calculation of motor starting time as first approximation _ eep

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Calculation of motor starting time as first Calculation of motor starting time as first approximation approximation Motor starting operations Motor starting operations The problems connected to motor starting operations are fundamentally linked to the The problems connected to motor starting operations are fundamentally linked to the type of motor which a determined type of motor which a determined motor operational torque “C motor operational torque “C M offers, to the starting offers, to the starting modality and to the modality and to the connected load connected load which has a determined which has a determined load torque “C ” load torque “C ”. The necessary The necessary starting torque “C starting torque “C a can be expressed as: can be expressed as: C a = C = C M – C – C L and shall be well calibrated to prevent it from being either too low, so as starting is and shall be well calibrated to prevent it from being either too low, so as starting is not too long and heavy – which causes not too long and heavy – which causes risks of temperature rise for the motor risks of temperature rise for the motor or from being too high on the joints or on the operating machines. or from being too high on the joints or on the operating machines. Calculation of motor starting time as first approximation | EEP http://electrical-engineering-portal.com/calculation-of-motor-starting-time... 1 of 6 8/4/2015 1:39 PM

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Page 1: Calculation of Motor Starting Time as First Approximation _ EEP

Calculation of motor starting time as firstCalculation of motor starting time as firstapproximationapproximation

Motor starting operationsMotor starting operations

The problems connected to motor starting operations are fundamentally linked to theThe problems connected to motor starting operations are fundamentally linked to the

type of motor which a determined type of motor which a determined motor operational torque “Cmotor operational torque “CMM”” offers, to the starting offers, to the starting

modality and to the modality and to the connected loadconnected load which has a determined which has a determined load torque “C ”load torque “C ”..

The necessary The necessary starting torque “Cstarting torque “Caa”” can be expressed as: can be expressed as:

CCaa = C = CMM – C – CLL

and shall be well calibrated to prevent it from being either too low, so as starting isand shall be well calibrated to prevent it from being either too low, so as starting is

not too long and heavy – which causes not too long and heavy – which causes risks of temperature rise for the motorrisks of temperature rise for the motor – –

or from being too high on the joints or on the operating machines.or from being too high on the joints or on the operating machines.

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Page 2: Calculation of Motor Starting Time as First Approximation _ EEP

A generic curve of the above mentioned quantities is shown in the A generic curve of the above mentioned quantities is shown in the Figure 1Figure 1 below. below.

The concept of The concept of motor starting time “ta”motor starting time “ta” can be associated to this concept of properly can be associated to this concept of properly

calibrated starting and can be evaluated making reference to concepts linked to thecalibrated starting and can be evaluated making reference to concepts linked to the

motion dynamics, but also by introducing simplifying hypotheses which allows, however,motion dynamics, but also by introducing simplifying hypotheses which allows, however,

an evaluation with a good approximation.an evaluation with a good approximation.

It is possible to relate the acceleration torque, expressed as a difference between theIt is possible to relate the acceleration torque, expressed as a difference between the

motor operational torque and the load torque, to the moment of inertia of the motor motor operational torque and the load torque, to the moment of inertia of the motor “J“JMM””,,

of the load of the load “J“JLL”” and to the motor angular speed, to obtain the following formula: and to the motor angular speed, to obtain the following formula:

where the expression of where the expression of “dω”“dω” assumes the following assumes the following

form:form:

and it is obtained byand it is obtained by

differentiating the well knowndifferentiating the well known

expression for the expression for the motor angularmotor angular

speedspeed ::

Through simple mathematical operations and applying Through simple mathematical operations and applying the methodthe method

of integral calculusof integral calculus, it is possible to make the unknown quantity, it is possible to make the unknown quantity

“ta” explicit by the following expression:“ta” explicit by the following expression:

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Page 3: Calculation of Motor Starting Time as First Approximation _ EEP

To express the value of the acceleration torque, it is necessary to introduceTo express the value of the acceleration torque, it is necessary to introduce

some simplifications:some simplifications:

The first oneThe first one consists in considering an average value for the motor operational consists in considering an average value for the motor operational

torque to be expressed as:torque to be expressed as:

CCMM = 0.45 x (C = 0.45 x (Css + C + Cmaxmax))

where where CCSS represents the represents the inrush torqueinrush torque and and CCmaxmax the the maximum torquemaximum torque;;

The second oneThe second one concerns the torque due to the load and which can be correct by concerns the torque due to the load and which can be correct by

applying the multiplying factor KL linked to the load typology as in Table 1 below.applying the multiplying factor KL linked to the load typology as in Table 1 below.

Table 1Table 1 – Values of factor K – Values of factor KLL

Type of comparable loadsType of comparable loads

Load CoefficientLoad Coefficient LiftLift FansFans Piston PumpsPiston Pumps FlywheelFlywheel

KKLL 11 0.330.33 0.50.5 00

In order to better understand In order to better understand the significance of the coefficient Kthe significance of the coefficient KLL we associate to the we associate to the

type of load indicated in the table the torque characterizing the type of load indicated in the table the torque characterizing the starting phase of thestarting phase of the

loadload by means of the following assumptions: by means of the following assumptions:

LiftLift = load torque constant during acceleration = load torque constant during acceleration

FansFans = load torque with square law increase during acceleration = load torque with square law increase during acceleration

Piston pumpsPiston pumps = load torque with linear increase during acceleration = load torque with linear increase during acceleration

FlywheelFlywheel = zero load torque. = zero load torque.

With these assumptions, With these assumptions, the acceleration torque can be expressed asthe acceleration torque can be expressed as::

These hypotheses allow to obtain theThese hypotheses allow to obtain the

motor starting time with the aid of themotor starting time with the aid of the

following formulafollowing formula

The starting time allows to define whether a normal or a heavy duty start mustThe starting time allows to define whether a normal or a heavy duty start must

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be realized and to choose correctly the protection and switching devices. Thebe realized and to choose correctly the protection and switching devices. The

above mentioned parameters relevant to the motor are given by theabove mentioned parameters relevant to the motor are given by the

manufacturer of the motor.manufacturer of the motor.

As an example, Table 2 below shows the values that these parameters can take forAs an example, Table 2 below shows the values that these parameters can take for

three-phase asynchronous motorsthree-phase asynchronous motors of common use and typically present on the market. of common use and typically present on the market.

Obviously the parameters relevant to the load characterize each single application andObviously the parameters relevant to the load characterize each single application and

must be known by the designer.must be known by the designer.

Table 2Table 2 – Typical values of some electrical and mechanical parameters of a three-phase – Typical values of some electrical and mechanical parameters of a three-phase

asynchronous motorasynchronous motor

Calculation of the starting time of a motorCalculation of the starting time of a motor

Making reference to the data of the above table, here is an example of calculation of theMaking reference to the data of the above table, here is an example of calculation of the

starting time of a motor, according to the starting time of a motor, according to the theoretical treatment previously developedtheoretical treatment previously developed..

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Three-phase asynchronous motor – 4 polesThree-phase asynchronous motor – 4 polesFrequencyFrequency

160 kW160 kW

FrequencyFrequency 50 Hz50 Hz

Rated speedRated speed 1500 rpm1500 rpm

Speed at full loadSpeed at full load 1487 rpm1487 rpm

Moment of inertia of the motorMoment of inertia of the motor JJMM = 2.9 Kgm = 2.9 Kgm22

Moment of inertia of the loadMoment of inertia of the load JJLL = 60 Kgm = 60 Kgm22

Load torqueLoad torque CCLL = 1600 Nm = 1600 Nm

Rated torque of the motorRated torque of the motor CCNN = 1028 Nm = 1028 Nm

Inrush torqueInrush torque CCss = 2467 Nm (C = 2467 Nm (Cs s = 2.4 x 1028)= 2.4 x 1028)

Max. torqueMax. torque CCmaxmax = 2981 Nm (C = 2981 Nm (Cmax max = 2.9 x= 2.9 x

1028)1028)

Load with constant torqueLoad with constant torque KKLL = 1 = 1

CCaccacc = 0.45 · ( C= 0.45 · ( CSS + C + Cmaxmax) – K) – KLL· C· CLL = 0.45 · (2467 + 2981) – (1 · 1600) = = 0.45 · (2467 + 2981) – (1 · 1600) = 851.6 Nm851.6 Nm

from whichfrom which

ttaa = (2 · π · 1500 · (2.9 + 60)) / 60 · 851.6 = = (2 · π · 1500 · (2.9 + 60)) / 60 · 851.6 = 11.6 s11.6 s

Load with quadratic rising torque Load with quadratic rising torque KKLL = 0.33 = 0.33

CCaccacc = 0.45 · ( C = 0.45 · ( CSS + C + Cmaxmax) – K) – KL L · C· CLL = 0.45 · (2467 + 2981) – (0.33 · 1600) = = 0.45 · (2467 + 2981) – (0.33 · 1600) = 1923.6 Nm1923.6 Nm

from whichfrom which

ttaa = (2 · π · 1500 · (2.9 + 60)) / 60 · 1923.6 = = (2 · π · 1500 · (2.9 + 60)) / 60 · 1923.6 = 5.14 s5.14 s

For both typologies of load, the esteemed motor starting time results to complyFor both typologies of load, the esteemed motor starting time results to comply

with the instruction given by the manufacturer with the instruction given by the manufacturer regarding the maximum timeregarding the maximum time

admitted for DOL startingadmitted for DOL starting. This indication can be also taken as a cue for a. This indication can be also taken as a cue for a

correct evaluation of the thermal protection device to be chosen.correct evaluation of the thermal protection device to be chosen.

Reference //Reference // Three-phase asynchronous motors: generalities and proposals for the Three-phase asynchronous motors: generalities and proposals for the

coordination of protective devices – ABBcoordination of protective devices – ABB

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Share with engineers //Share with engineers //

Recommended EE articles //Recommended EE articles //

About Author //About Author //

Edvard CsanyiEdvard Csanyi

EdvardEdvard - - Electrical engineer, programmer and founder of Electrical engineer, programmer and founder of EEPEEP..

Highly specialized for design of LV high power busbar trunkingHighly specialized for design of LV high power busbar trunking

(<6300A) in power substations, buildings and industry fascilities.(<6300A) in power substations, buildings and industry fascilities.

Designing of LV/MV switchgears.Professional in AutoCADDesigning of LV/MV switchgears.Professional in AutoCAD

programming and web-design.Present on programming and web-design.Present on Google+Google+

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