Iacdrive|Automation Solution

What is the best laptop for field work?

Dell D630 – it is the best laptop for field use I have used. And for some applications standard RS232 port is a must. We have Freja 300 test set which totally refuses to communicate with PC via widely available cheap USB-to-serial adapters. The only usable adapter I have found is semi-industrial type, costing about 50 Euro. Not that a price is so much concern, but it is not very convenient to deal with additional boxes, power supply units for them, etc. when commissioning at field.

But I do not expect you will have problems connecting Omicron via converters. We have been used CPC256 via various USB-RS232 converters without serious problems.
For communication with relay protections from Siemens and AREVA never had problems too. Cannot remember how it was with older ABB relays (last case we used them was 4 years ago), but newer ABB series are all with Ethernet communications.

So my advice will be – by special laptop for field work, not mix it with that for everyday office use. Load it with the minimal necessary software – MS Word, Excel, Adobe Reader, Omicron’s Test Universe and software for relays which will test.
For all these needs most older type laptops (4-5 years older) would be sufficient and you can buy for 200-300 Euro solid business class laptop. And also very important: look for non-glossy displays only!

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.

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.

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.

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

What need to be concerned to start a motor?

First, you need to know power (rated power and rated current) of your power source with whom you will supply your motor. For example, if you want to supply your motor by using low voltage synchronous generator (through high voltage power transformer), you need to know rated power and rated current of synchronous generator and rated power and rated currents of high voltage power transformer. This information is very important because if you don’t have powerful source for supplying your motor, there is possibility that you’ll never reach rated rotational speed during rated time which means that you’ll not start your motor.

Second, you need to know kind of your motor. Is that motor asynchronous motor with cage rotor or is that asynchronous motor with sliding rings? This information is very important because these kinds of asynchronous motors have different values of starting current: for asynchronous motors with cage rotor starting current is 6-8 times higher than rated current of mentioned kind of motor while for asynchronous motors with sliding rings starting current is 3-5 times higher than rated current of mentioned kind of motor. Also, too much higher starting current of your motor could be a reason for unallowed warming of windings of stator what it could lead to dangerous consequences, first all, for people in surrounding of motor and then also for equipment in surrounding of motor.

In relation with start of your motor with lower voltage because you will, on that way, reduce starting current 2 times and starting torque will be 4 times lower than rated torque of your motor. On that way, you will easily start your motor.

VFD replace mechanical gearbox to drive the load

Can an AC drive to replace the mechanical gearbox that used to decrease motor speed in conveyor application i.e to use a motor that will drive the load directly throw a coupling, belt or chain, without gearbox, motor rated up to 18.5 kw.

Theoretical is true as far the speed variation is concerned. Practically is not recommended for your application if the conveyor is required to be used with constant speed, on the other hand the gearbox also used for Torque purposes.

For light conveyors used on packing lines on which rate of production varies in accordance to some industrial parameters (Automation & PID control), direct coupled motor controlled by variable frequency drive may be feasible.

VFD is expensive (capital & running cost) its selectivity should be done carefully among the other available options.

By using a variable frequency drive we can change the speed of an AC Motor, and working for any time on any choosing speed, even in some case we can exceed the speed more than the normal one if the motor can withstand it. Noting that:
1- We should be careful when choosing the type of AC drive that should ne normally done according on the application “Conveyor, Fan, Pump, Compressor, ext ” to determine the torque’s level at running time.
2- In some special case when the motor runs at too low speed comparing by his normal one, maybe we need a forcing cooling for that motor.
3- Each VFD has a value of the Short Circuit’s level that can be withstanded, so, we should be careful of that point.

How to find the KA rating of Circuit breakers?

Before breaker’s selecting for your electrical system, you need to calculate value of expected short circuit current at the place of breaker’s installation. Then you need to calculate value of heat pulse and 1s current (expected value of current during one second). After that you need to calculate power of breaker and finally, after all, you can select appropriate breaker. Values of characteristics of selected breaker need to be higher from calculated values of characteristics of your power system.

1. The fault level of the upstream NW (Source) to be known, normally 500MVA or 250MVA.
2. Upstream impedance (reactance and resistor, capacitor to be ignored for Short cct calculation) can be determined accordingly.
3. The LV System starting from the secondary of the distribution transformer.
4. Short circuit percentage voltage for Transformer is known (normally 4% for 1000kVA and 6% for 1500 kVA) and hence reactance and impedance can also determined.
5. Impedance of Cables also can be determined from manufacturer TD sheet.
6. Subtotal impedance to be determined by conventional way (Submit if are in series/ (Z1+Z2+….Zn)/(Z1XZ2X…Zn) if are in parallel.
7. divide voltage by the Impedance up to the required location, will give you the fault current at that location.
8. Determine Maximum and minimum fault current. By the former you can decide the breaking capacity of CB and by the later the setting can be achieved.
9. verify the thermal constraints of the conductors(cables). ie

I²t ≤ S²K² , I short cct current, t time( < 5 s valid), K cable material Factor and S cable section area.
I²t Known as let through energy. accordingly breaking capacity of CB should be > than Circuit Maximum fault.

The MCB, MCCB, & ACB are all Low Voltage Circuit Breakers, where SF6 is a Non-active gaze used in Medium Voltage Circuit Breakers.

Now, to determine the value of Breaking Capacity of any circuit breaker, we should, by calculation, the Maximum Short Circuit Current Value ” Isc3max ” at the installation point of that circuit breaker, where we can calculate it by assuming a ” Short Circuit between 3 phases at that point “, then after knowing ” Isc3max ” we can determine the Breaking Capacity value that should be ” equal or bigger than Isc3max “.

Further:
1- The value ” 250 … 500MVA ” is the short circuit power at Medium voltage side for up to 36kV.
2- About the Short Circuit Voltage percentage value: we called ” Ucc or Usc ” and the value is ” 4& for up to 630kVA transformers “, and ” 6% for up to 2500kVA transformers “, but in all case, we can read it at the transformer’s name plate.
3- Sorry Mr. Omar, we can’t do, you mentioned, the sum of all Z, as these values aren’t on the same vector, so, we should first calculate ” R & X ” for each component, then do the sum of all R ” R total ” and all X ” X total “, then calculate the ” Z total “.
4- By knowing the Minimum Short Circuit Current value ” Iscmin “, we use it to determine the value of “Setting Value” of “Magnetic Protection or Short-time Protection”.

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.

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.