Many electric transportation vehicles such as locomotives, subways, electric cars, ships, etc. are jointly driven by multiple electric motors (Fig. 20.1). In these occasions and fields, there are optimization problems of how multiple motors or multiple engines output power and how to determine the number of operating units.

Fig. 20.1
Three photos of the electric vehicles, such as, a train, ship, and a car from the left to right, respectively.

Electric transportation vehicles

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20.1 Total Power Consumption of High-Speed Train Drive System

The work performed by transport vehicles is mainly to overcome friction. Take the EMU as an example. The carriages have the same speed. How to distribute the output of the motors in each carriage to minimize the total energy consumption?

The total force required by the EMU is Ft, the speed V0 of each power unit is equal, the output force Fi of each carriage is part of the total force, and the energy consumption expression of the EMU is:

$${P}_{t}=k3{\sum }_{i=1}^{n}\frac{{F}_{i}{V}_{i}}{{\eta }_{i}({F}_{i},{V}_{i})}=k3{F}_{t}{V}_{0}{\sum }_{i=1}^{n}\frac{{F}_{i}}{{F}_{t}}\times \frac{1}{{\eta }_{i}({F}_{i})}$$
(20.1)

where k3 is a constant.

The expression for energy consumption written as torque and speed is

$$\begin{aligned} P_{t} & = k3M_{t} n_{0} \sum _{{i = 1}}^{n} \frac{{M_{i} }}{{M_{t} }}\frac{1}{{\eta (M_{i} ,n_{0} )}} = k3M_{t} n_{0} \sum _{{i = 1}}^{n} \theta _{i} \frac{1}{{\eta (\theta _{i} M_{t} ,n_{0} )}} \\ & = P_{0} \sum _{{i = 1}}^{n} \theta _{i} \frac{1}{{\eta (\theta _{i} M_{t} )}} \\ \end{aligned}$$
(20.2)

where

$${\theta }_{i}=\frac{{M}_{i}}{{M}_{t}}$$
(20.3)
$${P}_{0}=k3{M}_{t}{n}_{0}$$

The total torque Mt required by the EMU is the total load, and the rotation speed of each power unit is equal to n0.

The efficiency curve of the first drive unit (including motor and reducer) at the speed n0 is shown in Fig. 20.2.

Fig. 20.2
A graph of eta versus the M. The dashed lines start at around eta m, remain horizontal and then decline vertically. The solid line begins at the origin and then follow a concave slope.

The efficiency curve of the first power unit

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In Fig. 20.2, M1m is the maximum output torque of the first power unit, M1e is the optimal output torque of the first power unit, and M1e corresponds to the maximum efficiencyηm.

20.2 Optimized Load Distribution Method for Drive Motors of High-Speed Trains

Assuming that Mt and n0 remain unchanged, let's find the optimal control method for the EMU. Since n0 is a constant, the operating efficiency of the speed-regulated motor of each power unit only considers the torque variable.

Consider minimizing the total power consumption of the drive system

$$ \begin{gathered} \min P_{t} \hfill \\ s.t.\,\theta_{i} > 0,\,\,i = 1,2, \ldots n \hfill \\ \sum\limits_{i = 1}^{n} {\theta_{i} = 1} \hfill \\ \end{gathered} $$
(20.4)

This problem can also be written as

$$ \begin{gathered} \min P_{0} \sum_{i = 1}^{n} \theta_{i} \frac{1}{{\eta_{i} \left( {\theta_{i} M_{0} } \right)}} \hfill \\ s.t.\,\theta_{i} > 0,i = {1,2},...n \hfill \\ \sum_{i = 1}^{n} \theta_{i} = 1 \hfill \\ \end{gathered} $$
(20.5)

We consider three cases:

  1. (1)

    n = 2

The system has two variables and has

$$ \begin{gathered} \theta_{1} + \theta_{2} = 1 \hfill \\ \theta_{1} > 0 \hfill \\ \theta_{2} > 0 \hfill \\ \end{gathered} $$
(20.6)

The objective function Pt can be expressed as

$$\begin{aligned} P_{t} & = P_{0} \left( {\theta _{1} \frac{1}{{\eta _{1} \left( {\theta _{1} M_{t} } \right)}} + \theta _{2} \frac{1}{{\eta _{2} \left( {\theta _{2} M_{t} } \right)}}} \right) \\ & = \,P_{0} \left( {\theta _{1} \frac{1}{{\eta _{1} \left( {\theta _{1} M_{t} } \right)}} + \left( {1 - \theta _{1} } \right)\frac{1}{{\eta _{2} \left( {\left( {1 - \theta _{1} } \right)M_{t} } \right)}}} \right) \\ \end{aligned}$$
(20.7)

The optimization condition is

$${{P}_{t}}{\prime}({\theta }_{1})=0$$
(20.8)

We have

$$\frac{1}{{\eta }_{1}({\theta }_{1}{M}_{t})}-{\theta }_{1}\frac{{{\eta }_{1}}{\prime}({\theta }_{1}{M}_{t}){M}_{t}}{{\eta }_{1}({\theta }_{1}{M}_{t}{)}^{2}}-\frac{1}{{\eta }_{2}((1-{\theta }_{1}){M}_{t})}+(1-{\theta }_{1})\frac{{{\eta }{\prime}}_{2}((1-{\theta }_{1}){M}_{t}){M}_{t}}{{\eta }_{2}((1-{\theta }_{1}){M}_{t}{)}^{2}}=0$$
(20.9)

Since all the motors are variable speed motors and the identical model, it is easy to see that

$$ \begin{gathered} \theta_{1} = \frac{1}{2} \hfill \\ \eta_{1}{\prime} \left( {\frac{{M_{t} }}{2}} \right) = \eta_{2}{\prime} \left( {\frac{{M_{t} }}{2}} \right) \hfill \\ \eta_{1} \left( {\frac{{M_{t} }}{2}} \right) = \eta_{2} \left( {\frac{{M_{t} }}{2}} \right) \hfill \\ \end{gathered} $$
(20.10)

is an optimization point. Then, we have

$${\theta }_{2}=1-{\theta }_{1}={\theta }_{1}=\frac{1}{2}=\frac{1}{n}$$
(20.11)

That is, the optimal control method is to keep

$${M}_{1}={M}_{2}=\frac{{M}_{t}}{2}=\frac{{M}_{t}}{n}$$
(20.12)

The minimum of the total power dissipation is

$${\mathit{minP}}_{t}={P}_{0}\frac{1}{{\eta }_{1}(\frac{{M}_{t}}{2})}={P}_{0}\frac{1}{{\eta }_{1}(\frac{{M}_{t}}{n})}$$
(20.13)
  1. (2)

    n = 3

The system has three variables, based on known conditions, we have

$${\theta }_{1}+{\theta }_{2}+{\theta }_{3}=1$$
(20.14)
$${\theta }_{1}>0$$
$${\theta }_{2}>0$$
$${\theta }_{3}>0$$

The Pt expression becomes

$${P}_{t}={P}_{0}({\theta }_{1}\frac{1}{{\eta }_{1}({\theta }_{1}{M}_{t})}+{\theta }_{2}\frac{1}{{\eta }_{2}({\theta }_{2}{M}_{t})}+{\theta }_{3}\frac{1}{{\eta }_{3}({\theta }_{3}{M}_{t})})$$
(20.15)

Assuming that θ1 is fixed and is an optimization point, and only θ2 and θ3 are variables, then we have

$${\theta }_{2}+{\theta }_{3}=1-{\theta }_{1}=constant$$
(20.16)

Based on the conclusion of n = 2 above, we have

$${\theta }_{2}={\theta }_{3}$$
(20.17)

to be the optimal point.

Assuming that θ2 is fixed and is an optimization point, and only θ1 and θ3 are variables, then we have

$${\theta }_{1}+{\theta }_{3}=1-{\theta }_{2}=constant$$
(20.18)

According to the conclusion of n = 2 above, we have

$${\theta }_{1}={\theta }_{3}$$
(20.19)

to be the optimal point.

Similarly, assuming that θ3 is fixed and is an optimization point, and only θ1 and θ2 are variables, then we have

$${\theta }_{1}+{\theta }_{2}=1-{\theta }_{3}=constant$$
(20.20)

According to the conclusion of n = 2 above, we have

$${\theta }_{1}={\theta }_{2}$$
(20.21)

to be is the optimal point.

So, we have the optimization point

$${\theta }_{1}={\theta }_{2}={\theta }_{3}=\frac{1}{3}=\frac{1}{n}$$
(20.22)

The optimal control method is to keep

$${M}_{1}={M}_{2}={M}_{3}=\frac{{M}_{t}}{3}=\frac{{M}_{t}}{n}$$
(20.23)

The minimum of the total power dissipation is

$$ minP_{t} = P_{0} \frac{1}{{\eta_{1} \left( {\frac{{M_{t} }}{3}} \right)}} = P_{0} \frac{1}{{\eta_{1} \left( {\frac{{M_{t} }}{n}} \right)}} $$
(20.24)
  1. (3)

    n = k

The system has n variables, and the above conclusion can be extended to the case of n = k, the optimal point is

$${\theta }_{1}={\theta }_{2}=...={\theta }_{k}=\frac{1}{k}$$
(20.25)

The optimal control method is to keep

$${M}_{1}={M}_{2}=...={M}_{k}=\frac{{M}_{t}}{k}$$
(20.26)

The minimum of the total power dissipation is

$${\mathit{minP}}_{t}={P}_{0}\frac{1}{{\eta }_{1}(\frac{{M}_{t}}{k})}$$
(20.27)

The optimal overall efficiency is

$${\mathit{max\eta }}_{t}({M}_{t})={\eta }_{1}(\frac{{M}_{t}}{k})$$
(20.28)

20.3 The Optimal Number of Operating Motors for High-Speed Trains

Assume that the EMU has m cars of the identical type, and all cars are driven by motors that can be adjusted in speed. The speed of each car is the same as n0, and the total torque is Mt. There are n car motors running with speed regulation, and n is less than is equal to m, the output torque of the i-th car motor is Mi, and the total power consumption is the minimum, there is

$$\begin{gathered} minP_{t} = P_{0} \frac{1}{{\eta _{1} \left( {\frac{{M_{t} }}{n}} \right)}} \hfill \\ n \le m \hfill \\ \frac{{M_{t} }}{n} \le M_{{1m}} \hfill \\ \end{gathered}$$
(20.29)

When the total torque Mt changes due to wind resistance, uphill, downhill, etc., we consider 2 cases:

1. Mt /n is equal to or less than m1e, which is the torque corresponding to the highest efficiency point, as shown in Fig. 20.3.

Fig. 20.3
A graph of eta versus the M. The line starts at around 0 and then follow a concave downward trend.

Mt/n is equal to or less than M1e

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In Fig. 20.3, we have

$$\mathit{max}\left({\eta }_{1}\left(\frac{{M}_{t}}{n-1}\right),{\eta }_{1}\left(\frac{{M}_{t}}{n}\right),{\eta }_{1}\left(\frac{{M}_{t}}{n+1}\right)\right)={\eta }_{1}\left(\frac{{M}_{t}}{n}\right)$$
(20.30)

When Mt increases, Mt/n also increases. If Mt/n > M1e and η1(Mt/n) < η1 (Mt/(n + 1)), then the optimal number of operating units becomes n + 1, and we should increase the number of running motors from n to n + 1, there are

$$\mathit{max}\left({\eta }_{1}\left(\frac{{M}_{t}}{n-1}\right),{\eta }_{1}\left(\frac{{M}_{t}}{n}\right),{\eta }_{1}\left(\frac{{M}_{t}}{n+1}\right)\right)={\eta }_{1}\left(\frac{{M}_{t}}{n+1}\right)$$
(20.31)

The optimal switching point is

$${\eta }_{1}\left(\frac{{M}_{t}}{n}\right)={\eta }_{1}\left(\frac{{M}_{t}}{n+1}\right)$$
(20.32)

When Mt decreases, Mt/n also decreases. If Mt/n > M1e and η1(Mt/n) < η1(Mt/(n − 1)), then the optimal number of running units becomes n − 1. We should reduce the number of running motors from n to n − 1, and then we have

$$\mathit{max}\left({\eta }_{1}\left(\frac{{M}_{t}}{n-1}\right),{\eta }_{1}\left(\frac{{M}_{t}}{n}\right),{\eta }_{1}\left(\frac{{M}_{t}}{n+1}\right)\right)={\eta }_{1}\left(\frac{{M}_{t}}{n-1}\right)$$
(20.33)

The optimal switching point is

$${\eta }_{1}\left(\frac{{M}_{t}}{n}\right)={\eta }_{1}\left(\frac{{M}_{t}}{n-1}\right)$$
(20.34)

2. Mt n is greater than M1e, as shown in Fig. 20.4

Fig. 20.4
A graph of eta versus the M. The dashed lines follow a vertical declines. The solid line starts at around 0 and then follow a concave trend.

Mt/n is greater than M1e

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In Fig. 20.4, there are

$$\mathit{max}\left({\eta }_{1}\left(\frac{{M}_{t}}{n-1}\right),{\eta }_{1}\left(\frac{{M}_{t}}{n}\right),{\eta }_{1}\left(\frac{{M}_{t}}{n+1}\right)\right)={\eta }_{1}\left(\frac{{M}_{t}}{n}\right)$$
(20.35)

When Mt increases, Mt/n also increases. If η1(Mt/n) < η1(Mt/(n + 1)), then the optimal number of running units becomes n + 1, and we should increase the number of running motors from n to n + 1, which yields

$$\mathit{max}\left({\eta }_{1}\left(\frac{{M}_{t}}{n-1}\right),{\eta }_{1}\left(\frac{{M}_{t}}{n}\right),{\eta }_{1}\left(\frac{{M}_{t}}{n+1}\right)\right)={\eta }_{1}\left(\frac{{M}_{t}}{n+1}\right)$$
(20.36)

The optimal switching point is

$${\eta }_{1}\left(\frac{{M}_{t}}{n}\right)={\eta }_{1}\left(\frac{{M}_{t}}{n+1}\right)$$
(20.37)

When Mt decreases, Mt/n also decreases, if η1(Mt/n) < η1(Mt/(n − 1)), then the optimal number of running units becomes n − 1, and we should reduce the number of running motors from n to n − 1, which yields

$$\mathit{max}\left({\eta }_{1}\left(\frac{{M}_{t}}{n-1}\right),{\eta }_{1}\left(\frac{{M}_{t}}{n}\right),{\eta }_{1}\left(\frac{{M}_{t}}{n+1}\right)\right)={\eta }_{1}\left(\frac{{M}_{t}}{n-1}\right)$$
(20.38)

The optimal switching point is

$${\eta }_{1}\left(\frac{{M}_{t}}{n}\right)={\eta }_{1}\left(\frac{{M}_{t}}{n-1}\right)$$
(20.39)