Multiple motors are used to drive long scraper conveyors or the wire section of paper machines, mainly to make long chains and long belts evenly stressed. This type of system does not have the problem of optimal switching of the number of drive motors. However, the system has the problem of how to distribute the load so that the system operates efficiently.

19.1 A Scraper Conveyor Driven by Two Different Motors

In coal mines, many scraper conveyors are very long, some exceeding 300 m. Most scraper conveyors are driven by more than one motor at the same time to ensure uniform force on the scraper conveyor. Scraper conveyors driven by two different motors are common. It has a traditional structure as shown in Fig. 19.1.

Fig. 19.1
An illustration of the scraper conveyor. It includes two motors.

The traditional structure of a scraper conveyor

Full size image

In Fig. 19.1, two different motors drive the scraper conveyor to ensure uniform force on the long scraper conveyor. There is a way to accomplish the same task that consumes the least amount of energy. In other words, there is a way to maximize the overall efficiency. However, there has not yet been a common conclusion on the efficiency optimization of such systems [1].

To achieve minimum energy, we should determine how much load each motor in the system carries.

This issue is becoming increasingly important with the increasing awareness of environmental protection and the need for energy-saving systems. Currently, optimization research on various scraper conveyors has been extensively studied [2,3,4,5,6,7,8,9]. Almost all such optimization methods require the establishment of an accurate model of the system. However, since accurate models of actual scraper conveyors are difficult to establish, these optimization methods quickly become difficult to apply in real systems. Solutions that avoid these complications would be highly beneficial, especially for business and industry.

By assuming a fixed total load, we propose an optimal control method similar to traditional static optimization [10], which does not rely on an accurate model of the scraper conveyor and is only based on the shape of the efficiency function. By varying the total load, we also propose a method similar to traditional dynamic optimization [11], which only relies on the efficiency function of the device at constant output. These methods can be used for systems consisting of the same type of motor and gear and having the same energy efficiency function.

However, in industrial applications, there are still a large number of scraper conveyors, and all of their motors do not have the same model. To address the optimization problem of such systems, we extend the discussion presented in [10, 11]. This chapter defines the normalized energy efficiency function of the motor and the scraper conveyor driven by multiple motors with the same normalized energy efficiency function. We can find the functional expression of the scraper conveyor power consumption and its minimum value. When the normalized efficiency function of each motor is the same, our approach can still be adopted [10, 11]. The minimum power consumption and optimal switching point of this type of scraper conveyor are determined by using a normalized energy efficiency function.

19.1.1 Characteristics of a Scraper Conveyor

In general, the efficiency function of a motor at constant speed n0 (rpm) is shown in Fig. 19.2. By exploiting the shape of the efficiency function, we are able to determine the total power consumption of a scraper conveyor.

Fig. 19.2
A graph of eta versus the M. The line starts at around 0 and then follow a concave downward trend. The dashed line follow a vertical decline at M e.

The efficiency function of a motor

Full size image

In Fig. 19.2, M is torque (N m), and η (M) is the efficiency function, Me is the torque at ηe which is the maximum efficiency. For a given motor, ηe and Me are constants. η (M) can be considered approximately a concave non-negative function through the origin, and we have

$$\begin{gathered} M \ge 0 \hfill \\ \eta (M) \ge 0 \hfill \\ \eta (0) = 0 \hfill \\ \end{gathered}$$
(19.1)

and

$$ \eta^{\prime\prime}(M) < 0 $$
(19.2)

Approximately, η (M) becomes

$$\eta (M)={\sum }_{i=0}^{\infty }{a}_{i}{M}^{i}=M{\sum }_{i=1}^{\infty }({a}_{i}{M}^{i-1})\approx M({a}_{1}+{a}_{2}M)\ge 0$$
(19.3)

where

$${a}_{0}=0$$
(19.4)

It follows that

$$ \begin{gathered} a_{1} + a_{2} M > 0 \hfill \\ a_{1} > 0 \hfill \\ a_{2} < 0 \hfill \\ \end{gathered} $$
(19.5)

The motor’s power has the form

$$P(M)=\frac{M{n}_{0}}{9550\eta (M)}=\frac{{P}_{0}}{\eta (M)}$$
(19.6)

and

$${P}_{0}=\frac{M{n}_{0}}{9550\eta }$$
(19.7)

where P(M) is the power consumption, M is the output torque, η (M) is the efficiency at point (M, n0), and P0 is the ideal work.

If a scraper conveyor consists of two motors which are running, the total power consumption function has the form

$${P_t}\left( {{M_t}} \right) = \frac{1}{{9550}}\sum\limits_{i = 1}^2 {\frac{{{n_i}{M_i}}}{{{\eta _i}\left( {{M_i}} \right)}}} $$
(19.8)

and

$${M_t} = \sum\limits_{i = 1}^2 {{M_i}} $$
(19.9)

where Pt(M) is the total power consumption function, Mi is the i-th motor’s output torque, ηi (Mi) is the ith motor’s efficiency at point (Mi, ni), Mt is the total load torque, and ni is the speed of the i-th motor.

In order to ensure the long scraper conveyor stress evenly, we should have

$${M}_{i}>0$$
(19.10)

If two motors are the same model in the scraper conveyor, their efficiency are the same, and the output speed of two motors is the same n0. The ηi (Mi) becomes η (Mi), We have the following optimal conclusion [10].

The optimal control method is to keep

$${M}_{1}={M}_{2}=\frac{{M}_{t}}{2}$$
(19.11)

The minimal value of the total power consumption is

$${\mathit{minP}}_{t}({M}_{t})={P}_{0}\frac{1}{\eta (\frac{{M}_{t}}{2})}$$
(19.12)

where P0 is the ideal work, it has the form

$${P}_{0}=\frac{{M}_{t}{n}_{0}}{9550}$$
(19.13)

The total optimal efficiency is

$${\mathit{max\eta }}_{t}({M}_{t})=\eta (\frac{{M}_{t}}{2})$$
(19.14)

19.1.2 Scraper Conveyor with Similar Efficiency

We define the torque rate γ as

$$\gamma =\frac{M}{{M}_{e}}$$
(19.15)

We define ηN(γ) as the normalized efficiency function of a motor. The normalized efficiency function ηN(γ) has a shape shown in Fig. 19.3.

Fig. 19.3
A graph of eta N versus the gamma. The line starts at around 0 and then follow a concave downward trend.

The normalized efficiency function ηN (γ)

Full size image

In Fig. 19.3, γ is a variable, ηN (γ) is a function of γ, and ηN (γ) has the following relationship with η (M).

$$\eta (M)=\eta (\gamma {M}_{e})={\eta }_{N}(\gamma )$$
(19.16)

If the normalized efficiency functions of two different motors are the same, we have

$${\eta }_{N1}(\gamma )={\eta }_{N2}(\gamma )$$
(19.17)

We define motor 1 and motor 2 as motors with similar efficiency. Scraper conveyors using motors with similar efficiencies are called scraper conveyors with similar efficiencies.

Suppose γi is the i-th motor’s torque rate and has the form

$${\gamma }_{i}=\frac{{M}_{i}}{{M}_{ie}}$$
(19.18)

For a scraper conveyor with similar efficiency driven by two motors with the same output speed n0, the total power consumption is of the form

$${P}_{t}({M}_{t})=\frac{{n}_{0}}{9550}{\sum }_{i=1}^{2}\frac{{M}_{i}}{{\eta }_{i}({M}_{i})}=\frac{{n}_{0}}{9550}{\sum }_{i=1}^{2}\frac{{\gamma }_{i}{M}_{ie}}{{\eta }_{N}({\gamma }_{i})}$$
(19.19)

19.1.3 Optimal Control of the Scraper Conveyor with Similar Efficiencies

In this section, we derive the expression for the total power consumption of scraper conveyors with similar efficiencies and determine their minimum power consumption.

Theorem

For the optimization problem of Pt (Mt), the minimization of the total power consumption.

$$ \begin{gathered} minP_{t} (M_{t} ) \hfill \\ s.t.\,\,\,\,\,\,\gamma_{i} > 0,i = {1,2} \hfill \\ \sum\limits_{i = 1}^{2} {\gamma_{i} M_{ie} = M_{t} } \hfill \\ \end{gathered} $$
(19.20)

is given by

$$min{P_t}\left( {{M_t}} \right) = \frac{{{P_0}}}{{{\eta _N}\left( {\frac{{{M_t}}}{{{M_{1e}} + {M_{2e}}}}} \right)}}$$
(19.21)

Proof

There are two variables. We have

$${\gamma }_{1}{M}_{1e}+{\gamma }_{2}{M}_{2e}={M}_{t}$$
(19.22)

where

$$\begin{array}{c}{\gamma }_{1}>0\\ {\gamma }_{2}>0\end{array}$$
(19.23)

The objective function Pt (Mt) is expressed as

$${P_t}\left( {{M_t}} \right) = \frac{{{n_0}}}{{9550}}\left( {\frac{{{\gamma _1}{M_{1e}}}}{{{\eta _N}\left( {{\gamma _1}} \right)}} + \frac{{{\gamma _2}{M_{2e}}}}{{{\eta _N}\left( {{\gamma _2}} \right)}}} \right)$$
(19.24)

The optimal condition is given for

$$P_t^\prime ({\gamma _1}) = 0$$
(19.25)

We have

$$\begin{aligned} &{\frac{{{M_{1e}}{\eta _N}\left( {{\gamma _1}} \right) - {\gamma _1}{M_{1e}}{\eta _N}\prime \left( {{\gamma _1}} \right)}}{{\eta _N^2\left( {{\gamma _1}} \right)}}} \\ & {\quad + \frac{{ - {M_{1e}}{\eta _N}\left( {\frac{{{M_t} - {\gamma _1}{M_{1e}}}}{{{M_{2e}}}}} \right) + \left( {{M_t} - {\gamma _1}{M_{1e}}} \right)\frac{{{M_{1e}}}}{{{M_{2e}}}}{\eta _N}\prime \left( {\frac{{{M_t} - {\gamma _1}{M_{1e}}}}{{{M_{2e}}}}} \right)}}{{\eta _N^2\left( {\frac{{{M_t} - {\gamma _1}{M_{1e}}}}{{{M_{2e}}}}} \right)}} = 0} \end{aligned} $$
(19.26)

It is then easily verified that

$${\gamma }_{1}={\gamma }_{2}=\frac{{M}_{t}}{{M}_{1e}+{M}_{2e}}$$
(19.27)

is an optimal point.

Checking the second derivative, we see

$${{P}_{t}}^{{\prime}{\prime}}({\gamma }_{1})>0$$
(19.28)

Therefore, the optimal point is the unique minimum.

The minimum of the total power consumption is then

$$min{P_t} = \frac{{{n_0}{M_t}}}{{9550}}\frac{1}{{{\eta _N}\left( {\frac{{{M_t}}}{{{M_{1e}} + {M_{2e}}}}} \right)}} = \frac{{{P_0}}}{{{\eta _N}\left( {\frac{{{M_t}}}{{{M_{1e}} + {M_{2e}}}}} \right)}}$$
(19.29)

where P0 is the ideal work.

Remark

If two motors have the same model, then the optimal point is.

$${\gamma }_{1}={\gamma }_{2}=\frac{{M}_{t}}{2{M}_{e}}$$
(19.30)
$${M}_{1}={M}_{2}=\frac{{M}_{t}}{2}$$

The minimum of the total power consumption is

$${\mathit{minP}}_{t}=\frac{{n}_{0}{M}_{t}}{9550}\frac{1}{{\eta }_{N}(\frac{{M}_{t}}{2{M}_{e}})}=\frac{{P}_{0}}{\eta (\frac{{M}_{t}}{2})}$$
(19.31)

19.1.4 Simulation

If a scraper conveyor is driven by two different motors, A and B, two motors run at the same speed 600 (rpm). The total torque is variable. The output torque of each motor is adjustable.

Motor A is smaller and its efficiency function with respect to torque M at speed 600 (rpm) is given by

$${\eta }_{1}(M)=0.0188M-0.000094{M}^{2}$$
(19.32)

where

$${M}_{1e}=100(N.M)$$
(19.33)

and

$${\eta }_{1e}=0.94$$
(19.34)

Motor B is larger and its efficiency function with respect to torque M at speed 600 (r.p.m) is

$${\eta }_{2}(M)=0.0094M-0.0000235{M}^{2}$$
(19.35)

where

$${M}_{2e}=200(N.M)$$
(19.36)

and

$${\eta }_{2e}=0.94$$

We see that A and B have the same normalization efficiency function, given by the following

$${\eta }_{N}(\gamma )=1.88\gamma -0.94{\gamma }^{2}$$
(19.37)

According to our previous conclusion, the optimal control method is to keep motors A and B at the same torque rate

$${\gamma }_{1}={\gamma }_{2}=\frac{{M}_{t}}{{M}_{1e}+{M}_{2e}}=\frac{{M}_{t}}{300}$$
(19.38)

If the total load torque Mt is 360 (N.M), then

$$\begin{gathered} \gamma _{1} = \gamma _{2} = \frac{{M_{t} }}{{M_{{1e}} + M_{{2e}} }} = 1.2 \hfill \\ M_{1} = \gamma _{1} M_{{1e}} = 120 \hfill \\ M_{2} = \gamma _{2} M_{{2e}} = 240 \hfill \\ \end{gathered}$$
(19.39)

19.1.5 Conclusion

The proof of the optimal control method given in this chapter is mainly based on the characteristics of the efficiency function. The efficiency function can be approximately considered as a concave non-negative function passing through the origin. Therefore, the optimal method is independent of the linearity or nonlinearity of the system and does not require a state equation or transfer function of the system. The optimal method has the following characteristics:

  1. (1)

    Includes linear and nonlinear systems,

  2. (2)

    No systematic mathematical model is required,

  3. (3)

    High versatility.

19.2 Wire Sections Driven by Three Different Motors in a Paper Machine

In the wire section of the paper machine, there is usually a very long felt belt. This kind of felt is often driven by multiple motors to ensure uniform felt tension. The structure of a piece of felt driven by three different motors is shown in Fig. 19.4.

Fig. 19.4
A structure of the 3 different motors drive in a paper machine. The shaded circle represents the motors 1, 2, and 3.

The structure of a felt driven by three motors

Full size image

In Fig. 19.4, three different motors drive a felt in a paper machine. There is a way to accomplish a task that consumes the least amount of energy, in other words, there is a way that maximizes overall energy efficiency.

The analysis method is basically the same as the idea in the previous section.

19.2.1 Optimal Control of the Wire Section with Similar Efficiencies

Theorem

For the optimization problem of Pt (Mt), the minimization of the total power consumption.

$$ \begin{gathered} minP_{t} (M_{t} ) \hfill \\ s.t.\,\,\,\,\,\,\gamma_{i} > 0,i = {1,}\,{2},\,3 \hfill \\ \sum_{i = 1}^{3} \gamma_{i} M_{ie} = M_{t} \hfill \\ \end{gathered} $$
(19.40)

is given by

$${\mathit{minP}}_{t}({M}_{t})=\frac{{P}_{0}}{{\eta }_{N}(\frac{{M}_{t}}{{M}_{1e}+{M}_{2e}+{M}_{3e}})}$$
(19.41)

Proof

We consider two kinds of situation.

  1. (1)

    n = 2

There are two variables. We have

$$ \begin{gathered} \gamma_{1} M_{1e} + \gamma_{2} M_{2e} = M_{t} \hfill \\ \gamma_{1} > 0 \hfill \\ \gamma_{2} > 0 \hfill \\ \end{gathered} $$
(19.42)

The objective function Pt is expressed as

$${P_t} = \frac{{{n_0}}}{{9550}}\left( {\frac{{{\gamma _1}{M_{1e}}}}{{{\eta _N}\left( {{\gamma _1}} \right)}} + \frac{{{\gamma _2}{M_{2e}}}}{{{\eta _N}\left( {{\gamma _2}} \right)}}} \right)$$
(19.43)

The optimal condition is

$${{P}_{t}}{\prime}({\gamma }_{1})=0$$
(19.44)

We have

$$ \begin{aligned} & \frac{{{M_{1e}}{\eta _N}\left( {{\gamma _1}} \right) - {\gamma _1}{M_{1e}}{\eta _N}\prime \left( {{\gamma _1}} \right)}}{{\eta _N^2\left( {{\gamma _1}} \right)}} \cr &\quad +\, \,\frac{{ - {M_{1e}}{\eta _N}\left( {\frac{{{M_t} - {\gamma _1}{M_{1e}}}}{{{M_{2e}}}}} \right) + \left( {{M_t} - {\gamma _1}{M_{1e}}} \right)\frac{{{M_{1e}}}}{{{M_{2e}}}}{\eta _N}\prime \left( {\frac{{{M_t} - {\gamma _1}{M_{1e}}}}{{{M_{2e}}}}} \right)}}{{\eta _N^2\left( {\frac{{{M_t} - {\gamma _1}{M_{1e}}}}{{{M_{2e}}}}} \right)}} = 0 \cr \end{aligned} $$
(19.45)

It is easy to see that

$${\gamma }_{1}={\gamma }_{2}=\frac{{M}_{t}}{{M}_{1e}+{M}_{2e}}$$
(19.46)

is an optimal point.

The minimal value of the total power consumption is

$${\mathit{minP}}_{t}=\frac{{n}_{0}{M}_{t}}{9550}\frac{1}{{\eta }_{N}(\frac{{M}_{t}}{{M}_{1e}+{M}_{2e}})}=\frac{{P}_{0}}{{\eta }_{N}(\frac{{M}_{t}}{{M}_{1e}+{M}_{2e}})}$$
(19.47)

P0 is the ideal work.

  1. (2)

    n = 3

There are three variables.

Based on known conditions, we have

$$ \begin{gathered} \gamma_{1} M_{1e} + \gamma_{2} M_{2e} + \gamma_{3} M_{3e} = M_{t} \hfill \\ \gamma_{1} > 0 \hfill \\ \gamma_{2} > 0 \hfill \\ \gamma_{3} > 0 \hfill \\ \end{gathered} $$
(19.48)

Pt expression becomes

$${P_t} = \frac{{{n_0}}}{{9550}}\left( {\frac{{{\gamma _1}{M_{1e}}}}{{{\eta _N}\left( {{\gamma _1}} \right)}} + \frac{{{\gamma _2}{M_{2e}}}}{{{\eta _N}\left( {{\gamma _2}} \right)}} + \frac{{{\gamma _3}{M_{3e}}}}{{{\eta _N}\left( {{\gamma _3}} \right)}}} \right)$$
(19.49)

Suppose γ1 is the fixed optimal point, only γ2 and γ3 are variables. We have

$${\gamma }_{2}{M}_{2e}+{\gamma }_{3}{M}_{3e}={M}_{t}-{\gamma }_{1}{M}_{1e}=cons\mathit{tan}t$$
(19.50)

Based on the above conclusions at n = 2, the optimal point is at

$${\gamma }_{2}={\gamma }_{3}$$
(19.51)

Suppose that γ2 is the fixed optimal point, only γ1 and γ3 are variables. We have

$${\gamma }_{1}{M}_{1e}+{\gamma }_{3}{M}_{3e}={M}_{t}-{\gamma }_{2}{M}_{2e}=cons\mathit{tan}t$$
(19.52)

According to the conclusion at n = 2, the optimal point is at

$${\gamma }_{1}={\gamma }_{3}$$
(19.53)

Similarly, assume that γ3 is fixed optimal point, only γ1 and γ2 are variables. We have

$${\gamma }_{1}{M}_{1e}+{\gamma }_{2}{M}_{2e}={M}_{t}-{\gamma }_{3}{M}_{3e}=cons\mathit{tan}t$$
(19.54)

According to the conclusion at n = 2, the optimal point is at

$${\gamma }_{1}={\gamma }_{2}$$
(19.55)

Thus, we have the optimal points

$${\gamma }_{1}={\gamma }_{2}={\gamma }_{3}=\frac{{M}_{t}}{{\sum }_{i=1}^{3}{M}_{ie}}$$
(19.56)

The minimum of the total power consumption is

$${\mathit{minP}}_{t}=\frac{{P}_{0}}{{\eta }_{N}(\frac{{M}_{t}}{{\sum }_{i=1}^{3}{M}_{ie}})}$$
(19.57)

Remark:

If three motors have the same model, then the optimal point is

$${\gamma }_{1}={\gamma }_{2}={\gamma }_{3}=\frac{{M}_{t}}{3{M}_{e}}$$
(19.58)
$${M}_{1}={M}_{2}={M}_{3}=\frac{{M}_{t}}{3}$$

The minimum of the total power consumption is

$${\mathit{minP}}_{t}=\frac{{n}_{0}{M}_{t}}{9550}\frac{1}{{\eta }_{N}(\frac{{M}_{t}}{3{M}_{e}})}=\frac{{P}_{0}}{\eta (\frac{{M}_{t}}{3})}$$
(19.59)

19.2.2 Conclusion

The proof of the optimal control method given in this chapter is mainly based on the characteristics of the energy efficiency function, which can be approximated as a concave non-negative function through the origin. Therefore, the optimal method is independent of the linearity or nonlinearity of the system and does not require a state equation or transfer function of the system.

In order to solve the optimization problem of general equipment, the authors have conducted extensive research [11,12,13,14,15,16,17,18]. The conclusions given in this chapter can also be applied to the optimization of pumps, fans, blowers, compressors, etc. We would like to thank Zhang Yanfang from Hebei Automation Company and Yao Bosheng from Bei**g IAO Technology Development Company for their valuable help and suggestions during the development and experiment of this theory.