Author: ABBdriveX

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AQ: Power Transformer power losses

Power losses of ferromagnetic core depend from voltage and frequency. In case where is no-load secondary winding, power transformer has a power losses in primary winding (active and reactive power losses) which are very small, due to low current of primary winding (less than 1% of rated current) and power losses of ferromagnetic core (active and reactive power losses) which are the highest in case of rated voltage between ends of primary winding…

Of course, we can give voltage between the ends of primary winding of power transformer (voltage who is higher from rated voltage), but we need include some limits before that:

1. if we increase voltage in the primary winding of power transformer (voltage who is higher from rated voltage), we need to set down frequency, otherwise ferromagnetic core of power transformer will come in area of saturation, where are losses to high, which has a consequence warming of ferromagnetic core of power transformer and finally, has a consequence own damage,

2. if we increase voltage in the primary winding of power transformer (voltage who is higher from rated voltage), also intensity of magnetic field and magnetic induction will rise until “knee point voltage”: after that point, we can’t anymore increase magnetic induction, because ferromagnetic core is in area of saturation…

In that case, current of primary winding of power transformer is just limited by impedance of primary winding… By other side, in aspect of magnetising current, active component of this current is limited by resistance of ferromagnetic core, while is reactive component of this current limited by reactance of ferromagnetic core.

There is a finite amount of energy or power that can be handled by various ferromagnetic materials used for core material. Current increases greatly with relatively small voltage increases when you are over the knee of the magnetization curve characterized by the hysteresis loop. Nickel/steel mix materials saturate at lower flux densities than silicon steel materials. 50ni/50fe materials saturate at about 12kG; 80Ni/20Fe as low as 6kG. Vanadium Permendur material saturates at levels as high as 22kGauss- Nano-crystallines- 12.5kG (type), Ferrites -typically over 4kG at room, decreasing as temperature rises. What causes saturation?: Exceeding material limits.

AQ: AC induction motor constant power

An AC induction motor is supposed to be a constant power motor, which implies it draws more current on low voltage. Consider a motor running a constant torque load at a particular speed. Suppose now the voltage is reduced, which should cause it to settle down at a lower speed supplying the same torque as per the new torque speed characteristic. If we consider the electrical side, higher slip will cause more current to be drawn that too at higher pf, which should maintain the power justifying the above theory. But on the mechanical side the new output power Torque x speed is supposed to be lesser now as speed is less now. Is it this contradiction?

The following guidelines prove there is no contradiction since your question about Motor under running condition:

1. Torque / Slip characteristic for Induction Motor has three Zones.
a)- Starting Torque @ S=1, selection of this torque depends on the application. The starting should be greater the system torque at time of starting.
b)- Unstable Zone during which acceleration and torque development took place. This zone up to the Max. Torque can be developed.
In this regard, it may be necessary to mention that the seventh harmonics to be considered otherwise crawling / clogging may occur.
c)- Normal Operating Zone. NOZ about which your query raised. NOZ ranged as ” 0 < S< 1″ ie up to the Max. Torque. It is worth mentioning that Max. torque always remain the same regardless to its location of occurrence.

2. The torque is directly proportional to rotor resistance “r2” & varies with slip “S”. hence increase of rotor resistance is the most practical method of changing the torque (ie wound rotor Slip ring Motors). Moreover, the Max torque achieved when rotor Resistance “r2” = The Stator impedance, At starting S=1.

3. Accordingly, the ration r2/x2 gives the location of the max. Torque w.r.t Slip (if the max. torque is required at starting (S=!) then r2/x2 should equal “1”.

4. load being constant. Mechanical output = Electrical input – losses.

5. Tmax Propotional to Sq(v). decrease of 50% of the supply voltage generate a reduction of 20% in the max. running Torque (zone c) , increase in slip and also Full load current and temperature raise increase while the full load speed decrease. the status of the above parameters will be opposite if the voltage increases by 10%.

Based on the above, in all cases since the Motor is running within the operating range will be no issue unless the supply voltage falls behind the above limits (-50%, +110%). Accordingly, variable frequency drives provided by under/Over voltage protection relay to avoid damage to insulation due to Heat/temperature rise that will be generated due to excessive current intend to composite load.

AQ: 480volt Solidly grounded system versus HRG system

High Resistance Ground will limit the current to about 5 amps. The good news is that it no longer be necessary to trip on a ground fault. The bad news is that you may not connect any single phase loads to that substation. If the single phase loads are an issue, it may be possible to support all those loads with one or two feeders. In that situation, an isolation transformer is added to create a separately derived ground.

High resistance grounding is an excellent option in systems where continuity of service is important. However it is important to understand that if a ground fault occurs, it needs to located and repaired. This can be at a time convenient to facility operation, but it must be chased down and fixed.

This troubleshooting is accomplished by pulsing the ground fault current between 5 and 8 amps. Then a hand held clamp on ammeter is used to search out the offending feeder. Most HRG manufacturers include this feature into their design.

480volt solid ground system the fault current L-g is limited to the fault rating of the system may be 50KA,and usually have a disastrous consequences during faults and system has to be properly selected and protected accordingly.

Presently for some emergency power systems in power plants there is a 480voltb 3ph supply is made with neutral grounded through a NGT ie to limit the L-g fault current within 10Amp, and this may result in overvoltage of the other two phases too.  Overall this becomes a more stable system.

I would be very hesitant to use HRG for a medium voltage system. While a fault is on the system the neutral will not be at earth potential. This is usually not so important because the neutral is not carried to any of the loads on a high resistance grounded system. But the effect of the neutral straying away from ground potential is the two unfaulted phases will have a higher than normal line-to-ground voltage. Low voltage systems have a lot of built in margin in their insulation so it is not a problem with most equipment connected to that system. However MV equipment does not have so much “spare” insulation so the effect of high resistance grounding of a MV system is that it significantly increases the chance of migrating the fault from a single-line-to-ground fault to a double-line-to-ground or line-to-line fault.

Also at the point of the fault, the energy of 5 amps flowing in a 480 volt or 400 volt system may be somewhat dangerous but in most cases will be dissipated easily. On a higher voltage system, whether it is 4.16kV, 6.6kV, or higher, the energy at the point of the fault is higher and is much more likely to damage the insulation on the adjacent conductors and quickly turn into a more severe fault.

So I would resist using a high resistance grounded design on a medium voltage system (over 2000 volts).

AQ: Resistance Grounding System

Low Resistance Grounding:
1. Limits phase-to-ground currents to 200-400A.
2. Reduces arcing current and, to some extent, limits arc-flash hazards
associated with phase-to-ground arcing current conditions only.
3. May limit the mechanical damage and thermal damage to shorted
transformer and rotating machinery windings.
4. Does not prevent operation of overcurrent devices.
5. Does not require a ground fault detection system.
6. May be utilized on medium or high voltage systems. GE offers low
resistance grounding systems up to 72kV line-to-line.
7. Conductor insulation and surge arrestors must be rated based on the lineto-
line voltage. Phase-to-neutral loads must be served through an
isolation transformer.

High Resistance Grounding:
1. Limits phase-to-ground currents to 5-10A.
2. Reduces arcing current and essentially eliminates arc-flash hazards
associated with phase-to-ground arcing current conditions only.
3. Will eliminate the mechanical damage and may limit thermal damage to
shorted transformer and rotating machinery windings.
4. Prevents operation of overcurrent devices until the fault can be located
(when only one phase faults to ground).
5. Requires a ground fault detection system to notify the facility engineer that
a ground fault condition has occurred.
6. May be utilized on low voltage systems or medium voltage systems up to
5kV. IEEE Standard 141-1993 states that “high resistance grounding
should be restricted to 5kV class or lower systems with charging currents
of about 5.5A or less and should not be attempted on 15kV systems, unless
proper grounding relaying is employed”.
7. Conductor insulation and surge arrestors must be rated based on the lineto-
line voltage. Phase-to-neutral loads must be served through an
isolation transformer.

Conclusion:
Resistance Grounding Systems have many advantages over solidly grounded systems including arc-flash hazard reduction, limiting mechanical and thermal damage associated with faults, and controlling transient overvoltages. High resistance grounding systems may also be employed to maintain service continuity and assist with locating the source of a fault.
When designing a system with resistors, the design/consulting engineer must consider the specific requirements for conductor insulation ratings, surge arrestor ratings, breaker single-pole duty ratings, and method of serving phase-to-neutral loads.

AQ: Directional Numerical over current relay

If current will flow in positive direction then the relay will behave as a Normal over current relay and if current will flow in negative direction then the relay will behave as a Directional over current relay…..Why the angle between healthy line voltage and fault current is required for sensing the direction??

Suppose you have purely resistive circuit with a voltage source connected to it. Now take any arbitrary node X in the network Look at the current flow from the node. Now when the source voltage is positive the current flows from the say upper node of the source to the lower node of source. Now if you look at this current flow from the arbitrary node X mentioned above, the current will be moving towards the node X from one side and it is moving away on the other side of the arbitrary node X node in one half cycle.
Now in the next half cycle the same thing repeats but with one difference, that the direction of current flow changes.

Now you want to operate your relay when the current is moving away from the arbitrary node in the first half cycle. place a CT at the node with primary P1 towards the node X1 and p2 away and take Secondary S1 to the relay . Now when current flow is from P1 to P2 current flow in the secondary will be say S1 to S2 thru relay.

Now we can say that, when the arbitrary node voltage is positive current flows from P1 to P2 and we take this as our direction required for the relay to operate.

Again when you look back from the node you will see that when the arbitrary node is at higher potential current is flowing towards you. Now place another CT with P1 towards the arbitrary node P2 on the other side and connect another relay at s2 side (now you can visualise 2 CTs on either side of the node with P1 towards the node X in both CTs). The current will be flowing from P2 to P1 in this CT and hence S1 will be negative with respect to S2 and current flow in this relay will be in reverse direction as that of the first relay.

In the next half cycle the current direction reverses and first relay current will be s2 to s1 thru relay and second relay current is from S1 to S2 thru second relay.

Now we want the relay no 1 to operate and relay no 2 not to operate. how do you achieve it when you connect the current alone to the relay which is changing the direction in every half cycle .
To achieve this now you connect a PT at the same arbitrary node X and connect the voltage to both the relays. The same point of the PT secondary voltage is connected to both the relays. Now find the phase angle between the current in the first relay and the voltage. You will see that when node voltage is positive, the current flow in the first CT will be from P1- P2 in primary and S1 to S2 in the secondary. In this case let us say that the phase angle between voltage and current is Zero in the first relay.

What happens in the second relay and CT /PT. The voltage is same as first relay which is positive and the current flow in the CT is P2 to P1 and in relay it is S2 to S1
– meaning opposite direction to first relay . As the current is in reverse direction with respect to voltage we can say that they are out of phase (180 degree).
Now you set both relays to operate when the voltage and currents are in phase. observe the result in first half cycle . Relay 1 operates (phase angle between v and I zeo ) and relay 2 no operation ( phase angle 180 deg ).

In the second half cycle observe the phase angle of relay 1 . Voltage at node is negative. (voltage phasor reversed ) Current flows from P2 to P1 and S2 to s1 in the relay 1 (current phasor also reversed) still the phase angle is zero and hence relay 1 operates Similarly relay 2 restrains.

Hence we found that, the relay 1 operates in both half cycle and relay 2 restrains. This is the importance of Voltage for directional relay.

AQ: Parallel connection transformers

AQ: Load Break Switch VS Circuit Breaker

There are two important different parameters as far as the disconnection/connection of Electrical utilities is concerned.
– Breaking capacity.
– Making Capacity.

Making capacity is more higher since it is considering the initial starting current, inrush and faults that might occur at switching on moment.

Accordingly, the Circuit breakers have both while the Isolator or LBS has only breaking capacity. Isolator therefore used to isolate/disconnect/break load. To be switched on only under no load.

Basically and LBS is only designed to make and break load currents. It can be closed onto a fault (has a making rating) and cannot break a fault current. A CB is designed to make and break fault currents and off course load currents. Need to be careful with switching long cables and long lines with LBS, due to its limitation with switching cable/line charging (highly capacitive) currents. Similar precaution with needed with switching reactive loads like large transformers. Normally a CB is recommended for the 2 latter cases. Check the manufacturer test specs.

In simple way the Load Break Switch used to cut off healthy circuits or to break / disconnect the load. As a precaution, normally the LBS / Isolator are to be switched on under no load, the connected load to be utilized after switching on the Isolator. This why it has no making capacity. Circuit Breakers are intended to operate under unnormal conditions in order to clear the fault & to isolate the defective circuits protecting its associated electrical equipment, therefore breaking and making capacities are considered as the most important criteria as far Circuit breaker are concerned.

The making current is not an RMS value it is the peak value ie. Impk = 2.5Irms. The peak value of 2.5 times RMS is the DC offset at point when the LBS closes on the fault and is taken as the worst case X/R ratio of the source (X/R of about 20). This peak decays to RMS value Ith (thermal withstand current) dependent on the X/R delay constant. The decay rate is exponential with time. There is a misconception that making is 2.5times breaking current, but making is normally quoted as a peak and breaking as RMS. The breaking current in a CB is an RMS value. Breaking fault current is far more difficult that making especially when the contacts open when current is not at zero crossing point on the sine wave. For HV systems 132KV and above, the restrike and TRV starts to become a major consideration in CB selection, especially for long cables and lines.

AQ: Ground fault detection in a Delta system

We have system which is connected to 16kV/2.4 utility transformer (delta on secondary) and we are using 2.kV/480V transformer for loads after 2000ft. Utility wants to protect against ground fault in the system. I am planning to select a ground over voltage relay using a broken delta PTs on secondary. I am having problem with calculating the 3Vo value, How much voltage will you set to trip the relay for SLG or LLG fault.

Let’s say you have a system 16kV/2.4kV with more than one, say 5 transformers T1, T2, T3….T5 interconnected transformers throughout your network, with the broken delta arrangement to detect the residual voltage on each of the transformer’s delta side, If you have an earth fault say on the LV of T1, the voltage displacement gets picked up on the delta side of all 5 transformers and there is a very high probability that all 5 transformers get taken out. This is because this scheme does not look for the earth fault current or where it exists, as long as it is on the interconnected system where the source is able to support the earth fault. The moment it notices a voltage displacement, bang goes your CB to clear the fault irrespective of it’s location.

A better scheme is to use the zig-zag transformer which offers a low impedance for zero sequence currents, generally used with a neutral grounding resistor to limit the current to more often than not the rated current (or lesser than that) of the transformer LV. So in a similar situation for an earth fault on T1 LV, the earth fault completes it’s path through the earthed NGR resistor and back to the fault point. A sensitive earth fault relay connected to a CT connected between the grounded resistor and the neutral point of the zig-zag transformer, designed to take out both the HV & LV CBs of T1 will do the job, without fear of taking out the other transformers.

Of course if this is the only transformer you are talking of, then the voltage displacement method should work in principle, however I would still go ahead and install the zig-zag arrangement described above. Let me know your thoughts, and then we can start discussing about the magnitude of 3Vo or 3Io as the case may be.

When we talk about detection of Earth Faults, that means we want to know it without tripping, so, we should absolutely use “IT Earthing system” for LV side, then with this system we can use “IMD – Injection Monitoring Devices” that monitor when it happen, and if wanted send the tripping order to the installed Circuit Breaker.

By the way:
1- As the transformer’s connection of secondary side is “Delta”, we use one of 3 phase to be connected via a special impedance to the earth to have “IT” system. Noting that for some of “IMD” we don’t need to use the special impedance as it’s integrated inside the IMD.
2- In the most of “IMD” we can adjust the value of “Insulation level” where above of this value an Alarm signal by auxiliary contact will be sent.
3- Some of “IMD” have 2 levels can be adjusted “1st level for Alarm” and “2nd level for tripping”.

AQ: Motor short circuit protection

In motor protection panel when 3 fuses are provided for short circuit protection, Is it always a condition that during short circuit minimum 2 fuses should be blown? If yes or no then why?

Because fuse is a type of low resistance resistor that acts as a sacrificial device to provide overcurrent protection, of either the load or source circuit.

Its essential component is a metal wire or strip that melts when too much current flows, which interrupts the circuit in which it is connected.
Short circuit, overloading, mismatched loads or device failure are the prime reasons for excessive current.
2 fuses are must & should blowing in motor control panel due to motor each winding sets are connected to 2 phase in delta connection (each winding set is works on 440 Volts power supply).

We need to identify the types of short circuits which can be experienced in a motor and if you are talking of 3 fuses for short circuit protection, it means you are dealing with a 3phase motor. All things being equal, a three phase motor should be balanced in operation hence the current in each phase must be the same.

If a short cct occurs, it could be a phase to frame ( L-E),or phase to phase(L-L)or even 3phase (3L) short cct. In each case, the fuse that ruptures will depend on its condition, rating, type and class. If a fuse has been subjected to various loading and abnormal conditions, the tendency to rupture faster exists. If I should view your question from the perspective that all fuses are of the same type, class and subjected to the same condition and motor windings are same and in the same healthy condition, then a L-E fault should not necessarily cause other fuses in the motor cct to rupture immediately. It should be the defaulted line. And if it happens, the motor will not be balanced which in turn causes the other fuses to rupture in turn due to increase in current. It is always better to use a circuit breaker to isolate all the lines in the event of a fault.

AQ: Motor fuses

Mainly Fuses are used for protection against short Circuits due its high rupurting capacity (breaking capacity) and fast response (less than 10ms).

As far the electrical drives are concerned, Fuses can be used to protect the feeders, while the Electrical Motors will be protected by Thermomagnetic CB to achieve Short Circuit as well as overload protection. At least thermal overload has to be provided for the Electrical Motors.

Accordingly, Plow of fuses depend on the type of Short Circuit, Single phase or 3 Phase fault (ie location of the fault) and the let through energy. In case one phase blown (say earth fault) -ve sequence and Zero sequence will be generated and subsequently the motor thermal overload will operate to protect the Motor.

It is worth to mention that, now a day proper protections for Electrical motors are commonly used, MCCB/MCB for overcurrent (short Circuit & overload), Single Phasing, Under voltage, Phase sequence relays… etc.
Thermal Protection can be achieved by many technique (ie Bi-metal, thermostat, resistance (NTC or PTC),,,etc.

The Fuse for Electrical Motor is efficient for O/C (Short Circuit, either L-G or L-L) or/and internal fault in the Motor windings. Taking into consideration the fuse rating considering the Starting current of the Motor. Therefore the fuse will not be effective for overload protection on the similar case.

However, I believe the motor was either subjected to an internal fault due to insulation failure (Humidity, water, bearing damages, high temperature rise, Stator/rotor gap,,etc) , subsequently the fuse blown or in prior the fuse blown due to an external factor and the thermal overload device associated with Motor Control panel not operated / out of order in the proper time -The heat rise impaired the winding accordingly or the Motor was subjected to stall current and prolong starting period.