Calculation and Selection of Motor Parameters in Automation System

Abstract

Motors are the most important executive components in electrical automation equipment, including AC motors, DC motors, AC servo motors, DC servo motors, stepping motors, etc. In the design of the automation system, the selection of the motor is a frequently encountered problem. You need to calculate the parameters of the required motor. Novices often do not know where to start. Below we give a simple calculation and selection method.

Motors are the most important executive components in electrical automation equipment, including AC motors, DC motors, AC servo motors, DC servo motors, stepping motors, etc. In the design of the automation system, the selection of the motor is a frequently encountered problem. You need to calculate the parameters of the required motor. Novices often do not know where to start. Below we give a simple calculation and selection method.

11.1 Determination of the Rated Torque N_e of the Motor

The selection of the motor is mainly based on the calculation of the rated torque, and the rated torque Ne of the motor should meet:

$${N}_{e}>{N}_{w}+{N}_{f}$$

(11.1)

Among them, N_W is the load torque, and N_f is the resistance torque of the equipment.

The load torque N_w includes the constant speed load torque N_c and the variable speed load torque N_ad. the variable speed load torque N_ad includes the acceleration load torque N_a and the deceleration load torque N_d. The variable speed load torque N_ad takes the maximum value of N_a and N_d.

$$ \begin{gathered} N_{w} = N_{c} + N_{ad} \hfill \\ N_{ad} = \max (N_{a} ,N_{d} ) \hfill \\ \end{gathered} $$

(11.2)

The equipment resistance torque N_f includes the equipment static resistance torque N_fs and the equipment motion resistance torque N_fw. Generally, the equipment static resistance torque N_fs is greater than the equipment movement resistance torque N_fw, and the resistance moment of the equipment is considered based on the static resistance moment of the equipment.

$${N}_{f}={N}_{fs.}$$

(11.3)

• 1. Calculation and determination of resistance torque

For the improvement of existing equipment, it is obtained by experimental methods. If it is new equipment, according to past experience or references, if no relevant data can be found, the static resistance torque of the equipment must be obtained by calculation methods. It is recommended that novices do not continue because the calculation process is complicated and the calculated results may not be correct. We suggest that if readers are inexperienced, they should not consider the resistance moment first, only the load moment, and then increase margin coefficient, which will make it easier to get started.

When obtaining the static resistance torque of the equipment by experiment, first remove the coupling between the motor and the connecting shaft. Three simple measurement methods are as follows:

• (1) Use a torque wrench (or torque electric wrench) to rotate the connecting shaft, and the torque value that makes the connecting shaft just rotate is the resistance torque value of the equipment. Write down the torque reading. The torque wrench is shown in Fig. 11.1.

  Fig. 11.1

  

  Torque wrench

• (2) Tie one end of a rod (such as a piece of wood or iron rod) to the connecting shaft with a nylon cable tie (or wire), or clamp a large wrench on the connecting shaft, as shown in Fig. 11.2.

  Fig. 11.2

  

  Test resistance torque

The rod is placed horizontally and perpendicular to the connecting shaft. The weight of the rod is W₁ (kg, kilogram force). An object (such as several nuts, or small pieces) is hung on the other end of the rod. The length of the connecting shaft to the object is L (m, m), increase the weight of the object to W₂ (kg), until the connecting shaft starts to rotate. If there is a tension scale, it can also be replaced by a tension scale. Read the kilogram force W₂ of the tension scale when the connection shaft just starts to rotate the conversion of kg into N needs to be multiplied by the acceleration of gravity g, then the resistance moment N_f (N · m) is equal to

$${N}_{f}={W}_{2}*9.81*L+{W}_{1}*9.81*L/2$$

(11.4)

• (3) If the method of fixing a rod on the connecting shaft is limited, a binding rope (or material tape) can also be wound on the connecting shaft. The radius of the connecting shaft is R (m), and the binding rope (or material tape) The other end is connected to a tension scale, as shown in Fig. 11.3.

  Fig. 11.3

  

  Test resistance torque

Pull the binding rope (or material belt) hard, and record the kilogram force reading W₂ (kg) of the tension scale when the connecting shaft starts to rotate, then the resistance torque N_f (N · m) is equal to

$${N}_{f}={W}_{2}*9.81*R$$

(11.5)

If it is inconvenient to remove the coupling between the motor and the connecting shaft, or it is inconvenient to fix the rod (or wrap the packing rope) on the connecting shaft. It is necessary to consider other shafts (connected with the connecting shaft) to measure. The transmission mode and the transmission ratio will affect the calculation of resistance torque, which will be explained later when we talk about load torque measurement.

• 2. Calculation and determination of constant speed load torque

  • (1) For rotating loads directly driven by the connecting shaft;

    • (1) If it is cutting head of cutting machine tool or digging head of mining excavation equipment, etc., as shown in Fig. 11.4.

      Fig. 11.4

      

      Cutting torque

The constant speed load torque N_c is equal to the cutting force F multiplied by the cutting head radius R₁.

$${N}_{c}=F*{R}_{1}$$

(11.6)

• (2) If it is a printing machine, rotary die-cutting machine or paper machine, etc., as shown in Fig. 11.5.

  Fig. 11.5

  

  Uniform speed load torque

The total pulling force F₁ minus the total pulling force F₂, the constant speed load torque N_c is

$${N}_{c}=({F}_{1}-{F}_{2})*{R}_{1}$$

(11.7)

where F₁ is the sum of the tensions of the various strips passing through the work roll 1, and the force opposite to the rotation direction of the coupling shaft. F₂ is the total amount of tension of strips and scraps after processing.

The force relationship in gear transmission or rubber roller transmission is shown in Fig. 11.6. The force at the edge of each gear is constant, but the torque of each gear is different.

Fig. 11.6

The relationship between gear transmission and force

• (2) For linear motion loads, such as CNC machine tools, machining centers, assembly machines, etc., the connecting shaft drives the ball screw, and the screw nut drives the worktable to perform linear motion. The pitch of the ball screw is h, as shown in Fig. 11.7.

  Fig. 11.7

  

  Linear motion load

The connecting shaft rotates once, and the movement angle is 2π radians. At the same time, the ball screw rotates once, and the screw nut drives the worktable to move linearly with a pitch h (m), and the thrust of the worktable is F (N). According to the principle of energy conservation, the uniform speed load moment N_c (N · m) is

$${N}_{c}=\frac{F*{h}_{1}}{2\pi }=F\frac{{h}_{1}}{2\pi }.$$

(11.8)

• 3. Calculation and determination of variable speed load torque

The calculation of acceleration torque and deceleration torque is the same, but the torque direction is opposite, so we only discuss the calculation of acceleration torque.

• (1) For rotating loads.

An object with a mass of m (kg) rotates around a radius r (m), its linear velocity is V (m/s), its angular velocity is ε, and its kinetic energy W is

$$W=\frac{1}{2}m{V}^{2}=\frac{1}{2}m(\varepsilon r{)}^{2}=\frac{1}{2}(m{r}^{2}){\varepsilon }^{2}=\frac{1}{2}J{\varepsilon }^{2}$$

(11.9)

The moment of inertia J(kg · m²) is defined as

$$J=m{r}^{2}$$

(11.10)

For a uniform disk with a diameter of R (m) and a total mass of m (kg) or a uniform circular shaft with a length of L, its moment of inertia J (kg · m²) is

$$J=\frac{1}{2}m{R}^{2}$$

(11.11)

An object with a moment of inertia J and an angular velocity ε passes through a reducer (or gear, synchronous belt), the reduction ratio is n, the angular velocity is n_ε, and the equivalent moment of inertia is J′. Regardless of friction, the kinetic energy on both sides is equal. Out of the equivalent moment of inertia J′ on the other side of the reducer,

$$ \begin{gathered} \frac{1}{2}J\varepsilon^{2} = \frac{1}{2}J^{\prime}(n\varepsilon )^{2} \hfill \\ J^{\prime} = \frac{J}{{n^{2} }} \hfill \\ \end{gathered} $$

(11.12)

All the k work rolls that are driven by the connecting shaft, the transmission ratio with the connecting shaft is n₁, n₂, … n_k. Their moments of inertia are J₁, J₂, … J_k. The moment of inertia converted to the connecting shaft on the motor side is J_1d, J_2d, … J_kd. The moment of inertia of the motor is J_m, then the total moment of inertia J converted to the motor side is

$$ J = J_{m} + J_{1d} + J_{2d} + + J_{kd} = J_{m} + \frac{{J_{1} }}{{n_{1}^{2} }} + \frac{{J_{2} }}{{n_{2}^{2} }} + \cdots + \frac{{J_{k} }}{{n_{k}^{2} }} $$

(11.13)

According to the speed requirements of the equipment, converted to the connecting shaft, the time for the angular velocity of the connecting shaft to accelerate from 0 to ε is t (s), and the torque required on the connecting shaft is N_ad (N · m), which is the speed change torque that the motor needs to provide.

$${N}_{ad}=F*r=(ma)r=m\frac{v}{t}r=m\frac{\varepsilon r}{t}r=m{r}^{2}\frac{\varepsilon }{t}=J\frac{\varepsilon }{t}$$

(11.14)

If there is a reducer between the connecting shaft and the motor, the connecting shaft torque N_ad (N · m) speed n₁ (rpm). After passing through the reducer with a reduction ratio of n, the torque converted to the motor side is N_ad'(N · m) and the speed is n₂ (rpm). Regardless of friction, the power on both sides of the reducer is equal, and the speed change torque $N_{ad^{\prime}}$ (N · m) on the motor side is obtained

$$ \begin{gathered} \frac{{N_{ad} n_{1} }}{9550} = \frac{{N_{{ad^{\prime}}} n_{2} }}{9550} \hfill \\ n_{2} = n_{1} n \hfill \\ N_{ad^{\prime}} = \frac{{N_{ad} }}{n}. \hfill \\ \end{gathered} $$

(11.15)

• (2) For linear motion loads.

The connecting shaft drives the ball screw, and the screw nut drives the worktable. The mass of the screw nut and the worktable is m1 (kg), and the worktable drags an object with a mass of m₂ (kg) to make a linear motion, as shown in Fig. 11.8.

Fig. 11.8

Linear motion load

The time for the object to accelerate from 0 to V_max (m/s) is t (s), and the force F (N) that the workbench needs to apply is

$$F=({m}_{1}+{m}_{2})a=({m}_{1}+{m}_{2})\frac{{v}_{{v}_{max}}}{t}$$

(11.16)

Regardless of the friction, the work done on the side of the connecting shaft and the side of the table is equal, and the acceleration torque N_ad1 converted to the side of the connecting shaft is given by

$${N}_{ad1}=F\frac{h}{2\pi }$$

(11.17)

The moment of inertia of the connecting shaft, lead screw and motor is J. When the speed of the object accelerates from 0 to V_max (m/s) within time t (s). The angular velocity of the connecting shaft and lead screw accelerates from 0 to ε_max (rad/s). The acceleration torque N_ad2 required on the connecting shaft

$$ \begin{gathered} \varepsilon_{max} = v_{max} \frac{2\pi }{h} \hfill \\ N_{ad2} = J\frac{{\varepsilon_{max} }}{t} \hfill \\ \end{gathered} $$

(11.18)

The total acceleration torque N_ad required by the motor

$${N}_{ad}={N}_{ad1}+{N}_{ad2.}$$

(11.19)

11.2 Determination of Motor Speed

The motor speed is determined according to the highest working speed required by the automation equipment.

For the load shown in Fig. 11.9, the maximum speed V (m/min) of the material belt corresponds to the maximum speed n_max (rpm) of the work roll 1.

Fig. 11.9

Tape load

The maximum speed of the connecting shaft and work roll 1 is the same, equal to

$$ \begin{gathered} v_{max} = n_{max} 2\pi R_{1} \hfill \\ n_{max} = \frac{{v_{max} }}{{2\pi R_{1} }} \hfill \\ \end{gathered} $$

(11.20)

For the load shown in Fig. 11.8, if the maximum linear motion velocity of the known object is V_max, the maximum angular velocity converted to the motor side of the connecting shaft is ε_max (rad/s). The maximum rotational speed is n_max (rpm).

$$ \begin{gathered} \varepsilon_{max} = v_{max} \frac{2\pi }{h} \hfill \\ n_{max} = \varepsilon_{max} \frac{60}{{2\pi }} \hfill \\ \end{gathered} $$

(11.21)

Select the rated speed of the motor n_e

$$ n_{e} \ge n_{\max } . $$

(11.22)

11.3 Determination of the Maximum Acceleration of the Servo Motor

According to the process conditions, general motion loads have an acceleration requirement. For example, the time required to increase the speed of the material belt (or paper) from 0 to the maximum speed V_max (m/s) does not exceed t (s). The acceleration of the material belt a (m/s²) for

$$a=\frac{{V}_{max}}{t}$$

(11.23)

Referring to the method in Sect. 11.2, calculate the maximum angular velocity ε_max (rad/s) of the motor connection shaft according to the maximum vehicle speed V_max (m/s). Then the angular acceleration ω (rad/s²) of the connection shaft is

$$\omega =\frac{{\varepsilon }_{max}}{t}$$

(11.24)

The maximum angular acceleration ω_max of the servo motor must satisfy

$${\omega }_{max}>\frac{{\varepsilon }_{max}}{t}$$

(11.25)

If a reducer is installed on the front of the servo motor, and the reduction ratio is n, the maximum angular acceleration ω_max of the servo motor must satisfy

$${\omega }_{max}>n\frac{{\varepsilon }_{max}}{t}$$

(11.26)

Otherwise, the driver of the servo motor will give an alarm during acceleration.

11.4 Determination of Motor Power

The power P(kw) of the motor is equal to

$$P\left(\text{kw}\right)=\frac{{N}_{e}\left(n\cdot m\right){n}_{e}\left(rpm\right)}{9550}.$$

(11.27)

11.5 Determination of Encoder Resolution

The selection of encoder resolution is determined according to the processing accuracy required by the automation equipment. For example, in Fig. 11.7, the positioning accuracy A of the material strip is required to be 0.1 mm. The diameter R1 of the work roll 1 is 50 mm, the circumference L of the work roll 1 is 2πR1 = 314.16 mm, and the connecting shaft of the work roll 1 is installed for the motor. The minimum resolution Re of the encoder should be L/A = 3140. In order to leave room for adjustment of the PID and feed-forward parameters of the positioning link, the resolution of the encoder should be increased by an order of magnitude, that is, greater than or equal to 31,400.

$${R}_{e}\ge 10\frac{L}{A}.$$

(11.28)

11.6 Servo Motor Inertia Ratio

Servo motor inertia ratio λ: refers to the ratio of the load's moment J of inertia to the motor's moment J_m of inertia.

$$\uplambda =\frac{J}{{J}_{m}}$$

(11.29)

For a fixed load, when λ increases, the response speed of the system will slow down and the control accuracy will decrease. λ is generally between 5:1 and 10:1. For some low-speed motion situations, λ can be selected between 3:1 and 5:1.

It is best to select this parameter according to the technical specifications provided by the servo motor manufacturer, otherwise it will seriously affect the control accuracy and dynamic response speed of the system.