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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: Phase rotation errors

Phase rotation errors are not as rare as they ought to be. I’ve seen more than one building with a systematic phase rotation error. This can be prevented by carefully following the color coding system (Yellow Orange Brown and Red Blue Black for 480 volt and 208 volt systems in the US for example) and tagging feeders at both ends to assure proper connections.

To check for proper phase rotation sequencing (ABC and not ACB) you can use a phase rotation meter. Without that you can bump a three phase motor that should be correctly connected to see if it turns in the right direction. If it’s wrong, reverse any two phase wires from the source to the distribution equipment. However, if you have a tie breaker and intend to operate the secondaries of two transformers in parallel by closing it that is not good enough. Both transformer distribution networks have to be connected correctly on all three phases. You have to check the voltage across each corresponding pair of terminals on the tie breaker and be certain they are all about zero volts. If you don’t and there is an error, closing the tie breaker if that is possible at all (some electronic breakers may lock you out) will result in a phase to phase bolted fault that can result in severe damage to your distribution equipment. Phase rotation errors are invariably the result of incompetent installation, inadequate specifications for feeder identification, and inadequate inspection.

There are times when the phase rotation error is made on the primary side of the transformer. If this happens it can be compensated for by reversing the phase rotation error from the secondary side. This is less desirable but it will work. If you have multiple phase rotation errors in the same distribution network you have quite a mess to clean up. It will be time consuming and expensive tracking all of them down to be certain you have eliminated them. False economies by cutting corners on the initial installation of substations and distribution equipment will result in necessitating very expensive and inconvenient repairs. If it is not corrected you risk severe damage to three phase load equipment.

AQ: High AC current inductors

There are several issues at work here. For high AC current inductors, you want to have low core losses, low proximity loss in the windings, and low fringing effects.

At normal frequencies, ferrites are by far the lowest core loss, much better than MPP and other so called “low-loss” materials. So you would like to use them from this aspect.

A toroid gives the greatest winding surface for the magnetic material, letting you use the least number of layers and hence minimizing proximity loss. The toroid also has the advantage of putting all the windings on the outside of the structure, facilitating cooling. This is very important.

However, you can’t easily gap a toroid of ferrite, it’s very expensive.

Some aerospace applications actually cut the ferrite toroid into segments and reassemble them with several gaps to solve the problem. The multiple gaps keep fringing effects low. It might be nice if you could buy a set of toroidal segments so you don’t have to do the cutting because that is a big part of the cost. I don’t know if that is a reasonable thing to do, maybe someone can comment.

Once you go to MPP, the core loss goes up, but the distributed gap minimizes the fringing losses.

The MPP lets you run somewhat higher on current before saturation, but if you have high ac you can’t take full advantage of that due to the core losses.

All these tradeoffs (and quite a few more not mentioned for brevity) are the reason that so many different solutions exist.

AQ: Moving data around within memory of an individual PLC

The first question would have to be – why do want to do it? If the data already exists in one location that is accessible by all parts of the program, why are you going to use up more PLC memory with exactly the same data?

Well, there are a couple of candidate reasons. One might be recipe data. You have an area of memory with a set of stored recipes for different products, and at an appropriate moment you want to copy a specific recipe from the storage area to the working area. The first thing to be said about that is that if your recipes are at all complex and you have a requirement to have a significant number of different recipes, then PLC memory is probably not the right place to be storing them. The ultimate, these days, of course, is that recipes are created by techies on PCs away from the production area, in nice quite, comfortable labs or whatever, and are stored on a SQL server. Only the recipe for today’s actual production run gets transferred to the PLC. But there are some applications where there is a limited number of different recipes required and the recipes themselves are quite simple, when it can be reasonable to store the recipes in PLC memory.

A second reason for copying memory areas within the same PLC is for procedures, sub-routines or whatever. But again, these days, all PLC languages have some sort of in-built facility for procedures – what Rockwell uniquely call Add On Instructions, what everyone else calls UDFBs – user defined function blocks. In any case, the point is that these facilities usually make all that memory management stuff transparent to the programmer. You just configure the UDFB and call it as required. The compiler takes care of all the memory data moves for you.

Another reason for copying memory, actually related to the previous, is a technique much used by PLC programmers where they use an area of memory as a ‘scratch-pad’. So they will copy some unprocessed data to the scratchpad area, all of the operations performed on the data take place using the scratchpad, and at the end, they copy the processed data back again. Again, it is questionable how much this technique is actually required these days, I would suggest that it most cases, there probably is a better way using a UDFB. But I have seen some programmers who routinely include a scratchpad area within any UDFBs they define.

AQ: Creepage in thermal substations

The term creepage distance is specifically associated with porcelain insulators used in the Air Insulated substations. Insulator surface attracts dust, pollution (in industrial areas) and salt (along the sea coast) and these form a conducting layer on the surface of the insulator body when the surface is wet. As long as this surface is dry, there is not much problem. But when it becomes wet during early morning or during winter season the outer surface forms a conducting path along the surface from high voltage terminal to earthed metal fitting at the end of metallic structure and may lead to surface conduction and finally external flash-over. The insulators are provided with Sheds to limit the direct exposer to mist or dew. The protected area of the sheds will not allow formation of continuous conducting layer along the surface of insulator as the part of surface which is under the sheds may not become wet due to mist or dew and this part (length along the bottom surface) of the insulator surface is called protected creepage.

Measurement of corona inception and extinction voltages give a fair idea about the possible flashover even with protected creepage. But these will change under different levels of pollution.
This problem is not present with Composite insulators as the Silicone rubber sheds surface does not allow formation of continuous wet conducting layer as the surface of these insulators is Hydrophobic. Hence higher creepage is not considered for composite insulators.

However air density is also a limiting factor for deciding the creepage of insulators, necessitating higher creepage in case of higher altitudes.
You may have to assess the level of pollution and altitude of substation and select the creepage accordingly.
Medium pollution levels may be 25mm/kV
Very high pollution areas like on the sea coast and chemical and pharmaceutical industrial areas 31mm/kV where the insulators may become expensive alternatively periodic hot line washing is also another solution for cleaning of pollution on insulators.
In case of very high pollution levels GIS may be safe solution (if cost is not an issue)

Thermal substations where there are no electrostatic precipitators may also experience equipment failures due to pollution. Pressurized equipments like SF6 gas circuit breaker experienced external flash-overs during winter months in Northern India The utility was not accepting the theory of insulation failure due to pollution initially but they had to accept the cause of failure as pollution when they had similar failure in the consecutive year during the same winter months and they have resorted to hot line washing since then and there are no more such failures. Sometimes these deposits may not be seen glaringly but failure may happen.

AQ: Sub-transmission network

Q:
What factors determine current distribution between two 33kV feeders feeding a 33/11kV Substation within a sub-transmission network.

A:
Try the voltage divider rule. Take the R + X of each feeder (resistance and reactance) and find the Z. Remember that square root of R² + X² = Z. Now that you have the Z of each feeder, now find the Z of the two in parallel. To do this we have Z = 1/(1/Z1 +1/Z2). So, if Z1 = 2.16 ohms and Z2 = 1.67 ohms, then our Z of the two in parallel is 0.94 ohms. Now we pass the current of the entire substation between these two feeders. Let’s say that the current is 240 amps. Now we have 240Ax0.94ohms = 226 volts (IxZ=V). And since voltage divided by impedance gives us current (V/Z = I) we now take the voltage drop across the two feeders in parallel and divide each of the feeder impedances to get the separate feeder current. So we get 226V/2.16ohms = 105 amps (feeder 1) and 226V/1.67ohms = 135 amps (feeder 2). I have not tried this with your exact situation. Having different voltages from two different substations will change things, but at least this way you have a good start on the problem.

Since one end is tied together and the two other ends are from different substations, then you will have the classic voltage sending and receiving formula. Since the load is the one substation, then their will only be one power factor of the one load, so I would think this formula would apply: Es = Square Root of ((ErCosƟ + IR)² +(ErSinƟ +IX)²), which is square root of ((Receiving voltage times the cosine of the current phase angle plus current times resistance of the line)² + (Receiving voltage times Sine of the current phase angle plus current times reactance of the line)²). The voltage drop across each line would be VD=I(RcosƟ +XsinƟ) where R is the line resistance and X is the line reactance and the Ɵ is the phase angle of the load.