electrical calculation guide

The total Power Dissipated in the busbar is dependent on the resistance of the bar, it's length andthe square of the RMS current flowing through it... 7 Busbar CalculationsL.M.Photonics

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L.M.Photonics Ltd 2006

Electrical Calculations

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Products that are referred to in this document may be either trademarks and/or registered trademarks of the

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Printed: January 2006 in Christchurch New Zealand

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3 Power dissipated in Enclosure

L.M.Photonics Ltd 2006

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I

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Busbars Busbar Power dissipation for given currents are also calculated.

The Power Factor Correction calculations provide for an accurate sizing of static power factorcorrection of AC Induction motors Most selection tables are highly inaccurate as the variations inindividual motor designs result in a wide variation of magnetizing current

The Motor Starter Selection calculations allows the correct starter to be matched to any specificmotor and load provided the speed torque curves for the motor and load are available

Metric to imperial and imperial to metric conversions are included for many of the commonly usedunits in the electrical industry under the topics of Area, Length, Mass, Pressure, Torque and

Volume More conversions will be added in later releases of this software

This software is under constant development If you have any comments of suggestions, please post these at:

http://www.lmphotonics.com/contact.php or post them on our forum at http://www.lmpforum.com/forumdisplay.php?fid=71

For discussions and announcments, watch http://www.lmpforum.com/forumdisplay.php?fid=60

P.O Box 13 076

Christchurch NEW ZEALAND.

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II

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6 Electrical Calculations

The Busbar voltage drop is the expected resistive voltage drop on a busbar circuit, based on thelength and cross sectional area of the bar There may be an additional voltage drop due to theinductance of the bar This can become particularly important at high frequencies and highcurrents Where there are a number of bars in parallel, assume the bar width is the actual widthmultiplied by the number of bars in parallel i.e 5 bars of 50 x 6 mm in parallel would give the same

resistive voltage drop as a single bar of 50 x 30mm

To calculate the resistive voltage drop of a length of busbar, enter in the width, length and

thickness of the bar Select the units as either metric or imperial and the current passing throughthe bar The circuit configuration also needs to be specified "Single bar" refers to the voltage dropalong a single length of bar, while "Single Phase" refers to the voltage drop of two equal lengths ofbar, one in the active circuit and one in the neutral circuit "Three Phase" calculates the voltagedrop between the supply and a three phase load where three equal bars are used for the threephase circuits Enter the ambient temperature around the bar as Celsius or Fahrenheit and theprogram will check the suitability of the bar for that current The program displays the resistivevoltage drop for both an aluminium bar of these dimensions and a copper bar of these dimensions

The total Power Dissipated in the busbar is dependent on the resistance of the bar, it's length andthe square of the RMS current flowing through it

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7 Busbar Calculations

The power dissipated in the busbar is proportional to the square of the current, so if the busbarhas a cyclic load, the current should be the RMS current rather than the average If the maximumcurrent flows for a considerable period of time, this must be used as the current to determine themaximum busbar temperature, but the power dissipation is based on the square root of the

maximum current squared times the period for which it flows plus the lower current squared timesthe period it flows all divided by the square root of the total time For example, a busbar carries acurrent of 600 Amps for thirty seconds, then a current of 100 amps for 3000 seconds, then zerocurrent for 3000 seconds The power dissipation is based on an RMS current of sqrt(600x600x30 +100x100x3000 + 0 x 3000)/sqrt(30 + 3000 + 3000) = 82.25 Amps

To calculate the Power Dissipation of a busbar, enter in the width, length and thickness of the bar,and the RMS Current passing through it Select the units as either metric or imperial The programdisplays the Power Dissipated in both an aluminium bar of these dimensions and a copper bar ofthese dimensions Enter the ambient temperature around the bar in either Celsius or Fahrenheitand the program will check the suitability of the bar for this application

Optimal ratings are achieved when the bar runs horizontally with the face of the bar in the verticalplane i.e the bar is on its edge There must be free air circulation around all of the bar in order toafford the maximum cooling to its surface Restricted airflow around the bar will increase thesurface temperature of the bar If the bar is installed on its side, (largest area to the top) it will run

at an elevated temperature and may need considerable derating The actual derating requireddepends on the shape of the bar Busbars with a high ratio between the width and the thickness,are more sensitive to their orientation than busbars that have an almost square cross section Vertical busbars will run much hotter at the top of the bar than at the bottom, and should be

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derated in order to reduce the maximum temperature within allowable limits

Maximum Busbar ratings are not the temperature at which the busbar is expected to fail, rather it

is the maximum temperature at which it is considered safe to operate the busbar due to otherfactors such as the temperature rating of insulation materials which may be in contact with, orclose to, the busbar Busbars which are sleeved in an insulation material such as a heatshrinkmaterial, may need to be derated because of the potential aging and premature failure of theinsulation material

The Maximum Current rating of Aluminium Busbars is based on a maximum surface temperature

of 90 degrees C (or a 60 degree C temperature rise at an ambient temperature of 30 degrees C) If

a lower maximum temperature rating is desired, increase the ambient temperature used for thecalculations i.e If the actual ambient temperature is 40 degrees C and the desired maximum bartemperature is 80 degrees C, then set the ambient temperature in the calculations to 40 + (90-80)

= 50 degrees C

The Maximum Current rating of Copper Busbars is based on a maximum surface temperature of

105 degrees C (or a 75 degree C temperature rise at an ambient temperature of 30 degrees C)

The Busbar Width is the distance across the widest side of the busbar, edge to edge

The Busbar Thickness is the thickness of the material from which the Busbar is fabricated If thebusbar is manufactured from a laminated material, then this is the overall thickness of the barrather than the thickness of the individual elements

The Busbar Length is the total length of busbar used

The Busbar Current is the maximum continuous current flowing through the busbar The powerdissipated in the busbar is proportional to the square of the current, so if the busbar has a cyclicload, the current should be the RMS current rather than the average If the maximum current flowsfor a considerable period of time, this must be used as the current to determine the maximumbusbar temperature, but the power dissipation is based on the square root of the maximum currentsquared times the period for which it flows plus the lower current squared times the period it flowsall divided by the square root of the total time For example, a busbar carries a current of 600Amps for thirty seconds, then a current of 100 amps for 3000 seconds, then zero current for 3000seconds The power dissipation is based on an RMS current of sqrt(600x600x30 + 100x100x3000

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9 Busbar Calculations

+ 0 x 3000)/ssqrt30 + 3000 + 3000) = 82.25 Amps

The Ambient Temperature is the temperature of the air in contact with the busbar If the air is in anenclosed space, then the power dissipated by the busbar will cause an increase in the ambienttemperature within the enclosure

To calculate the rating of a busbar, enter in the width and thickness of the bar, and the ambienttemperature around the bar Select the units as either metric or imperial, and the temperature asCelsius or Fahrenheit The program displays both the current rating of an aluminium bar of thesedimensions and a copper bar of these dimensions

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11 Cable Calculations

Cable ratings are based on the resistance of the cable, the surface area of the cable, and thetemperature rating of the insulating material The rating applied to a cable is at a given ambienttemperature Variation in ambient temperature will result in a variation of the cable rating

The resistance of the cable is a function of the material from which the conductor is manufactured,i.e copper or aluminium, and the cross sectional area of the conductor

The Cable size can be nominated in either square millimeters, or in the US American Wire Gauge.The European metric ratings are based on figures from VDE 0100 and the American Wire Gaugefigures are taken from the National Electrical Code (NEC)

Ambient Temperature The air temperature around the cable

Ventilation Free air movement around the cable will allow more cooling than cables enclosed inconduit or trunking

To calculate the current rating of a cable, select the cable size and the ambient temperaturearound the cable in degrees Celsius or Fahrenheit and the ventilation ("In Free Air" or "Enclosed"

in conduit or trunking.) The current rating of a copper cable is displayed

The Cable voltage drop is the expected voltage drop on a cable circuit, based on the length andcross sectional area of the bar Where there are a number of cables in parallel, assume the cablecross sectional area is the actual area multiplied by the number of cables in parallel i.e 5 cables of

6 mm in parallel would give the same resistive voltage drop as a single cable of 30mm

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To calculate the voltage drop of a length of cable, select the cable size and the current passingthrough the cable The circuit configuration also needs to be specified "Single Cable" refers to thevoltage drop along a single length of cable, while "Single Phase" refers to the voltage drop of twoequal lengths of cable, one in the active circuit and one in the neutral circuit "Three Phase"

calculates the voltage drop between the supply and a three phase load where three equal cablesare used for the three phase circuits The program displays the voltage drop for a copper cable

The total Power Dissipated in the cable is dependent on the resistance of the cable, it's length and

the square of the RMS current flowing through it

The power dissipated in the cable is proportional to the square of the current, so if the cable has acyclic load, the current should be the RMS current rather than the average If the maximum currentflows for a considerable period of time, this must be used as the current to determine the

maximum cable temperature, but the power dissipation is based on the square root of the

maximum current squared times the period for which it flows plus the lower current squared timesthe period it flows all divided by the square root of the total time For example, a cable carries acurrent of 600 Amps for thirty seconds, then a current of 100 amps for 3000 seconds, then zerocurrent for 3000 seconds The power dissipation is based on an RMS current of sqrt(600x600x30 +100x100x3000 + 0 x 3000)/sqrt(30 + 3000 + 3000) = 82.25 Amps

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IV

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In three phase applications, Delta connected circuits can be replaced by star connected circuitswith the same resultant impedance The Delta impedances are Z12, Z23 and Z31 These can bereplaced by star impedances Z10, Z20 and Z30

For the calculations, the impedance Zxx is expressed as R + JX where R is the resistive

component and X is the reactive component

For an inductor, X = 2 x pi x F x L and

for a capacitor, X = -1/(2 x pi x F x C) NB X value for capacitor is negative!!

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15 Circuits

In three phase applications, Star connected circuits can be replaced by Delta connected circuitswith the same resultant impedance The Star impedances are Z10, Z20 and Z30 These can bereplaced by Delta impedances Z12, Z23 and Z31

For the calculations, the impedance Zxx is expressed as R + JX where R is the resistive

component and X is the reactive component

For an inductor, X = 2 x pi x F x L and

for a capacitor, X = -1/(2 x pi x F x C) NB X value for capacitor is negative!!

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17 Constants

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VI

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19 Conversions

To make a conversion, from the "Conversions" menu, select the required conversion

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Enter a value (Greater then 1) in the left hand window and select the units to convert from Selectthe units to convert to on the right hand side and the result is displayed in the right hand window

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VII

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The temperature rise within the enclosure is directly proportional to the thermal resistance of theenclose and the total power dissipated within the enclosure The thermal resistance is a function ofthe shape and size of the enclosure, and also to the amount of exposed surface area Given thedimensions of the enclosure, and the environment within which it is mounted, it is possible toapproximate the enclosure thermal resistance for a sealed enclosure

The addition of ventilation grills will reduce the thermal resistance, but not by a significant amount.(Typically less than 10%)

If the thermal resistance of the enclosure needs to be reduced significantly, the only way to

accurately control the temperature rise is by the addition of forced ventilation fans With forcedventilation, the amount of airflow required for a given temperature rise is proportional to the totalpower dissipated

When fans are used, adequate open area for air input and air output must be provided If the openarea equals the size of the fan, the velocity of the air flow will be equal to the velocity of the airthrough the fan If the open area is too small, the airflow velocity will increase, the pressure acrossthe fan will increase and the airflow will reduce

Inlet and exhaust ports can provide a much higher resistance to air flow, and thereby restriction inventilation than is generally appreciated Wherever possible, it is preferable to keep the smallestdimension of the open area at no less than 10 mm If the smallest dimension is reduced below thisfigure, the boundary effect around the edges of the opening will reduce the effective open area Agood rule is to try for twice the area of the fans in open area for the inlet and exhaust ports with theminimum dimension of the opening at 10mm or greater

If the smallest dimension is halved, then allow for double the open area

A further reduction in the minimum opening dimension should be compensated for by and

increase in the open area Failure to do this will result in reduced air flow and increased

temperature rise One solution where very fine mesh is employed is to double the fan capacity Airfilters can severely restrict the air flow, but will often be accompanied by flow/pressure curves Where these curves are available, they can be superimposed on to the fan curves and the actualexpected flow can be easily predicted NB Small box fans very quickly lose airflow when a

constrictive filter is applied

Typical Air flow figures for small fans:

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23 Enclosure Ventilation and Cooling

A sealed enclosure containing electrical apparatus will have an internal temperature rise that isdependent on the thermal resistance of the enclosure and the total power dissipated within thatenclosure

The thermal resistance of the enclosure is a function of the size of the enclosure, the shape of theenclosure and the exposed surface area of the enclosure If the enclosure is free standing with airable to freely circulate over all vertical surfaces, then it will have a lower thermal resistance than anenclosure where the air flow over one or more surfaces is restricted Likewise, increasing width isfar more effective in reducing the thermal resistance of an enclosure than increasing height Adding ventilation louvers without fans will only slightly reduce the thermal resistance of theenclosure (usually less than 10% reduction in thermal resistance or temperature rise!)

The thermal resistance can be calculated for a given height, width and depth of enclosure

provided the installation conditions are specified From the thermal resistance and total powerdissipated within the enclosure, the temperature rise can be calculated

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The total Power dissipated in the enclosure is the sum of all power dissipated by all componentsmounted within the enclosure

In order to calculate/approximate the ventilation required for an enclosure, the power dissipatedwithin the enclosure must be known

Here are some guidelines for calculating the total power dissipated in the enclosure

Soft Starter Allow 4.5 Watts per amp i.e MSX 0175 operating at 150 amps, allow 150 x 4.5 =

675 Watts

Speed Drive Allow 20 Watts per amp

Contactor AC3 If dissipation not known, allow 0.15 Watts per Amp contact dissipation, plus 0.1Watts per amp coil dissipation

Contactor AC1 If dissipation not known, allow 0.6 Watts per Amp contact dissipation, plus 0.1Watts per amp coil dissipation

Thermal overload If not known, allow:

10 watts if Ie < 32A

18 watts if 32A < Ie < 70A

22 watts if 70A < Ie < 500A

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25 Enclosure Ventilation and Cooling

50 watts if Ie > 500A

Fans (Not part of starter i.e cabinet ventilation fans)

Allow 10 watts for 25mm x 120 mm

Lamps Allow 2.5 watts

Isolators If not known, allow 0.6 watts per line amp

Power factor correction capacitors Refer to data sheet and/or supplier

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VIII

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27 Induction Motor Starting

To engineer the system, it is important to firstly establish the starting torque requirements of thedriven load Next the starting characteristics of the induction motor should be analyzed in order toestablish the start current required by the motor to develop the required starting torque A startercan now be designed/selected to meet the start current requirement, and an appropriate supplyconnected

Induction motors exhibit a very low impedance at speeds less than their rated speed This results

in a very high start current when Direct On Line started The Direct On Line starting current isindependent of the motor load and is dependent only on the motor design, rotor speed and theapplied voltage Variations in motor loading will affect the start duration only Typically, the Direct

On Line starting current falls somewhere between 550% Full Load Current and 900% Full LoadCurrent The actual start current of a given design is determined primarily by the design of therotor Shallow bar rotor designs are generally referred to as Design 'A' rotors and are characterized

by a high start current (650% - 900%) and a low starting torque (60% - 150%) Design 'B' rotors aredeeper bar rotors and typically exhibit a starting current of (550% - 650%) and a starting torque of(150% - 300%)

In many installations, the maximum starting torque is not required, and the very high startingcurrent places stress on the supply causes voltage disturbances and interference to other users onthe supply Reduced voltage starting is a means of reducing the start current, however a reduction

in the start voltage will also reduce the starting torque

In order to achieve a useful start at a reduced starting current, it is important that the motor is able

to develop sufficient torque at all speeds up to full speed to exceed the load torque at those

speeds If the reduced torque developed by the motor is less than the load torque at any speed,the motor will not accelerate to full speed Stepping the starter to full voltage at less than full speedwill result in a high current and little if any advantage over using a Direct On Line starter Theselection of a start voltage that is too low will result in an inferior start characteristic

Star/Delta (Wye/Delta) starters are open transition When the transition is made from the reducedvoltage to full voltage, there is a period of time when the motor is effectively open circuited from thesupply During this period, the motor is effectively acting as a generator at a frequency proportional

to it's actual shaft speed When the starter reconnects the motor to the supply in Delta, there is avery high transient current and resulting transient torque which is much more severe and damagingthan the Direct On Line starting conditions

Other reduced voltage starters commonly employed are the Autotransformer Starter and the

Solid State Soft Starter

The induction motor has two major components: The Rotor and The Stator In most motors, theStator is in the outer part of the motor and comprises a stack of steel laminations and two or morewindings The inner part of the stator is hollow, and the windings are distributed around the innersurface of the stator imbedded in a number of slots The windings are organized to form two or

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more electromagnetic poles

The Rotor is a solid cylindrical stack of laminations with a series of conducting bars imbeddednear the surface The ends of these bares are shorted together by shorting rings

When the supply is connected to the stator windings, a magnetic field is created which is rotating

at the supply frequency The field in a two pole machine will do one complete revolution per cycle

of the supply A Four pole machine requires two cycles for a complete revolution and a Six polemachine requires three cycles for a complete revolution

The rotating magnetic field developed by the stator, causes a current to flow in the short circuitedrotor winding in the same manner as the secondary current is caused to flow in a transformer -infact the motor emulates a transformer with a short circuited secondary

The rotor current in turn develops a rotating magnetic field which interacts with the stator field todevelop a rotating torque field in the direction of the stator field rotation The strength of the torquefield is dependent on the interaction of the two magnetic fields, and is therefore dependent on themagnitude of the fields and their relative phase angle

The full voltage start current and start torque curves vary tremendously between different motordesigns due to the variations in rotor designs

In designing a motor starting system, it is important to base the design on the actual motor beingused A design based on "typical" curves can yield very erroneous results

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29 Induction Motor Starting

The induction motor is used to convert electrical energy into mechanical energy The driven loadpresents a mechanical load to the shaft of the motor As the motor is started, it accelerates thedriven load from zero speed to the rated full load speed of the motor As the load accelerates, thetorque presented to the motor shaft will vary depending on the design of the machine

Generally, the load torque is expected to be higher at full speed than at lower speeds Someapplications such as loaded conveyors may require a high breakaway torque to get the load tobegin to move from zero speed

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