Motor Starting

Motor Starting Contents [hide]

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1 Introduction

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1.1 Why do the calculation?

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1.2 When to do the calculation?

2 Calculation Methodology

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2.1 Step 1: Construct System Model and Collect Equipment Parameters

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2.2 Step 2: Calculate Equipment Impedances

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2.2.1 Network Feeders

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2.2.2 Synchronous Generators

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2.2.3 Transformers

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2.2.4 Cables

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2.2.5 Standing Loads

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2.2.6 Motors

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2.3 Step 3: Referring Impedances

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2.4 Step 4: Construct the Equivalent Circuit

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2.5 Step 5: Calculate the Initial Source EMF

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2.6 Step 6: Calculate System Voltages During Motor Start

3 Worked Example

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3.1 Step 1: Construct System Model and Collect Equipment Parameters

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3.2 Step 2: Calculate Equipment Impedances

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3.3 Step 3: Referring Impedances

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3.4 Step 4: Construct the Equivalent Circuit

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3.5 Step 5: Calculate the Initial Source EMF

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3.6 Step 6: Calculate System Voltages During Motor Start

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4 Computer Software

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5 What Next?

Introduction

High voltage motor (courtesy of ABB)

This article considers the transient effects of motor starting on the system voltage. Usually only the largest motor on a bus or system is modelled, but the calculation can in principle be used for any motor. It's important to note that motor starting is a transient power flow problem and is normally done iteratively by computer software. However a static method is shown here for first-pass estimates only.

Why do the calculation? When a motor is started, it typically draws a current 6-7 times its full load current for a short duration (commonly called the locked rotor current). During this transient period, the source impedance is generally assumed to be fixed and therefore, a large increase in current will result in a larger voltage drop across the source impedance. This means that there can be large momentary voltage drops system-wide, from the power source (e.g. transformer or generator) through the intermediary buses, all the way to the motor terminals. A system-wide voltage drop can have a number of adverse effects, for example: 

Equipment with minimum voltage tolerances (e.g. electronics) may malfunction or behave aberrantly

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Undervoltage protection may be tripped

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The motor itself may not start as torque is proportional to the square of the stator voltage, so a reduced voltage equals lower torque. Induction motors are typically designed to start with a terminal voltage >80%

When to do the calculation? This calculation is more or less done to verify that the largest motor does not cause system wide problems upon starting. Therefore it should be done after preliminary system design is complete. The following prerequisite information is required: 

Key single line diagrams

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Preliminary load schedule

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Tolerable voltage drop limits during motor starting, which are typically prescribed by the client

Calculation Methodology This calculation is based on standard impedance formulae and Ohm's law. To the author's knowledge, there are no international standards that govern voltage drop calculations during motor start. It should be noted that the proposed method is not 100% accurate because it is a static calculation. In reality, the voltage levels are fluctuating during a transient condition, and therefore so are the load currents drawn by the standing loads. This makes it essentially a load flow problem and a more precise solution would solve the load flow problem iteratively, for example using the Newton-Rhapson or Gauss-Siedel algorithms. Notwithstanding, the proposed method is suitably accurate for a first pass solution. The calculation has the following six general steps: 

Step 1: Construct the system model and assemble the relevant equipment parameters

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Step 2: Calculate the relevant impedances for each equipment item in the model

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Step 3: Refer all impedances to a reference voltage

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Step 4: Construct the equivalent circuit for the voltage levels of interest

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Step 5: Calculate the initial steady-state source emf before motor starting

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Step 6: Calculate the system voltages during motor start

Step 1: Construct System Model and Collect Equipment Parameters The first step is to construct a simplified model of the system single line diagram, and then collect the relevant equipment parameters. The model of the single line diagram need only show the buses of interest in the motor starting calculation, e.g. the upstream source bus, the motor bus and possibly any intermediate or downstream buses that may be affected. All running loads are shown as lumped loads except for the motor to be started as it is assumed that the system is in a steady-state before motor start. The relevant equipment parameters to be collected are as follows: 

Network feeders: fault capacity of the network (VA), X/R ratio of the network

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Generators: per-unit transient reactance, rated generator capacity (VA)

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Transformers: transformer impedance voltage (%), rated transformer capacity (VA), rated current (A), total copper loss (W)

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Cables: length of cable (m), resistance and reactance of cable (

)

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Standing loads: rated load capacity (VA), average load power factor (pu)

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Motor: full load current (A), locked rotor current (A), rated power (W), full load power factor (pu), starting power factor (pu)

Step 2: Calculate Equipment Impedances Using the collected parameters, each of the equipment item impedances can be calculated for later use in the motor starting calculations.

Network Feeders Given the approximate fault level of the network feeder at the connection point (or point of common coupling), the impedance, resistance and reactance of the network feeder is calculated as follows:

Where

is impedance of the network feeder (Ω)

is resistance of the network feeder (Ω) is reactance of the network feeder (Ω) is the nominal voltage at the connection point (Vac) is the fault level of the network feeder (VA) is a voltage factor which accounts for the maximum system voltage (1.05 for voltages 1kV) is X/R ratio of the network feeder (pu)

Synchronous Generators The transient resistance and reactance of a synchronous generator can be estimated by the following:

Where (Ω)

is the transient reactance of the generator

is the resistance of the generator (Ω) is a voltage correction factor (pu) is the per-unit transient reactance of the generator (pu) is the nominal generator voltage (Vac) is the nominal system voltage (Vac) is the rated generator capacity (VA) is the X/R ratio, typically 20 for generators with nominal voltage

100MVA, 14.29 for

100MVA, and 6.67 for all

1kV

is a voltage factor which accounts for the maximum system voltage (1.05 for voltages 1kV) is the power factor of the generator (pu)

Transformers The impedance, resistance and reactance of twowinding transformers can be calculated as follows:

Where

is the impedance of the

transformer (Ω) is the resistance of the transformer (Ω) is the reactance of the transformer (Ω) is the impedance voltage of the transformer (pu) is the rated capacity of the transformer (VA) is the nominal voltage of the transformer at the high or low voltage side (Vac) is the rated current of the transformer at the high or low voltage side (I) is the total copper loss in the transformer windings (W)

Cables Cable impedances are usually quoted by manufacturers in terms of Ohms per km. These need to be converted to Ohms

based on the length of the cables:

Where

is the

resistance of the cable {Ω) is the reactance of the cable {Ω) is the quoted resistance of the cable {Ω / km) is the quoted reactance of the cable {Ω / km) is the length of the cable {m)

Standing Loads Standing loads are lumped loads comprising all loads that are operating on a particular bus, excluding the motor to be started. Standing loads for each bus need to be calculated. The impedance, resistance and reactance of the standing load is calculated by:

W h er e is th e i m p e d a nc e of th e st a n di n g lo a d { Ω) is the resistance of the standing load {Ω) is the reactance of the standing load {Ω) is the standing load nominal voltage (Vac) is the standing load apparent power (VA) is the average load power factor (pu)

M ot or s

Th e m ot or' s tra nsi en t im pe da nc e, re sis ta nc e an d re ac ta nc e is cal cul at ed as fol lo ws :

Where

transien

impedan

the mot is transient resistance of the motor (Ω) is transient reactance of the motor (Ω) is ratio of the locked rotor to full load current is the motor locked rotor current (A) is the motor nominal voltage (Vac) is the motor rated power (W) is the motor full load power factor (pu) is the motor starting power factor (pu)

Step Refer Impe

Where t

multiple

levels, t

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calculat

need to

to a refe voltage

HV side

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single e circuit.

The win

transfor

calculat

Where

transfor

is the transformer nominal secondary voltage at the principal tap (Vac) is the transformer nominal primary voltage (Vac) is the specified tap setting (%)

Using th

impedan

resistan

can be r

(HV) sid

by the f

Where

referred (Ω) is the impedance at the secondary (LV) side (Ω) is the transformer winding ratio (pu)

Convers

equatio

referred

Step Equiv

"Near" Th

The equ

a voltag

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form th circuit, voltage

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power s

generat used in

the initi

Assump

conditio 

The source point of common coupling (PCC) is at its nominal voltage



The motor is switched off



All standing loads are operating at the capacity calculated in Step 2



All transformer taps are set at those specified in Step 2



The system is at a steady-state, i.e. there is no switching taking place throughout the system

Since w

the PCC

by volta

Where is the nominal voltage (Vac)

is the source impedance (Ω) is the equivalent load impedance with the motor switched off (Ω)

Step Durin

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the initi

constan

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the pow

Next, w

is suppl period.

circuit d

impedan

starting

The cur

Where is the initial source emf (Vac) is the equivalent load impedance during motor start (Ω) is the source impedance (Ω)

The volt

Where is the initial source emf (Vac) is the system current supplied by the source (A) is the source impedance (Ω)

The dow

simple a

the mot

that the

action n

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levels a

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(S1) an

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initial ge

Vac

Step

Now we current

Next, th

Vac (or 87.98% of nominal voltage)

The volt

Vac (or 87.92% of nominal voltage)

The volt

Vac, then referred to the LV side = 359.39Vac (or 86.60% of nominal voltage)

Any oth

Suppose voltage

the syst

Comp

Motor s

this calc

softwar

What

If the re

levels w

softwar 

Reduce the motor starting current, e.g. via soft-starters, star-delta starters, etc



Reduce the source impedances, e.g. increase the size of the generator, transformer, supply cables, etc

The calc

Sizing Gen-Sets For Large Motor Starting Feb 1, 2008 12:00 PM, By Larry A. Bey, Cummins Onan Corp.

Remember, an on-site engine-generator set is a limited source of power, both in horsepower available from the engine, and kVA available from the generator. As such, it must be large enough to start as well as run connected motor loads. You've lost normal power. Your engine-generator set (gen-set) starts up and reaches speed. Now, you want to start some large motors key to your operation. Suddenly, starter holding coils drop out, starter contacts chatter, and a few motors stall due to insufficient torque for acceleration. Can this happen to you? It sure can, if you haven't sized your gen-set properly. We all know that motors draw a high inrush current during starting:typically six times full load current. But, inrush currents for the highefficient motors specified today are almost double that amount. Motors with high inertia loads can also require up to three times rated power during starting. Yes, it's common for motor starting kVA requirements to determine the size of the set. However, the following factors also play a key role in sizing gen-sets: •

Harmonics caused by variable frequency drives.

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Use of high-efficiency motors.

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Sequential starting of motors.

What's involved. When starting motors, large voltage and frequency dips may occur if the generator set isn't sized properly. Other loads connected to the generator output may be more sensitive to voltage and frequency dips than the motor or motor starter, and this may cause problems. For example, a rate of change greater than 1 Hz/sec in generator frequency may cause some static UPS units to malfunction. If the load on the generator set is a single large motor, particularly one requiring high starting torque, a number of problems can occur. They include: sustained low-voltage operation that can cause overheating; extended load acceleration times; opening of circuit breakers or motor protective devices; engine-generator protection shutdowns; and more. Your gen-set's ability to start large motors without excessive voltage and frequency dip is a function of the complete system. This includes: •

The engine power available;

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The generator's capacity;

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The response of the generator excitation system;

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The energy stored in the rotating inertia of the gen-set; and

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The acceleration of the motor and its load.

You must consider all these factors for proper gen-set sizing. Here's a simple rule for estimating the size of an engine-generator set for motor starting: 1kW of generator set rating per each 3/4 to 1 hp of motor nameplate. Let's look more closely at a motor start. Induction motors have typical starting characteristics. The curve of motor current versus speed shows that during starting, the motor draws approximately six times its full load current; this current remains high until the motor reaches about 80% of speed. This high inrush current causes a dip in generator voltage. The electric power initially required by the motor (with the motor at standstill) is about 150% of rated power. The power required by the motor peaks at about 300% of rated power and 80% of speed with full voltage applied. But, the generator set supplies less than 300% power because starting voltage is lower than full voltage during acceleration, and because the generator set's rotating inertia transfers energy to the motor. The motor must develop greater torque than required by the load. The motor's torque curve at full voltage is above the load's torque curve. The difference between the torque developed by motor and the torque required by the load determines the rate of acceleration. Since torque is proportional to voltage, any reduction in voltage means a proportional reduction in torque. A properly sized generator set will support the high starting kVA requirements of the motor, and maintain sufficient output voltage for the motor, so it can develop adequate torque to accelerate the load to rated speed. All standby gen-sets use synchronous generators with exciters. Many are available with permanent magnet generator (PMG) excitation systems. The PMG provides excitation power independent of the generator terminal voltage. As such, it can maintain full excitation:even during transient loading, such as motor starting. Full excitation power results in a less extensive voltage dip and improved recovery times. Using reduced-voltage starting. Though a voltage dip often causes various problems, a controlled reduction in voltage at motor terminals can be beneficial, but only when reduction in motor torque is acceptable. Reducing motor starting kVA can reduce the required size of the gen-set, reduce the voltage dip, and provide a softer start for the motor loads. When sizing gen-sets, you must first determine the acceptable level of motor torque required during starting, or the loads will accelerate slowly, or even fail to reach full speed:ultimately causing motor damage. Using solid-state starters. Solid-state starters can adjust the starting torque, acceleration ramp time, and current limit for controlled acceleration of a motor when it starts. For the purpose of sizing a gen-set, the current limit adjustment reduces the inrush current and may be used to reduce the starting kW and kVA requirement on the generator. The range of available current limit settings is typically from 150% to 600% of full-load current. A 600% current limit setting on the solid-state starter results in a gen-set sizing that's the same as an across-the-line starting. A 300% current limit setting reduces starting kVA by 50%. Use of the current limit setting also reduces motor torque available to the load. From a gen-set sizing perspective, an extended acceleration ramp time and low current limit setting (if appropriate for the motor and the mechanical load) would result in the least voltage and frequency excursions. One downside to using solid-state motor starters is their integral SCRs (silicon-controlled rectifiers) will cause voltage distortion. To compensate, you'll have to oversize the generator. The recommendation: two times the running kW load, except where you're using an automatic bypass. If the solid-state starter does have an automatic bypass, the SCRs are only in the circuit during starting. Once the motor is running, the bypass contactor closes and shunts the SCRs. In this case, you can ignore the voltage distortion during starting, and you don't have to add generator capacity. VFDs require larger generators. All versions of variable frequency drives (VFDs) are current limiting and reduce starting kW and kVA. The current drawn by these drives is nonlinear (having harmonics), which causes a distorted voltage drop across the reactance of the generator. Since VFDs are nonlinear, you must include an additional generator capacity sizing factor to keep voltage distortion to a reasonable level of approximately 15% total harmonic distortion (THD) or less. The larger the generator, the greater the reduction in impedance of the power source (generator), which in turn, reduces the effects caused by harmonic current distortion.

For six-pulse VFDs, a typical generator sizing factor would be twice the running kW of the drive. This offsets any reduction in starting kW and kVA. If it is the pulse width modulated (PWM) type (or includes an input filter to limit current distortion to less than 10%), then you can reduce the sizing factor down to 1.4 times the running kW of the drive. Using a step starting sequence. The starting sequence of loads can have a significant effect on the size of a gen-set. One commonly used approach is to assume all connected loads will start in a single step. This results in the largest gen-set selection. Unless you do something to add load incrementally (such as multiple transfer switches with staggered time delays, or a step load controller), then you should use a single-step load for sizing purposes. In multiple step applications, you start the largest motor first, to minimize the gen-set size. Once placing all loads on line with the genset, you can stop and start load equipment with automatic controls. Here, you'll have to size the gen-set by assuming the largest motor starts last, with all other connected loads already on line. Examples of sizing gen-sets. You can size a gen-set with manual calculations (using a worksheet) or with PC software available from most major gen-set manufacturers. The basic process is the same. It's always best to use actual data (if known). If this information isn't available, using PC software is the best option, since much of the required information on typical load characteristics is available as default information. If you use the manual sizing procedure, it should result in a recovery voltage of at least 90% of rated voltage and a starting instantaneous voltage dip of approximately 20% to 40%. The instantaneous voltage dip and frequency dip will likely vary from manufacturer to manufacturer, based on equal ratings of gen-sets. For a closer estimation of transient (starting instantaneous voltage) performance, use the manufacturer's sizing software. Using the manual sizing procedure. Step 1: Gather information. You'll need to know the following for each motor load: •

Nameplate hp,

• •

Running kilowatts (RkW),

•

Running motor power factor (PF),

•

Starting motor PF, and

•

Locked rotor kVA/hp.

Running kilovolt-amperes (RkVA),

You can use the following equation to calculate RkW and RkVA for motors: RkW = [(Nameplate hp) x (0.746kW/hp)] / Efficiency (eq. 1) RkVA = RkW / Running motor PF (eq. 2) To calculate starting kilovolt-amperes (SkVA) and starting kilowatts (SkW) for motors, use these equations: SkVA = (Nameplate hp) x (Locked rotor kVA/hp) (eq. 3) SkW = (SkVA) x (Starting motor PF) (eq. 4) Step 2: Total the RkW, RkVA, SkW, and SkVA numbers for all the loads. Step 3: Select the gen-set by comparing the RkW, RkVA, SkW, and SkVA to the ratings on the manufacturer's specification sheets (after appropriate derating for ambient temperature and altitude). Example One calculation. Determine gen-set size for three loads started across-the-line in a single step. Here's pertinent information: •

Two 200 hp motors, Code G, 92% running efficiency, 0.25 starting PF, 0.91 running PF.

•

Total 100kVA of fluorescent lighting, starting PF of 0.95, and running PF of 0.95 (Note: We're using the terms starting and running PF for the lighting load here for clarification when adding the motor loads. Actually, the ballast for the lighting load has a constant PF of 0.95.)

Step 1: Information gathering and calculations. 200 HP motor: RkW = (200 hp x 0.746 kW/hp) / 0.92 = 162.2kW RkVA = 162.2kW / 0.91 PF = 178.2kVA SkVA = 200 hp x 5.9 kVA/hp41180kVA SkW = 1180kVA x 0.25 PF = 295kW Florescent Lighting: RkW = 100kVA x 0.95 PF = 95kW RkVA = 100kVA SkVA = 100kVA

SkW = 100kVA x 0.95 PF = 95kW Step 2: Totals. Load.......... | RkW | RkVA | SkW | SkVA 200hp Motor | 162.2 | 178.2 | 295 | 1180 200hp Motor | 162.2 | 178.2 | 295 | 1180 Lighting....... | 95.... | 100... | 95. | 100 Totals (kVA). | 420... | 457.. | 685 | 2460 Step 3: Selection. At a minimum, you'll have to size the gen-set to supply the maximum starting (surge) demands and the steady-state running loads of the connected load equipment. In this example (using one manufacturer's published data), you would select a 750kW generator set with 2944 SkVA available at 90% recovery voltage to supply the total load SkVA of 2460. The load totals for RkW, RkVA, and SkW are well within the rating of the 750kW (938kVA) gen-set you selected. The running kilowatt load of 420kW is 56% of the 750kW gen-set standby rating. Example Two calculation. Assume you have the same three loads as in Example One, but now you're using an autotransformer type reduced voltage starter that is set at the 65% starting voltage to start the two motors. This tap setting will reduce the starting kVA by the square of the voltage (0.65)squared, or 0.42 times the starting kVA. Step 1: Calculations 200 HP motor: RkW = (200 hp x 0.746 kW/hp) / 0.92 = 162.2kW RkVA = 162.2kW / 0.91 PF = 178.2kVA SkVA = 200 hp x 5.9 kVA/hp = 1180 x (0.65)squared = 495kVA SkW = 495kVA x 0.25 PF = 124kW Florescent Lighting: RkW = 100kVA x 0.95 PF = 95kW RkVA = 100kVA SkVA = 100kVA SkW = 100kVA x 0.95 PF = 95kW Step 2: Totals Load.......... | RkW.. | RkVA | SkW | SkVA 200hp Motor | 162.2 | 178.2. | 124. | 495 200hp Motor | 162.2 | 178.2. | 124. | 495 Lighting...... | 95..... | 100... | 95... | 100 Totals (kVA) | 420... | 457... | 343. | 1090 Step 3: Selection. Using one manufacturer's published data, you would select a 450kW gen-set to supply the required starting kVA. The running kilowatt load of 420kW is 93% of the gen-set's standby rating. So, if you want a margin for future load additions, you would select a 500kW gen-set running at 84% of rated standby power.

Sidebar: Here's What Causes Dip in Starting Voltage When you start a motor across-the-line with a gen-set, the motor represents a low impedance load while at locked rotor or stalled condition. This causes a high inrush current. The high motor inrush current (I ms) flows through the generator armature windings and is affected by the reactance. This causes a drop in generator voltage. Impedance controls the flow of current in AC circuits. But, the generator armature reactance is such a large part of its total impedance that resistance is ignored. The generator terminal voltage drops instantaneously when the motor starter contacts close at time t40, as a function of the subtransient reactance (X"d). Generally, the larger the generator, the lower its reactance. So, one way to minimize the instantaneous voltage dip is to increase the generator size.

The generator terminal voltage may drop further, depending on response of the generator's automatic voltage regulator and the power capability of the excitation system. (Most gen-set automatic voltage regulators include underfrequency protection.) During momentary overloads, the engine speed may also dip. If it does, the automatic voltage regulator reduces excitation power to the main field, which lowers the generator terminal voltage. This, in turn, reduces the load on the engine, allowing it to recover to rated speed. Typically, a maximum generator terminal voltage dip of 30% will not cause coils to drop out. (This allows for approximately 5% additional voltage drop in the conductors between the generator and the motor). Although the voltage dip, due to under frequency protection, may extend the voltage recovery time, it also allows the engine to be sized closer to the steady-state running load rather than starting load. This is particularly important with diesel engines, which should not run for an extended duration at less than 30% of rated load. (Extended light-load operation of a diesel engine can result in the accumulation of unburned fuel in the exhaust system, due to incomplete combustion from low combustion temperatures, called wet stacking. Light load operation can also result in engine damage from fuel and water contaminating lubricating oil.) After the initial voltage dip, it's important the generator restore voltage to a minimum of 90%-rated value while supplying the motor starting kVA. At least 90% recovery voltage is necessary for the motor to develop adequate torque to accelerate its load to rated speed. A motor starting a high starting torque load, such as a loaded compressor, requires higher recovery voltage than one starting an unloaded compressor. As the motor comes up to speed, the voltage will rise, as the starting kVA input decreases. Once the motor is up to speed, the voltage should return to rated value, if the gen-set is sized properly.

Sidebar: How Inertia Affects Gen-Set Sizing The moment of inertia of a rotating mass offers resistance to acceleration. The load connected to the motor shaft has its moment of inertia, and in practical situations for specific equipment, this may or may not be available information. Fortunately, for the purpose of sizing a gen-set, or more specifically to determine the engine power needed to start and accelerate a rotating motor load, the motor load's moment of inertia need only be broadly categorized as low or high inertia. High inertia loads are characterized by high breakaway torque requiring prolonged acceleration. Low inertia loads are characterized by low starting torque at standstill, with increasing torque as motor speed increases resulting in rapid acceleration to rated speed. Starting low inertia loads will reduce the normal starting kW needed. Look for more information on this is in the sample calculations within this article.

Sidebar: Examples of High and Low Inertia High inertia loads include: •

Single- and multi-cylinder pumps

•

Single -and multi-cylinder compressors without unloading valves

•

Crushers

•

Hydraulic elevators without unloading valves

Low inertia loads include: •

Fans, centrifugal and blower

•

Compressors starting unloaded

•

Centrifugal pumps

•

Motor-generator elevators

Note: Pumps starting into high head pressure and large diameter fans or fans starting into high restriction areas should be classified as high inertia lo

I

n our example in Voltage Calculation - Part 2, we have only considered a single cable and checked only the voltage drop at starting.

In actual practice, we shall be comparing several cable sizes, selecting the cable that provides optimum design consideration. In this example, we shall be using several cables from 35 mm2 up to 120 mm2. Likewise, we shall be considering, not only the starting condition but the running condition of the motor as well. You might be asking, why we selected a motor in our example. The reason, a motor is a dynamic load. A motor circuit load varies from the starting to the normal running condition, thus there will be multiple considerations when selecting the cable for a motor circuit. In Table 2 Cable Selection, we have considered 5 cables sizes, all cables except one satisfy our design consideration which are the following: Voltage drop during starting < 15% Voltage drop during running < 5% Again we will be using values from our previous example which are the following: Voltagesending end = 400 volts Motor Running: Irunning = 180 A PFrunning = cos Ørunning = 0.87 sin Ørunning = 0.493 Motor Starting Istarting = 6 * 180 = 1080 A PFstarting = cos Østarting = 0.25 sin Østarting = 0.968

Using the formula

Table 2. Cable Selection Size

r50

x50

R

X

%Vdrunning

%Vdstarting

35

0.674

0.0867

0.0809

0.0104

5.9%

14.2%

50

0.499

0.0858

0.0599

0.0103

4.5%

11.7%

70

0.344

0.0850

0.0413

0.0102

3.2%

9.4%

95

0.271

0.0825

0.0325

0.0099

2.6%

8.3%

120

0.214

0.0808

0.0257

0.0097

2.1%

7.4%

In Table 2. Cable Selection, the 35 mm2 cable passes the motor starting voltage drop but fails the motor running voltage drop which makes this cable to be eliminated from our selection list. The 70 mm2, 90 mm2 and 120 mm2 cables satisfy all our design conditions, however, selecting any of these cables will make the installation more expensive than required. The 50 mm2 cable satisfy all conditions and cost of the project. Cable selection considerations does not stop here, there are more that need to be considered such as operating temperature and short-circuit withstand. In our next article, we shall be discussing about cable operating temperatures.