## Designing Mechatronic Systems

Question:

Discuss About The Measures Public Library Networked Services?

This assignment focuses on the clutches and brakes as well as electric motors and control. Mechatronics design gives the skills and knowledge to the student and also it provides exposure to the designs process of the mechatronic system.it also gives the appreciation of the mechatronics component system like actuators, sensors, the principle fundamental operation of the components, their weaknesses and strength and their characteristics in terms of operation. This leads to the process of design of integrated interactive, system divided into a subsystem, component sizing and selection, and the inclusion of some considerations to the justified design that is qualified. The subject also gives the wider knowledge on the background of mechatronics, exposing the student to the current challenges of the art state (Devdas Shetty, 2013).

The following questions were answered according to the performance of the AC motor curve below. The curve represents an AC motor (Janschek, 2011). The table below illustrates AC motor speed at the power of 60Hz

Poles number |
Speed of the full load in (rpm) |
Synchronous speed in (rpm) |

2 |
3450 |
3600 |

4 |
1725 |
1800 |

6 |
1140 |
1200 |

8 |
850 |
900 |

10 |
690 |
720 |

12 |
575 |
600 |

The synchronized speed of the six pole motor type if the frequency is at 60Hz is calculated below:

From the table above, for the 6 pole motor at a speed of 60HZ, the full load speed is, N that is 1140 in the column where the pole number 6 is.

The velocity or the speed if the frequency is 60Hz is given by ? = (2πN/60)

N is the full load speed of 6 poles = 1140

Frequency = 60Hz

Π = 22/7

Hence the speed will be = (6*22/7*1140)/60

= 358.29rad/sec

The torque that a six pole motor type can exert ta a start load when it is rated as 559.3 + 0.PW is calculated below:

P = last two digits of the student id number:

559.3 + 0.05 = 559.35W

Torque is given by power/angular velocity/speed

Power is 559.35W and angular velocity is 358.29rad/sec

T= P/?

Torque= 595.35W/358.29rads/sec

Torque =1.56N.m

Hence the torque that a six motor type can exert to start a load when it is rated as 559.35 and power velocity or speed of 358.39 is 1.56N/M.

Therefore the motor produces a torque of 1.56N.m

Amount of the torque that a motor can develop to start a load is calculated as below:

To start a load, it is known that starting torque is 150% of the full load torque

## AC Motor Speed and Torque Calculations

T_{S}= 150 percent of the full load torque

= 150/100*1.56

= 2.34N.m

Therefore the torque that is developed by the motor to start the load is 2.34N.m

The breakdown torque for the motor is calculated below:

The breakdown torque is 350% of the full load torque

T_{b}=350%*torque of the full load

350/100*1.56

5.46Nm

The computation of the dimensions of the annular plate types that give the braking torque of 33.9N.m, spring provides a normal force of 1423.4 + 0.Q between the frictions surface, the friction coefficient of 0.25 and the stopping load of the 750rpm are done below (Kevin C. Craig, 2015)

Where Q is the student identification: 1423.4 + 0.05 = 1423.45N

The required mean radius of the annular plate’s type’s _{brakes} is calculated by:

R _{m=}T_{f}/fN where the coefficient friction is f and friction of the torque on the brake is T_{f and} N is the normal force.

Substitute 33.9N.m in for the T_{f}, 0.25 for f and 1423.45N for N

R_{m}=33.9/ (0.25*1423.45)

0.09526m which equals to 0.1 when rounded off.

Hence the annular plate type brake requires the mean radius (R_{m}), is 0.1m

Let inner and outer radius of the annular plate type brake be R_{1} and R_{0}

Take the (R_{o}/R_{i}) of the desired ratio of approximately 1.50

(R_{o}/R_{i})= 1.50

R_{o}= 1.50R_{i}

The mean radius which is R_{m} can be expressed in terms of the inner radius and outer radius of the annular plate type brake

R_{m}= (R_{o}+R_{i})/2

1.05R_{i} can be substituted for R_{o}

R_{m}= (1.50R_{i}+R_{i})/2

2.50R_{i}/2

1.25R_{i}

0.09526is substituted for the R_{m} in the equation so that the inner radius of annular plate type brake can be calculated

0.1m=1.25R_{i,} both sides are divided by 1.25 so that we can get R_{i}

R_{i}=0.09526m/1.25

=0.07621m

Hence the radius of the inner annular plate type brake is 0.07621m

0.07621m is substituted for R_{i }in the equation so that the outer radius of the annular plate type brake (Ro) is calculated as:

R_{o}=1.50*0.07621

0.1143

Hence the radius of the outer annular plate type brake is 0.1143m

The area of the friction surface (A) is calculated by:

A=π {(R_{o}^{2})-(R_{i}^{2})}, 0.1143 is substituted for R_{o }and 0.07621 is substituted for R_{i}

A=π {(0.1143^{2})-(0.07621^{2})}

A=22/7(0.0132-0.005808)

A=22/7*0.007392

A=0.0228m^{2,} hence the frictions surface area is 0.0228m^{2}

The frictional power absorbed can be calculated as follows:

P=T_{f }n/63000, the speed in rpm is represented by n. 750rpm is substituted for n and 33.9N.m is substituted for Tf.

## Annular Plate Type Brake Design

P= (33.9N.m) (750rpm)/63000

P=0.4036hp, hence the absorbed frictional power is 0.4036hp.

The wear ratio is calculated by (WR)

WR=P_{f}/A, 0.4036 is substituted for P_{f} and 0.0228m^{2} is substituted for A.

WR= 0.4036hp/0.0228m2

17.69hp/m^{2}, hence the wear ratio (WR) is acceptable because its values is too low.

A band brake is designed that exert the braking torque of 81.36 + 0.PN.m while the drum is slowed from 120rpm.

Where P is the student identification which is 505

81.36 + 0.05 = 81.41Nm

From the coefficient of friction table, friction materials like the design and woven asbestos are selected for 25psi maximum pressure (Pmax) approximately, the value of coefficient friction (f) is considered as 0.25. The trials of the values for the radius (r) of the brake drum are used as 6.0, in.2.5 width (w) and (?) as 210degrees.

The maximum band tension is calculated by:

Pi =p_{max}rw, for the maximum pressure, 25psi is substituted in, 6 in. for radius and 2.5in. For width

Pi= (25psi) (6in) (2.4in)

=375lb

The minimum bad tension can be calculated by (P_{2})

P_{2}=Pi/e^{f?}, 375 is substituted for Pi, 0.25 for f and 210 degrees for ?

P_{2}=375lb/ e^{ (0.25) (210)}

=357lb/e^{ (0.25) (210*πrad/180)}

^{ }=375lb/e^{ (0.25) (3.66rad)}

^{ }=150lb

The friction torque can be calculated by:

T_{f}= (P_{i}-P_{2}) r, 375lb is substituted in Pi, 150lb for P2 and 6in. for radius

Tf= (375-150) (6)

=225*6

=1350lb.in.

The friction torque value is 1350lb.in. Is very high so some of the adjustments can be made in radius to be 5.5in, width 2.0in and ? 210degrees.

Maximum band tension is calculated as (P_{i})

P_{i}=P_{max}rw, 25psi is substituted for the maximum pressure, 5.5 in. for the radius and 2in for the width.

Pi= (25psi) (2in) (5.5in)

=275lb

The minimum band tension is calculated by (P_{2})

P_{2}=P_{i}/e^{f?}, 275lb is substituted in Pi, 0.25 for f and 210 degrees for the ?

P_{2}=275lb/e^{ (0.25) (210)}

^{ } =275lb/e^{ (0.25) (210*πrad/180)}

^{ } =275lb/e^{ (0.25) (3.66rad)}

^{ }=110lb

The friction torque can be calculated by (T_{f})

T_{f}= (P_{i}-P_{2}) r, 275lb is substituted in P_{i}, 5.5 in radius and 110lb for the P_{2}

Tf= (275lb-110lb) (5.5in)

=908lb.in.

The friction torque value of 908lb.in is satisfactory.

The actuation force (W) is calculated by W=P_{2}^{a}/L

L is the lever length and a is the distance, the value of the distance is considered as 5.5in and the lever length is 12.0 in.110lb is substitute for P2, 12.0 in for L and 5.5 in for a

## Band Brake Design

W= (110lb)^{5.5in}./12in.

=50.4lb, hence the (W), actuation force is 50.4lb.

The friction power absorbed (P) is calculated as P=T_{f}n/63000, and the speed in rpm is n.

120rpm is substituted in n and 908lb in. is substituted in Tf

P= (908lb.in.) (120rpn)/63000

=108960/63000

=1.7295hp, hence the absorbed friction power is 1.7295hp.

The area projected of the brake shoe (A) is calculated by A=2πrw?/360.

5.5in. is substituted in radius. 2in. for width and 210 degrees for?

A=2π (5.5in.) (2in) 210/360

=40.3in^{2}, hence the area of the brake projected is 40.3in^{2}.

The wear ratio is calculated by, WR=Pf/A1.7295hp is substituted In Pf and 40,3in^{2} is substituted in A’

WR=1.7295hp/40.3in^{2}

=0.04292hp/in^{2}

Hence the wear ratio is 0.04292, service average is reasonable and also the band brake geometry is acceptable.

Conclusion

The paper covers the mechatronic systems and its examples like clutches and brakes and electric motor control. Clutches were developed by Thomson and the clutch brakes are used in many applications like medicine, robots, automation, material handling, office machine, and manufacturing industries. This assignment focuses on the clutches and brakes as well as electric motors and control. Mechatronics design gives the skills and knowledge to the student and also it provides exposure to the designs process of the mechatronic system.

It also gives the appreciation of the mechatronics component system like actuators, sensors, the principle fundamental operation of the components, their weaknesses and strength and their characteristics in terms of operation. This leads to the process of design of integrated interactive, system divided into a subsystem, component sizing and selection, and the inclusion of some considerations to the justified design that is qualified. The subject also gives the wider knowledge on the background of mechatronics, exposing a student to the current challenges of the art state.

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