## Calculate Size of Contactor / Fuse / CB / OL Relay of Star-Delta Starter

• Calculate Size of each Part of Star-Delta starter for 10HP, 415 Volt Three Phase Induction Motor having Non Inductive Type Load, Code A, Motor efficiency 80%, Motor RPM 600, Power Factor 0.8. Also Calculate Size of Overload Relay if O/L Relay Put in the wingdings (overload is placed after the Winding Split into main and delta Contactor) or in the line (Putting the overload before the motor same as in DOL).

## Basic Calculation of Motor Torque & Current:

• Motor Rated Torque (Full Load Torque) =5252xHPxRPM
• Motor Rated Torque (Full Load Torque)=5252x10x600=88 lb-ft.
• Motor Rated Torque (Full Load Torque) =9500xKWxRPM
• Motor Rated Torque (Full Load Torque)=9500x(10×0.746)x600 =119 Nm
• If Motor Capacity is less than 30 KW than Motor Starting Torque is 3xMotor Full Load Current or 2X Motor Full Load Current.
• Motor Starting Torque=3x Motor Rated Torque (Full Load Torque).
• Motor Starting Torque==3×119=356 Nm.
• Motor Lock Rotor Current =1000xHPx figure from below Chart/1.732×415
 Locked Rotor Current Code Min Max A 1 3.14 B 3.15 3.54 C 3.55 3.99 D 4 4.49 E 4.5 4.99 F 5 2.59 G 2.6 6.29 H 6.3 7.09 I 7.1 7.99 K 8 8.99 L 9 9.99 M 10 11.19 N 11.2 12.49 P 12.5 13.99 R 14 15.99 S 16 17.99 T 18 19.99 U 20 22.39 V 22.4
• As per above chart Minimum Locked Rotor Current =1000x10x1/1.732×415=14 Amp
• Maximum Locked Rotor Current =1000x10x3.14/1.732×415=44 Amp.
• Motor Full Load Current (Line) =KWx1000/1.732×415
• Motor Full Load Current (Line) = (10×0.746)x1000/1.732×415=13 Amp.
• Motor Full Load Current (Phase) ==13/1.732=7 Amp.
• Motor Starting Current (Star-Delta Starter) =3xFull Load Current.
• Motor Starting Current (Line)=3×13=39 Amp

### (1) Size of Fuse:

 Fuse  as per NEC 430-52 Type of Motor Time Delay Fuse Non-Time Delay Fuse Single Phase 300% 175% 3 Phase 300% 175% Synchronous 300% 175% Wound Rotor 150% 150% Direct Current 150% 150%
• Maximum Size of Time Delay Fuse =300% x Full Load Line Current.
• Maximum Size of Time Delay Fuse =300%x13= 39 Amp.
• Maximum Size of Non Time Delay Fuse =1.75% x Full Load Line Current.
• Maximum Size of Non Time Delay Fuse=1.75%13=23 Amp.

### (2) Size of Circuit Breaker:

 Circuit Breaker as per NEC 430-52 Type of Motor Instantaneous Trip Inverse Time Single Phase 800% 250% 3 Phase 800% 250% Synchronous 800% 250% Wound Rotor 800% 150% Direct Current 200% 150%
• Maximum Size of Instantaneous Trip Circuit Breaker =800% x Full Load Line Current.
• Maximum Size of Instantaneous Trip Circuit Breaker =800%x13= 104 Amp.
• Maximum Size of Inverse Trip Circuit Breaker =250% x Full Load Line Current.
• Maximum Size of Inverse Trip Circuit Breaker =250%x13= 32 Amp.

### Thermal over Load Relay (Phase):

• Min Thermal Over Load Relay setting =70%x7= 5 Amp
• Max Thermal Over Load Relay setting =120%x7= 9 Amp

### Thermal over Load Relay (Line):

• For a star-delta starter we have the possibility to place the overload protection in two positions, in the line or in the windings.
• If O/L Relay Placed in Line: (Putting the O/L before the motor same as in DOL).Supply>Over Load Relay>Main Contactor
• If Over Load Relay supply the entire motor circuit and are located ahead of where the power splits to the Delta and Star contactors, so O/L Relay size must be based upon the entire motor Full Load Current.
• Thermal over Load Relay setting =100%x13= 13 Amp
• Disadvantage: O/L Relay will not give Protection while Motor runs in Delta (Relay Setting is too High for Delta Winding)
• If O/L Relay Placed In the windings: (overload is placed after the Winding Split into main and delta Contactor).Supply>Main Contactor-Delta Contactor>O/L Relay
• If overload is placed after the Point where the wiring Split into main and delta Contactor, Size of over load relay at 58% (1/1.732) of the motor Full Load Current because we use 6 leads going to the motor, and only 58% of the current goes through the main set of conductors (connected to the main contactor).
• The overload then always measures the current inside the windings, and is thus always correct. The setting must be x0.58 FLC (line current).
• Thermal over Load Relay setting =58%x13= 8 Amp.

• ### Main and Delta Contactor:

• The Main and Delta contactors are smaller compared to single contactor used in a Direct on Line starter because they Main and Delta contactors in star delta starter are controlling winding currents only. The currents through the winding are 1/√3 (58%) of the current in the line. These two contactors (Main contactor and Delta Contactor) are close during run. These rated at 58% of the current rating of the motor.
• ### Star Contactor:

• The third contactor is the star contactor and that only carries star current while the motor is connected in star in starting. The current in star winding is 1/√3= (58%) of the current in delta, so this contactor can be rated at 1/3 (33%) of the motor rating. Star contactor can be selected smaller than the others, providing the star contactor pulls first before the main contactor. Then no current flows when third contactor pulls.
• In star connection at start, the motor draws and delivers 1/3 of its full rated power.
• When the starter switches over to Delta, the motor draws full power, but since the contactors and the overload relay are usually wired within the Delta, you need to use contcators and relay which are only rated 1/√3 =58% of the full rated power of the motor.
 Application Contactor Making Cap Non-Inductive or Slightly Inductive ,Resistive Load AC1 1.5 Slip Ring Motor AC2 4 Squirrel Cage Motor AC3 10 Rapid Start / Stop AC4 12 Switching of Electrical Discharge Lamp AC5a 3 Switching of Electrical Incandescent Lamp AC5b 1.5 Switching of Transformer AC6a 12 Switching of Capacitor Bank AC6b 12 Slightly Inductive Load in Household or same type load AC7a 1.5 Motor Load in Household Application AC7b 8 Hermetic refrigerant Compressor Motor with Manual O/L Reset AC8a 6 Hermetic refrigerant Compressor Motor with Auto O/L Reset AC8b 6 Control of Restive & Solid State Load with opto coupler Isolation AC12 6 Control of Restive Load and Solid State with T/C Isolation AC13 10 Control of Small Electro Magnetic Load ( <72VA) AC14 6 Control of Small Electro Magnetic Load ( >72VA) AC15 10
• As per above Chart
• Type of Contactor= AC1
• Making/Breaking Capacity of Contactor= Value above Chart x Full Load Current (Line).
• Making/Breaking Capacity of Contactor=1.5×13= 19 Amp.
• Size of Star Contactor (Starting Condition) = 33%X Full Load Current (Line).
• Size of Star Contactor =33%x13 = 4 Amp.
• Size of Main Contactor (Starting-Transition-Running) = 58%X Full Load Current (Line).
• Size of Main Contactor =58%x13 = 8 Amp.
• Size of Delta Contactor (Running Condition) = 58%X Full Load Current (Line).
• Size of Delta Contactor =58%x13 = 8 Amp.

### Summary:

•  Type of Contactor= AC1
• Making/Breaking Capacity of Contactor=19 Amp.
• Size of Star Contactor =4 Amp.
• Size of Main Contactor = 8 Amp.
• Size of Delta Contactor =8 Amp.

## Effects of unbalanced Electrical Load (Part:2)

• ### Harmonics in system by UPS:

• UPS or inverter supplies also perform with poor efficiency and inject more harmonic currents in case of unbalances in the system
• ### Decrease Life cycle of Equipment:

• Unbalanced Voltage increase I2R Losses which increase Temperature. High temperatures, exceeding the rated value of a device, will directly decrease the life cycle of the device and speed up the replacement cycle for the device, and significantly increase the costs of operation and maintenance.
• ### Relay malfunction

• Unbalanced Voltage flows Negative and Unbalanced Voltage of Voltage or Current.
• The high zero-sequence current in consequence of voltage imbalance may bring about malfunctions of relay operation or make the ground relay less sensitive. That may result in serious safety problems in the system.
• ### Inaccurate Measurement

• Negative and zero-sequence components of voltages or currents will give rise to inaccurate measurements in many kinds of meters.
• The imprecise measured values might affect the suitability of settings and coordination of relay protection systems and the correctness of decisions by some automated functions of the system.
• ### Decrease Capacity of transformers, cables and lines

• The capacity of transformers, cables and lines is reduced due to negative sequence components. The operational limit is determined by the RMS rating of the total current, due to ‘useless’ non-direct sequence currents the capacity of equipment is decrease.
• ### Increase Distribution Losses

• Distribution network losses can vary significantly depending on the load unbalance.
• Unbalance load increase I2R Losses of distribution Lines.
• ### Increase Energy Bill by increasing Maximum Demand

• Unbalanced Load increase maximum Demand of Electrical supply which is significantly effects on energy bill. By load balancing we can reduce energy bill.
• For Energy Consumption Energy Supply Company does not charge on kVA but on kW for Residential customers. This means that they are charged for the “actual” energy used and not charged for the “total” energy supplied. Thus the power factor and Maximum Demand do not impact residential customers.
• But Commercial, Industrial and H.T Connection charged by its maximum demand . We have to specify the maximum “demand“(in kVA) at the time of connection. During the month if you exceed your maximum “demand” you have to pay penalty (or extra price) for the same. That is the MDI penalty that appears on electricity bills.
• Let’s assume That Two Company has same approved load of 40 KW and runs 30KW for 100 hours.
• Electricity charge = 65 Rs per kWh
• Demand charge = 210Rs per kW
• Example 1: Company A runs a 30 KW loads continuously for 100 hours but It’s Maximum Demand is 50KW
• 30 KW x 100 hours = 3,000 KWh
• Energy Consumption Charge =3000×65=195000Rs
• Demand difference = 50 KW-40KW=10KW
• Demand Charges = 10X210=2100Rs
• Total Bill:  195000+2100=197100Rs
• Example 2: Company A runs a 30 KW loads continuously for 100 hours but It’s Maximum Demand is 40W
• 30 KW x 100 hours = 3,000 KWh
• Energy Consumption Charge =3000×65=195000Rs
• Demand difference = 40 KW-40KW=0KW
• Demand Charges = 0X210=00Rs
• Total Bill:  195000+0=195000Rs
• ### Failure of Transformer

• Three-phase voltage with high unbalanced may cause the flux inside the transformer core to be asymmetrical.
• This asymmetrical flux will cause extra core loss, raise the winding temperature and may even cause transformer failure in a severe case.
• Ideally any distribution transformer gives best performance at 50% loading and every electrical distribution system is designed for it. But in case of unbalance the loading goes over 50% as the equipments draw more current.
• For a distribution transformer of 200KVA rating, the eddy currents accounts for 200W but in case of 5% voltage unbalance they can rise up to 720W.
• ### Bad / Loose connection of neutral wire

• In balance Load condition Bad connection of Neutral wire does not make more impact on distribution System but in unbalance load condition such type of Bad neutral connection make worse impact on distribution.
• The Three Phase power supplies a small a three-floor building. Each floor of this three-floor building is serviced by a single-phase feeder with a different phase. That is the first, second and third floor are serviced by phase R, Y and B. The external lighting load is connected only on R Phase.
• The supply transformer is rated at 150 kVA and connected delta-grounded wye to provide for 430/220 V three-phase four-wire service.
• This Transformer has a loose or Bad Neutral connection with the earth.
• The transformer delivers a load of 35 kVA at 220 V with 0.9 power factor lagging to each floor.
• During the daytime on, most of the Load of the Building are distributed equally over the three floors which is R Phase=30A, Y Phase =32A, B Phase=38A.
• In Daytime The Bad connection of Neutral does not effected the Distribution system due to equal load distribution of the System
• However it is not case in Nighttime. the Load on Y Phase and B Phase are negligible but R Phase Load is high compare to Y and B Phase.
• In R Phase due to High Electrical Load and The fluorescent lamps flash frequently during the Nighttime of external Lighting Load
• In Night time a bad electrical contact of the neutral wire of the supply makes the high contact resistance between the neutral wire and connector .which was about 15 kΩ.
• This extra high impedance caused an unusually high voltage drop in the phase a circuit. In this case, the voltage of phase a dropped from the normal 220V to 182.5V, about 17% based on the nominal voltage. If the contact impedance goes higher than 20 kΩ, it may result in more serious conditions such as extinguishing all lamps.
• This problem can be removed by fixing the bad connection and keeping the contact impedance near to zero.
 Neutral Wire Contact Resistance Voltage across  bad Connection Point Voltage across  Transformer Secondary Side Day Time Night Time Day Time Night Time R Y B R Y B R Y B R Y B Proper Connection (0Ω) 0v 0v 0v 0v 0v 0v 220v 220v 220v 220v 220v 220v Bad Connection (15Ω) 0v 0v 0v 40v 0v 0v 220v 220v 220v 182v 220v 220v
• ### Neutral wire broken

• The effect of a broken neutral makes voltage imbalance in a Three Phase Four Wire System.
• For a Three Phase Four Wire System, high neutral wire impedance might enlarge a voltage imbalance (Some Phase Voltage increase while some Phase Voltage decreases).
• High Voltage damage the equipment connected and even destroy on other hand low voltage effect operation of equipments.
• The Three Phase star connected lighting loads are fed by a 430 V balanced three-phase voltage source. The fluorescent lamps are all rated at 220 V, 100 W each. The lamps are not equally in R Phase 5 No’s of Bulbs are connected, in Y Phase 3 No’s of Bulbs are connected and on B Phase 3 No’s of Bulbs are Connected. And, the normal impedance of the neutral wire is 1Ω
• In unbalanced three phase load arrangement, high neutral wire impedance will enlarge the voltage across the neutral wire. The voltages of phases B and C at the load terminal raised to 255 V and 235 V, respectively, and gaining 16.15% and 5.77% based on rated voltage. These abnormally high phase voltages might damage the lamps in phase B and C.
• On the other hand, the voltage in phase A was reduced from 220V to 185V. That might cause the lamps to flash.
• If the broken neutral line problem is fixed, then the three phase voltages will go back to normal in near balanced status .however, if the loads are distributed equally to the three phases this problem can also be removed or minimized.
 Conditions Voltage across the neutral wire Voltage at  the load terminal R Y B R Y B Normal Condition 1v 1v 1v 220v 220v 220v Neutral Broken 0v 0v 0v 182v 255v 235v
• ### Unsuitable capacitor bank installation

• For reducing energy loss, utilities always force their customers to maintain the power factor within a limit. Penalty will be applied to the customers if their loads’ power factors run outside the limits.
• Installation of shunt capacitor banks is the most common and cheapest manner to improve the power factor. However, unsuitable installation (single Phase Capacitor instead of Three Phase Capacitor ) may make it worse.
• The supply transformer is rated at 150 kVA, 11kV/430 V, and supplies a three-phase load of 105 kVA with power factor 0.7 lagging.
• A single-phase 20KVAR capacitor bank is connected to B phase to improve system power. The impedance of the shunt capacitor bank is 1.805Ωper phase.
• This kind of single phase Capacitor installation should make the system unbalanced. This unsuitable installation consumes extra real power of 44355 W.
• The extra real power consumption = 1.732x2XV(RB) / 4xXc =(1.732x2x430) / (4×1.805) =44355W
• This case shows that the system balance should be considered when installing a capacitor bank to correct the system power factor for a three-phase power distribution system.

## Remedial Action to prevent unbalances Load:

• All the single phase loads should be distributed on the three phase system such that they put equal load on three phases.
• Replacing the disturbing equipments i.e. with unbalanced three phase reactance.
• Reducing the harmonics also reduces the unbalance, which can be done by installing reactive or active filters. These filters reduce the negative phase sequence currents by injecting a compensating current wave.
• In case the disturbing loads cannot be replaced or repaired, connect them with high voltage side this reduces the effects in terms of percentage and even controlled disturbance in low voltage side.
• Motors with unbalanced phase reactance should be replaced and re-winded.
• Distribution of single-phase loads equally to all phases.
• Single-phase regulators have been installed that can be used to correct the unbalance but care must be exercised to ensure that they are controlled carefully not to introduce further unbalance.
• Passive network systems and active power electronic systems such as static var compensators and line conditioners also have been suggested for unbalance correction.
• Use of passive networks and static VAR compensators.
• Equipment that is sensitive to voltage unbalance should not be connected to systems which supply single-phase loads.
• Effect of voltage unbalance on ac variable speed drives can be reduced by properly sizing ac side and dc link reactors
• Tight all Neutral Connections of the System.
• Install Proper size of Capacitor Bank to the System.
• Load Scheduling, where the loads in an electrical network are scheduled in a way to turn on and off at precise times to prevent the overloading of any one phase.
• Manual Load Shifting, where an electrician opens a breaker panel and physically removes the loads from one phase and inserts them onto another phase.
• Load Shedding, where the loads in an electrical network are immediately turned off in order to instantly “rebalance” the phases. This is usually done by ranking the loads in a network by how long they can be turned off before it affects operations

## Introduction:

• Generally, three phase balance is the ideal situation for a power system and quality of delivered Electrical Power. However Voltage unbalance may makes worse effect on Power quality of Electrical Power at distribution level.
• The voltages are quite well balanced at the generator and transmission levels. but the voltages at the utilization level can become unbalanced due to the unequal system impedances, the unequal distribution of single phase loads, asymmetrical three-phase equipment and devices (such as three-phase transformers with open star-open delta connections), unbalanced faults, bad connections to electrical connectors.
• An excessive level of voltage unbalance can have serious impacts on power quality. In the system the level of current unbalance is several times the level of voltage unbalance. Such an unbalance in the line currents can lead to excessive line losses, losses in the stator and rotor of Motor Malfunctioning of Relay, unsymmetrical measuring of Meters. Voltage unbalance also has an impact on ac variable speed drive systems where the front end converter consists of three-phase rectifier systems
• Phase balancing is very important and usable to reduce distribution feeder losses and Improve system stability and security

## What is unbalance Voltage

• Any deviation in voltage and current waveform from perfect sinusoidal, in terms of magnitude or phase shift is termed as unbalance
• In ideal conditions the phases of power supply are 120 degree apart in terms of phase angle and magnitude of their peaks should be same. On distribution level, the load imperfections cause current unbalance which travel to transformer and cause unbalance in the three phase voltage. Even minor unbalance in the voltage at transformer level disturbs the current waveform significantly on all the loads connected to it
• If three phase voltages have the same magnitude and are in exactly 120deg phase displacement, then the Three-phase voltage is called balanced, otherwise, it is unbalanced.
• There are no negative- and zero-sequence voltages in a balanced system, only positive-sequence components of balanced three-phase voltage exist. On the contrary, if the system is unbalanced, negative-sequence components or zero-sequence components or both may exist in the system.

## Causes of unbalance Voltage

• Switching of three phase heavy loads results in current and voltage surges which cause unbalance in the system.
• Unequal impedances in the power transmission or distribution system cause differentiating current in three phases.
• Any large single phase load, or a number of small loads connected to only one phase cause more current to flow from that particular phase causing voltage drop on line
• With continuous operation of motor’s in various environment cause degradation of rotor and stator windings. This degradation is usually different in different phases, affecting both the magnitude and phase angel of current waveform
• A three phase equipment such as induction motor and Transformer with unbalance in its windings. If the reactance of three phases is not same, it will result in varying current flowing in three phases and give out system unbalance.
• A current leakage from any phase through bearings or motor body provides floating earth at times, causing fluctuating current.
• Unbalanced incoming utility supply
• Unequal transformer taps settings
• Large single phase distribution transformer on the system
• Open phase on the primary of a 3 phase transformer on the distribution system
• Faults or grounds in the power transformer
• Open delta connected transformer banks
• A blown fuse on a 3 phase bank of power factor improvement capacitors
• Unequal impedance in conductors of power supply wiring
• Unbalanced distribution of single phase loads such as lighting
• Heavy reactive single phase loads such as welders

## How to calculate unbalance

• %voltage unbalance= 100x (maximum deviation from average voltage) / (average voltage)
• Example: With phase-to-phase voltages of The System is 430V, 435V, and 400V.
• The average Voltage=(430+435+400)/3=421V.
• The maximum Voltage deviation from Average Voltage=435-421=14V
• %voltage unbalance=14×100/421=3.32%
• The permissible limit in terms of percentage of negative phase sequence current over positive sequence current is 1.3% ideally but acceptable up to 2%.

## Effects of unbalance Voltage on System and Equipment:

• The factors for voltage unbalances can be classified into two categories: normal factors and abnormal factors.
• Voltage imbalances due to normal factors, such as single-phase loads and three-phase transformer banks with open star-open delta connections, can generally be reduced by properly designing the system and installing suitable equipment and devices.
• Abnormal factors include series and shunt faults of circuits, bad electrical contacts of connectors or switches, asymmetrical breakdown of equipment or components, asynchronous burnout of three phase power fuses, single-phase operation of motors, etc. The abnormal factors just mentioned above might result in critical damage of systems and equipment.
• ### Increase Neutral Return Current

• The unequal distribution of loads between the three phases of the system cause the flow of unbalanced currents in the system, that produce unbalanced voltage drops on the electric lines. This increase in neutral current which cause line losses.
• If the system has balanced phase then Neutral current flow will be less on a system. We can save thousands to millions of rupees money by reduce losses be the reducing the neutral current flow in the system
• Thus unbalance in LV distribution network resulting in increase of neutral current.
• ### Voltage or Current Shift

• If the system is unbalanced, negative-sequence components or zero-sequence components or both may exist in the system.
• The resistance for negative sequence current is 1/6th of the positive sequence current, which means a small unbalance in voltage waveform will give more current and thus losses.
• ### Excessive power loss

• The unbalance Voltage always causes extra power loss in the system. The higher the voltage unbalance is the more power is dissipated means higher power bills.
• The imbalance of current will increase the I2R Losses
• Let’s look at a simple exercise, In balance System The Load current in R Phase=200A, Y Phase=200A,B Phase=200A and in Unbalance System The Load current in R Phase=300A, Y Phase=200A,B Phase=100A,Consider Resistance of line are same in both case and all phases.
• In Balanced System:
• Total Load current =R+Y+B = 200+200+200=
• Total Losses =R(I2R)+Y(I2R)+B(I2R)=40000+40000+40000=120,000Watt.
• In Un Balanced System:
• Total Load current =R+Y+B = 300+200+100=
• Total Losses =R(I2R)+Y(I2R)+B(I2R)=90000+40000+10000=140,000Watt.
• Here Total Load current is same in both case but Losses in unbalance system is more than balance system.
• An unbalance of 1% is acceptable as it does not affect the cable. But above 1% it increases linearly and at 4% the de-rating is 20%. This means – 20% of the current flowing in the cable will be unproductive and thus the copper losses in the cable will increase by 25% at 4% unbalance.
• ### Motor failure

• In general, a three-phase motor fed by a balanced three-phase voltage with only positive-sequence component which produces only positive-sequence torque.
• Reduce Motor life by heating: Extra loss due to voltage imbalance will heat the motor windings, by increasing the operating temperature of Motor leads to the breakdown of winding insulation and might finally in motor failure. This may also decompose the grease or oil in the bearing and de-rate the motor windings. The voltage unbalance of 3% increases the heating by 20% for an induction motor.
• Winding insulation life is reduced by one-half for each 10°C increase in operating temperature
• Vibration of Motor: The negative-sequence voltage caused by voltage imbalance produces opposite torque and leads to motor vibration and noise. Severe voltage imbalance may even result in motor collapse.
• Reduce Motor Life: Heat generated by Unbalance Voltage may also reduce the Motor life
• Reduce Efficiency: In induction motors connected to unbalanced supply, the negative sequence currents flow along with positive sequence current resulting in decreased percentage of productive current and poor motor efficiency. Any unbalance above 3% hampers the motor efficiency.
 Motor Efficiency % Motor Load % Full Voltage Unbalance Nominal 1% 2.5% 100 94.4 94.4 93.0 75 95.2 95.1 93.9 50 96.1 95.5 94.1
• Assume that the 100-HP motor tested was fully loaded and operated for 800 hours per year with an unbalanced voltage of 2.5%. With energy priced at 23Rs/KWH. the annual energy and cost savings calculation are
• With Normal Voltage
• Annual Energy Consumption=100HPx0.746X800X(100/94.4)x23=1454068Rs
• With Unbalanced Voltage
• Annual Energy Consumption=100HPx0.746X800X(100/93)x23=1475957Rs
• Annual Cost Savings = 1475957-1454068=21889Rs
• Overall savings may be much larger because an unbalanced supply voltage may power numerous motors and other electrical equipment.
• Tripping of Motor: Negative phase sequence current flowing due to unbalance can cause faults in the motor, resulting in, tripping or permanent damage of the electrical equipment
• Reduce Capacity: For motors, an unbalance of 5% will result in capacity reduction by 25%.
• Tripping of VFD Drives: The variable frequency or speed drives connected to an unbalanced system can trip off. VFD treats high level unbalances as phase fault and can trip on earth fault or missing phase fault.