AQ: Bushing insulation testing
In bushing insulation test there are three major current elements which any of those could affect the test result. These current elements are Capacitive current
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
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: 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: How to select a drive between motor and machine?
We should select a drive (direct/flexible, chain, flat/vee/ribbed belt, gearbox, soft start). The motor/starter/drive characteristics should match that of the load. Design and factors to be considered in selection.