ࡱ> ^`YZ[\] bjbj@@ "e1g"e1g%hhhhh|||8X|+b,#*"***G8G,GaaaaaaadsgnahLcAGLLahh**aVVVLh*h*aVLaVVZ`ha*[]N`aa0+b`4gQg aaghaGHVmI|IGGGaa!V|GGG+bLLLLgGGGGGGGGGX .: STARTERS & MOTORS BACK TO BASICS MOTOR LONG TERM STORAGE MOTOR VOLTAGE VARIATION TEMPERATURE RISE AND INSULATION CLASS Issued by C.J. Hewetson Date 7 Jan 02 This back to basics information covers basic starter information we have forgotten or have not been taught. Basic is not intended to provide in depth technical training, merely simple motor starting in its basic form, the starter that is found in general use around industry covering, DOL, Star Delta, Primary Resistor, Auto Transformer and Secondary resistor starters. 1) (DOL) DIRECT ON LINE STARTERS DOL motor starting provides maximum torque, but starting current under these conditions can be as high as 7 times normal full load current of the motor in question. DOL starters generally provide the lowest cost and highest starting torque. Certain motors such as double cage rotors are available which maintain maximum torque and provide reduced starting currents. These motors are more costly and are usually only considered for outputs over 75 KW or where maximum starting torque is essential with some reduction of starting current. Should these special motors not be available then some form of reduced voltage starting which in turn limits starting current, and also starting torque must be considered. The following factors must also be taken into account when selecting the type of starting to be used. Motor design, the driven machine torque requirements, incoming power supply (transformer may have insufficient capacity to withstand the large current requirements). Last but not least the Local supply authorities may have a maximum allowable KW/KVA or inrush current which they will allow for a particular site. The following factors normally determine whether or not a reduced voltage or DOL starter should be used. Any limitation in the supply transformer, existing distribution or incoming fuses and wiring . The incoming power supply transformers etc, may have insufficient capacity to withstand the full load inrush current of full voltage starting (DOL), which result in large voltage drops, transients. Driven machinery torque requirements or acceleration requirements. Medium tension voltage e.g. above 4.5Kv will require DOL starting in most cases. 2) AUTOMATIC STAR DELTA STARTER The most common method of reduced voltage starting uses the Star Delta connection. With this method all six ends of the motor windings are connected to the starter so that in the starting position the windings are connected in Star and the voltage per phase is (3 or 57.7% of line voltage. When the motor has reached sufficient acceleration in Star (at least 90% of synchronous speed) the starter is switched so that the motor windings are connected in Delta and full voltage applied to each phase. It should be noted that not all motors with 6 leads or terminals are suitable for Star-Delta starting as some Continental motors having 6 terminals are wound for dual voltage (220/38Ov) and are not suitable for starting Star-De1ta except on 22Ov 3 phase. Star Delta starters are available in "Closed" or " Open transition". On open transition during the changeover period from star to delta tappings, the squirrel-cage motor is disconnected from the supply and as the secondary winding (rotor) is short circuited, the magnetic field does not collapse immediately. When the motor is reconnected to the supply in delta, the phase position of the residual voltage may move with respect to the supply voltage and cause a large transient current to flow in the switchgear and motor windings at the same time the driven equipments speed stalls. Depending on the phase displacement of the two voltages, the current transient may exceed 20 times the full load current of motor at the instant of changing over from star to delta running position and transient voltages have been measured up to 8000 volts. The condition is similar to that of switching an alternator on to the supply without synchronising. Closed transition starters incorporate a set of transition contactors and resistors which prevent the motor from stalling and creating a transient spike, by holding in the star contactors until the line contactor has made. Closed transition starters are more expensive than open transition but should always be used for large reciprocating or rotary displacement compressors. All automatic Star-Delta starters use a timing device for switching from Star to Delta and the timer therefore should be carefully set for each individual application. The star delta is not easily converted to a closed transition starter, and even the closed transition (Wanchop) star delta starter still has the problem that the start voltage can not be altered. If there is insufficient torque available in star, then it will go DOL. The star delta starter does get around the regulations in some countries where there is a requirement for a reduced voltage-current starter, but in reality, in many situations results in more severe transients than DOL. The main benefits of the star delta starter are that it puts more money in the pockets of the switchgear supplier, and it is politically correct. 2.1) STARTING TORQUE This can vary according to type and design of motor but generally the starting torque in Star is approximately 33 1/3% of the maximum value obtained by full voltage D.O.L. starting. Double cage squirrel-cage motors can produce values of starting torque approximately twice that of single cage squirrel-cage motors with reduced starting current. To calculate any motor starting torque use the following formulae HPx 5250/RPM or Kw x 9544/RPM. For reciprocating or Screw compressors a speed torque curve can be provided by the manufacturer and it is easy to check the motor has sufficient torque to break away the compressor and accelerate it. For every quotation, the motor torque must be checked. 2.3) STARTING CURRENT As for starting torque approximately 33 1/3% of the maximum value obtained by full voltage D.O.L. starting. For example a 7.5KW motor having a full load current of 15 amps would draw about 90 amps (6 times F.L.C.) on D.O.L. starting but would be about 30 amps ( 1/3 x 90) starting current in Star. An approximate guide for starting current in Star is 2 3 x the full load current of the motor. However, depending on type and design of motor, starting currents in Star can vary quite considerably. Most motor manufacturers usually list the D.O.L. and Star starting currents and torques. 2.4) APPLICATIONS One disadvantage of Star-Delta starting is the much reduced starting torque in Star, and is therefore only suitable for unloaded or lightly loaded motors on starting. The motor must be capable of reaching at least 90% of synchronous speed in Star before changeover to Delta occurs. Such loads as fans, blowers etc., may be suitable applications but it may be necessary to restrict the airflow during starting to allow the motor to accelerate freely. Most refrigeration compressors are not suitable for Star Delta starting, unless some method of mechanical unloading is employed and or the suction and discharge gas density ratio and pressures are equalized and even then, care must be taken to ensure that the motor has sufficient torque in Star to overcome both the compressor and motor inertia, break away the compressor and accelerate it up to 90% speed within the allowed time period. Many sales people increase the motor KW rating in order to provide adequate starting torque, but this is incorrect as at full load, the motor output will far exceed the compressor shaft power required and the full load efficiency and power factor will drastically reduce which increases the operational running costs. 2.5) ACCELERATION PERIOD BS587-1957 specifies that the maximum starting time of a Star-Delta starter as 9 + KW/3 in seconds, with a maximum of 45 seconds for motors up to 132KW. As an example a 11KW motor must be capable of reaching at least 90% of synchronous speed in Star before changeover to Delta in a maximum of 14 seconds (9 +15/3=14) If this rate of acceleration of speed cannot be obtained, the load is outside a Star Delta application and some means of reducing the load or use of reduced voltage starting must be employed. Failure to achieve this acceleration can result in severe stresses to the motor, possible switching difficulties of the starting equipment, tripping on the overload and in severe cases failure of the compressor to run up to speed. 2.6) SHORT CIRCUIT PROTECTION HRC fuses motor starting rated will provide the best protection for all motor starters. 3) AUTOMATIC AUTO TRANSFORMER STARTER Another method of reducing starting current of a squirrel-cage motor is to employ an auto- transformer having tappings at usually 51%, 65% and 80% of line voltage. Other values can be used for special starting conditions. Being an auto- transformer it has a single winding per phase so that part of the winding is common to both primary and secondary. Some type auto- transformer starters use a two-coil transformer up to and including 45KW and three coil over that size to reduce any imbalance in line current. At the instant of start the Transformer Star Contactor closes first, which connects the ends of the transformer windings in "Star". The closing of the Transformer Star Contactor also closes the Transformer Line Contactor which energises the transformer. The voltage to the motor terminals then depends on the tapping used. Simultaneously the timing device is energised and after a pre-determined time, sufficient for acceleration, the timer opens the Transformer Star contactor which in turn closes the Motor Line Contactor and applies full voltage to the motor. During changeover the supply is not disconnected from the motor so avoiding any switching transients. Most Auto-Transformer Starters are wired in this way, which is known as the KORNDORFER connection. 3.1) STARTING TORQUE Each auto- transformer is provided with several sets of voltage taps making it an easy matter to adjust the motor starting torque and current to suit the particular load conditions. Taps which are clearly marked on the auto-transformer nameplate generally provide for 50%, 65% and 80% of line voltage at starting. Starting torque is proportional to the square of the voltage so that the starting torque on these taps will be 25%, 42% and 64% respectively of the maximum value obtained by full voltage starting. 3.2) STARTING CURRENT Neglecting the magnetising current of the auto- transformer and any saturation of the motor, the starting current is proportional to the square of the voltage. With tappings of 50%, 65% and 80% of full voltage the starting current wi11 be 25%, 42% and 64% respectively of the maximum value obtained by full voltage starting. Both starting torque and current may vary from these figures, depending on type and design of motor. 3.3) APPLICATION These starters are suitable for most applications such as pumps, blowers, conveyors, compressors, fans and other machines where it is necessary to maintain maximum starting torque with a minimum of current and power losses drawn from the line. They are designed for general purpose industrial applications where heavy duty reliable starting equipment is of paramount importance. 3.4) STARTING DUTY British Standards lists the number of permissible starts per hour. Care should be taken in specifying the correct duty starter for each application, as exceeding the maximum number of starts per hour can have a serious effect on the transformer windings.  Minimum Subsequent Normal Maximum Cooling Period Number of Number of for starts as Starts per Starts in any Col.3 if made Duty Rating Hour. 15 minutes Consecutively  Ordinary 2 2 60 minutes Intermittent 15 6 15 minutes Frequent 40 12 15 minutes  3.5) ACCELERATING PERIOD In accordance with same standard the period required to accelerate the motor shall be assumed not to exceed the following values,-  KW Ordinary Duty - On 75% tap or higher: 6 + 5 seconds with a maximum of 30 seconds.  KW On 60% tap or lower: 9 + 3 seconds with a maximum of 45 seconds. Intermittent Duty - 1 minute in each 15 minutes. Frequent Duty - 2 minutes in each 15 minutes. Most manufactures of Auto-Transformer starters have a factory set on the 65% tapping and it may be necessary when starting heavy loads to increase to the 80% tapping to achieve the above accelerating periods. It is important to set the timer correctly for each individual application so that the starter does not remain in the reduced voltage position after the motor has ceased to accelerate, otherwise excessive currents can be drawn by the auto-transformer and motor, thus shortening their working life. 3.6) OVERLOAD RELAY The overload protection is as for D.O.L. starting and should be set to rating of the motor. 3.7) SHORT CIRCUIT PROTECTION H.R.C. fuses motor starting will provide the best protection for all motor starter types. 3.8) TRANSFORMER PROTECTION As an auto-transformer is a relatively costly item, it is recommended that transformer protection is well worth the small added expense. If the Auto-Transformer is provided with pockets to accommodate Microtherms, which when specified, are wired into the Control Circuit to protect the Transformer Windings against overheating caused by an excessive number of starts or possible failure of a timer or contactor. 3.9) GENERAL The Auto- Transformer starter is the most expensive of the three described and for this reason is not used as often as it should be. Much time and money is wasted on installing and repairing Star-Delta starters that are used incorrectly on totally unsuitable loads requiring a large starting torque, where an Auto-Transformer starter would have been entirely satisfactory and cheaper in the long run. The great advantage of the Auto- Transformer starter is the ease with which the starting torque and current can be adjusted to suit each application by changing the voltage taps on the transformer. As mentioned previously, auto-transformers can be obtained with higher or lower voltage taps than normal to suit any special starting conditions. 4) AUTOMATIC PRIMARY RESISTOR STARTERS Primary Resistor Starting consists of connecting the motor to the line through a series resistor in each phase. The motor is started from a push button or other two wire pilot device, which energises the accelerating contactor. The closing of this contactor supplies reduced voltage through the starting resistor. Simultaneously, the time delay relay is energised, and after a predetermined time sufficient for acceleration, this timing relay energises the Line Contactor, connecting the Motor Directly to the Line. Unlike Star-Delta starting, at no time is the supply disconnected from the Motor. Primary Resistor starting can provide very smooth acceleration since, as the motor speed increases, the current fails and the voltage drop across the resistor is reduced, resulting in a rise in voltage at the motor terminals with an increase in torque during acceleration. 4.1) STARTING TORQUE Standard resistors are designed to allow approximately 60% of line voltage to be impressed on the motor terminals at starting. Starting torque is proportional to the square of the voltage so that a standard Primary Resistor would give approximately 36% of the maximum value obtained by full voltage D.O.L. starting. 4.2) STARTING CURRENT The design of standard resistors are based on the average motor (which takes 6 to 8 times full load current on D.O.L. starting), and will allow 60% of line voltage at the motor terminals on starting. The starting current is directly proportional to the voltage, so that the standard resistor would reduce the starting current to approximately 60% of the maximum value obtained by full voltage D.O.L. starting. Both Starting Torque and Starting Current values can vary considerably depending on the type and design of motor. Likewise the value of resistor can be varied to give a greater or lesser voltage to the motor at instant of start and although the standard is 60% of full voltage based on the average motor, any variation of this can be made to suit a particular application. 4.3) APPLICATIONS Application is particularly suitable for fans, blowers, centrifugal pumps, compressors and general purpose applications where smooth acceleration is desirable together with a constantly increasing torque, as the motor accelerates. As with Star-Delta starting, care should be taken that initial low starting torque is satisfactory, as some of the above applications can, because of their size, weight and high inertia require a large starting torque. Where a smooth start is required to prevent mechanical damage to couplings etc. from a motor with too much starting torque, a step of primary resistance may often provide a simple solution. This resistance is sometimes left in circuit permanently on small motors with a short duration of load cycle. 4.4) STARTING DUTY The number of starts per hour is limited by the rating of the resistor and as the resistor must carry the stalled rotor current of the motor, the price and physical dimensions tend to limit its use to the lower horsepower range, generally below 75KW.  Duty NORMAL NO. MAXIMUM AGGREGATE PERMISSIBLE OF STARTS STARTING PERIOD CONSECUTIVE WITHOUT COOLING STARTS  Ordinary 5 per hour Up to 37KW.  KW. 8 + 4 = seconds (with min.of 10 secs) 3 starts Above 37 up to 450 KW. KW  18 + 20 = seconds  Intermittent 15 per 1 minute in 15 minutes Repeated Hour starting for 2 minutes  Frequent 40 per 2 minutes in l5 minutes Repeated Hour starting for 4 minutes  Some Standards gives somewhat similar figures except that ordinary duty is limited to 2 starts per hour with a maximum of 2 consecutive starts. 4.5) ACCELERATING PERIOD As seen from above (Column 3) that a definite maximum starting period is specified, in which time the motor should reach full speed. There is nothing to be gained by allowing the motor to run at reduced voltage after the motor has ceased to accelerate and the timer should be adjusted to switch the motor to full voltage at this point. 4.6) OVERLOAD RELAY The Overload Protection used should be the same is as for D.O.L. 4.7) SHORT CIRCUIT PROTECTION H.R.C. fuses motor starting will provide the best protection for all motor starter types. 4.8) RESISTOR PROTECTION A Thermostat mounted near the resistances and wired into the control circuit can be fitted to provide protection of the resistors against excessive number of starts or abuse. Where two wire remote control is used, it is usually necessary to add an additional control circuit relay to prevent the motor from automatically re-starting due to the Thermostat reclosing after operation. 4.9) GENERAL The greatest advantage of the Primary Resistor starter is its ability to provide an increase in torque during acceleration. Unlike Star-Delta starting which has a fixed value of voltage at the Motor terminals, the Primary Resistor allows the voltage at the motor to increase as the motor accelerates. As the supply to the Motor is not interrupted during changeover there are no great switching transients as can happen with open transition Star-Delta. It will be noted that the initial starting torque is somewhat similar to Star-Delta but the starting current is much greater with Primary Resistor which can limit its applications. 5) AUTOMATIC SECONDARY RESISTOR STARTER The function of the Secondary Resistor Starter is to control the starting requirements of the Wound Rotor or Slip- ring motor. The slipring motor is used for starting high inertia loads or where a high starting torque with minimum starting current is required. The method of starting is to introduce resistance into the rotor circuit during starting, which has the effect of reducing proportionally the inrush stator current and torque developed to any desired figure within the capacity of the motor. Acceleration is provided by suitable adjustable timing relays ensuring uniform acceleration which is not dependent on human element. The accelerating relays reduce by specific steps the amount of resistance connected in the rotor circuit, until finally no resistance is left and the slip -rings are short-circuited with the performance of the motor similar to that of a standard squirrel-cage. Each Secondary Resistance starter is individually designed to suit a particular motor and the design is dependent on two major factors: 1 . The rotor characteristics of the particular motor, ROTOR VOLTS at standstill, and FULL LOAD ROTOR AMPS. 2. The starting characteristics of the drive which will dictate the initial starting torque necessary, and the number of accelerating steps required to limit the accelerating step peak currents to the permissible value. The selection of the correct rating of Contactors, together with the design of the starting resistance, and the number of accelerating steps all depend on the above requirements and data. 5.1) GENERAL There are few problems with a well designed Secondary Resistance starter but the following points are worth noting. When installing a new starter it is important to adjust each timer, as there is no advantage to be gained by dwelling on a given step after acceleration has ceased, as this only creates unnecessary heating in the starting resistor. It is generally found that the initial steps require a longer period and the time can be shortened as the motor builds up speed. Resistance protection by means of a Thermostat fitted near the resistances is well worth the small additional cost, to prevent resistor damage due to an excessive number of starts or possible timer malfunction. The size of cables between the starter and motor should be carefully checked as the rotor current may be three or four times the stator current, so that where the starter is some distance from the motor the voltage drop and heating in the rotor cables may be serious. When replacing an old worn out Secondary Resistance starter, it so often happens that the rotor characteristics are no longer legible on the motor nameplate and of course a new starter cannot be designed without these essential figures. When the rotor voltage is known the approximate current can be calculated from the equation. Rotor Current KW x 1000 x 1. 1 (3x Rotor Voltage The Rotor voltage is the open circuit voltage at standstill and can be measured across any two slip-rings when the starter is connected to the supply with the brushes lifted, or insulated from the slip-rings and the rotor held to prevent rotation. Slip-ring motors should not be started on full voltage (D.O.L.) with the slip-rings short-circuited . Under this condition the starting current is higher than that of the equivalent squirrel-cage motor, while the starting torque is not only lower, but it can vary with the position of the rotor in respect to the stator winding. The rotor can actually lock (known as cogging) and will stay in this position with the associated high starting current until either turned by hand, or the H.R.C. protective fuses have blown. Under emergency conditions the possibility of cogging can be overcome by introducing a small amount of permanent resistance into the rotor circuit, such as shorting the slip-rings with resistance wire, but with a loss of efficiency and slight decrease in ultimate speed. Unless as a short term emergency requirement, it is not recommended to start a slip-ring motor, other than with its correct Secondary Resistance starter. The many uses for Secondary Resistance starters and slip-ring motors are very complex and a number of text books have been written covering the subject. Such applications as Inching, Variable Speed, Slip Resistors, Load sharing and Synchronised Drives. As there seems to be some difficulty in arriving at the actual starting torque and starting currents for the various reduced voltage starters following examples are given. An average 4 pole 7.5Kw motor could have the following performance data.  KW FULL LOAD FULL LOAD DIRECT STARTING % SPEED R.P.M. AMPS OF FULL LOAD AT 40OV TORQUE CURRENT  7.5 1430 15 200 600  Taking D.O.L. starting, the starting torque will be 200% and the starting current  15 x 600 = 90 amps. 100   REDUCED VOLTAGE FULL LOAD STARTING STARTER STARTING TORQUE CURRENT  Star-Delta 33 x 200 = 66% 33 x 90 (D.O.L.)  100 100 = 29.7 amps   Primary Resistor 36 x 200 = 72% 60 x 90 100 100 = 54 amps  Auto-transformer 25 x 200 = 50% 25 x 90  50% Tap 100 100 = 22.5 amps 42 x 200 = 84% 42 x 90  65% Tap 100 100 = 37.8 amps 80% Tap 64 x 200 =128% 64x 90 100 100 = 57.6 amps  BASIC MOTOR FORMULAES & CALCULATIONS The formulas and calculations which appear below should be used for estimating purposes only. It is the responsibility of the customer to specify the required motor Hp, Torque, and accelerating time for his application. The salesman may wish to check the customers specified values with the formulas in this section, however, if there is serious doubt concerning the customers application or if the customer requires guaranteed motor/application performance, the Product Department Customer Service group should be contacted. Rules Of Thumb (Approximation) At 1800 rpm, a motor develops a 3 lb.ft. per hp At 1200 rpm, a motor develops a 4.5 lb.ft. per hp At 575 volts, a 3-phase motor draws 1 amp per hp At 460 volts, a 3-phase motor draws 1.25 amp per hp At 230 volts a 3-phase motor draws 2.5 amp per hp At 230 volts, a single-phase motor draws 5 amp per hp At 115 volts, a single-phase motor draws 10 amp per hp Mechanical Formulas Torque in lb.ft. = HP x 5250  INCLUDEPICTURE "http://www.reliance.com/nav_elements/4pxlblack.gif" \* MERGEFORMATINET  rpm  INCLUDEPICTURE "http://www.reliance.com/nav_elements/1pxltrn.gif" \* MERGEFORMATINET  HP = Torque x rpm  INCLUDEPICTURE "http://www.reliance.com/nav_elements/4pxlblack.gif" \* MERGEFORMATINET  5250  INCLUDEPICTURE "http://www.reliance.com/nav_elements/1pxltrn.gif" \* MERGEFORMATINET  rpm = 120 x Frequency  INCLUDEPICTURE "http://www.reliance.com/nav_elements/4pxlblack.gif" \* MERGEFORMATINET  No. of Poles Temperature Conversion Deg C = (Deg F - 32) x 5/9 Deg F = (Deg C x 9/5) + 32 High Inertia Loads t =WK2 x rpm  INCLUDEPICTURE "http://www.reliance.com/nav_elements/4pxlblack.gif" \* MERGEFORMATINET  308 x T av. INCLUDEPICTURE "http://www.reliance.com/nav_elements/1pxltrn.gif" \* MERGEFORMATINET WK2 = inertia in lb.ft.2 t = accelerating time in sec. T = Av. accelerating torque lb.ft..T =WK2 x rpm  INCLUDEPICTURE "http://www.reliance.com/nav_elements/4pxlblack.gif" \* MERGEFORMATINET  308 x t inertia reflected to motor = Load Inertia  INCLUDEPICTURE "http://www.reliance.com/prod_elements/2lparen.gif" \* MERGEFORMATINET Load rpm  INCLUDEPICTURE "http://www.reliance.com/nav_elements/4pxlblack.gif" \* MERGEFORMATINET  Motor rpm INCLUDEPICTURE "http://www.reliance.com/prod_elements/2rparen.gif" \* MERGEFORMATINET 2 Synchronous Speed, Frequency And Number Of Poles Of AC Motors ns =120 x f  INCLUDEPICTURE "http://www.reliance.com/nav_elements/4pxlblack.gif" \* MERGEFORMATINET  P INCLUDEPICTURE "http://www.reliance.com/nav_elements/1pxltrn.gif" \* MERGEFORMATINET f =P x ns  INCLUDEPICTURE "http://www.reliance.com/nav_elements/4pxlblack.gif" \* MERGEFORMATINET  120 INCLUDEPICTURE "http://www.reliance.com/nav_elements/1pxltrn.gif" \* MERGEFORMATINET P =120 x f  INCLUDEPICTURE "http://www.reliance.com/nav_elements/4pxlblack.gif" \* MERGEFORMATINET  ns Relation Between Horsepower, Torque, And Speed HP =T x n  INCLUDEPICTURE "http://www.reliance.com/nav_elements/4pxlblack.gif" \* MERGEFORMATINET  5250 INCLUDEPICTURE "http://www.reliance.com/nav_elements/1pxltrn.gif" \* MERGEFORMATINET T =5250 HP  INCLUDEPICTURE "http://www.reliance.com/nav_elements/4pxlblack.gif" \* MERGEFORMATINET  n INCLUDEPICTURE "http://www.reliance.com/nav_elements/1pxltrn.gif" \* MERGEFORMATINET n =5250 HP  INCLUDEPICTURE "http://www.reliance.com/nav_elements/4pxlblack.gif" \* MERGEFORMATINET  T Motor Slip % Slip =ns - n  INCLUDEPICTURE "http://www.reliance.com/nav_elements/4pxlblack.gif" \* MERGEFORMATINET  nsx 100 CodeKVA/HPCodeKVA/HPCodeKVA/HPCodeKVA/HPA0-3.14F5.0 -5.59L9.0-9.99S16.0-17.99B3.15-3.54G5.6 -6.29M10.0-11.19T18.0-19.99C3.55-3.99H6.3 -7.09N11.2-12.49U20.0-22.39D4.0 -4.49I7.1 -7.99P12.5-13.99V22.4 & UpE4.5 -4.99K8.0 -8.99R14.0-15.99Symbols I=current in amperesE=voltage in voltsKW=power in kilowattsKVA=apparent power in kilo-volt-amperesHP =output power in horsepowern=motor speed in revolutions per minute (RPM)ns=synchronous speed in revolutions per minute (RPM)P=number of polesf=frequency in cycles per second (CPS)T=torque in pound-feetEFF=efficiency as a decimalPF=power factor as a decimal Equivalent Inertia In mechanical systems, all rotating parts do not usually operate at the same speed. Thus, we need to determine the "equivalent inertia" of each moving part at a particular speed of the prime mover. The total equivalent WK2 for a system is the sum of the WK2 of each part, referenced to prime mover speed. The equation says: WK2EQ = WK2part  INCLUDEPICTURE "http://www.reliance.com/prod_elements/2lparen.gif" \* MERGEFORMATINET  Npart  INCLUDEPICTURE "http://www.reliance.com/nav_elements/4pxlblack.gif" \* MERGEFORMATINET  Nprime mover  INCLUDEPICTURE "http://www.reliance.com/prod_elements/2rparen.gif" \* MERGEFORMATINET 2 This equation becomes a common denominator on which other calculations can be based. For variable-speed devices, inertia should be calculated first at low speed. Let's look at a simple system which has a prime mover (PM), a reducer and a load. WK2 = 100 lb.ft.2WK2 = 900 lb.ft.2 (as seen at output shaft)WK2 = 27,000 lb.ft.2PRIME MOVER  3:1 GEAR REDUCER  LOAD The formula states that the system WK2 equivalent is equal to the sum of WK2parts at the prime mover's RPM, or in this case: WK2EQ = WK2pm + WK2Red.  INCLUDEPICTURE "http://www.reliance.com/prod_elements/2lparen.gif" \* MERGEFORMATINET  Red. RPM  INCLUDEPICTURE "http://www.reliance.com/nav_elements/4pxlblack.gif" \* MERGEFORMATINET  PM RPM  INCLUDEPICTURE "http://www.reliance.com/prod_elements/2rparen.gif" \* MERGEFORMATINET 2 + WK2Load  INCLUDEPICTURE "http://www.reliance.com/prod_elements/2lparen.gif" \* MERGEFORMATINET  Load RPM  INCLUDEPICTURE "http://www.reliance.com/nav_elements/4pxlblack.gif" \* MERGEFORMATINET  PM RPM  INCLUDEPICTURE "http://www.reliance.com/prod_elements/2rparen.gif" \* MERGEFORMATINET 2 Note: reducer RPM = Load RPM WK2EQ = WK2pm + WK2Red.  INCLUDEPICTURE "http://www.reliance.com/prod_elements/2lparen.gif" \* MERGEFORMATINET  1  INCLUDEPICTURE "http://www.reliance.com/nav_elements/4pxlblack.gif" \* MERGEFORMATINET  3  INCLUDEPICTURE "http://www.reliance.com/prod_elements/2rparen.gif" \* MERGEFORMATINET 2 + WK2Load  INCLUDEPICTURE "http://www.reliance.com/prod_elements/2lparen.gif" \* MERGEFORMATINET  1  INCLUDEPICTURE "http://www.reliance.com/nav_elements/4pxlblack.gif" \* MERGEFORMATINET  3  INCLUDEPICTURE "http://www.reliance.com/prod_elements/2rparen.gif" \* MERGEFORMATINET 2 The WK2 equivalent is equal to the WK2 of the prime mover, plus the WK2 of the load. This is equal to the WK2 of the prime mover, plus the WK2 of the reducer times (1/3)2, plus the WK2 of the load times (1/3)2. This relationship of the reducer to the driven load is expressed by the formula given earlier: WK2EQ = WK2part  INCLUDEPICTURE "http://www.reliance.com/prod_elements/2lparen.gif" \* MERGEFORMATINET  Npart  INCLUDEPICTURE "http://www.reliance.com/nav_elements/4pxlblack.gif" \* MERGEFORMATINET  Nprime mover  INCLUDEPICTURE "http://www.reliance.com/prod_elements/2rparen.gif" \* MERGEFORMATINET 2 In other words, when a part is rotating at a speed (N) different from the prime mover, the WK2EQ is equal to the WK2 of the part's speed ratio squared. In the example, the result can be obtained as follows: The WK2 equivalent is equal to: WK2EQ = 100 lb.ft.2 + 900 lb.ft.2  INCLUDEPICTURE "http://www.reliance.com/prod_elements/2lparen.gif" \* MERGEFORMATINET  1  INCLUDEPICTURE "http://www.reliance.com/nav_elements/4pxlblack.gif" \* MERGEFORMATINET  3  INCLUDEPICTURE "http://www.reliance.com/prod_elements/2rparen.gif" \* MERGEFORMATINET 2 + 27,000 lb.ft.2  INCLUDEPICTURE "http://www.reliance.com/prod_elements/2lparen.gif" \* MERGEFORMATINET  1  INCLUDEPICTURE "http://www.reliance.com/nav_elements/4pxlblack.gif" \* MERGEFORMATINET  3  INCLUDEPICTURE "http://www.reliance.com/prod_elements/2rparen.gif" \* MERGEFORMATINET 2 Finally: WK2EQ = lb.ft.2pm + 100 lb.ft.2Red + 3,000 lb.ft2Load WK2EQ = 3200 lb.ft.2 The total WK2 equivalent is that WK2 seen by the prime mover at its speed. Electrical Formulas To FindAlternating CurrentSingle-PhaseThree-PhaseAmperes when horsepower is knownHP x 746  INCLUDEPICTURE "http://www.reliance.com/nav_elements/4pxlblack.gif" \* MERGEFORMATINET  E x Eff x pfHP x 746  INCLUDEPICTURE "http://www.reliance.com/nav_elements/4pxlblack.gif" \* MERGEFORMATINET  1.73 x E x Eff x pfAmperes when kilowatts are knownKw x 1000  INCLUDEPICTURE "http://www.reliance.com/nav_elements/4pxlblack.gif" \* MERGEFORMATINET  E x pfKw x 1000  INCLUDEPICTURE "http://www.reliance.com/nav_elements/4pxlblack.gif" \* MERGEFORMATINET  1.73 x E x pfAmperes when kva are knownKva x 1000  INCLUDEPICTURE "http://www.reliance.com/nav_elements/4pxlblack.gif" \* MERGEFORMATINET  EKva x 1000  INCLUDEPICTURE "http://www.reliance.com/nav_elements/4pxlblack.gif" \* MERGEFORMATINET  1.73 x EKilowattsI x E x pf  INCLUDEPICTURE "http://www.reliance.com/nav_elements/4pxlblack.gif" \* MERGEFORMATINET  10001.73 x I x E x pf  INCLUDEPICTURE "http://www.reliance.com/nav_elements/4pxlblack.gif" \* MERGEFORMATINET  1000KvaI x E  INCLUDEPICTURE "http://www.reliance.com/nav_elements/4pxlblack.gif" \* MERGEFORMATINET  10001.73 x I x E  INCLUDEPICTURE "http://www.reliance.com/nav_elements/4pxlblack.gif" \* MERGEFORMATINET  1000Horsepower = (Output)I x E x Eff x pf  INCLUDEPICTURE "http://www.reliance.com/nav_elements/4pxlblack.gif" \* MERGEFORMATINET  7461.73 x I x E x Eff x pff  INCLUDEPICTURE "http://www.reliance.com/nav_elements/4pxlblack.gif" \* MERGEFORMATINET  746I = Amperes; E = Volts; Eff = Efficiency; pf = Power Factor; Kva = Kilovolt-amperes; Kw = Kilowatts Locked Rotor Current (IL) From Nameplate Data Three Phase: IL =577 x HP x KVA/HP  INCLUDEPICTURE "http://www.reliance.com/nav_elements/4pxlblack.gif" \* MERGEFORMATINET  E INCLUDEPICTURE "http://www.reliance.com/nav_elements/1pxltrn.gif" \* MERGEFORMATINET  HYPERLINK "http://www.reliance.com/mtr/flaclcmn.htm" \l "kvahp#kvahp" See: KVA/HP ChartSingle Phase: IL =1000 x HP x KVA/HP  INCLUDEPICTURE "http://www.reliance.com/nav_elements/4pxlblack.gif" \* MERGEFORMATINET  E EXAMPLE: Motor nameplate indicates 10 HP, 3 Phase, 460 Volts, Code F. IL = 577 x 10 x (5.6 or 6.29)  INCLUDEPICTURE "http://www.reliance.com/nav_elements/4pxlblack.gif" \* MERGEFORMATINET  460 IL = 70.25 or 78.9 Amperes (possible range)  Effect Of Line Voltage On Locked Rotor Current (IL) (Approx.) IL @ ELINE = IL @ EN/P xELINE  INCLUDEPICTURE "http://www.reliance.com/nav_elements/4pxlblack.gif" \* MERGEFORMATINET  EN/P EXAMPLE: Motor has a locked rotor current (inrush of 100 Amperes (IL) at the rated nameplate voltage (EN/P) of 230 volts. What is IL with 245 volts (ELINE) applied to this motor? IL @ 245 V. = 100 x 254V/230V IL @ 245V. = 107 Amperes Basic Horsepower Calculations Horsepower is work done per unit of time. One HP equals 33,000 ft-lb of work per minute. When work is done by a source of torque (T) to produce (M) rotations about an axis, the work done is: radius x 2  INCLUDEPICTURE "http://www.reliance.com/prod_elements/pi.gif" \* MERGEFORMATINET x rpm x lb. or 2  INCLUDEPICTURE "http://www.reliance.com/prod_elements/pi.gif" \* MERGEFORMATINET TMWhen rotation is at the rate N rpm, the HP delivered is: HP = radius x 2  INCLUDEPICTURE "http://www.reliance.com/prod_elements/pi.gif" \* MERGEFORMATINET x rpm x lb.  INCLUDEPICTURE "http://www.reliance.com/nav_elements/4pxlblack.gif" \* MERGEFORMATINET  33,000 = TN  INCLUDEPICTURE "http://www.reliance.com/nav_elements/4pxlblack.gif" \* MERGEFORMATINET  5,250 For vertical or hoisting motion: HP =W x S  INCLUDEPICTURE "http://www.reliance.com/nav_elements/4pxlblack.gif" \* MERGEFORMATINET  33,000 x EWhere: W = total weight in lbs. to be raised by motor S = hoisting speed in feet per minute E = overall mechanical efficiency of hoist and gearing. For purposes of estimating E = .65 for eff. of hoist and connected gear. For fans and blowers: HP = Volume (cfm) x Head (inches of water)  INCLUDEPICTURE "http://www.reliance.com/nav_elements/4pxlblack.gif" \* MERGEFORMATINET  6356 x Mechanical Efficiency of Fan Or HP = Volume (cfm) x Pressure (lb. Per sq. ft.)  INCLUDEPICTURE "http://www.reliance.com/nav_elements/4pxlblack.gif" \* MERGEFORMATINET  3300 x Mechanical Efficiency of Fan Or HP = Volume (cfm) x Pressure (lb. Per sq. in.)  INCLUDEPICTURE "http://www.reliance.com/nav_elements/4pxlblack.gif" \* MERGEFORMATINET  229 x Mechanical Efficiency of Fan For purpose of estimating, the eff. of a fan or blower may be assumed to be 0.65. Note: Air Capacity (cfm) varies directly with fan speed. Developed Pressure varies with square of fan speed. Hp varies with cube of fan speed.For pumps: HP = GPM x Pressure in lb. Per sq. in. x Specific Grav.  INCLUDEPICTURE "http://www.reliance.com/nav_elements/4pxlblack.gif" \* MERGEFORMATINET  1713 x Mechanical Efficiency of Pump Or HP = GPM x Total Dynamic Head in Feet x S.G.  INCLUDEPICTURE "http://www.reliance.com/nav_elements/4pxlblack.gif" \* MERGEFORMATINET  3960 x Mechanical Efficiency of Pump  where Total Dynamic Head = Static Head + Friction Head For estimating, pump efficiency may be assumed at 0.70. Accelerating Torque The equivalent inertia of an adjustable speed drive indicates the energy required to keep the system running. However, starting or accelerating the system requires extra energy. The torque required to accelerate a body is equal to the WK2 of the body, times the change in RPM, divided by 308 times the interval (in seconds) in which this acceleration takes place: ACCELERATING TORQUE = WK2N (in lb.ft.)  INCLUDEPICTURE "http://www.reliance.com/nav_elements/4pxlblack.gif" \* MERGEFORMATINET  308t Where: N = Change in RPM W = Weight in Lbs. K = Radius of gyration t = Time of acceleration (secs.) WK2 = Equivalent Inertia 308 = Constant of proportionality Or TAcc = WK2N  INCLUDEPICTURE "http://www.reliance.com/nav_elements/4pxlblack.gif" \* MERGEFORMATINET  308t The constant (308) is derived by transferring linear motion to angular motion, and considering acceleration due to gravity. If, for example, we have simply a prime mover and a load with no speed adjustment: Example 1 PRIME LOADER  LOAD WK2 = 200 lb.ft.2WK2 = 800 lb.ft.2The WK2EQ is determined as before: WK2EQ = WK2pm + WK2LoadWK2EQ = 200 + 800WK2EQ = 1000 ft.lb.2If we want to accelerate this load to 1800 RPM in 1 minute, enough information is available to find the amount of torque necessary to accelerate the load. The formula states: TAcc = WK2EQN  INCLUDEPICTURE "http://www.reliance.com/nav_elements/4pxlblack.gif" \* MERGEFORMATINET  308t or 1000 x 1800  INCLUDEPICTURE "http://www.reliance.com/nav_elements/4pxlblack.gif" \* MERGEFORMATINET  308 x 60 or 1800000  INCLUDEPICTURE "http://www.reliance.com/nav_elements/4pxlblack.gif" \* MERGEFORMATINET  18480  TAcc = 97.4 lb.ft. In other words, 97.4 lb.ft. of torque must be applied to get this load turning at 1800 RPM, in 60 seconds. Note that TAcc is an average value of accelerating torque during the speed change under consideration. If a more accurate calculation is desired, the following example may be helpful. Example 2 The time that it takes to accelerate an induction motor from one speed to another may be found from the following equation: t = WR2 x change in rpm  INCLUDEPICTURE "http://www.reliance.com/nav_elements/4pxlblack.gif" \* MERGEFORMATINET  308 x T Where: T = Average value of accelerating torque during the speed change under consideration. t = Time the motor takes to accelerate from the initial speed to the final speed. WR2 = Flywheel effect, or moment of inertia, for the driven machinery plus the motor rotor in lb.ft.2 (WR2 of driven machinery must be referred to the motor shaft). The Application of the above formula will now be considered by means of an example. Figure A shows the speed-torque curves of a squirrel-cage induction motor and a blower which it drives. At any speed of the blower, the difference between the torque which the motor can deliver at its shaft and the torque required by the blower is the torque available for acceleration. Reference to Figure A shows that the accelerating torque may vary greatly with speed. When the speed-torque curves for the motor and blower intersect there is no torque available for acceleration. The motor then drives the blower at constant speed and just delivers the torque required by the load. In order to find the total time required to accelerate the motor and blower, the area between the motor speed-torque curve and the blower speed-torque curve is divided into strips, the ends of which approximate straight lines. Each strip corresponds to a speed increment which takes place within a definite time interval. The solid horizontal lines in Figure A represent the boundaries of strips; the lengths of the broken lines the average accelerating torques for the selected speed intervals. In order to calculate the total acceleration time for the motor and the direct-coupled blower it is necessary to find the time required to accelerate the motor from the beginning of one speed interval to the beginning of the next interval and add up the incremental times for all intervals to arrive at the total acceleration time. If the WR2 of the motor whose speed-torque curve is given in Figure A is 3.26 ft.lb.2 and the WR2 of the blower referred to the motor shaft is 15 ft.lb.2, the total WR2 is: 15 + 3.26 = 18.26 ft.lb.2,And the total time of acceleration is: WR2  INCLUDEPICTURE "http://www.reliance.com/nav_elements/4pxlblack.gif" \* MERGEFORMATINET  308  INCLUDEPICTURE "http://www.reliance.com/prod_elements/2lbracket.gif" \* MERGEFORMATINET  rpm1  INCLUDEPICTURE "http://www.reliance.com/nav_elements/4pxlblack.gif" \* MERGEFORMATINET  T1 + rpm2  INCLUDEPICTURE "http://www.reliance.com/nav_elements/4pxlblack.gif" \* MERGEFORMATINET  T2 + rpm3  INCLUDEPICTURE "http://www.reliance.com/nav_elements/4pxlblack.gif" \* MERGEFORMATINET  T3 + - - - - - - - - - + rpm9  INCLUDEPICTURE "http://www.reliance.com/nav_elements/4pxlblack.gif" \* MERGEFORMATINET  T9  INCLUDEPICTURE "http://www.reliance.com/prod_elements/2rbracket.gif" \* MERGEFORMATINET  Or t = 18.26  INCLUDEPICTURE "http://www.reliance.com/nav_elements/4pxlblack.gif" \* MERGEFORMATINET  308  INCLUDEPICTURE "http://www.reliance.com/prod_elements/2lbracket.gif" \* MERGEFORMATINET  150  INCLUDEPICTURE "http://www.reliance.com/nav_elements/4pxlblack.gif" \* MERGEFORMATINET  46 + 150  INCLUDEPICTURE "http://www.reliance.com/nav_elements/4pxlblack.gif" \* MERGEFORMATINET  48 + 300  INCLUDEPICTURE "http://www.reliance.com/nav_elements/4pxlblack.gif" \* MERGEFORMATINET  47 + 300  INCLUDEPICTURE "http://www.reliance.com/nav_elements/4pxlblack.gif" \* MERGEFORMATINET  43.8 + 200  INCLUDEPICTURE "http://www.reliance.com/nav_elements/4pxlblack.gif" \* MERGEFORMATINET  39.8 + 200  INCLUDEPICTURE "http://www.reliance.com/nav_elements/4pxlblack.gif" \* MERGEFORMATINET  36.4 + 300  INCLUDEPICTURE "http://www.reliance.com/nav_elements/4pxlblack.gif" \* MERGEFORMATINET  32.8 + 100  INCLUDEPICTURE "http://www.reliance.com/nav_elements/4pxlblack.gif" \* MERGEFORMATINET  29.6 + 40  INCLUDEPICTURE "http://www.reliance.com/nav_elements/4pxlblack.gif" \* MERGEFORMATINET  11  INCLUDEPICTURE "http://www.reliance.com/prod_elements/2rbracket.gif" \* MERGEFORMATINET   t = 2.75 sec.Figure A Curves used to determine time required to accelerate induction motor and blower Accelerating TorquesT1 = 46 lb.ft.T4 = 43.8 lb.ft.T7 = 32.8 lb.ft.T2 = 48 lb.ft.T5 = 39.8 lb.ft.T8 = 29.6 lb.ft.T3 = 47 lb.ft.T6 = 36.4 lb.ft.T9 = 11 lb.ft.  INCLUDEPICTURE "http://www.reliance.com/mtr/images/fcfiga.gif" \* MERGEFORMATINET  Duty Cycles Sales Orders are often entered with a note under special features such as:  INCLUDEPICTURE "http://www.reliance.com/nav_elements/1pxltrn.gif" \* MERGEFORMATINET "Suitable for 10 starts per hour" Or  INCLUDEPICTURE "http://www.reliance.com/nav_elements/1pxltrn.gif" \* MERGEFORMATINET "Suitable for 3 reverses per minute" Or  INCLUDEPICTURE "http://www.reliance.com/nav_elements/1pxltrn.gif" \* MERGEFORMATINET "Motor to be capable of accelerating 350 lb.ft.2" Or  INCLUDEPICTURE "http://www.reliance.com/nav_elements/1pxltrn.gif" \* MERGEFORMATINET "Suitable for 5 starts and stops per hour" Orders with notes such as these can not be processed for two reasons. The appropriate product group must first be consulted to see if a design is available that will perform the required duty cycle and, if not, to determine if the type of design required falls within our present product line. None of the above notes contain enough information to make the necessary duty cycle calculation. In order for a duty cycle to be checked out, the duty cycle information must include the following: Inertia reflected to the motor shaft. Torque load on the motor during all portions of the duty cycle including starts, running time, stops or reversals. Accurate timing of each portion of the cycle. Information on how each step of the cycle is accomplished. For example, a stop can be by coasting, mechanical braking, DC dynamic braking or plugging. A reversal can be accomplished by plugging, or the motor may be stopped by some means then re-started in the opposite direction. When the motor is multi-speed, the cycle for each speed must be completely defined, including the method of changing from one speed to another. Any special mechanical problems, features or limitations. Obtaining this information and checking with the product group before the order is entered can save much time, expense and correspondence. Duty cycle refers to the detailed description of a work cycle that repeats in a specific time period. This cycle may include frequent starts, plugging stops, reversals or stalls. These characteristics are usually involved in batch-type processes and may include tumbling barrels, certain cranes, shovels and draglines, dampers, gate- or plow-positioning drives, drawbridges, freight and personnel elevators, press-type extractors, some feeders,presses of certain types, hoists, indexers, boring machines,cinder block machines, keyseating, kneading, car-pulling, shakers (foundry or car), swaging and washing machines, and certain freight and passenger vehicles. The list is not all-inclusive. The drives for these loads must be capable of absorbing the heat generated during the duty cycles. Adequate thermal capacity would be required in slip couplings, clutches or motors to accelerate or plug-stop these drives or to withstand stalls. It is the product of the slip speed and the torque absorbed by the load per unit of time which generates heat in these drive components. All the events which occur during the duty cycle generate heat which the drive components must dissipate. AL SALEM YORK SERVICES LTD PROCEDURES FOR EXTENDED STORAGE OF MOTORS 1) GENERAL 1.1. Motors must be kept in their original packaging (or with equivalent protection). In addition, they must be stored in a warehouse free from extremes in temperature, humidity and corrosive atmosphere. 1.2. If unusual vibrations exist at the storage location, the motor should be protected with isolation pads. 1.3. All breathers and drains are to be operable while in storage and/or the moisture drain plugs should be removed. The motors must be stored so the drain is at the lowest point. Storage Preparation Improper storage of electric machines will result in seriously reduced reliability of that equipment. For example, the following items can occur to an electric motor that does not experience regular usage while exposed to normally humid atmospheric conditions: Corrosion of shaft, stator, rotor & bearings. Rust particles from surrounding surfaces may contaminate the bearings. The electrical insulation may absorb an excessive amount of moisture leading to the motor winding failing to ground. Brinelling of the bearings due to vibration frequencies in excess of the natural frequency of the motor housing and bearings. 1.4. Minimize condensation in and around the motor by use of desiccants or other humidity control methods. 1.5. Motor space heaters, when fitted, should have a tempory power supply connected and be energized when there is a possibility that the ambient storage conditions will reach the dew point. 1.6. Coat all external machined surfaces including couplings with a material to prevent corrosion. An acceptable product for this purpose is Exxon Rust Ban #392 (or equivalent). 1.7. Measure and record the electrical resistance of the winding insulation with a megger or insulation resistance meter. Minimum accepted megohm level is the insulation kv rating + 1 megohm. If levels fall below this value, contact Al Salem York Office. The recorded data will be needed when the motor is removed from storage. 2) Extended Storage of AC Motors, Ship or Rig board Motors 2.1. Where the motor is not shipped as part of a drive line assembly, a shipping brace must be attached to the shaft to prevent damage during transportation. The shipping brace, if provided, must be removed and stored for future use. Before the motor is moved, the brace must be reinstalled to hold the shaft firmly in place against the bearing. Large AC motors should not be moved without the shipping brace in place. 2.2. When placing the motor into extended storage (greater than 3 months), the motors with regreasable bearings must be greased. The motor shaft must then be rotated a minimum of 15 times after greasing. Non-regreasable motors with a Do Not Lubricate nameplate should also be rotated 15 times to redistribute grease within the bearing and prevent Brinelling of the bearings. 2.3. Before lubricating the motor, remove the grease drain plug (opposite the grease fitting) on the bottom of each end bracket. Replace the plug after greasing. 2.4. Motors with oil lubricated bearings and or lube oil system, should have the oil reservoir filled with the appropriate oil, for providing lubrication during storage. 2.5. Under no circumstances should motor shafts be rotated without first ensuring the bearings are greased or lubricated as appropriate ( ex non greasable bearings). 2.6. Shafts on non-regreasable motors should be rotated 15 revolutions every 3 months. 2.7. All breather drains should be fully operable while in storage. The motors must be stored so the drain is at the lowest point. All breathers and automatic T drains must be operable to allow breathing at points other than through the bearing fits. 2.8. Space heaters, when fitted, are to be connected and operable while in storage. 2.9. Windings must be meggered at the time equipment is put in storage 2.10. When the motor is removed from storage, the insulation resistance must not have dropped more than 50% from the initial reading. Any drop below this point necessitates electrical or mechanical drying. Refer to motor drying procedure. 2.11. When motors are not stored in the original containers, but are removed and mounted on other pieces of machinery, the mounting must be such that the drains/breathers and space heaters are fully operable. In this respect, the drains must be kept at the lowest point in the motor so that all condensation can automatically drain out. TEMPERATURE RISE AND CLASS OF INSULATION FOR AC MOTORS ISSUED BY C.J.HEWETSON 10/1/03 The correct class of insulation is essential in terms of a satisfactory operational life of AC motors, particularly in Middle East high ambient operation. The motor operational life is dependant upon a number of factors The winding insulation class. The ambient temperature. The motor efficiency. The Copper/Iron mass of the stator and windings. Voltage fluctuation & phase imbalance. Winding temperature. Bearing operating temperature and lubrication specification. The load on the motor. The winding operating temperature directly effects the life of winding insulation materials plus the bearings. Winding temperature rise is directly related to the efficiency of the motor, which is the power output at the shaft divided by the power input at the supply side. In general, motors are 85% efficient unless high efficiency motors are requested, in which case the efficiency would be 90-95%. High efficiency motors have more Copper/Iron by design; therefore the losses and the equivalent heat generation will be less. The difference between power input and output is called the loss and this power loss is transformed into heat, which warms up the motor windings and must be expelled from the motor to avoid excessive temperature rise. If the efficiency of the motor is 85% then 15% of the KW input is generated in winding heat with an equivalent winding temperature rise based upon the mass of Copper and Iron and the specific heat thereof. The hottest part of the windings is in the centre of the slots where the heat is generated as a result of the losses. The heat generated is dissipated to the ambient air through the external surfaces of the motor and assistance of air forced over the surface in the case of TEFC or through the stator/rotor air gap in the case of ODP motors, via a fan mounted on the rotor shaft end. For every 10degf (5.5degC) winding temperature rise above the insulation rating of the motor, the motor operating life will be reduced by 50%. Motor insulation temperature rise allowable for each class of insulation is always based upon an ambient of 72degf (40degC) which is the air on temperature to the motor for cooling purposes. The ambient design should be taken as 50 deg C for the Eastern Province (unless the motor is located indoors in an air conditioned or ventilated environment in which case the design ambient may be lower). As motor insulation class is based on 40DegC ambient, and design ambient is 50DegC, the winding temperature will be 10degC (18degf) above the allowable temperature rise and likely to exceed the maximum temperature for the windings. For TEAO or TEFC motors the bearings provided will relate to the expected motor winding temperature. As most of the York product designs are based on N.American markets, the insulation class specified will generally be class B and the bearings fitted will reflect this. For Saudi Arabia the winding temperature is likely to fall into class F or H and if you consider that lubricants begin to break down and lose their lubricating properties at 85-90degC, it can be readily understood that the bearings may well fail due to a reduction in grease viscosity and leakage or vapourisation before the motor windings. Indeed as the bearing lubrication is gradually lost, the bearing elements will pick up, resulting in higher current draw which in itself may cause the motor windings to fail. Most motors supplied by York have no service factor or thermal reserve as described by NEMA or IEC standards, which allows for overloading of the motor for short or continuous periods without serious damage of overheating. The service factors that should be specified are 1.10, 1.15 or 1.2 which provides a 10,15 or 20% overload; sales people should not accept a service factor of 1.0. It is very easy to overload motors especially during pull down, increased air flow on AHU's due to lower ESP, changes in air density/humidity or due to power supply voltage fluctuations or phase imbalances. Under such conditions a thermal reserve will allow the motor to operate without exceeding their class rating by using some or all of their thermal reserve. The temperature rise for motors with 1.0 service factor is as follows. INSULATION CLASSAEBFHF WITH CLASS B RISETEMPERATURE RISE OF WINDING6075 80 10012580AMBIENT TEMPERATURE 404040404040ALLOWANCE FOR HOT SPOTS5510151515THERMAL RESERVE 0000020TOTAL WINDING TEMPERATURE105120130155180155 For all motors be they single or 3 phase for operation in Saudi Arabia to ensure trouble free operation and a normally expected life of 25,000hrs, the winding class should be class F with a service factor of 1.1 as a minimum especially in view of the likelihood of voltage imbalances and voltage reduction general to SA. In the Eastern Province we are frequently faced with changing condenser fan motors on rooftop packages, split condensing unit condensers and air cooled chillers or to change the bearings. Motor failure due to incorrect specification of winding insulation class causes customer dissatisfaction with our products even though he may not appreciate the reasons for the failure. This in turn may result in lost sales in the future where customers do not perceive our units as being reliable or suited for high ambient operation. ASAC REFRIGERATION DIVISION EFFFECTS OF MOTOR VOLTAGE & FREQUENCY VARIATION C.HEWETSON All motors are designed to operate within limited voltage and frequency variations: voltage variation at rated frequency must be within 10%, and frequency variations at rated voltage must be 5%. The combined variation of voltage and frequency must be limited to the arithmetic sum of 10%. Variations are expressed as deviation from motor nameplate values, not necessarily system nominal values. The allowable 10% voltage variation is based upon the assumptions that horsepower will not exceed nameplate rating and that motor temperature may increase. For instance, a 230-Volt motor operating at 207-Volts (90% of rated) loses any service factor indicated on the nameplate, and could run hotter than at rated voltage. The following conditions are likely to occur with variations in voltage: An increase or decrease in voltage may result in increased heating at rated horsepower load. Under extended operation this may accelerate insulation deterioration and shorten motor life. An increase in voltage will usually result in a noticeable decrease in power factor. Conversely, a decrease in voltage will result in an increase in power factor. Locked-rotor and breakdown torque will be proportional to the square of the voltage. Therefore, a decrease in voltage will result in decreased torque availability. An increase of 10% in voltage will result in a reduction of slip of approximately 17%. A voltage reduction of 10% would increase slip by about 21%.  HYPERLINK "http://oge.apogee.net/pd/dmvf01.htm"  The following conditions occur with variations in frequency: Frequency greater than rated frequency normally improves power factor but decreases locked-rotor and maximum torque. This condition also increases speed, and the accompanying friction and windage losses. Conversely, a decrease in frequency will usually lower power factor and speed while increasing locked-rotor maximum torque and locked rotor current.  HYPERLINK "http://oge.apogee.net/pd/dmvf02.htm"  The important parameter relative to all induction motors is flux density. This is the ratio of line volts over line frequency. V/Hz for a 460 Volt motor operating at 60Hz is 7.67 V/Hz or for a 380Volt 60Hz motor 6.33V/Hz, these are the common operating voltages and frequencies in Saudi. On many occasions we face a situation where York supply 460Volt 60Hz motors designed for a V/Hz ratio of 7.67V/Hz to operate on 380Volt 60Hz which has a V/Hz ratio of 6.33V/Hz. The % between 7.67 & 6.33V/Hz is 17.47% which is outside the allowable combined voltage/Hz tolerance of +/-10% and should not therefore be used as the motor will overheat, also its rated output, PF and torque will be different. The situation would be far worse if the Saudi 380Volt power supply was 10% which it often is, where the ratio would then be 380-10%/60 = 5.7V/Hz. On some occasions we are supplied with 415Volt 50Hz motors, here the flux density is 8.3V/Hz so again the motor cannot be used on 380Volts. However a 415Volt 50Hz motor would run satisfactorily without overheating operating on 460Volt 60Hz as the flux density ratio difference is 7.5% which is within the +/- 10% of the nameplate volts.     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