Category: Blog

Home  ›  Blog  ›  Page 14

AQ: The effect of power failure on VFD

Q: I am planning to put some variable frequency drives on non-critical section of factory where there will be planned interruption of 30 seconds but 2 times a day.

A: Why are you going to use VFD, variable frequency drives are expensive. What is the application?
If the application require fix speed / rate Soft Start device is required.

However, if there might be frequent start stop (Power OFF/ON) AC Contractor Duty AC3 are recommended to be used to bypass the Soft Starter or Static device once the required motor speed is reached and then Start/stop have no impact on the installation.

The technique of using AC3 Contactors, is not applicable for VFD if the VFD is necessary for application. In this case the other advantage associated with VFD no longer will be valid (Protection, control and monitoring).

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: Parallel operation of autotransformers

Q:
We have 2 no 160MVA 220/132/11 kV transformers with short circuit impedance 46.06 ohm, and one 160MVA transformer, 220/132/11 kV with % impedance 15.02%. Can we parallel these three transformers?

A:
Indeed, the vector group is an important (mandatory) consideration when connecting transformers in parallel. And also important if a transformer is going to close a loop in either the HV or LV sides.

But it is perfectly OK to parallel transformers with different impedances. All it is going to happen is an uneven distribution of the power flow, among the parallel transformers. The unit with the lowest impedance would carry a larger share of the load.

Regarding “same ratio”: are you talking about the transformer ratio, such as 138/230 kV? Or are you talking about tap positions? Within certain constraints, it is possible to parallel transformers with different ratios (let’s think, for a second, of identical transformers at different tap positions). This is not recommended, though, because of reactive power circulation.

So, without disagreeing with the factors that you have listed, I would like to re-order, if you will, the conditions you have described:
1) Mandatory: making sure that the vector group and nominal voltages of transformers being considered for parallel operation are indeed adequate and compatible with the intended parallel operation
2) Desirable: ability to operate parallel transformers at the same tap positions or as close as possible, to minimize reactive current circulation
3) Almost indifferent: identical impedances on the parallel transformers simplify things a bit, but this is not a “show stopper” for parallel operation of these transformers. Actually, it is more realistic to expect some differences in impedances, even for otherwise “identical” transformers (same manufacturer, same nameplate ratings, etc.)

For the “Y-Y- Delta” transformers operated in parallel, there exist two kinds of the circulating currents between the tanks and between the banks of the delta side. As the circulating current between the tanks is 90 degree out of phase of the load current, it is estimated by decomposing the line current into the component 90 degree out of phase of the load current. The circulating current between the banks in the delta side is estimated from the delta winding current and the line currents.

The estimated circulating current depends on the power factor of the system even with the same tank currents. This characteristic is derived from the view point of the active and reactive power. Also, it needs the voltage as well as the tank and the load currents.

AQ: Parallel connection transformers

AQ: What happen if we put a magnet near digital energy meter?

In the “olden” days when there were only moving disk meters, I heard that people drilled small holes into the Bakelite cases and tried to get spiders to make a web inside the meter and slow the meter down. It probably wasn’t true, but there have always been people trying to get something for nothing.
I also heard that some people were using a welder and found that their moving disk meter went backwards, but it depended where they positioned the welder, and how strong the welding current was.

Back to electronic meters, if there are transformers inside the electronic meter, placement of a magnet as close to this transformer as possible could cause over fluxing every half a cycle, this could cause a diode like affect in the meter electronics, and if the electronics are designed to eliminate harmonics for calculating energy usage, then the magnet has let this person pay less for electricity, i.e. steal electricity.

Of course the meter may also have a detection circuit for high harmonics and send a message back to the utility to say the harmonic level is too high and a serviceman may then discover this magnet.
I do know that some electronic meter IC manufacturers have added a bump circuit into their ICs so I am sure they have thought about this sort of trickery too.

I like everyone paying full dollar for their electricity, otherwise most of us are carrying the small number of people doing these sorts of things.

“Meters should offer compliance to requirements of CBIP-304 and its amendments for tampering using external magnets. The meter should be immune to tamper using external magnets. The meters should be immune to 0.2T of A.C. magnetic fields and 0.5 T of D.C. magnetic fields, beyond which it should record as tamper if not immune.”
The above statement is a requirement during the manufacturing of digital energy meter. Hence we shall assume that digital meters are tamper proof using Magnets.

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: Reactive consumptions in AC power system

There are two types of reactive consumptions in AC power system, inductive and capacitive reactances. We can not call them losses. The loss of a transmission line is the active power consumed by the line resistance which is determined by the current on the line. Reactive power can adjust the power factor and control the apparent power, then the current and losses on the line.

The minus reactive power means capacitive load is higher than the inductive load, which happens when the transmission line has no load or with pure resistive load because the capacitive load along the TL dominates the reactive load. In this situation the voltage at the end of the line should be higher than the one at the beginning (you should get it when you get the negative reactive power).

When the load (80% of the industry load is inductive) increases, the reactive power will be positive as the inductive load will dominate the reactive power consumption, and then voltage will lower than that at the beginning. So the optimized choice for the reactive load is that in power plant generating less reactive power (reducing the losses on the line) and generating the compensating reactive power (negative reactive power) at consumer side by using capacitor banks or synchronizing motor, which can increase the power factor of the consumption and regulate the voltage (if the transformer has no taps), and then efficiency (save money) as well.

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: Electrical equipment in hazardous areas

With regards to hazardous areas, Electrical equipment to be installed in those areas should comply with the zone classification. I believe the location where you are intending to install this motor would have been classified according to your local classification standards or IEC 60079 for Liquid/gas/vapour explosives OR IEC 61241 for dusts. Therefore your motor should is to be certified to be installed in those areas, to verify this information you can ask the manufacturer or supplier to provide the Certificate of conformity.

Other information to be looked at, when installing the hazardous motors with variable frequency drive etc, the IEC requirements state that
– both motor & VFD to be certified and type tested together
– IP ratings, protection technique, temp class, gas group to comply with zone classification

It is critical to remember that the starting torque is reduced by the square, as the voltage is reduced. So at 70% voltage, the torque is down to 50%. That is where I have experienced the most trouble with soft starts.

It’s probably important to model or have someone model your load versus the motor torque on the soft starter, to make sure the motor will start, and that it doesn’t take so long to accelerate the load that it causes excessive heating, or trips overloads.

AQ: Earthing conductors size calculation

1. As per IEE/BS7671, The Minimum cross-sectional area of protective conductor in relation to the cross-sectional area of associated line conductor (Say =S mm2) by taking into consideration that are of same material, as follows:

If S ≤ 16 then the Minimum cross-sectional area of the corresponding earth conductor = S mm2

If 16 < S ≤ 35 then the Minimum cross-sectional area of the corresponding earth conductor = 16 mm2

if S > 16 , then the Minimum cross-sectional area of the corresponding earth conductor = S/2 mm2

2. It may be necessary to verify the same by using the following equation
I²t ≤ S²K

I earth fault current and t tripping time.

while the following equation is applicable for bonding conductor

Zs < 50/Ia where Zs = Earth loop impedance and Ia is protective device operating current