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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: 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.

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.