Frequently asked Questions

How do I size a generator to start an induction motor driving a pump?

Motor 3KW 15A 230V single phase 40A start

If you wish to size the gen set correctly, there are several things that you need to consider:

Firstly, the gen set is made up of two major components, the engine and the alternator. The engine is loaded by KW and the alternator is loaded by amps or KVA.

Secondly, an induction motor draws a significant overload current when starting, typically in the order of 6 - 9 times the rated curent of the motor when started at full line voltage. The actual start current is dependant on the motor design and the line voltage and is independant of the motor load. If a reduced votage starter is employed, the current will reduce, and the minimum current at which the motor will start is a function of motor desing and load torque requirements. In the case of a single phase motor, I will assume that it is going to be full voltage started.

If the rated start current is 40 Amps, then the alternator needs to be able to supply this for about 5 seconds. Typically, you would be looking at a 25 to 30 amp rated alternator for this, but check with the alternator supplier.

If we assume that the KW load is the rated motor shaft load plus the copper loss during start, then you would be looking at a KW load of 3KW plus Cu losses. A motor of this rating would typically have a copper loss in the order of 15% of the motor rating at full load. (Say 0.45KW) During start, the current is higher, in this case quoted as 40 A / 15 A so the copper loss would be about 0.45 x 40/15 x 40/15 = 3.2KW

As this is a submersible pump with a high rated current, then it is probable that the Cu losses would be higher than 15% at rated load and so the 3.2 is probably on the light side!

My suggestion would be for an engine rated at at least 4KW continuous and 6.5KW intermittent coupled to an alternator rated at at least 15 A contiuous and 40Amp intermittant

This would probably be an alternator rated at about 7KVA for 240 V single phase operation.

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Can I Bridge out the slip rings and use a soft starter on a slip ring motor?

A slip ring motor uses resistors in the rotor circuit to modify the starting characteristics of the slip ring motor. Increasing the resistance in the rotor circuit has two effects:
1. It reduces the start current
2. It increases the slip at which maximum torque occurs.

If the slip ring motor has been employed to provide a very high starting torque across the entire speed range during start, then the slip ring or secondary resistance starter can not be replaced. In this case, the first stage of the resistors would be selected to provide a high torque at 100% slip (zero speed) and a number of stages are then employed, each with reducing resistance to move the Slip point in steps from 100% towards 0%. The effect of this is to provide maximum torque at all speeds and at a reduced start current. (typically 200 - 300%)

Shorting out the slip rings and attempting any form of reduced voltage start in the stator supply, will result in a much reduced start torque at a much higher start current. Effectively, the motor could exhibit a Locked Rotor Current in excess of 1000% and a Locked Rotor Current less than 100%. If we reduce the start current down to say 400%, then the start torque would be less than 100 x (400/1000) x (400/1000) or less than 16%!

If the driven load does not require a high start torque, then the slip ring motor can be set up to emulate a standard cage motor by applying rotor resistance that will cause a full voltage start current of about 550%. A reduced voltage starter can now be applied, and the rings should be shorted out once the machine reaches full speed. If you do not short the rings at full speed, the slip will be higher than ideal and the motro efficiency will be reduced. There will be a high power dissipation in the resistors.

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Can I Bridge out the slip rings and use an inverter on a slip ring motor?

Yes. Infact it is best to remove the brush gear and short directly onto the rings. This reduces maintenance and the potential for arcing on the ring gear damaging the inverter. You may have problems with some loads if the start torque is over 120%, depending on the output capacity of the inverter.

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Why does a Current Transformer Saturate?

The equivalent circuit of a CT consists of the following -
- An Ideal transformer with a turns ratio to match the current ratio.
- A variable inductance connected across the secondary of the ideal CT (This is the magnetising or excitation current)
- CT secondary resistance connected in series with the secondary load

The magnetising current of the CT varies with the secondary voltage. The magnetizing curve can be measured by applying a variable voltage, at the operating frequency, to the secondary terminals with the primary open circuited. The resulting curve shows that as you increase the applied voltage, the magnetising current increases gradually up to the kneepoint voltage, at which point the CT iron saturates and the magnetising current increases rapidly.

The Magnetizing current is subtracted from the ideal CT output resulting in a reduced output current when the CT saturates, effectively reducing the CT ratio.

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What happens if I operate a motor in Star instead of Delta?

If the motor is designed to be operated in delta on your local supply voltage, then operating continuously in star can cause damage to the motor. When a delta motor is connected in star, the voltage across the windings is reduced by the square root of three resulting in reduced flux in the iron. This will reduce the magnetising current, and will also reduce the torque capacity of the motor. If you operate at light loads, there will be no problem, however if you operate at high loads, the slip of the motor will be increased dramatically and it may stall. The increased slip will result in a dramatic increase in the power dissipated in the rotor. If the motor begins to stall, the stator will also suffer excess heating causing a motor failure.

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Can we OverSpeed a Motor on a VFD?

We are having an induction motor WITH FAN feeding air to one of our AIR HANDLING UNIT.name plate details of the 3phase induction motor is 55 amp,415v,50hz,985rpm,make ABB.wecan vary the speed with the help of VFD.
Consider a case, to meet our CFM requirement we want the motor to run on say 1300rpm. WHETHER IT IS ADVISABLE TO RUN THE MOTOR on say 60hz if it is drawing the current less than full load current ? or what is your view on this?

You can certainly overspeed a motor driven by a VFD provided that you are careful not to overload the motor and also not to overspeed the bearings. A six pole motor would typically use bearings capable of much higher speeds so this should not be a problem. Induction motors operating off a VFD need to be re rated depending on the speed that they are operating and the cooling applied. At speeds below the rated speed of the motor, they should be rated on the rated full speed torque. As torque x speed = power, the effective power capacity of the motor reduces with speed. If you continually operate the motor at it's rated torque at less than full rated speed, the reduced cooling will cause a temperature rise that could damage the motor. Additional cooling may be required. At speeds higher than the rated speed of the motor, the flux in the iron reduces due to the output voltage being limited to the supply voltage. This causes a reduction in torque capacity with speed. The result is that the motor capacity is power limited to the rated power of the motor.
In you case, provided the power requirements of the driven load do not exceed the rated power of the motor at the higher speed, then there should be no problem in operating the motor at the increased speed.

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How do you set up a thermal over load? Do you set it for the current drawn or for the motor rating?

Strictly speaking, you would normally set the overload protection to the rating of the motor, but if the motor always operates below its rating, then setting the overload closer to the actual operating current will afford a higher level of protection, provided that the motor is still able to start without tripping the overload.

The method that I normally advocate for setting a thermal overload is to operate the motor at maximum load for an extended period of time, then slowly adjust the overload down until it trips, then set the relay a small margin higher. The problem with thermal overloads is that the calibration is coarse to say the least, and this method ensures that you have protection against a change in the load characteristics of motor condition. If there are occasional nuisance trips, then the setting can be increased, but should not exceed the motor rating.

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Under what circumstances would semi-conductor (or high speed fuses) be used on a soft starter application? What purpose do they serve?

Semiconductor fuses are an energy limiting fuse using specially shaped silver elements in a silicon sand environment. Because of their energy limiting characteristics, they are able to protect semiconductor devices from excess energy let through under fault current conditions.
A soft starter uses reverse parallel connected SCRs or thyristors in series with the supply to the motor. If a short circuit occurs between the starter and the motor, or in the motor itself, there is no limiting impedance to restrict the current flow and so the current is essentially the short circuit current of the supply. In most cases, this is enough to damage or destroy the SCRs. It is not practical to turn an SCR off once current has begun to flow, so electronic protection is not practical. Current will stop at the next zero crossing which could be half a cycle later. The semiconductor fuse is able to interupt the current flow in less than half a cycle, and the rupture time is dependant on the total energy flow.
SCRs have a maximum short term energy rating that is usually called I squared t (Current squared times time) and provided the maximum let through I2t (or total clearing I2t) of the fuse at the operating voltage of the supply, is less than the I2t of the SCRs, then the fuse should fail before the SCRs.
Semiconductor fuses can be a bit of a problem at times because their pre arcing I2t is well below their total clearing I2t and they can be stressed by the normal operation of the starter and fail prematurely. It is important to compare the overload time current curve of the fuse with the operation of the starter.

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Under what circumstances would semi-conductor (or high speed fuses) be used on a soft starter application? What purpose do they serve?

Semiconductor fuses are an energy limiting fuse using specially shaped silver elements in a silicon sand environment. Because of their energy limiting characteristics, they are able to protect semiconductor devices from excess energy let through under fault current conditions.
A soft starter uses reverse parallel connected SCRs or thyristors in series with the supply to the motor. If a short circuit occurs between the starter and the motor, or in the motor itself, there is no limiting impedance to restrict the current flow and so the current is essentially the short circuit current of the supply. In most cases, this is enough to damage or destroy the SCRs. It is not practical to turn an SCR off once current has begun to flow, so electronic protection is not practical. Current will stop at the next zero crossing which could be half a cycle later. The semiconductor fuse is able to interupt the current flow in less than half a cycle, and the rupture time is dependant on the total energy flow.
SCRs have a maximum short term energy rating that is usually called I squared t (Current squared times time) and provided the maximum let through I2t (or total clearing I2t) of the fuse at the operating voltage of the supply, is less than the I2t of the SCRs, then the fuse should fail before the SCRs.
Semiconductor fuses can be a bit of a problem at times because their pre arcing I2t is well below their total clearing I2t and they can be stressed by the normal operation of the starter and fail prematurely. It is important to compare the overload time current curve of the fuse with the operation of the starter.

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Can I run a 60Hz rated motor on 50Hz?

Induction motors use an iron core and require flux in the iron to operate. In order to achieve the commercial goals of smasllest size and lowest price at best efficiency, induction motors are designed to operate at a high level of flux in the iron. The flux is determined by the turns, voltage and frequency. In a modern motor, if the flux is increased by a small amount, the iron losses increase and the iron tends towards saturation. At saturation, the inductance begins to fall and the current increases further. To reduce the flux at a given voltage and frequency, the turns on the stator are increased. This reduces the Iron loss, but a longer length of thinner wire is used and the copper loss increases. Design becomes a balancing act between copper loss and iron loss and so the design is optimised for a given voltage and frequency.

If the voltage applied to the motor is held constant and the frequency is increased, the inductive reactance increases and so the flux reduces. This effectively reduces the maximum torque capacity of the motor and so the motor power rating at the higher frequency remains the same.

If the voltage applied to the motor is held constant and the frequency is reduced, the current will increase and in theory, the torque will also increase. The motor should be able to deliver the same power also, BUT the flux in the iron is now too high resulting in excessive iron loss, and the motor will fail prematurely. Above a very low frequency, (5 - 10Hz) the impedance of the magentising circuit of the motor is primarily inductive and so in order to keep the flux within limits, it is important to keep a linear V/F ratio (Voltage to Frequency ratio). If the frequency is reduced by 10%, the voltage must also be reduced by 10%. Because the flux in the iron remains the same, the torque capacity remains the same and so the power rating of the motor also drops by 10%.

Provided the voltage is dropped by the same proportion as the frequency, it is OK to run a 60Hz motor on 50Hz. The speed will be reduced by the reduction in frequency and the power capacity will also reduce by the reduction in frequency.

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