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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

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: How to connect 3 phase motor?

Making a connection of 3 phase motor the nameplate shows different voltages for delta it is 380-400 volt and 660-690 volt for star, what option should be selected? the supply Line to Line voltage is 380-400.

Each stator winding of the motor can withstand 380-400 V.
Thus, if you connect your motor (the stator of your motor) in delta, it should be connected to 380-400 V line-to-line.

On the other hand, if you connect the stator winding of your motor in Y, you’d be able to connect your motor to line-to-line voltage that is sqrt(3) x 380-400 V = 660-690 V.

The actual output power (for a standard squirrel cage 3-phase AC motor) is not determined by the motor itself, but by the load it is driving. The motor will attempt to run at a speed near its synchronous speed, and to deliver the power required by the driven machinery at that speed. This means that the current taken up by the motor at any given voltage, will be almost the same whether it is star, or delta connected. If you therefore connect the motor in star while supplying it by the voltage it is designed for when delta connected, the current through each winding will be sqrt(3) times the winding is designed for. This again means that the heat dissipation in the winding will be approximately 3 times what it is designed for, and therefore it will burn out if you load the motor with its nominal load.

We should be aware that the motor power as mentioned on its nameplate, in relation with the available power of the MCC panel to which it is connected, are the important factors in choosing the type of starting of the motor. Take into account the fact that starting the motor direct in Delta connection (which is the correct one based on your network voltage) the currents may be up to 8xInomianl of motor and if your MCC doesn’t have the capacity to withstand this current (by decreasing its supply voltage ) you may fail with DOL Delta starting type. Is that why, based on the power of motors, in order to avoid high currents during the starting time, it is recommended the Y/D connection. Limitations in starting currents by Y/D are considerable by decreasing the current first with sqrt3 because the feeding voltage is not 660V (you feed the motor with 380-400V) and the current initially in Y is sqrt3<I delta, so it is 3 times less than Delta DOL. Y/D is not the single one, there are a lot of solution to start AC motors.

AQ: 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.

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.

AQ: Automation solution

Automation is a solution:
1. To reduce the manpower & the CTC due to them.

2. Few skilled technicians can run the automated machines smoothly, with much lesser number of errors & faults (as human is not directly controlling every thing & is not burdened with multitasking challenge for extended duration which causes fatigue and hence errors/faults)

3. The power consumption & time to market can be estimated & reduced as machines will be operated on time & can work for longer durations than human beings & they don’t ask for tea/coffee/lunch breaks nor they ask for incentives. (Care for peoples who are maintaining them as well as care for machines which are earning profits for you, by regular maintenance & regular proper inspection of their conditions). Now days very good automatic power mgmt processors/controllers are available which can maintain the power as per the defined conditions as per the load & real time necessity.

4. Train the operators / technicians regularly to keep them up to date with tricks / methods / operations / principals to handle most situations by their own (will reduce the cost of a nonsense manager who is kept to yell, threat & discriminate subordinates and know only one slogan: “do it properly, otherwise, i will ….”). A training department which actually hold the capability to technically train employees from labor to talented engineers is a necessity of this age, as things are not remains just a lifting boulder & digging holes. We are living in an advance age in which we are having many expectations, competition & external pressures.

5. Finance bugs cries for expenses on NRE costs, salaries & treats this investments like invested in a share/equity/debt fund, but, earning from a business & financial mgmt capability must be inline with level & operations performed by the company. Instead of keeping low minds in tech industries, hire the engineers who has reached to an expertise level in automation industry & know the in depth issues occurring in between & underneath to estimate & expect correct values & timelines. Qualified project managers are much more realistic in their approaches, thoughts, assumptions & mentality.

AQ: System with difference neutral

Q:
I have one system with two source. One from genset and the other from PLN (national power supply company) that each system has neutral.
The question is
1. Is there any problem if I connect both neutral directly?
2. Is there any spark when I connect both neutral?
3. How is the best solution to connect both neutral?

A:
1. I understand that the Genset is dedicated for essential load as an Emergency power supply which will be operated by hand (only in Manual Mode).

2. The Control Philosophy for a Generator that intended to be connected to PLN as emergency power source depends on the local service provider regulations.

3. Usually in your case there should be Electrical as well as Mechanical interlocks between the mains incomer & genset main breaker. ie both Sources will never be in Synchronism ( will not feeding the same load simultaneously).This measures will ensure that there will be only neutral point to the system.

AQ: Cross regulation for multiple outputs

Cross regulation is a very important component of multiple outputs. This can be done in several ways: transformer coupling, mutually coupled output filter chokes (forward-mode) and/or shared output sensing voltages/currents. All, of which, are impossible to model. I have tried them all.

I have sort of written of the first two off, since it is under the control of external vendors, which make their own decisions as to their most cost effective solutions. At best, transformer solutions yield a +/- 5 percent regulation, and can be many times much worse. .Coupled inductors yield a much better cross regulation, but the turns ratio is critically important. If you are off by one turn, you loose a percentage of efficiency.

Shared current/voltage cross sensing is so much more common sense. First, choose the respective weighing of the percentage of sense currents from each outputs approximately in proportion to their respective output powers. Keep in mind that, without cross-sensing, the unsensed outputs can be as much +/- 12 % out of regulation. Decide your sense current through you lower sense resistor. Then multiply your percentages by this sense current from your positive outputs. Calculate each output’s resistance to provide that respective current. Try it, you will be amazed. The negative outputs will also improve immensely.

One can visualize this by, if one senses only one output, only the load of that output influences the feedback loop, which, for example, increases the pulsewidth for each increase in load of the heavily loaded output. The lighter, unsensed loads go crazy. By cross sensing, the lighter loads are more under control and the percent of regulation of the primary load is loosened somewhat.
By sharing the current through the lower sense resistor, you can improve the regulation of every output voltage in a multiple output power supply.

AQ: Soft start motor tripped in fuel oil suction and discharge

First of all check all the component i.e.CB, CT, Heat Element, and the O/L setting then megger the motor to be shore that there is no problem with the motor winding insulation.
After that let the mechanical check the vibration analyses during the start-up also measure the startup currant of the motor and diffidently you will find where is the problem.
It could be a relay setting; or problem in the insulation; or even a problem in the motor itself.

On the other hand, check the motor on No Load condition and tune it to the Soft starter before coupling it to the pump.
Auto Tuning feature is generally inbuilt to Advanced soft starters.
If the No load startup of the motor is perfect, 2 causes arise:
1) Improper design.
2) Viscosity _ this can be tackled if you can make some temporary arrangement for pre-heating to confirm if this is the culprit.

As using soft starter could result in reducing torque of the motor. Soft starter normally reduces starting current by reducing starting voltage. However, decreasing voltage will lead to starting toque reduction. Hence, the motor may take longer time, especially when driving high-inertia load, with somewhat high current until it reach its full speed. Using an inverter will help you get full starting torque or even boost up it to 150-200% while keeping starting current at 150-200% of full load. Installation of heat tracing might also help and economic.

Assuming it is an electrical problem. On a motor of this size it has separate overload protection from the ground fault and short circuit protection. There are tolerance levels for motor that you may not be within. However a megger will not answer all the possibilities with motors unless you are ready to perform polarization index test etc….A power analyzer will allow you to see the operation in real world application. Assuming you have confirmed this is an electrical problem your next step would be to use a power analyser. You should be able to confirm by the signature and different placements of the analyzer the problem. Analyzer should be around all three phases.