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ES2C6 Week 4 Examples - WECdwCWC

WECdwCWC

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Electromechanical System Design (ES2C6)

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The University of Warwick

Academic year: 2023/2024

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Zaki Ibrahim Johare

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ES2C6 Week 4 Examples: Assessing,

sizing and analysing an electromechanical

system

Lecture 4: Rotary Transmission Systems (with efficiency from

lecture 4).

Question 1: A simple gear chain consisting of two meshed gears is shown below (A), along with the

number of teeth on the gears, the efficiency of the gearbox, and the input angular velocity and

torque. Find the input power, output power, output angular velocity in rad/s and output torque.

1

2

15

60 1000 rpm

0 Nm

0. i

N

T





=

Solution 1: This is similar to an example from lecture 4, but with the added complication of an

efficiency, and the input velocity is given in rpm. The first step is to convert the input speed to rad/s:

2 1000 104 rad/s

60

i 

 = =

Now the input and output power can be calculated as:

104 0 1 W

0 1 0 W

i i i o i

P T

P P





= =  =

Remembering that the efficiency influences the torque transfer, not the velocity transfer, the output

torque and velocity can be found as:

60 0 0 0 Nm

15 1 15

104 26 Nm

60

o i o i

T T N

N



 

= =   =

= =  =

Question 2: A compound gear chain using intermediate shafts is below, along with the number of

teeth on each gear, the efficiency between each driven and driver gear connection, and the input

velocity and torque. Calculate the product of the teeth on the driven gears, and the product of the

teeth on the driver gears. Use this to find the total gear ratio. Calculate the total efficiency of the

system. Find the output velocity and toque, and the velocity and torque of shaft A.

1 2 3 4 5 50 10 40 N N N N = = = = 5 6 8 20 400 rad/s 0 Nm i i N N T  = = = = 1 2 3 4 5 6 0. 0. 0. N N N N N N    → → → = = = Solution 2: This is similar to an example in Lecture 4 but with added efficiencies. Here, an efficiency is given for each of the driven/driver gear pairs. First, the gear ratio is found by dividing the product of the teeth on the driven gears by the product of the teeth on the driver gears:

####### ( ) ( )

2 4 6 3 1 3 5 50 40 20 1 1 100 5 10 8 N N N n N N N N = − = − = − The total efficiency of the whole compound gear chain is the product of the efficiencies of all the efficiencies between each driven and driver gear connection: 1 2 3 4 5 6 3 tot =  N →N  N →N  N →N = 0 =0. Note that this is a disadvantage of a compound gear chain, i. they can have lower efficiencies that single stage gear chains. Now the output velocity and torque can be found: 400 4 rad/s 100 0 0 100 4 Nm i o o tot i N T T N    = = − = − = = −   = − Finally, the velocity and torque of shaft A can be found. 1 2 5 400 40 rad/s 50 A i N N  = −  = − = 1 2 2 1 50 0. 0 0 Nm 5 A i N N N T T N = −  → = −  = − Question 3: A worm and wheel system is shown below. The worm has 3 starts, an axial pitch of 4𝑚𝑚, and a diameter of 15𝑚𝑚. The number of teeth on the wheel is 90 , and the efficiency is 0. If the worm is rotating at an angular velocity of 250 𝑟𝑎𝑑/𝑠 with a torque of 0 𝑁𝑚, what is the angular velocity and torque of the wheel? Find the value of tan 𝜆. Solution 3: The lead and gear ratio must first be calculated as follows:

( )

3 L N p 1 a 3 4 10 0 m = =   − = N = N 2 / N 1 = 90 3 = 30

maximum velocity and acceleration have been found. The parameters of a potential gearbox to be used with the system are given. Assess whether the gearbox is suitable for this system: Simple Geared Transmission System

####### ( )

2 L 0 kgm 0 Nm B 0 Nm/ rad/s L L J T = = = Motion Profile Applied to the Load max 2 1 s 2 s 1 s 0 s 1 s 3 s 5 s 4 rad/s 3 rad/s ACC S DEC a b c d t t t t t t t   = = = = = = = = = Gearbox Datasheet Values 12 5 N x = = max continuous 100 rad/s 50 rad/s   = = max continuous 0 Nm 0 Nm T T = = Solution 5: This problem relates to the selection of a suitable gearbox for a transmission system, given the system’s parameters and motion profile applied to the load. The values of 𝑇𝑒𝑞𝑢𝑖𝑣 and 𝜔𝑚𝑒𝑎𝑛 must be found, whose equations are given below: x x x x ACC ACC ACC S S S DEC DEC DEC ACC ACC S S DEC DEC equiv mean ACC ACC S S DEC DEC ACC S DEC t T t T t T t t t T t t t t t t          

- - - = =
- - - Each quantity will now be found, which is relatively simple for a trapezoidal motion profile with constant torque load:

(  ) ( ) ( )

(  ) ( )( )

(  ) ( ) ( )

1 1 , 3 1 2 rad/s 2 2 , 3 1 4 rad/s 1 1 , 3 1 2 rad/s 2 2 trap a b ACC ACC trap b c S b trap c d DEC DEC t t t t t t t t t t t t           = = = =  = = = =  = = = =

3 rad/s 2 0 rad/s 2 3 rad/s 2  ACC = = S =  DEC= − = −

####### ( ) ( ) ( ) ( )

####### ( )( ) ( ) ( )

####### ( ) ( ) ( ) ( )

2, 2 2, 2 2, 2 0 3 0 2 0 0 Nm 0 0 0 4 0 0 Nm 0 3 0 2 0 0 Nm ACC L ACC ACC L S L S S L DEC L DEC DEC L T J B T T J B T T J B T       = + + = + + = = + + = + + = = + + = − + + = Now the values of 𝑇𝑒𝑞𝑢𝑖𝑣 and 𝜔𝑚𝑒𝑎𝑛 can be found

####### ( )( )( ) ( )( )( ) ( )( )( )

####### ( )( ) ( )( ) ( )( )

5 5 5 5 2 1 0 4 2 0 2 1 0. 0 Nm 2 1 4 2 2 1. x x x x ACC ACC ACC S S S DEC DEC DEC equiv ACC ACC S S DEC DEC t T t T t T T t t t      

- =
- = =

####### ( )( ) ( )( ) ( )( )

####### ( ) ( ) ( )

2 1 4 2 2 1. 3 rad/s 1 2 1. ACC ACC S S DEC DEC mean ACC S DEC t t t t t t    

- =
- = =
- Now follow the process for choosing (or in this case assessing the suitability of) a gearbox.

Choose a gearbox whose rated torque is larger than 𝑇𝑒𝑞𝑢𝑖𝑣.
The gearbox meets the requirement
Ensure that the maximum torque is greater than the max torque (generally at the end of 𝑇𝑎𝑐𝑐.)
Because there is no damping and the load torque is constant, the max torque occurs through 𝑡𝐴𝑐𝑐 and is thus 0 𝑁𝑚.
So, the gearbox meets the requirement.
Divide the maximum speed of the gearbox by 𝜔𝑆 to determine maximum possible gear ratio.
𝜔𝑚𝑎𝑥 ⁄𝜔 𝑆= 100 4⁄ = 23.
So the gearbox meets the requirement
Select a standard gear ratio 𝑁 that is below this value.
In this case we are given N = 12;
Calculate the input mean velocity as N × 𝜔𝑚𝑒𝑎𝑛.
𝜔𝑖,𝑚𝑒𝑎𝑛 = N × 𝜔𝑚𝑒𝑎𝑛 = 12 × 3 = 37 𝑟𝑎𝑑/𝑠
So the gearbox meets the requirement
Calculate the input peak velocity as N × 𝜔𝑆.
𝜔𝑖,𝑝𝑒𝑎𝑘 = N × 𝜔𝑆 = 12 × 4 = 50 𝑟𝑎𝑑/𝑠
So the gearbox meets the requirement

####### ( )

2 2 1 0 1 3 10 3 Nm/ rad/s 0 4 L L M B B  N − →     =   =   =     

####### ( )

0 3 10 3 0 Nm/ rad/s BT = +  − = Finally, the torque load referred to the motor is 1 0 1 0 Nm 0 4 L L M T T  N →     =   =  =     Question 7: For the lead and screw system shown below along with its parameters, find the total inertia, damping and torque load referred to the motor. Assume the rotation of the motor causes a positive velocity linear velocity of the load.

####### ( )

2 2 0 kgm 0 kgm 0 Nm/ rad/s 10 kg 1 kg M S M L N J J B M M = = = = = 0 m/rad 30 degrees 0. 5 N 0. L P L F    = = = = = Solution 7: The total inertia referred to the motor is: J tot = J M + J S +J L →M

####### ( ) 2 ( 10 1 0) 2 3 2

1 10 kgm 0. L N L M M M J L  − →

- = = =  3 2 J tot J M JS J L M 0 0 1 10 0 kgm −  = + + → = + +  = The total damping is just 𝐵𝑀 For the load torque referred to the motor, the forces applied to the load must be found:

####### ( ) ( ) ( )

L M 5 N (given) cos sgn( ) 0 10 1 9 cos(30)(1) 4 N sin 10 1 9 sin 30 53 N 5 53 4. T 0 1 Nm 0. P f L L N L g L N P f g F F M M g V F M M g F F F L     → = = + =  +   = = + = +   =  + +   + +   =   =   =     Question 8: For the conveyor belt system shown below along with its parameters, find the total inertia, damping and torque load referred to the motor. Assume the rotation of the motor causes a positive velocity linear velocity of the load.

####### ( )

2 2 (1,2,3) (1,2,3) ( , ) 0 kgm 0, 0, 0 kgm 0 Nm/(rad/s) 5 N 0, 0, 0 m 1,10 kg 30 degrees 0. 0. M P M p P B L L J J B F D M    = = = = = = = = = Solution 8: The total inertia referred to the motor is:

####### ( ) ( )

1 2 3 2 2 1 2 2 2 2 3 1 3 2 3 3 2 2 1 3 2 0 0. 0 kgm 0 0. 0 0. 9 10 kgm 0 0. 1 10 0. 8 10 kgm 2 0 2 T M P P M P M L M P P P M P P P P M P B L P L M T J J J J J J J D J D J D J D M M D J J    → → → → − → − → = + + + +     =   =   =         =   =   =     

  +   =   =   =       = 0 + 0 + 0 + 9  10 − 3 + 8  10 − 3 =0 kgm 2 The total damping is just 𝐵𝑀 For the load torque referred to the motor, the forces applied to the load must be found. For simplicity, the forces are the same as those found in question 7. However, their conversion to torque applied to the motor is different:

####### Lecture 4: Efficiency and RMS Torque

We have been applying the efficiency to the transmission systems throughout this examples sheet, so for this lecture there will just be a simple example of finding the RMS torque. Question 10: A transmission system has a motor torque load time as shown below. Calculate the RMS motor torque: Solution 10: In this example, we are given all of the values needed to calculate the RMS torque using the following equation:

( ) ( )

( )( ) ( )( ) ( ( ) )( )

2 2 2 2 2 , , , , , 2 2 2 2 2 1 1 2 2 1 1 4 5 0 5 5 7 4 0. 2 2 5. 5 Nm M ACC P ACC ACC P S S M DEC P DEC DEC RMS M T T t T t T T t T t

- - - =
- - − += End of Examples Mark Dooner 2021

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