We divide the energy system into three parts: generation, transmission and consumption. When generating electricity, improve energy efficiency and maximize power generation. In terms of power transmission and energy consumption, we need to improve energy efficiency and minimize energy consumption.

2.1 Energy Form of the Power Station

The power station converts raw energy into electrical energy, which can be expressed as

$$ {\text{W}}_{t} = {\text{W}}_{0} {\upeta }_{0} $$
(2.1)

where W0 represents the original energy, that is, the ideal work, η0 represents the overall energy efficiency of the power station, and Wt is the power generation of the power station. In the case of a certain ideal work W0, the higher the η0, the greater the Wt, and the more energy output by the power station.

2.1.1 Convert Potential Energy to Electrical Energy

A hydropower station or energy storage power station has n hydroelectric generating units. The working head of each hydro-generator set in the parallel system is the same. The potential energy of water in a river or reservoir is used to generate electricity, expressed as

$$ W_{t} = W_{0} \eta_{o} = \frac{{Q_{t} H}}{{k_{1} }}\eta_{0} = \frac{H}{{k_{1} }}\mathop \sum \limits_{i = 1}^{n} Q_{i} \eta_{i} = \frac{{Q_{t} H}}{{k_{1} }}\mathop \sum \limits_{i = 1}^{n} \frac{{Q_{i} }}{{Q_{t} }}\eta_{i} = W_{0} \mathop \sum \limits_{i = 1}^{n} \theta_{i} \eta_{i} $$
(2.2)
$$ W_{0} = \frac{{Q_{t} H}}{{k_{1} }} $$

where k1 is a constant, W0 is the potential energy of water flowing through the hydro-generator unit, Wt is the electric energy output by the power station, n is the total number of hydro-generator units. Qt is the total water flow rate through the hydro-generator unit. H is the water head, η0 is the overall energy efficiency of the hydropower station, ηi is the operating energy efficiency of the i-th hydro-generator set, Qi is the flow rate through the i-th hydro-generator set. θi is the load rate of the i-th hydro-generator set, as the load rate of the i-th hydroelectric generating unit, expressed as

$$ Q_{t} = \mathop \sum \limits_{i = 1}^{n} Q_{i} $$
(2.3)
$$ \theta_{i} = \frac{{Q_{i} }}{{Q_{t} }} $$
$$ \mathop \sum \limits_{i = 1}^{n} \theta_{i} = 1 $$
$$ \eta_{0} = \mathop \sum \limits_{i = 1}^{n} \theta_{i} \eta_{i} $$

For a fixed Qt and H, the higher the η0, the larger the Wt.

2.1.2 Convert Heat Energy to Electricity

Thermal power plants consist of steam boilers and turbo generators that convert the thermal energy of coal, natural gas or oil into electricity. The electrical energy produced by a thermal power plant is expressed as

$$ W_{t} = \mathop \sum \limits_{i = 1}^{n} (W_{i} \eta_{i} ) = W_{t0} \mathop \sum \limits_{i = 1}^{n} \frac{{W_{i} }}{{W_{t0} }}\eta_{i} = W_{t0} \mathop \sum \limits_{i = 1}^{n} \theta_{i} \eta_{i} = W_{t0} \eta_{o} $$
(2.4)
$$ W_{t0} = \mathop \sum \limits_{i = 1}^{n} W_{i} $$

where n is the total number of generating units in thermal power plants, Wt0 is the thermal energy contained in coal, natural gas or oil used in thermal power plants, Wt is the total electric energy output by thermal power plants. η0 is the comprehensive energy efficiency of thermal power plants, and ηi is the i-th generator operating efficiency, θi is the load rate of the i-th generator, the expression is

$$ \theta_{i} = \frac{{W_{i} }}{{W_{t0} }} $$
(2.5)
$$ \mathop \sum \limits_{i = 1}^{n} \theta_{i} = 1 $$
$$ \eta_{0} = \mathop \sum \limits_{i = 1}^{n} \theta_{i} \eta_{i} $$

2.1.3 Wind Power Hydrogen Production System

Due to the involvement of chemical reactions, fission reactions and other factors, some systems cannot directly convert energy. Such systems can only maximize the output as the optimal goal of the overall energy efficiency of the system. For example, electrolyzing water to produce hydrogen, you cannot say that all the hydrogen energy produced is converted from the electricity consumption of the electrolyzer.

Assuming that there are m hydrogen generators in the hydrogen production station, the total input power is Wt, the electric energy input by the i-th device is Wi, and the overall efficiency of the system is ηt. The overall energy efficiency expression of the system is

$$\begin{array}{c}{\eta }_{t}=\sum\limits_{i=1}^{m}\frac{{W}_{i}}{{W}_{t}}{\eta }_{i}({W}_{i})= \sum\limits_{i=1}^{m}{\theta }_{i}{\eta }_{i}({\theta }_{i}) \\ \begin{array}{c}s.t. \sum\limits_{i=1}^{m}{W}_{i}={W}_{t} >0 \\ {W}_{{i}{\max}}\ge {W}_{i}>0\end{array}\end{array}$$
(2.6)

where Wimax is the maximum load of the hydrogen generator.

If expressed in terms of hydrogen production, this optimization problem can also be transformed into

$$ W_{t} = \mathop \sum \limits_{i = 1}^{n} W_{i} = k_{0} \mathop \sum \limits_{i = 1}^{n} \frac{{Q_{i} }}{{\eta_{i} }} = k_{0} Q_{t} \mathop \sum \limits_{i = 1}^{n} \frac{{Q_{i} }}{{Q_{t} }} \cdot \frac{1}{{\eta_{i} }} = k_{0} Q_{t} \mathop \sum \limits_{i = 1}^{n} \theta_{i} \frac{1}{{\eta_{i} }} = k_{0} Q_{t} \frac{1}{{\eta_{o} }} $$
(2.7)
$$ \eta_{0} = \frac{1}{{\mathop \sum \nolimits_{i = 1}^{n} \theta_{i} \frac{1}{{\eta_{i} }}}} $$

where k0 is an energy conversion constant, Wt is the energy consumption of the system, Wi is the energy consumption of the i-th hydrogen generator, Qt is the total hydrogen production of the system. Qi is the hydrogen production provided by the i-th hydrogen generator, n represents the total number of hydrogen generators in the system. η0 represents the overall energy efficiency of the system, ηi represents the operating efficiency of the i-th hydrogen generator, and θi represents the load rate of the i-th hydrogen generator, the expression is

$$ Q_{t} = \mathop \sum \limits_{i = 1}^{n} Q_{i} $$
(2.8)
$$ \theta_{i} = \frac{{Q_{i} }}{{Q_{t} }} $$
$$ \mathop \sum \limits_{i = 1}^{n} \theta_{i} = 1 $$

For a fixed Qt, the higher the η0, the smaller the Wt.

2.2 Power Dispatch and Distribution

The high-voltage power generated by power stations needs to be transmitted to customers through many transmission lines and a large number of transformers. So we need to consider the issue of power dispatch and distribution.

2.2.1 Power Distribution

There are n transformers in the power supply system, and all transformers are connected in parallel. For a parallel system, the line voltage of each transformer is U0. The total apparent power St and the total output current I0 of all transformers, the line current of the i-th transformer is Ii, and the total energy consumption Wt of n transformers is expressed as

$$ \begin{aligned} W_{t} &= \sqrt 3 \mathop \sum \limits_{i = 1}^{n} \frac{{U_{0} I_{i} }}{{\eta_{i} }} = \sqrt 3 U_{0} \mathop \sum \limits_{i = 1}^{n} \frac{{U_{0} I_{i} }}{{\eta_{i} }} = \sqrt 3 U_{0} I_{0} \mathop \sum \limits_{i = 1}^{n} \frac{{I_{i} }}{{I_{0} }} \cdot \frac{1}{{\eta_{i} }} = \sqrt 3 U_{0} I_{0} \mathop \sum \limits_{i = 1}^{n} \theta_{i} \frac{1}{{\eta_{i} }} \\ & = S_{t} \mathop \sum \limits_{i = 1}^{n} \theta_{i} \frac{1}{{\eta_{i} }} = \frac{{S_{t} }}{{\eta_{0} }} \end{aligned} $$
(2.9)
$$ I_{0} = \mathop \sum \limits_{i = 1}^{n} I_{i} $$
$$ S_{t} = \sqrt 3 UI_{0} $$
$$ \eta_{0} = \frac{1}{{\mathop \sum \nolimits_{i = 1}^{n} \theta_{i} \frac{1}{{\eta_{i} }}}} $$

Among them, η0 represents the overall efficiency of the power supply system, θi represents the load rate of the i-th transformer, expressed as

$$ \mathop \sum \limits_{i = 1}^{n} \theta_{i} = 1 $$
(2.10)
$$ \theta_{i} = \frac{{I_{i} }}{{I_{0} }} $$

2.2.2 Power Dispatch

There are n transmission lines supplying power in an area, and the power grids are connected in parallel. The total apparent power required is St, and the line voltage U of each transmission line is the same. The total current I0, the resistance and current of the i-th line are Ri and Ii respectively, then the total energy consumption Wt of the n transmission lines is expressed as

$$ W_{t} = 3\mathop \sum \limits_{i = 1}^{n} I_{i}^{2} R_{i} = 3I_{0}^{2} \mathop \sum \limits_{i = 1}^{n} \left( {\frac{{I_{i} }}{{I_{0} }}} \right)^{2} R_{i} = 3I_{0}^{2} \mathop \sum \limits_{i = 1}^{n} \theta_{i}^{2} R_{i} $$
(2.11)
$$ I_{0} = \mathop \sum \limits_{i = 1}^{n} I_{i} $$
$$ S_{t} = \sqrt 3 UI_{0} $$

where θi represents the load rate of the i-th transmission line, expressed as

$$ \mathop \sum \limits_{i = 1}^{n} \theta_{i} = 1 $$
(2.12)
$$ \theta_{i} = \frac{{I_{i} }}{{I_{0} }} $$

2.3 Energy Consumption System

Energy consumption system transforms electricity energy into target energy, ideal work W0, Wt is the total energy consumption of the system, Wt is expressed as

$$ {\text{W}}_{t} = {\text{W}}_{0} \frac{1}{{{\upeta }_{0} }} $$
(2.13)

where W0 represents the target energy which is the ideal work, η0 denotes the overall efficiency of the energy consumption system. It is seen that for a fixed ideal work W0, the higher η0 is, the smaller Wt will be, so that the system needs less energy.

2.3.1 Gaining Potential Energy

In pumping stations and energy storage power stations, pumps are used to obtain the potential energy of water and work in parallel or in series.

  1. 1.

    Working in Parallel

A pumping station or a storage power station has n pumps. Each pump in a parallel system gets the same head. Using electric energy to obtain the potential energy of water. Expressed as follows:

$$ W_{t} = W_{0} \frac{1}{{\eta_{0} }} = \frac{{Q_{t} H}}{{k_{1} }} \cdot \frac{1}{{\eta_{0} }} = \frac{H}{{k_{1} }}\mathop \sum \limits_{i = 1}^{n} Q_{i} \frac{1}{{\eta_{i} }} = \frac{{Q_{t} H}}{{k_{1} }}\mathop \sum \limits_{i = 1}^{n} \frac{{Q_{i} }}{{Q_{t} }} \cdot \frac{1}{{\eta_{i} }} = W_{0} \mathop \sum \limits_{i = 1}^{n} \theta_{i} \frac{1}{{\eta_{i} }} $$
(2.14)
$$ W_{0} = \frac{{Q_{t} H}}{{k_{1} }} $$
$$ \eta_{0} = \frac{1}{{\mathop \sum \nolimits_{i = 1}^{n} \theta_{i} \frac{1}{{\eta_{i} }}}} $$

where k1 is a constant, W0 is the obtained water potential energy, Wt is the electric energy consumption of the pumping station or energy storage power station, n is the total number of pumps. Qt is the total flow rate of water delivered, H is the obtained water head. η0 is the overall energy efficiency of the pumping station or energy storage power station, ηi is the operating efficiency of the i-th pump. Qi is the flow rate of the i-th pump, θi is the load rate of the i-th pump. Expressed as follows:

$$ Q_{t} = \mathop \sum \limits_{i = 1}^{n} Q_{i} $$
(2.15)
$$ \theta_{i} = \frac{{Q_{i} }}{{Q_{t} }} $$
$$ \mathop \sum \limits_{i = 1}^{n} \theta_{i} = 1 $$

When Qt and H are fixed, the higher η0 is, the smaller Wt is.

  1. 2.

    Running in Series

There are n pumps in a pumping station, and the flow rate of each pump in the series system is the same, and the potential energy of water is obtained by using electric energy, which is expressed as

$$ W_{t} = W_{0} \frac{1}{{\eta_{0} }} = \frac{{QH_{t} }}{{k_{1} }} \cdot \frac{1}{{\eta_{0} }} = \frac{Q}{{k_{1} }}\mathop \sum \limits_{i = 1}^{n} H_{i} \frac{1}{{\eta_{i} }} = \frac{{QH_{t} }}{{k_{1} }}\mathop \sum \limits_{i = 1}^{n} \frac{{H_{i} }}{{H_{t} }} \cdot \frac{1}{{\eta_{i} }} = W_{0} \mathop \sum \limits_{i = 1}^{n} \theta_{i} \frac{1}{{\eta_{i} }} $$
(2.16)
$$ W_{0} = \frac{{QH_{t} }}{{k_{1} }} $$
$$ \eta_{0} = \frac{1}{{\mathop \sum \nolimits_{i = 1}^{n} \theta_{i} \frac{1}{{\eta_{i} }}}} $$

where k1 is a constant, W0 is the potential energy of water obtained, Wt is the power consumption of the pumping station, n is the total number of water pumps, Q is the flow rate of water delivered by the pumps. Ht is the total water head obtained. η0 is the overall efficiency of the pumping station, ηi is the operating efficiency of the i-th pump, Hi is the lift of the i-th pump, θi is the load rate of the i-th pump, the expression is

$$ H_{t} = \mathop \sum \limits_{i = 1}^{n} H_{i} $$
(2.17)
$$ \theta_{i} = \frac{{H_{i} }}{{H_{t} }} $$
$$ \mathop \sum \limits_{i = 1}^{n} \theta_{i} = 1 $$

For a fixed Q and Ht, the higher the η0, the smaller the Wt.

2.3.2 Provide Pressure Energy

Fans, compressors and blowers are used to obtain pressure energy of air or other gases. Take a fan station as an example, there are n fans in the station. Each fan in a parallel system provides the same pressure. Using electric energy to obtain the pressure energy of air, the expression is

$$ W_{t} = W_{0} \frac{1}{{\eta_{o} }} = \frac{{Q_{t} P}}{{k_{2} }} \cdot \frac{1}{{\eta_{0} }} = \frac{P}{{k_{2} }}\mathop \sum \limits_{i = 1}^{n} Q_{i} \frac{1}{{\eta_{i} }} = \frac{{Q_{t} P}}{{k_{2} }}\mathop \sum \limits_{i = 1}^{n} \frac{{Q_{i} }}{{Q_{t} }} \cdot \frac{1}{{\eta_{i} }} = W_{0} \mathop \sum \limits_{i = 1}^{n} \theta_{i} \frac{1}{{\eta_{i} }} $$
(2.18)
$$ {\text{W}}_{0} = \frac{{Q_{t} P}}{{k_{2} }} $$
$$ \eta_{0} = \frac{1}{{\mathop \sum \nolimits_{i = 1}^{n} \theta_{i} \frac{1}{{\eta_{i} }}}} $$

where k2 is a constant, and W0 represents the pressure energy of the output air of the fan station. Wt is the power consumption of the fan station, Qt is the air flow rate provided by the fan station, P is the air pressure, n is the total number of fans, and η0 is the overall energy efficiency of the fan station. ηi represents the operating energy efficiency of the i-th fan, Qi represents the flow rate of the i-th fan, θi represents the load rate of the i-th fan, the expression is

$$ Q_{t} = \mathop \sum \limits_{i = 1}^{n} Q_{i} $$
(2.19)
$$ \theta_{i} = \frac{{Q_{i} }}{{Q_{t} }} $$
$$ \mathop \sum \limits_{i = 1}^{n} \theta_{i} = 1 $$

For a fixed Qt and P, the higher the η0, the smaller the Wt.

2.3.3 Provide Cold and Heat Energy

  1. 1.

    Central Air Conditioning System

There are many central air-conditioning systems in public buildings, which consume a lot of energy. There are n chillers in the central air-conditioning system to provide cold energy. The total power consumption is expressed as

$$\begin{aligned} W_{t} &= \mathop \sum \limits_{i = 1}^{n} W_{i} = k_{3} \Delta t\mathop \sum \limits_{i = 1}^{n} \frac{{Q_{i} }}{{\eta_{i} }} = k_{3} \Delta tQ_{t} \mathop \sum \limits_{i = 1}^{n} \frac{{Q_{i} }}{{Q_{t} }} \cdot \frac{1}{{\eta_{i} }} = k_{3} \Delta tQ_{t} \mathop \sum \limits_{i = 1}^{n} \theta_{i} \frac{1}{{\eta_{i} }} \\ & = k_{3} \Delta tQ_{t} \frac{1}{{\eta_{o} }} \end{aligned}$$
(2.20)
$$ \eta_{0} = \frac{1}{{\mathop \sum \nolimits_{i = 1}^{n} \theta_{i} \frac{1}{{\eta_{i} }}}} $$

where Wt is the power consumption of the system, k3 is a constant, Δt is the temperature difference between the inlet and outlet fluids of the system, Wi is the power consumption of the i-th chiller. Qt is the total cooling or heat required by the building, Qi is the cooling or heat provided by the i-th chiller, n represents the total number of chillers in the central air-conditioning system, η0 represents the overall energy efficiency of the system, ηi represents the operating efficiency of the i-th chiller, and θi represents the load rate of the i-th chiller, the expression is

$$ Q_{t} = \mathop \sum \limits_{i = 1}^{n} Q_{i} $$
(2.21)
$$ \theta_{i} = \frac{{Q_{i} }}{{Q_{t} }} $$
$$ \mathop \sum \limits_{i = 1}^{n} \theta_{i} = 1 $$

For a fixed Qt, the higher the η0, the smaller the Wt.

  1. 2.

    Central Heating System

There are many boilers in the city district heating system to provide heat energy. There are n boilers in the system, which are used to provide heat energy, and the total energy consumption is expressed as

$$ W_{t} = \mathop \sum \limits_{i = 1}^{n} W_{i} = k_{4} \Delta t\mathop \sum \limits_{i = 1}^{n} \frac{{Q_{i} }}{{\eta_{i} }} = k_{4} \Delta tQ_{t} \mathop \sum \limits_{i = 1}^{n} \frac{{Q_{i} }}{{Q_{t} }} \cdot \frac{1}{{\eta_{i} }} = k_{4} \Delta tQ_{t} \mathop \sum \limits_{i = 1}^{n} \theta_{i} \frac{1}{{\eta_{i} }} = k_{4} \Delta tQ_{t} \frac{1}{{\eta_{o} }} $$
(2.22)
$$ \eta_{0} = \frac{1}{{\mathop \sum \nolimits_{i = 1}^{n} \theta_{i} \frac{1}{{\eta_{i} }}}} $$

where Wt is the energy consumption of the system, k4 is a constant, Δt is the temperature difference between the inlet and outlet fluids of the system, Wi is the energy consumption of the i-th boiler. Qt is the total flow rate of the system, Qi is the flow rate provided by the i-th boiler, n is the total number of boilers in the system. η0 is the overall energy efficiency of the system, ηi indicates the operating energy efficiency of the i-th boiler, θi indicates the load rate of the i-th boiler, the expression is

$$ Q_{t} = \mathop \sum \limits_{i = 1}^{n} Q_{i} $$
(2.23)
$$ \theta_{i} = \frac{{Q_{i} }}{{Q_{t} }} $$
$$ \mathop \sum \limits_{i = 1}^{n} \theta_{i} = 1 $$

For a fixed Qt, the higher the η0, the smaller the Wt.

2.3.4 Motion System

Many motion systems, such as high-speed trains, subway trains, electric vehicles, ships, long conveyors, and paper machines, may be driven by multiple motors.

  1. 1.

    High-Speed Trains and Subway Trains

It is driven by n motors to overcome friction, air resistance and weight potential energy. The speed of each carriage is the same. For a given vehicle speed V0, the power consumption of a high-speed train can be expressed as

$$ W_{t} = W_{0} \frac{1}{{\eta_{0} }} = k_{5} \mathop \sum \limits_{i = 1}^{n} \frac{{F_{i} V_{0} }}{{\eta_{i} }} = k_{5} V_{0} \mathop \sum \limits_{i = 1}^{n} \frac{{F_{i} }}{{\eta_{i} }} = k_{5} V_{0} F_{t} \mathop \sum \limits_{i = 1}^{n} \frac{{F_{i} }}{{F_{t} }} \cdot \frac{1}{{\eta_{i} }} = W_{0} \mathop \sum \limits_{i = 1}^{n} \theta_{i} \cdot \frac{1}{{\eta_{i} }} $$
(2.24)
$$ W_{0} = k_{5} F_{t} V_{0} $$
$$ \eta_{0} = \frac{1}{{\mathop \sum \nolimits_{i = 1}^{n} \theta_{i} \frac{1}{{\eta_{i} }}}} $$

where k5 is a constant, V0 is the speed of the high-speed train, Ft is the total traction force of the high-speed train, W0 is the ideal work, Wt is the total power consumption of the high-speed train. n is the total number of motors, and η0 is the overall energy efficiency of the high-speed train. ηi represents the operating energy efficiency of the i-th motor, Fi represents the traction force of the i-th motor, θi represents the load rate of the i-th motor, the expression is

$$ F = \mathop \sum \limits_{i = 1}^{n} F_{i} $$
(2.25)
$$ \theta_{i} = \frac{{F_{i} }}{F} $$
$$ \mathop \sum \limits_{i = 1}^{n} \theta_{i} = 1 $$

For a fixed Ft and V0, the higher the η0, the smaller the Wt.

V0 corresponds to the rotational speed n0 of the motor, and Ft corresponds to the total torque Mt of all the motors. The power consumption of high-speed trains can also be expressed as

$$ W_{t} = W_{0} \frac{1}{{\eta_{0} }} = k_{6} n_{0} \mathop \sum \limits_{i = 1}^{n} \frac{{M_{i} }}{{\eta_{i} }} = k_{6} n_{0} M_{t} \mathop \sum \limits_{i = 1}^{n} \frac{{M_{i} }}{{M_{t} }} \cdot \frac{1}{{\eta_{i} }} = W_{0} \mathop \sum \limits_{i = 1}^{n} \theta_{i} \cdot \frac{1}{{\eta_{i} }} $$
(2.26)
$$ W_{0} = k_{6} n_{0} M_{t} $$
$$ \eta_{0} = \frac{1}{{\mathop \sum \nolimits_{i = 1}^{n} \theta_{i} \frac{1}{{\eta_{i} }}}} $$

where k6 is a constant, n0 is the speed of each motor, the same, Mt is the total torque of all motors, Mi is the output torque of the i-th motor, θi is the load rate of the i-th motor, the expression is

$$ M_{t} = \mathop \sum \limits_{i = 1}^{n} M_{i} $$
(2.27)
$$ \theta_{i} = \frac{{M_{i} }}{{M_{t} }} $$
$$ \mathop \sum \limits_{i = 1}^{n} \theta_{i} = 1 $$
  1. 2.

    Electric Vehicles

It is driven by n motors to overcome friction, air resistance and weight potential energy. Each motor has the same speed. If it is driven by a motor, we can replace this motor with n small motors.

Similarly, for a given vehicle speed V0 (corresponding to the rotational speed n0 of the electric motors), the total tractive force Ft corresponds to the total torque Mt of all electric motors. The electric energy consumption of electric vehicles can also be expressed as

$$ W_{t} = W_{0} \frac{1}{{\eta_{0} }} = k_{6} n_{0} \mathop \sum \limits_{i = 1}^{n} \frac{{M_{i} }}{{\eta_{i} }} = k_{6} n_{0} M_{t} \mathop \sum \limits_{i = 1}^{n} \frac{{M_{i} }}{{M_{t} }} \cdot \frac{1}{{\eta_{i} }} = W_{0} \mathop \sum \limits_{i = 1}^{n} \theta_{i} \cdot \frac{1}{{\eta_{i} }} $$
(2.28)
$$ W_{0} = k_{6} n_{0} M_{t} $$
$$ \eta_{0} = \frac{1}{{\mathop \sum \nolimits_{i = 1}^{n} \theta_{i} \frac{1}{{\eta_{i} }}}} $$

where n0 is the speed of each motor, the same, W0 is the ideal work, Wt is the total electric energy consumption of the electric vehicle, Mt is the total torque of all motors, Mi is the output torque of the i-th motor, and η0 represents the electric vehicle’s overall energy efficiency, ηi represents the operating energy efficiency of the i-th motor, θi represents the load rate of the i-th motor, the expression is

$$ M_{t} = \mathop \sum \limits_{i = 1}^{n} M_{i} $$
(2.29)
$$ \theta_{i} = \frac{{M_{i} }}{{M_{t} }} $$
$$ \mathop \sum \limits_{i = 1}^{n} \theta_{i} = 1 $$

For a fixed Mt and n0, the higher the η0, the smaller the Wt.

  1. 3.

    Electric Boat

It is driven by n motors to overcome friction, water resistance and wind resistance. Each motor has the same speed. If it is driven by a motor, we can replace this motor with n small motors.

Similarly, the power consumption of an electrical boat can be expressed as

$$ W_{t} = W_{0} \frac{1}{{\eta_{0} }} = k_{6} n_{0} \mathop \sum \limits_{i = 1}^{n} \frac{{M_{i} }}{{\eta_{i} }} = k_{6} n_{0} M_{t} \mathop \sum \limits_{i = 1}^{n} \frac{{M_{i} }}{{M_{t} }} \cdot \frac{1}{{\eta_{i} }} = W_{0} \mathop \sum \limits_{i = 1}^{n} \theta_{i} \cdot \frac{1}{{\eta_{i} }} $$
(2.30)
$$ W_{0} = k_{6} n_{0} M_{t} $$
$$ \eta_{0} = \frac{1}{{\mathop \sum \nolimits_{i = 1}^{n} \theta_{i} \frac{1}{{\eta_{i} }}}} $$

where n0 is the speed of each motor, W0 is the ideal work, Wt is the total power consumption of the ship, Mt is the total torque of all motors, Mi is the output torque of the i-th motor. η0 is the overall energy efficiency of the ship, ηi represents the operating efficiency of the i-th motor, θi represents the load rate of the i-th motor, the expression is

$$ M_{t} = \mathop \sum \limits_{i = 1}^{n} M_{i} $$
(2.31)
$$ \theta_{i} = \frac{{M_{i} }}{{M_{t} }} $$
$$ \mathop \sum \limits_{i = 1}^{n} \theta_{i} = 1 $$

For a fixed Mt and n0, the higher the η0, the smaller the Wt.

  1. 4.

    Conveyor Belt and Scraper Conveyor

In order to make the conveyor belt evenly stressed, it is generally necessary to use multiple motors to drive a long conveyor belt, this type of optimization does not present the problem of choosing the best number of motors to run.

There is a forming wire in the wire section of the paper machine, which is jointly driven by n motors. The speed of each motor must be the same (regardless of the reducer), only the torque of each motor is adjusted. This situation is similar with the high-speed train above.

There are many long scraper conveyors in coal mines, usually driven by two motors at both ends. The speed of each motor must be the same, just adjust the torque of each motor so that the force is even. This situation is also similar with high-speed rail.

Ports have many long conveyor belts for transporting raw materials. For uniform force, one belt is driven by two motors.

Similarly, the power consumption can be expressed as

$$ W_{t} = W_{0} \frac{1}{{\eta_{0} }} = k_{6} n_{0} \mathop \sum \limits_{i = 1}^{n} \frac{{M_{i} }}{{\eta_{i} }} = k_{6} n_{0} M_{t} \mathop \sum \limits_{i = 1}^{n} \frac{{M_{i} }}{{M_{t} }} \cdot \frac{1}{{\eta_{i} }} = W_{0} \mathop \sum \limits_{i = 1}^{n} \theta_{i} \cdot \frac{1}{{\eta_{i} }} $$
(2.32)
$$ W_{0} = k_{6} n_{0} M_{t} $$
$$ \eta_{0} = \frac{1}{{\mathop \sum \nolimits_{i = 1}^{n} \theta_{i} \frac{1}{{\eta_{i} }}}} $$

where n0 is the speed of each motor, which are the same as each other, W0 is the ideal work, Wt is the total power consumption of the belt. Mt is the total torque of all motors, Mi is the output torque of the i-th motor. η0 is the overall energy efficiency of the system. ηi is the operating energy efficiency of the i-th motor, θi is the load rate of the i-th motor, the expression is

$$ M_{t} = \mathop \sum \limits_{i = 1}^{n} M_{i} $$
(2.33)
$$ \theta_{i} = \frac{{M_{i} }}{{M_{t} }} $$
$$ \mathop \sum \limits_{i = 1}^{n} \theta_{i} = 1 $$

2.3.5 Manpower Scheduling

The manpower scheduling mentioned here is aimed at completing tasks while keeping everyone in the team relaxed and not left behind.

  1. 1.

    A Group of n People

There are n individuals in a military operation. Since one must eat even when not working, there is no question of choosing the optimal number of people. When the machine is turned off, it consumes no energy. In this respect, humans are different from machines. The total weight of the items to be carried is Qt, and Qi is the weight of the items carried by the i-th person. The total energy consumption is expressed as

$$ W_{t} = W_{0} \frac{1}{{\eta_{0} }} = k_{7} Q_{t} \mathop \sum \limits_{i = 1}^{n} \frac{{Q_{i} }}{{Q_{t} }} \cdot \frac{1}{{\eta_{i} }} = W_{0} \mathop \sum \limits_{i = 1}^{n} \theta_{i} \frac{1}{{\eta_{i} }} $$
(2.34)
$$ W_{0} = k_{7} Q_{t} $$
$$ \eta_{0} = \frac{1}{{\mathop \sum \nolimits_{i = 1}^{n} \theta_{i} \frac{1}{{\eta_{i} }}}} $$

where k7 is a constant, W0 represents the ideal work, Wt represents the total body energy consumption of the group, n represents the total number of people. η0 represents the overall energy efficiency of the group, ηi represents the operation energy efficiency of the i-th person, θi represents the i-th person’s load rate, expressed as

$$ Q_{t} = \mathop \sum \limits_{i = 1}^{n} Q_{i} $$
(2.35)
$$ \theta_{i} = \frac{{Q_{i} }}{{Q_{t} }} $$
$$ \mathop \sum \limits_{i = 1}^{n} \theta_{i} = 1 $$

For a fixed Qt, the higher the η0, the smaller the Wt.

  1. 2.

    Teams of Two

Two people use a pole to carry an object from point A to point B. The total weight of things to be moved is Qt, and Qi is the weight of things carried by the i-th person. The total energy consumption is expressed as

$$ W_{t} = W_{0} \frac{1}{{\eta_{0} }} = k_{7} \mathop \sum \limits_{i = 1}^{2} Q_{i} \frac{1}{{\eta_{i} }} = k_{7} Q_{t} \mathop \sum \limits_{i = 1}^{2} \frac{{Q_{i} }}{{Q_{t} }} \cdot \frac{1}{{\eta_{i} }} = W_{0} \mathop \sum \limits_{i = 1}^{2} \theta_{i} \frac{1}{{\eta_{i} }} $$
(2.36)
$$ W_{0} = k_{7} Q_{t} $$
$$ \eta_{0} = \frac{1}{{\mathop \sum \nolimits_{i = 1}^{2} \theta_{i} \frac{1}{{\eta_{i} }}}} $$

where k7 is a constant, W0 represents the ideal work, Wt is the total energy consumption of the group, n represents the total number of people. η0 represents the overall efficiency of the group, ηi represents the work efficiency of the i-th individual, and θi represents the load rate of the i-th individual, expressed as

$$ Q_{t} = \mathop \sum \limits_{i = 1}^{2} Q_{i} $$
(2.37)
$$ \theta_{i} = \frac{{Q_{i} }}{{Q_{t} }} $$
$$ \mathop \sum \limits_{i = 1}^{2} \theta_{i} = 1 $$

For fixed Qt and the road conditions of two points A and B, the higher η0 is, the smaller Wt is.

2.4 Overall Energy Efficiency and Weighted Energy Efficiency

The η0 represents the overall operating energy efficiency of the system.

In power stations, it is expressed as

$$ \eta_{0} = \mathop \sum \limits_{i = 1}^{n} \theta_{i} \eta_{i} $$
(2.38)

In the energy consumption system, it can be expressed as

$$ \eta_{0} = \frac{1}{{\mathop \sum \nolimits_{i = 1}^{n} \theta_{i} \frac{1}{{\eta_{i} }}}} $$
(2.39)

The η0 is also called the weighted energy efficiency of the system.

2.5 Efficiency Function

In nature, both humans and artificial machines have efficiency functions. In general, the shape of the efficiency function η is shown in Fig. 2.1.

Fig. 2.1
A graph of eta versus beta. The dashed line remains horizontal and then follow a vertical decline. The solid line begins at around 0 and then follow a concave slope.

Efficiency function η

Full size image

In Fig. 2.1, β is the load, and η is the efficiency function. ηe is the efficiency at βe, which is the maximum efficiency, βm is the maximum load, and

$$ \beta_{m} \ge \beta \ge 0 $$
(2.40)
$$ \eta \left( \beta \right) \ge 0 $$
$$ \eta \left( 0 \right) = 0 $$
$$ \eta^{\prime \prime } \left( \beta \right) < 0 $$

Efficiency can be expressed as

$$ \eta \left( \beta \right) = \beta \mathop \sum \limits_{i = 1}^{\infty } \left( {a_{i} \beta^{i - 1} } \right) $$
(2.41)

In the energy consumption system, when β = 0, the energy consumption Wt is not necessarily zero. Wt = W0/η = 0/0, which can be derived by the L’Hopital’s Law.

The efficiency function is a concave function.

2.6 Unification of Optimization of Power Generation and Energy Consumption

The total power generation Wt of the generalized power station can also be expressed as

$$ W_{t} = \mathop \sum \limits_{i = 1}^{n} W_{i} $$
(2.42)

where n is the total number of generators, Wt is the total power generation of the power station, and Wi is the power generation of the i-th generator.

The total input energy Wt0 consumed by the generalized power station is expressed as

$$ W_{t0} = \mathop \sum \limits_{i = 1}^{n} W_{i0} = \mathop \sum \limits_{i = 1}^{n} \frac{{W_{i} }}{{\eta_{i} }} = W_{t} \mathop \sum \limits_{i = 1}^{n} \frac{{W_{i} }}{{W_{t} }}\frac{1}{{\eta_{i} }} = W_{t} \mathop \sum \limits_{i = 1}^{n} \theta_{i} \frac{1}{{\eta_{i} }} = W_{t} \frac{1}{{\eta_{0} }} $$
(2.43)
$$ \theta_{i} = \frac{{W_{i} }}{{W_{t} }} $$
$$ \eta_{0} = \frac{1}{{\mathop \sum \nolimits_{i = 1}^{n} \theta_{i} \frac{1}{{\eta_{i} }}}} $$
$$ \mathop \sum \limits_{i = 1}^{n} \theta_{i} = 1 $$

where Wi0 represents the energy consumed by the i-th generator, η0 represents the overall energy efficiency of generalized power station. θi represents the load rate of the i-th generator, and ηi represents the operating energy efficiency of the i-th generator.

For the maximization problem of the total power generation Wt

$$ \max W_{t} $$
(2.44)

equivalent to the minimization problem of the total consumed energy Wt0

$$ \min W_{t0} $$
(2.45)

The optimization of these systems is to solve two problems, one is to determine the optimal number of operating units, and the other is how to distribute the load of each unit.

2.7 Not Working Is Different from Shutting Down

Regarding equipment optimization and manpower optimization, there are several interesting phenomena that need to be noticed. When everyone travels together, it is the most labor-saving arrangement to carry items evenly, and there is no problem of changing the number of people. If you arrange a group of people to complete a job, you need to consider the number of people, because even if the people sent out do not work, they still need to eat. This is the difference between people and machines. When the machine is turned off, it will no longer consume energy; in addition, the machine does not work (such as the pressure-holding operation when the water supply system has zero flow) is not the same as the machine does not work when it is turned off. Energy is still consumed when the machine is on and not working.