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Home , Jerk (physics), Leadscrew

10/15/2016

MSD Mechanics Basics

Dr. Lutfi Al-Sharif Dr. Ashraf Saleem Dr. Ibrahim Al-Naimi

Introduction

• In preparation for the study of motor selection and drive, a quick review is provided of some basic concepts in mechanics that are essential for control systems. • Mechanics consists of three : , Statics and Dynamics. Kinematics involves the study of the motion of bodies in a frame of reference (without any consideration of the causing the motion). Statics involves the study of the equilibrium of bodies under the effect of a number of forces. Dynamics is the study of motion of bodies under the effect of a number of forces.

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Translational and Rotational Systems • When using motors and drives to control loads, in the majority of cases there are two types of motion: – Rectilinear translational. – Rotational. • For example a motor could be used to lift a load upwards via a box and a sheave. In this case the motor, its to the gearbox and the within the gearbox, as well as the sheave will be moving in a rotational manner, while the load will be moving in a rectilinear translational manner.

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Translational and Rotational Systems • These two systems are shown in a graphical format in the next Figure. In the first diagram, the m is moving in a straight line with non-zero , v(t), and non-zero , a(t). In the rotational diagram, an inertia I is rotating around a central axis at a non-zero angular speed of ω(t) and non-zero α(t).

Translational and Rotational Systems

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Translational and Rotational Systems • The table below shows a comparison of a number of quantities between the two systems (rectilinear translational, rotational). This table is useful in two aspects: – It provides a clear description of the analogies between the translational and rotational systems. – Any equation that applies to the translational system (that we are familiar with) can be translated into the rotational system just by replacing the suitable variables as shown in the table.

Translational and Rotational Systems

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Translational and Rotational Systems • The of acceleration in the rotational system is called the jerk and this is an important quantity in the passenger transport systems such as lifts () and trains, as they are considered to be an indicator of passenger comfort. The of acceleration is called the jounce (or snap). This is used in fairground ride design. Although these two quantities do have equivalents in the rotational system, they are not used in practice and thus have not been included in the table.

Rectilinear Translational Motion

• Let us take a body that is moving under constant

acceleration between the ti (initial time) and tf (final time). The speed time relationship (speed profile) for such a body is shown in following Figure. Notice that the shaded under the curve is in fact the travelled.

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Rectilinear Translational Motion

Rectilinear Translational Motion

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Rotational Motion

The Effect of a Mass on a Shaft

• In order to accelerate a mass in a linear horizontal direction, we need to apply a as follows:

• In order to accelerate the mass in a vertical direction upwards, the force needs to first overcome gravity (the of the mass), as follows:

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The Effect of a Mass on a Shaft • This represents the tension in the (or the rope) that is pulling the mass upwards.

The Effect of a Mass on a Shaft • If this belt (or rope) is wound over a pulley (or sheave) then must be supplied to the pulley (or sheave) in order to achieve this acceleration. Multiplying both sides by the radius of the pulley (or sheave) gives:

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The Effect of a Mass on a Shaft

• As can be seen from the last equation above the mass suspended from the belt (or rope) has two different effects on the rotational shafts as follows: – A torque at the shaft that is subtracted in this case from the motor torque, leaving a net torque that can be to accelerate the system. This torque is referred to as an out of balance torque meaning that it is a torque that is not balanced by another torque. It is equal to the product of the weight of the mass (m·g) multiplied by the radius of the pulley (or sheave).

The Effect of a Mass on a Shaft

• An inertia I that is equal to the mass multiplied by the square of the radius of the pulley (or sheave). We could think of this as if the mass was removed and replaced by a flywheel fitted to the pulley (or sheave) shaft, the value of which is I.

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Motor Lifting a Load • An example of a mechatronics system is a hoisting application where a motor is used to lift and lower a load (e.g., crane, lift). Let us take the case where a motor is lifting a load vertically against gravity. For simplicity we have assumed that the motor does not require a gearbox (i.e., that the motor speed is suitable). The geared case is discussed later in this Chapter. The motor shaft has certain inertia and is connected to the load via a pulley, whereby the load is connected via a belt. [an alternative to the pulley and belt could be a sheave and a rope, or a and a chain].

Motor Lifting a Load

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Motor Lifting a Load

• The diameter of the sheave is ds. A load of mass m is lifted by the pulley (or sheave) via a belt (or ropes). The weight of the mass m is producing a force in the belt (or rope) of its weight (m·g). The

torque from the motor is Td (where the subscript d is used to denote the drive). The torque caused by the load is . If we assume that the shaft

is subject to a and windage torque of Tfw, then the total torque acting on the shaft is:

Motor Lifting a Load • The total inertia at the shaft is the sum of the inertia of the shaft (including the pulley, sheave or sprocket) and the inertia caused by the load m which is • The total inertia is:

• Combining the two equations, gives the important result:

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Motor Lifting a Load

• Obviously this notation assumes that the load is being accelerated in the upward direction and thus the out of balance torque caused by the mass is negative. Where the drive torque is less than the torque caused by the weight of the load, the resultant acceleration will be negative, hence the load will accelerate downwards.

Motor Lifting a Load

• Example (1): A shaft is rotating at a speed of 1375 rpm. The effect of friction and windage (air resistance) is equivalent to a torque of magnitude 3 N.m (and can be assumed to be speed independent for simplicity). The shaft has a of 0.25 kg m2. How long will it take the shaft to come to a standstill?

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Motor Lifting a Load

• Example (2): A motor has a moment of inertia of 1.2 kg m2, and can provide a torque of 83 Nm, and has a top speed of 1450 rpm. If it is fitted with a pulley that has a diameter of 600 mm and the pulley is fitted with a rope to lift a mass of 20 kg. Calculate the time it will take for it to achieve 1450 rpm from standstill (assuming constant torque from standstill to top speed) and what this equates to in terms of linear speed for the load. Ignore the mass of the rope for simplicity.

The Effect of Gearing • When a gearbox is fitted between the motor and the load, it has an effect on the torque, speed and inertia. In general, a gearbox is used to match the drive to the load. Most gearboxes are used as reduction gearbox (i.e., to reduce the rotational speed). Thus the speed will be reduced, and the torque will be increased. This is very useful in practice; the rotational speed of most motors is too high to directly drive the load, and the torque is too low. The gearbox matches the drive to the load.

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The Effect of Gearing • A step-down gearbox reduces the rotational speed from the drive and increases the torque. We shall refer to the gearbox reduction ratio as

rg. We shall refer to the drive side shaft as the high speed shaft (H.S.S.) and the load side shaft of the gearbox as the low speed shaft (L.S.S.). • In an ideal world, the efficiency of the gearbox would be 100% and hence no loss of will result from the conversion. Under such a scenario:

The Effect of Gearing • By definition the relationship between the input speed and the output speed is:

But

Thus

showing the increase in the output torque.

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The Effect of Gearing • In reality, the efficiency will not be 100%. Denoting the forward efficiency of the gearbox as

ηf , we can rewrite the equations above with the effect of the efficiency as follows (note that the efficiency does not affect the speed relationships, as these remain unchanged; the efficiency will affect the torque and the power):

The Effect of Gearing • The relationship between the remains unchanged:

• The equations above show clearly how the efficiency affects the value of the low speed shaft torque but not the low speed shaft speed.

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The Effect of Gearing

• The reason for using the forward efficiency above is that in certain gearboxes the forward and reverse efficiencies are unequal. The term ‘forward’ and ‘reverse’ refer to the direction of flow of the power. The power is flowing in the forward direction if it is flowing from the high speed shaft to the low speed shaft; the power is considered to be flowing in reverse if it is flowing from the low speed shaft to the high speed shaft.

The Effect of Gearing

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The Effect of Gearing

• The gearing will also have an effect on the inertia when it is ‘referred’ to the high speed shaft. At the beginning, the inertia of the gearbox itself should be considered in our calculation. The inertia when referred from the low speed shaft to high speed shaft is divided by the square of the

gearbox ratio, rg. When an inertia from the high speed shaft is referred to the low speed shaft it is multiplied by the square of the gearbox speed reduction ratio.

The Effect of Gearing • The reason for using the square of the gearing ratio (rather than just the gearing ratio) can be understood by looking at the principle of conservation of . Let us assume that we

have a flywheel on the low speed shaft of ILSS rotating at a speed ωLSS and we want to find its equivalent inertia at the HSS. We have a gearing

ratio of rg between the two shafts. Then the kinetic energy must be the same whether the flywheel is fitted on the HSS or the LSS.

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The Effect of Gearing

• So, if the efficiency of the gearbox is 100%, then:

The Effect of Gearing

• The practical implication of this result is that if we wish to add more inertia to a system it is better to fit the extra flywheel to the high speed shaft as it will appear much larger when referred to the low speed shaft. For this reason, flywheels are always in practice fitted to the high speed shaft (usually the motor shaft).

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The Effect of Gearing

Counter weighting • In many mechanical systems the concept of dead load and live load exists. Dead load is always present in the system and does not provide any useful contribution. An example of this is the mass of the car that transports passengers. In each journey we move the mass of the whole vehicle back and forth. This movement does not provide any useful contribution and is only necessary in order to carry the passenger and provide safety. Live load on the other hand is not always present, and the purpose is to move the live load from one location to another.

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Counter weighting

• The concept of counter weighting is to compensate for the dead load in the system. This can be done by adding an additional mass that compensates for some or all of the dead load in the system. In a geared hoisting system this is done by adding another mass on the other side of the pulley (or sheave or sprocket). As shown in the figure below, a second mass is fitted to the other side of the

pulley. The two will be denoted as m1 and m2.

Counter weighting

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General Equation for Geared Hoisting Systems

Friction and Traction

• When a force pushes two surfaces against each other, a frictional force develops that opposes the direction of motion of the body. The magnitude of the frictional force is equal to the product of the perpendicular force pushing the two surfaces against each other multiplied by the coefficient of friction of the two surfaces. This is shown in the following Figure. It is important to emphasise that the coefficient of friction is measured for each pair of surfaces.

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Friction and Traction

Friction and Traction

• Friction can be used to an advantage in transport by preventing tyres and wheels from slipping against the surface on which they are running. For example rubber against certain surfaces has a high coefficient of friction (1.0-2.0). In this case this phenomenon is called Traction, and in fact it works in this case in the same direction as the force moving the body (see the following Figure).

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Friction and Traction

Flywheels

• A flywheel is a wheel that is fitted to a shaft to increase its inertia. Flywheels are needed for two reasons: 1. Reducing shock and : A flywheel can act like a capacitor in an electrical circuit where it can filter out ripples in voltage. Flywheels can reduce the effect of mechanical shock.

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Flywheels

2. Storage of energy: A flywheel can be used to store energy to be used later on. For example, let us say that the electrical supply is sometimes lost for 2-3 , using a flywheel can continue supplying energy during that interruption. The kinetic energy (in ) stored in a flywheel is equal to:

Where I is the moment of inertia (mass) in kg m2 and ω is the angular in rad/s.

Flywheels

• The first Figure shows an example of flywheel that has been fitted to the high speed shaft of a motor-gearbox assembly that is driving a heavy duty escalator. The flywheel is in fact in two halves to facilitate fitting and removal. In some cases the flywheel is much smaller and in effect is only used as hand-wheel (to rotate the shaft when needed). An example from a lift machine is shown in the second Figure.

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Flywheels

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Flywheels

• Flywheels are used extensively on many drive systems especially where a variable speed drive has not been used (e.g., two speed and single speed lift installations). They help reduce the shock of initial starting, and the value of initial acceleration. They also reduce the variations between full load and no load.

Torque and Moment of Inertia Calculation • Even though the calculation of complex mechanical devices seems to be very complicated at first glance, the math behind it is very much straight forward as long as there is an understanding that most components are derived from basic shapes (mostly a cylinder). Also helpful is the knowledge that speed transmissions are mostly accomplished using two cylinders (tooth wheels, pulleys, etc.) and that any has an on acceleration, inertia and torque that is being reflected to the motor.

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Moment of Inertia Calculation • The most complex mechanical devices are calculated based on three “basic” components: 1. Solid Cylinder 2. Hollow Cylinder 3. Solid Rectangular

Moment of Inertia Calculation

• Inertia of a Solid cylinder

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Moment of Inertia Calculation

• Inertia of a hollow cylinder

Moment of Inertia Calculation

• Inertia of Solid Rectangular Pillar

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Moment of Inertia Calculation

Torque and Moment of Inertia Calculation • Disk • • Coupling • Gear/Gearbox • Belt Pulley • Conveyor • Leadscrew • Nip Roll • Rack and • Rotary Table

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Torque and Moment of Inertia Calculation • Disk: The disk is generally the most used mechanical component when it comes to inertia calculations. A disk can be used to resemble other mechanical components such as pulleys, , , gears, , etc. Naturally, the disk is nothing else but a solid (or hollow) cylinder and thus the equations for a disk and solid (hollow) cylinder are the same.

Disk

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Torque and Moment of Inertia Calculation • Chain Drive: The chain drive consists of two tooth wheels and a chain. In the following the disk on the motor side is called the driver sprocket, the disk on the load side is called the driven Sprocket. Both are calculated like a regular disk (cylinder) or, if necessary, a hollow cylinder.

Chain Drive

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Torque and Moment of Inertia Calculation • Chain Drive: - The transmission rate of the chain sprocket device is derived from either the sprocket diameters or the number of teeth on both sprockets.

Torque and Moment of Inertia Calculation • Chain Drive: - The total inertia of the chain sprocket device includes the driver sprocket inertia, the chain inertia reflected to the motor side and the inertia of the driven sprocket reflected to the motor side.

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Torque and Moment of Inertia Calculation • Coupling: The coupling is nothing else but a solid (or hollow) cylinder and thus the equations for a coupling and solid (hollow) cylinder are the same. Usually the coupling is calculated like a hollow cylinder.

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Torque and Moment of Inertia Calculation

• Gear/Gearbox: Gears/Gearbox are basically two tooth wheels where one wheel (motor side) drives the other one (load side). Both wheels are calculated like a regular disk (cylinder) or, if necessary, a hollow cylinder.

Gear/Gearbox

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Torque and Moment of Inertia Calculation • Gear/Gearbox: - The transmission rate of the gear device is derived from either the gear diameters or the number of teeth on both gears.

Torque and Moment of Inertia Calculation

• Gear/Gearbox: - The total inertia of the gear device includes the driver gear inertia and the inertia of the driven gear reflected to the motor side.

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Torque and Moment of Inertia Calculation

• Belt-Pulley: A Belt Pulley (or timing belt) consists of two disks (pulleys) and a belt. The disk on the motor side is called the motor side pulley, the disk on the load side is called the load side pulley. Both pulleys are calculated like a regular disk (cylinder) or, if necessary, a hollow cylinder.

Belt-Pulley

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Torque and Moment of Inertia Calculation • Belt-Pulley: - The transmission rate of the belt pulley is derived from the pulley diameters.

Torque and Moment of Inertia Calculation

• Belt-Pulley: - The total inertia of the belt pulley includes the motor side pulley inertia, the belt inertia reflected to the motor side and the inertia of the load side pulley reflected to the motor side.

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Torque and Moment of Inertia Calculation

• Conveyor: A Conveyor consists of two disks (pulleys) and a belt. The disk on the motor side is called the motor side pulley, the disk on the load side is called the load side pulley. Both pulleys are calculated like a regular disk (cylinder) or, if necessary, a hollow cylinder.

Conveyor

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Torque and Moment of Inertia Calculation • Conveyor: - The transmission rate of the conveyor is derived from the pulley diameters.

Torque and Moment of Inertia Calculation

• Conveyor: - The total inertia of the conveyor includes the motor side pulley inertia, the belt inertia reflected to the motor side and the inertia of the load side pulley reflected to the motor side.

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Torque and Moment of Inertia Calculation • Ballscrew and Leadscrew The is a mechanical element that converts a rotational motion into a or torque to thrust with high accuracy and efficiency.

Ballscrew and Leadscrew

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Ballscrew and Leadscrew

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Torque and Moment of Inertia Calculation • Ballscrew and Leadscrew

Leadscrew Ballscrew

Friction High Extremely low

Efficiency Low (1/3 of Ballscrew High (more than 90%) efficiency) Required High low force Wear High low

Noise Quieter (No re-circulating Fairly high balls) Inertia Low High

Cost Low High

Ballscrew and Leadscrew Applications

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Torque and Moment of Inertia Calculation

• Ballscrew and Leadscrew In order to calculate a leadscrew inertia the screw is considered a disk, thus the inertia of the screw is calculated like a disk. A leadscrew’s transmission ratio is called a pitch (rev/in) or lead (in/rev). The total inertia of a leadscrew includes the reflected load inertia (in this case the load contains of the table weight and the actual load weight) and the screw inertia.

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Leadscrew

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Torque and Moment of Inertia Calculation

• The preload force in leadscrew - The Preload is a static loading force created by using tensioning spring in the leadscrew mechanism to overcome or reduce the effect. - If the value of preload force is not indicted in the leadscrew datasheet, it can be roughly calculated by using the following equation:

Preload Force = (Fp + Fg + Ffr)/3

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Torque and Moment of Inertia Calculation

• Nip Roll: A nip roll consists basically two rollers where one roller (motor side) drives the other one (load side, support roller). Both rollers are calculated like a regular disk (cylinder) or, if necessary, a hollow cylinder.

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Nip Roll

Torque and Moment of Inertia Calculation • Nip Roll: - The nip roll is a load device, i.e. there is no other load attached to the nip roll. For that reason the transmission ratio is not significant for speed calculation, but it is needed to calculate the driven rollers inertia that is reflected to the motor. The transmission rate of the nip roll is derived from the roller diameters.

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Torque and Moment of Inertia Calculation

• Nip Roll - The total inertia of the nip roll includes the driver roller inertia and the inertia of the driven roller reflected to the motor side.

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Torque and Moment of Inertia Calculation

• : A rack pinion mechanism consists of a rotating disk (pinion) and a rectangular block (rack). The pinion inertia can be calculated using the equations of a disk

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Rack and Pinion

Torque and Moment of Inertia Calculation • Rack and Pinion: Since the rack is being used for linear (not rotary) motion we are only interested in its weight, which will then be used to calculate the inertia.

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Torque and Moment of Inertia Calculation

• Rotary Table: A rotary table consists of a rotating disk and a gear, which may be a right- gear. The gear’s data used to calculate the rotary table are the gear inertia, transmission ratio, friction coefficient and efficiency.

Rotary Table

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Torque and Moment of Inertia Calculation • Example (3):

A flywheel is manufactured with R2= 200mm and R1= 150mm, W= 30mm. If it is manufactured from Aluminium, calculate its inertia. [Density for Aluminium is 2720 kg/m3].

Torque and Moment of Inertia Calculation • Example (4): Repeat the previous example, if the flywheel was: (a) Completely solid. What do you notice? (b) Made of carbon-steel (The density of carbon steel is 7850 kg/m3.

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Speed Curve Profiling

• Most variable speed drive systems are closed loop feedback systems. They will monitor the speed of the motor. They will then compare the actual value of the speed to the required value (sometimes called the reference value) of the speed, and will drive or brake the motor according to the relative value of these two signals

Speed Curve Profiling

• In order to know what the speed should be at any point in time, a profile of the desired speed of the motion control system has to be either available or generated. This profile represents the value of the speed at which the load should travel at any point in time; it does not necessarily mean that that is the speed at which it has actually travelled. The actual value from the speed feedback device represents the speed at which the load is actually travelling. In an ideal situation, the two profiles should be identical.

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Optimum Speed Profiling • When a motion control system attempts to move the load from one point to another, one important criterion is time: To move the load between the two points in the minimum possible time. Hypothetically speaking, the load

would attain top speed, vmax in zero time, would travel to the destination at vmax and then would be brought to a halt in zero time. We will refer to this as Case I. This case is the optimum for minimising travelling time, but requires theoretically infinite acceleration, which is impossible. This is shown graphically in the following

Figure, where it takes time t1 to move the load. Obviously this is not possible, as we cannot have infinite acceleration, for three reasons:

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Optimum Speed Profiling 1. A high value of acceleration requires excessively high values of torque from the motor which are not possible. This also necessitates high values of current that would trip the power supply protection. 2. High acceleration values lead to high shock forces on the equipment and could lead to mechanical failure. 3. Human safety and comfort: The human body cannot be subjected to high values of acceleration for safety and comfort reasons.

Optimum Speed Profiling

• So in order to avoid infinite (or extremely large) values of acceleration we must allow the load to

accelerate towards the top speed, vmax. This attains a maximum value of acceleration during the acceleration phase, which we will be denoted

as amax. This case will be denoted as Case II, in which the top speed is attained via a finite value

of acceleration, amax.

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Optimum Speed Profiling

• This has removed the problem of high acceleration values, and has thus removed the requirement for high torque values, high starting currents and shock forces on the mechanical equipment. This is thus suitable for moving inanimate loads. The time

taken in this case to move the load has risen to t2 where t2>t1. This case is shown graphically in the following Figure.

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Optimum Speed Profiling

• However, it can be noticed that infinite values of jerk are experienced under case II (see the jerk values in the previous Figure). When moving human this abrupt change from zero acceleration to maximum acceleration causes a noticeable level of discomfort. It is thus desirable to have a gradual change in the value of

acceleration by limiting the top value of the jerk to jmax. This can be done by ‘filleting’ the speed-time profile at the ‘corner’ points in Case III. This is shown graphically in the following Figure. This is sometimes referred to as the S-curve facility, and is an important feature of many of the modern variable speed drive systems.

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Optimum Speed Profiling • The price paid for the improved riding comfort is

the increased journey time, t3. Note that t3>t2>t1. The selection of the values of the acceleration and jerk is a compromise between minimising travel time and maximising comfort. Typical values for human comfort in elevators for example are 1 m/s2 for acceleration and 1 m/s3 for jerk. Japanese manufacturers use lower values (around 0.5 m/s2 for acceleration) while Americans use higher values (around 1.5 m/s2 for acceleration). Europeans use intermediate values.

Optimum Speed Profiling • A real-life trace of a speed curve taken from a lift with a variable voltage AC drive is shown in the following Figure. It shows the reference speed- time curve, the actual speed-time curve measured using a tacho-generator coupled to the motor, as well as the drive and brake signals generated inside the drive.

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Optimum Speed Profiling • It is worth noticing that the curve in the Figure has a low speed approach phase towards the end. This is when the load is gradually approaching the target . The speed profiles explained previously have been shown for profiles which do not have a levelling phase; however, the same principles apply in these profiles as well.

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Speed profile: Applications in Elevators • The relationships between a fictitious speed profile for an , the distance covered, the acceleration and the jerk are shown in the following Figure. These values are for an 8 trip, at a top speed of 1.6 m/s. These curves are fictitious because of the sharp edges of the speed profile, which would generate high values of jerk as shown. As discussed in the last section, high values of jerk are not acceptable in modern systems, as they have a detrimental effect on riding comfort. However, jerk is not controlled directly; but it is controlled indirectly by ensuring that the speed profile curve is smooth and filleted.

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Speed profile: Applications in Elevators • The “filleting” effect is the responsibility of the reference curve generator within the variable speed drive. The jerk value is limited by introducing a radius in the speed profile curve when changing from rest to acceleration; or from acceleration to constant top speed; or from constant top speed to deceleration; or from deceleration to standstill. This is shown for the speed profile in the following Figure, where all the transitions have been filleted. The larger the radii of these transition, the lower the value of jerk and the better the riding comfort. The limiting case is when an “S” or an “inverted S” shape form when the two curves of two consecutive transitions meet.

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Speed profile: Applications in Elevators • Notice that the time taken to achieve the same journey has increase from 6.6 seconds in the case of infinite jerk in the first Figure to 7 seconds in the case of finite jerk in the second Figure. This is the price paid for improving the ride comfort.

Speed profile: Applications in Elevators • Example (5): The graph below shows the time profile of the jerk for a body. Draw the acceleration and velocity time profiles on the diagram below. Assume that the initial velocity and acceleration are equal to zero. Find the equations (as a function of time) of the acceleration and velocity for t=0 to t=8 seconds.

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Concepts of Drives

• Speed torque curves • Stability Criterion • Efficiency • 4 Quadrant Operation • Peak and effective torque • Variable speed drive

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Concepts of Drives

• Regardless of the method of implementing a variable speed drive, there are a number of general guidelines and principles which have to be followed to ensure successful operation. These principles are discussed in this section. The term drive is usually used to refer to the combination of the power electronic variable speed drive and the motor, although in some instances it is also used to refer to the power electronic variable speed drive.

Speed-Torque Characteristics

• A speed torque characteristic is a curve which describes the relationship for a load or a drive between the speed and torque at various points. • It effectively answers the question: “If the load (or the drive) is running at a certain speed, what would be the required (or supplied) torque?”

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Speed-Torque Characteristics

• Speed torque characteristics are important for matching the load and the drive. The following Figure shows a block diagram of a drive supplying a load, with the speed-torque curves for both shown. The drive is made up of the variable speed drive and the motor, and is coupled mechanically to the load. Thus, the speed and torque of both should always be equal. However, there are certain conditions for stability.

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Speed-Torque Characteristics

When the s-t curve for the load is placed on the same diagram as the s-t curve of the drive, the following rules apply: 1- Operating point: The operating point will be the intersection point of the two curves, at which the speed of both the load and the drive are equal, and the of both are equal.

Speed-Torque Characteristics 2- Stability: The slope of the load s-t (speed-torque) curve at the intersection point should be more than the slope of the drive curve, to ensure stability. Otherwise, if the slope of the load is less than that of the drive, then any slight increase in speed, would raise the drive torque above the load torque causing the system to run-away, and any slight decrease in the speed, would lower the drive torque below the load torque and cause the system to stall. This stability

criterion can be expressed as follows (where TL is the load torque, TD is the drive torque and N is the speed):

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Speed-Torque Characteristics

3- For ideal operation, the intersection should be as near orthogonal as possible (i.e., take place at right ). 4- To achieve good regulation, the s-t curve of the drive should be as near to the vertical as possible. This ensures that any changes in the load torque are met without any significant change in speed.

Speed-Torque Characteristics

• An example of a stable system is shown in the following Figure, where the slope of the drive s-t curve is less than the slope of the load s-t curve at the point of intersection. A slight increase in speed will cause a reduction in drive torque and an increase in load torque which will slow the system back to its original operating point. A slight decrease in speed would cause the drive torque to rise above the load torque and thus speed up the system and return it back to its operating point.

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Speed-Torque Characteristics • On the other hand, the next Figure shows an example of an unstable system, because the slope of the drive’s s-t curve is larger than that of the load at the point of intersection. A slight increase in the speed would cause the drive torque to exceed the load torque. This difference in torque will cause the system to speed up, causing a further rise in drive torque against load torque...and so on. This will cause the drive to runaway, until it arrives at a stable point. In the same way, a small reduction in speed would cause the drive torque to drop below the load torque, which will cause the system to slow down, which will further increase the difference leading to further slowing down...and so on. The system will eventually stall.

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Speed-Torque Characteristics

• The next Figure shows a good s-t curve set-up for drive and load, for an elevator. The curves intersect at right angles, and the drive s-t curves are nearly vertical. The drives changes the s-t curve needed to achieve the required speed (S1, S2 and S3).

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Constant/Variable Torque Systems

• Systems differ in the manner in which their torque requirements vary with the speed. Fans and pumps for example, require a higher torque as the speed increases, while lifts, escalators and conveyors requires a nearly constant torque throughout their speed range, for a specific load. The torque requirements of fans and pumps follow the so-called square law, by which the torque required is proportional to the square of the speed, and the torque approaches zero, as the speed approaches zero. The torque requirements for lifts and escalators are mainly constant, with a slight dependence on speed.

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Constant/Variable Torque Systems

• The speed torque characteristic of a driven system is important, as it is vital to match the speed torque characteristic of the driven system to that of the motor and the drive. “Good speed control requires that there should be a large geometrical angle (ideally 90°) between the drive characteristic T-N

and the load characteristic TL-N. This is to prevent excessive speed variation arising from small changes of load torque”

Constant/Variable Torque Systems

• The Figure below shows an example of a speed-torque curve for a hoist, at the two extremes of full load and no load. The slight speed dependence is due to the speed dependent friction element.

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Constant/Variable Torque Systems

• The Figure below shows a speed-torque curve of a fan or pump which varies with the square of the speed.

Constant/Variable Torque Systems

• The Figure below shows the speed torque curves for a compressor, which is a polynomial dependent on speed.

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Constant/Variable Torque Systems

• In an analytical form, these could be represented as follows, assuming that the speed is represented by N, and the torque by T:

- For a lift or an escalator, TL=k0 + k1N 2 - For a fan or pump, TL=kN 2 - For a compressor, TL=k0 + k1N+k2N +.....

Efficiency • The efficiency of any system is the ratio of the output power over the input power. In some systems the efficiency depends on the direction of flow of power (or energy). This is evident in gearboxes (especially worm-gearboxes) where the reverse efficiency is lower that the forward efficiency. • In certain worm gearboxes, the reverse efficiency is as low as zero and this causes the gearbox to lock when driven in the reverse direction. This is sometimes used as a feature of the gearbox where it prevent a runaway situation in the load.

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Efficiency

• Example (6): A 3-phase squirrel cage induction motor (SCIM) has an output power of 75kW. The line current is 138 A line and the line voltage is 400 V line voltage. The power factor is 0.85 lagging. If the motor has 4 poles, the supply is 50 Hz, and it delivers an output torque of 500.8 Nm, calculate its efficiency. Assume the value of slip is 5%.

Four Quadrant Operation

• One of the expressions often used in drive terminology is the expression "four quadrant operation". This refers to the ability to motor the load and brake it (or regenerate, if applicable) in both the forward and reverse directions. The term four quadrant refers to the four quadrants in the speed-torque plane. These four modes are illustrated in the following Figure, and are explained below.

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Four Quadrant Operation

• 1st quadrant: In this quadrant, the speed is positive and the torque is positive. Whenever the speed and torque are in the same direction, the motor is in a motoring mode. The fact that the speed or torque are positive, means that they are anti-clockwise (or clockwise, depending on the convention). This quadrant is thus called "Forward motoring".

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Four Quadrant Operation

• 2nd quadrant: In the second quadrant the speed is negative and the torque positive. Whenever the speed and torque are in opposite senses, the motor is in a braking mode. The braking can either be dynamic (energy usually dissipated as heat in the motor or in external resistors) or regenerative (where energy is returned back to the main supply). As the speed is negative in this quadrant, it is referred to as "Reverse braking".

Four Quadrant Operation

• 3rd quadrant: In this quadrant, both torque and speed are negative, and thus is referred to as "Reverse motoring" mode. • 4th quadrant: In this quadrant, the speed is positive, and the torque is negative, and thus is referred to as "Forward braking".

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Four Quadrant Operation

• Although most hoisting systems have drives capable of four quadrant operations, most still use reversing contactors to reverse the sense of rotation rather than using the reversing capability of the variable speed drive. • The four quadrant operation can be applied to a hoisting system (a motor using a sheave or drum to hoist a load or lower it). This is shown in the following Figure.

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Four Quadrant Operation

• The various parts of a hoisting journey for a load are also related to the four quadrant operation as shown in the next Figure. • Note that in quadrants 1 and 3 the power is positive indicating a power flow from the drive to the load. While in quadrants 2 and 4 the power is negative indicating a power flow from the load to the drive.

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Calculation for Peak and Effective Torque The basic required data to select a motor are: • Load maximum (intermittent) torque • Load RMS (rated) torque • Load inertia • Load maximum speed/acceleration • Load speed profile • Load s-t curve

Calculation for Peak and Effective Torque • Load Calculation: The load of a motor is determined by its own rotor inertia, the total inertia reflected from the mechanical set up, the constant torque from the mechanical set up, and the maximum speed and maximum acceleration of the application. • As demonstrated in the following picture any calculation of the load parameters starts with the last mechanical component in the set up.

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Calculation for Peak and Effective Torque • The speed, acceleration, inertia and constant torque is “reflected” from mechanism to mechanism until it reaches the motor. Each component will add its own inertia and constant torque. Speed transmission components, like, for instance, a gear, will also transform the inertia, speed and acceleration from the previous component according to the transmission ratio.

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Calculation for Peak and Effective Torque • The total (system) inertia and the maximum acceleration will determine the acceleration torque. The total torque, i.e. the required peak (intermittent) torque is the sum of acceleration torque and constant torque. • The required RMS torque result from the entire motion cycle and the acceleration and constant torque of each motion segment.

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Peak Torque

• The calculation of the maximum (intermittent) torque is fairly easy, since it depends mainly on the maximum acceleration. However, the maximum torque contains of two components: 1. Constant torque as inflicted by the mechanical setup. This is the torque due to all other non-inertial forces such as gravity, friction, preloads and other push-pull forces. 2. Acceleration torque as inflicted by the inertia from the mechanical setup plus the maximum acceleration as required by the load cycle.

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The equations used to calculate the maximum torque are:

Peak Torque

• Practically, the required peak torque is calculated by multiplying the sum of load torque and acceleration torque by the safety factor.

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RMS Torque

• While the calculation of the maximum (intermittent) torque is fairly easy, the calculation of the RMS load torque is a bit more complex. The RMS torque (“Root Mean Squared”) represents the average torque over the entire duty cycle. • To calculate the RMS torque the following parameters are required: – Torque during acceleration – Torque during constant speed – Torque during deceleration – Acceleration time – Time of constant speed – Deceleration time – Holding time

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• The calculation of the acceleration and deceleration torque is achieved the same way as during the peak torque calculation, however, in this case the calculation has to be accomplished for each individual motion segment.

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The equation to calculate the RMS torque is as shown below:

Principle of Variable Speed Drives

• The speed torque characteristic of the induction motor fed directly from the main supply does not allow any flexibility in varying the speed. Moreover, as the load changes, the speed also changes. • In order to vary the speed, two methods are used: - Variable voltage drives - Variable frequency drives.

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Principle of Variable Speed Drives • Variable voltage drives vary the voltage applied to the motor terminals thus changing the speed torque characteristic as shown in the following Figure. This method of speed control is not very satisfactory, especially when low speeds are required (e.g., 10% of the synchronous speed). This is due to the fact that the intersection of the drive speed-torque characteristic and the load speed-torque characteristic is not perpendicular. As can also be seen from the figure, the range of speed control is very limited. • Note that the synchronous speed is still the same for the different voltages.

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Principle of Variable Speed Drives

• A much better method of speed control is to vary the frequency of the voltage supplied to the terminals of the motor. So if the output frequency of the voltage applied to the motor terminals is 25 Hz, the synchronous speed is effectively half the rated synchronous speed (assuming the main frequency is 50 Hz). The frequency can generally be varied between 5 Hz up to 65 Hz (i.e., 15 Hz above the rated main frequency).

Principle of Variable Speed Drives

• Using this method, the speed-torque characteristic of the drive is changed to a new one that has a different synchronous speed, as shown in the next Figure. Under this method, better speed control is achieved, as the intersection of the drive and load speed-torque characteristics is nearly perpendicular. Moreover, the range of speed control is very large. • It is also worth noting that the synchronous speed is changing with each different frequency.

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• Example (7): A squirrel cage induction motor (SCIM) is used to hoist a load of mass 148.2 kg upwards. The load suffers from speed dependent friction described as: Frictional torque (in N·m)= 72.7+(0.2181·n) where n is the rotational speed in rpm. The SCIM has the speed torque curve shown in the following Figure. It also has 6 poles, runs from a 50 Hz supply and has a rated torque of 727 N·m.

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• Example (7): Use the following assumptions: - The diameter of the pulley is 600 mm. - Acceleration due to gravity is 9.81 m/s2 - Assume that counter-clockwise (CCW) torque is positive, and that upward speed is positive.

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• Example (7): Answer the following questions: - Is the hoisted load in this case a constant torque load or a variable torque load (ignoring the friction)? - What is the value of the starting torque of the motor in N·m? - What is the initial acceleration when the motor starts accelerating the load? - Find the speed-torque curve of the load and its friction as an equation. - Draw the curve for the speed-torque equation of the load on the Figure (include the frictional effect as well as the load itself).

Example (7): Answer the following questions: - What is the speed in rpm at which the system will settle? - What quadrant will the system be in? - What is the direction of the power flow?

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• Example (8):

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