There are many hydropower stations (HPS) around the world, which consist of hydropower units that convert the potential energy of water in rivers or reservoirs into electrical energy (Fig. 15.1). Global hydropower generation in 2022 totaled 4,334.190 billion kilowatt-hours, a year-on-year increase of 1.1%. The proportion of global hydropower generation in total global power generation was 16.0% in 2022.

Fig. 15.1
Two parts. On the left is an illustration of a dam with a turbine and water flow. On the right is an aerial view photo of a large dam and its reservoir surrounded by several buildings.

Hydropower stations

Full size image

In a hydropower station, under a certain water head and capacity, there is a maximum power supply. To achieve the maximum energy output, we should decide how many generators should be used in the system and how much capacity should be allocated to each generator. This issue is becoming increasingly important with the increasing awareness of environmental protection and the need for energy-saving systems.

At present, various optimization methods have been widely studied, such as linear programming [1], recursive quadratic programming [2], stochastic dynamic programming [3], Lagrangian relaxation [4], genetic algorithm [5], simulated annealing [6] etc.], particle swarm optimization [7], ant colony optimization [8], quasi-Newton [9], neural network [10], evolutionary strategy [11], equal increment principle [12], etc. [13,14,−15].

Almost all of these optimization methods require the establishment of mathematical models of the system. However, since accurate models of real systems are difficult to establish, these optimization methods quickly become difficult to apply in real systems, considering the complexity of the models and algorithms, the curse of dimensionality, and the computational time.

Optimization solutions that avoid these problems would be very useful for business and industry.

15.1 Energy Efficiency of a Hydroelectric Generating Set

A hydroelectric generating set must be accelerated to the rated speed n0 before it can be connected to the grid. The energy efficiency of a hydroelectric generating set is shown in Fig. 15.2.

Fig. 15.2
A line graph of eta versus Q plots a concave-down increasing trend of eta of Q that rises from Q 0 and ends at Q m, with the projection lines of eta e and Q e.

Energy efficiency of a hydroelectric generating set

Full size image

In Fig. 15.2, Q is the flow rate through the hydroelectric generating set, η is the efficiency function of the hydroelectric generating set at rated speed n0 and given water head H0, Q0 is the zero-load flow rate of the hydroelectric generating set, and ηe is the maximum efficiency value, Qe is the flow rate corresponding to the maximum efficiency value ηe, Qm is the maximum flow rate, and

$${Q}_{m}\ge Q\ge {Q}_{0}$$
(15.1)
$${\eta }_{e}=\text{max}\left(\eta \right)=\eta \left({Q}_{e}\right)$$
$$\eta (Q)\ge 0$$
$${\eta }^{{\prime}{\prime}}\left(Q\right)<0$$

The energy efficiency function η can be expressed as

$$\eta \left(Q\right)=\sum_{i=0}^{\infty }{a}_{i}{Q}^{i}$$
(15.2)

15.2 HPS Power Generation

The water heads of each hydroelectric generating set in a hydropower station are identical. The electrical energy generated by the hydropower station is expressed as

$$\begin{array}{ccccc}{W_t}&= \mathop \sum \limits_{i = 1}^n (9.81{H_0}{Q_i}{\eta _i}) = 9.81{Q_t}{H_0}\mathop \sum \limits_{i = 1}^n \frac{{{Q_i}}}{{{Q_t}}}{\eta _i}\\& = 9.81{Q_t}{H_0}\mathop \sum \limits_{i = 1}^n {\theta _i}{\eta _i} = 9.81{Q_t}{H_0}{\eta _0} = {W_0}{\eta _o}\end{array}$$
(15.3)
$${W}_{0}=9.81{Q}_{t}{H}_{0}$$

where n is the total number of hydroelectric generating sets in the hydropower station, W0 is the potential energy of water flowing through the hydropower station, kW, Wt is the total electric energy output by the hydropower station, kW, and Qt is the total flow rate of water flowing through the hydroelectric generating sets, m3/s, H0 is the water head, m, η0 is the overall energy efficiency of the hydropower station, ηi is the operating efficiency of the i-th hydroelectric generating set, Qi represents the flow through the i-th hydroelectric generating set, Qi0 denotes the zero-load flow of ith hydroelectric generating set, θi represents the load rate of the i-th hydroelectric generating set, θi0 denotes the load rate of the i-th hydroelectric generating set at Qi0, expressed as

$${Q}_{t}={\sum }_{i=1}^{n}{Q}_{i}>{\sum }_{i=1}^{n}{Q}_{i0}$$
(15.4)
$${\theta }_{i}=\frac{{Q}_{i}}{{Q}_{t}}$$
$${\theta }_{i0}=\frac{{Q}_{i0}}{{Q}_{t}}$$
$${\sum }_{i=1}^{n}{\theta }_{i}=1$$
$${\eta }_{0}={\sum }_{i=1}^{n}{\theta }_{i}{\eta }_{i}$$

15.3 Optimal Control in a Hydropower Station

Assuming that all hydroelectric generating sets are the identical model, that is

$${Q}_{10}={Q}_{20}=\dots ={Q}_{n0}={Q}_{0}$$
(15.5)
$${\eta }_{1}\left(Q\right)={\eta }_{2}\left(Q\right)=\dots ={\eta }_{n}\left(Q\right)=\eta (Q)$$

Theorem For the optimization problem of Wt at the fixed head H0 and Qt, the maximization of the total electricity energy output Wt of the power station

$$\underset{\mathit{ }\begin{array}{c}{s.t.\mathit{ }{Q}_{i}>Q}_{0}\\ {\sum }_{i=1}^{n}{Q}_{i}={Q}_{t}\\ \end{array}}{\text{max}}{W}_{t}$$
(15.6)

is given by

$$\text{max}{W}_{t}={W}_{0}\eta (\frac{{Q}_{t}}{n})$$
(15.7)

The maximization of the overall efficiency η0 of the hydropower station

$$\underset{\mathit{ }\begin{array}{c}{s.t.\mathit{ }{Q}_{i}>Q}_{i0}\\ {\sum }_{i=1}^{n}{Q}_{i}={Q}_{t}\\ \end{array}}{\text{max}}{\eta }_{0}$$
(15.8)

is given by

$$\text{max}{\eta }_{0}=\eta (\frac{{Q}_{t}}{n})$$
(15.9)

Proof

We begin our inductive proof by considering the case where n = 2.

The constraint condition then becomes

$${Q}_{1}+{Q}_{2}={Q}_{t}$$
(15.10)

where

$${Q}_{1}>{Q}_{0}$$
(15.11)
$${Q}_{2}>{Q}_{0}$$

The objective function Wt is expressed as

$${W}_{t}={9.81H}_{0}({Q}_{1}\eta ({Q}_{1})+{Q}_{2}\eta ({Q}_{2}))$$
(15.12)

The optimal condition is given for

$$W'_{t}({Q}_{1})=0$$
(15.13)

We have

$$\eta \left({Q}_{1}\right)-\eta ({Q}_{t}-{Q}_{1})+{Q}_{1}{\eta }{\prime}\left({Q}_{1}\right)-\left({Q}_{t}-{Q}_{1}\right){\eta }{\prime}\left({Q}_{2}\right)=0$$
(15.14)

It is then easily verified that

$${Q}_{1}={Q}_{2}=\frac{{Q}_{t}}{2}$$
(15.15)

is an optimal point.

We then check the second derivative,

$${W}_{t}^{{\prime}{\prime}}<0$$
(15.16)

So, the optimal point is only maximum.

The maximal value of the total electricity energy output Wt of the power station is

$$\text{max}{W}_{t}={W}_{0}\eta (\frac{{Q}_{t}}{2})$$
(15.17)

The maximal value of the overall efficiency η0 of the power station is

$$\text{max}{\eta }_{0}=\eta (\frac{{Q}_{t}}{2})$$
(15.18)

We then assume that this holds for n = k. The above conclusion is readily extended to the case of n = k, and the optimal point is then

$${Q}_{1}={Q}_{2}=\dots ={Q}_{k}=\frac{{Q}_{t}}{k}$$
(15.19)

The maximal value of Wt is

$$\text{max}{W}_{t}={W}_{0}\eta (\frac{{Q}_{t}}{k})$$
(15.20)

The maximal value of the overall efficiency η0 of the power station is

$$\text{max}{\eta }_{0}=\eta (\frac{{Q}_{t}}{k})$$
(15.21)

Our inductive case is then given by n = k + 1. For the total electricity energy output Wt of the power station we have

$${W}_{t}=9.81{H}_{0}({\sum }_{i=1}^{k}{Q}_{i}\eta ({Q}_{i})+{Q}_{k+1}\left.\eta ({Q}_{k+1})\right)$$
(15.22)

and the maximum of the first item is

$$\text{max}9.81{H}_{0}{\sum }_{i=1}^{k}{Q}_{i}\eta ({Q}_{i})=9.81{H}_{0}({Q}_{t}-{Q}_{k+1})\eta (\frac{{Q}_{t}-{Q}_{k+1}}{k})$$
(15.23)

where

$${Q}_{1}={Q}_{2}=\dots ={Q}_{k}$$
(15.24)

The expression for Wt becomes

$${W}_{t}=9.81{H}_{0}(({Q}_{t}-{Q}_{k+1})\eta (\frac{{Q}_{t}-{Q}_{k+1}}{k})+{Q}_{k+1}\left.\eta ({Q}_{k+1})\right)$$
(15.25)

Based on the above conclusion for n = 2, the optimal point is

$${Q}_{k+1}=\frac{{Q}_{t}-{Q}_{k+1}}{k}$$
(15.26)

The solution is

$${Q}_{k+1}=\frac{{Q}_{t}}{k+1}$$
(15.27)

and

$${Q}_{1}={Q}_{2}=\dots ={Q}_{k}=\frac{{Q}_{t}-{Q}_{k+1}}{k}=\frac{{Q}_{t}}{k+1}$$
(15.28)

The optimal point is then

$${Q}_{1}={Q}_{2}=\dots ={Q}_{k}={Q}_{k+1}=\frac{{Q}_{t}}{k+1}$$
(15.29)

and the maximal value of the total electricity energy output Wt of the power station is

$$\text{max}\,{W}_{t}={W}_{0}\eta \left(\frac{{Q}_{t}}{k+1}\right)$$
(15.30)

The maximal value of the overall energy efficiency η0 of the power station is

$$\text{max}{\eta }_{0}=\eta \left(\frac{{Q}_{t}}{k+1}\right)$$
(15.31)

Load distribution theorem: In a hydropower station which consists of n hydroelectric generating sets that are the identical model, the optimal control method is to keep each hydroelectric generating set to have the same load.

15.4 Optimization Discriminants

In a hydropower station with the identical model hydroelectric generating sets, if n is the optimal, as shown in Fig. 15.3, there must be

Fig. 15.3
A line graph of eta versus Q plots a concave-down increasing trend of eta of Q that rises from Q 0 and ends at Q t over n minus 1, with the projection lines of Q t over n + 1 and Q t over n on the x-axis.

The n is the optimal

Full size image
$${W}_{0}\eta \left(\frac{{Q}_{t}}{n}\right)=\text{max}({W}_{0}\eta \left(\frac{{Q}_{t}}{n-1}\right),{W}_{0}\eta \left(\frac{{Q}_{t}}{n}\right),{W}_{0}\eta \left(\frac{{Q}_{t}}{n+1}\right))$$
(15.32)

Namely

$$\eta \left(\frac{{Q}_{t}}{n}\right)=\text{max}\left(\eta \left(\frac{{Q}_{t}}{n-1}\right),\eta \left(\frac{{Q}_{t}}{n}\right),\eta \left(\frac{{Q}_{t}}{n+1}\right)\right)$$
(15.33)

15.5 Optimal Switch in the Hydropower Station

Now we consider the optimal switch point for a hydropower station. Suppose the hydropower station has M-unit hydroelectric generating sets in total. Then we take n to be less than or equal to M and the optimum, the total electricity energy output Wt of the power station is the maximum.

$$\underset{\mathit{ }\begin{array}{c}{s.t.\mathit{ }{{Q}_{m}\ge Q}_{i}>Q}_{0}\\ {\sum }_{i=1}^{n}{Q}_{i}={Q}_{t}\\ n\le M\\ \end{array}}{\text{max}}{W}_{t}={W}_{0}\eta \left(\frac{{Q}_{t}}{n}\right)$$
(15.34)

The optimal overall efficiency is then

$$\eta \left(\frac{{Q}_{t}}{n}\right)=\text{max}\left(\eta \left(\frac{{Q}_{t}}{n-1}\right),\eta \left(\frac{{Q}_{t}}{n}\right),\eta \left(\frac{{Q}_{t}}{n+1}\right)\right)$$
(15.35)

If the total flow Qt varies, the optimal n may change also. The optimal switch point is dependent upon whether Qt is increasing or decreasing.

Theorem

In a hydropower station with the identical model hydroelectric generating sets, if n is the optimal, when Qtis increasing, the optimal switch point is at

$$\eta \left(\frac{{Q}_{t}}{n}\right)=\eta \left(\frac{{Q}_{t}}{n+1}\right)$$
(15.36)

when Qt is decreasing, the optimal switch point is at

$$\eta \left(\frac{{Q}_{t}}{n}\right)=\eta \left(\frac{{Q}_{t}}{n-1}\right)$$
(15.37)

Proof

In Figs. 15.4 and 15.5, the ηe is the maximum efficiency. Qe is the flow rate at ηe. The η (Qt/(n−1)) is less than η(Qt/n) on the right side of Qe, the η (Qt/(n + 1)) is less than η (Qt/n) on the left side of Qe.

From Figs. 15.4 and 15.5, we see that

Fig. 15.4
A line graph of eta versus Q plots a concave-down increasing trend of eta of Q that rises from Q 0 and ends at Q t over n minus 1, with the projection lines of Q t over n + 1, Q t over n, Q e, and Q t over n minus 1 on the horizontal axis.

Qt/n is equal or less than Qe

Full size image
Fig. 15.5
A line graph of eta versus Q plots a concave-down increasing trend of eta of Q that rises from Q 0 and ends at Q t over n minus 1, with the projection lines of Q t over n + 1, eta e, Q e, Q t over n, and Q t over n minus 1.

Qt/n is greater than the flow Qe

Full size image
$$\eta \left(\frac{{Q}_{t}}{n}\right)=\text{max}\left(\eta \left(\frac{{Q}_{t}}{n-1}\right),\eta \left(\frac{{Q}_{t}}{n}\right),\eta \left(\frac{{Q}_{t}}{n+1}\right)\right)$$
(15.38)

As seen in Fig. 15.4, if Qt/n < Qe and Qt increases, then η (Qt/n) will continue to increase until Qt/n reach Qe, at which point immediately after it will begin to decrease. At the same time, η (Qt/(n + 1)) will increase until it will eventually become greater than η(Qt/n) and will be the new maximum.

$$\eta \left(\frac{{Q}_{t}}{n+1}\right)=\text{max}\left(\eta \left(\frac{{Q}_{t}}{n-1}\right),\eta \left(\frac{{Q}_{t}}{n}\right),\eta \left(\frac{{Q}_{t}}{n+1}\right)\right)$$
(15.39)

This change of the maximum efficiency defines the optimal switch point at

$$\eta \left(\frac{{Q}_{t}}{n}\right)=\eta \left(\frac{{Q}_{t}}{n+1}\right)$$
(15.40)

If Qt/n > Qe, then η (Qt/n) will decrease with the increase of Qt while η(Qt/(n + 1)) will increase as can be inferred from Fig. 15.5. Ultimately, η(Qt/(n + 1)) will increase to such a point that it will become the new maximum

$$\eta \left(\frac{{Q}_{t}}{n+1}\right)=\text{max}\left(\eta \left(\frac{{Q}_{t}}{n-1}\right),\eta \left(\frac{{Q}_{t}}{n}\right),\eta \left(\frac{{Q}_{t}}{n+1}\right)\right)$$
(15.41)

and the optimal switch point is at

$$\eta \left(\frac{{Q}_{t}}{n}\right)=\eta \left(\frac{{Q}_{t}}{n+1}\right)$$
(15.42)

In a manner akin to our arguments for an increasing Qt, we begin by looking at Qt/n > Qe. If Qt is decreasing, then Fig. 15.5 shows that η (Qt/n) will increase until Qt/n reaches the value of Qe, where it will immediately begin to decrease. Simultaneously, η(Qt/(n−1)) will increase. As before, there will be point in which η (Qt/(n−1)) overtakes η(Qt/n) as the new maximum.

$$\eta \left(\frac{{Q}_{t}}{n-1}\right)=\text{max}\left(\eta \left(\frac{{Q}_{t}}{n-1}\right),\eta \left(\frac{{Q}_{t}}{n}\right),\eta \left(\frac{{Q}_{t}}{n+1}\right)\right)$$
(15.43)

This is our optimal switch point and is given by

$$\eta \left(\frac{{Q}_{t}}{n}\right)=\eta \left(\frac{{Q}_{t}}{n-1}\right)$$
(15.44)

As seen in Fig. 15.4, for a given Qt/n < Qe, η(Qt/n) will decrease as Qt decreases. We observe that η (Qt/(n−1)) will increase to the point where it will be the maximum efficiency.

$$\eta \left(\frac{{Q}_{t}}{n-1}\right)=\text{max}\left(\eta \left(\frac{{Q}_{t}}{n-1}\right),\eta \left(\frac{{Q}_{t}}{n}\right),\eta \left(\frac{{Q}_{t}}{n+1}\right)\right)$$
(15.45)

The optimal switch point is then

$$\eta \left(\frac{{Q}_{t}}{n}\right)=\eta \left(\frac{{Q}_{t}}{n-1}\right)$$
(15.46)

15.6 Operation at the Maximum Efficiency

For the total flow rate Qt(m3/s) of the river, the optimal run-unit of a hydropower station is n as described above. The maximal value of the overall efficiency η0 of the power station is

$$\text{max}{\eta }_{0}=\eta \left(\frac{{Q}_{t}}{n}\right)$$
(15.47)

and

$$\eta \left(\frac{{Q}_{t}}{n}\right)\le {\eta }_{e}$$
(15.48)

If a hydropower station is located at a reservoir which has a certain storage capacity, then the hydropower station can operate at the rate efficiency ηe. As shown in Fig. 3.1, ηe is the maximum, and greater than any efficiency value.

The total flow of each day in the river is 24*3600*Qt, and can be resolved into t1*Qt1 and t2*Qt2 as following

$$24\times 3600\times {Q}_{t}={t}_{1}\times {Q}_{t1}+{t}_{2}\times {Q}_{t2}$$
(15.49)
$$24\times 3600={t}_{1}+{t}_{2}$$

where t1 denotes running time (s) under run-unit n1, Qt1 denotes the flow rate (m3/s) through all hydroelectric generating sets during t1 under run-unit n1, t2 denotes time (s) under run-unit n2, Qt2 denotes the flow rate (m3/s) through all hydroelectric generating sets during t2 under run-unit n2. n1and n2 satisfy the following relationship

$$\eta \left(\frac{{Q}_{t}}{{n}_{1}}\right)={\eta }_{e}$$
(15.50)
$$\eta \left(\frac{{Q}_{t}}{{n}_{2}}\right)={\eta }_{e}$$

Thus, the hydropower station can work under the maximum efficiency ηe.

15.7 Conclusion

By supposing the total flow Qt fixed value, we propose an optimal control method, this method does not depend on the exact model of a hydropower station. By changing the total flow Qt, we also propose an optimal switch method, this method only depends on the efficiency function of the generator at constant head.

The proof of the optimal control and switch method given by this chapter are mainly based on the efficiency function characteristics which can be approximately considered a concave, non-negative function. Thusly, this optimal method has the following features:

  1. (1)

    Both liner and non-liner systems are included,

  2. (2)

    The system's mathematical model is not needed,

  3. (3)

    The method is high universal.