm{d}y}=0 $$

(30)

The constrained problem is solved by finding a function y and constant C such that Eqs. (28) and (30) hold.