Electrical Tariff in brief

Tariff is of these types:
1. Simple Tariff: Uniform cost / unit of energy consumed. Simple 2 comprehend by consumer but doesn’t facilitate usage of electricity and provides for no discrimination among the type of consumers
2. Flat rate tariff: Uniform cost/ unit of energy consumed set differently for different types of consumers. So a consumer with power load will be charged at lesser rate than lighting loads. But it doesn’t discriminate in costs within a particular type of consumers irrespective of how much energy they consume.
3. Block rate tariff: Energy supplied in blocks with each block having a particular number of units of energy. The first few blocks are charged highest followed by the next…then the next and so on. This helps in maintaining a good load factor (avg. load/peak load) by providing the said incentive
4. 2 part tariff: Fixed+Running charges. Fixed -> Based on maximum demand in kW (charged on /kW basis) & Running -> Charges / unit energy consumed in kWhr. Maximum demand estimated on basis of rateable value (total connected load)
5. Maximum Demand Tarriff: Similar to 2 part tariff except that maximum demand is obtained from MDI installed at consumers premises. Normally used for big consumers only.
6. Power factor tariff:
(a) kVA tariff: Fixed charges are based on kVA instead of kW to include the pf component as kVA is inversely proportional to pf
(b) kW & kVAR tariff: Charged on the basis of kW & kVAR units consumed
(c) Average pf method: A reference pf is set. If pf> ref, discount is provided to consumer else he is penalised at a predetermined amount
7. 3 part tariff: Includes the 2 part tariff+ fixed charges made during each billing period to include labour charges, interest. Depreciation on the basis of secondary power distribution

Foreign Internships...Interested??

This is for students currently in third year to help them choose wisely where to apply for their summer internships as well as when to start applying! Do check it out if ur seriously planning to apply for projects abroad!
Follow this link...
http://www.4shared.com/document/rI0zNIxB/Foreign_Internships.html

The 3rd harmonic Problem...& how delta configuration eliminates it...

Harmonics are multiples of the fundamental frequency of a wave. In a balanced 3-phase wyesystem, the currents going into the neutral node are supposed to cancel, so that there is no current on the neutral wire. Harmonic waves may not behave this way, so we should investigate… Harmonics can be modeled as current sources as we have seen, but equivalently, they can be modeled as voltage sources. Lets look at the 2nd & 3rd Harmonics…

The system will still be balanced in the presence of the 2ndharmonic (except negative sequence) and the current from the 2ndharmonic sources will still cancel in the neutral wire! It can be shown that this is true with ANY even harmonic. There is something interesting that is immediately visible aboutthe third harmonic voltage equations we just found. All the waves are in phase with each other.
This is the 3rd harmonic problem
In a similar derivation as the previous slide, with current instead of voltage we can get in the neutral wire:
Ian=Ibn=Icn for 3rd harmonic
In the 2ndharmonic, the current still cancels in the neutral wire, but the 3rd harmonic current adds, producing a very large neutral current. So we need to get rid of the neutral wire. By using a delta configuration, we can trap the current in the delta loop and also take advantage of one useful property. can easily see that the power across the load in a delta loopdue to the third harmonic is zero.

Source: Derek Grant

Braking in AC Motors

• Generator Action (AC Motor)
If an induction motor is forced to run at speeds in excess of the synchronous speed, the load torque exceeds the machine torque and the slip is negative, reversing the rotor induced EMF and rotor current. In this situation the machine will act as a generator with energy being returned to the supply.
If the AC supply voltage to the stator excitation is simply removed, no generation is possible because there can be no induced current in the rotor.
o Regenerative braking
Thus in traction applications, regenerative braking is not possible below synchronous speed in a machine fed with a fixed frequency supply. If however the motor is fed by a variable frequency inverter then regenerative braking is possible by reducing the supply frequency so that the synchronous speed becomes less than the motor speed.
AC motors can be microprocessor controlled to a fine degree and can regenerate current down to almost a stop whereas DC regeneration fades quickly at low speeds.
o Dynamic Braking
Induction motors can be brought rapidly to a stop (and / or reversed) by reversing one pair of leads which has the effect of reversing the rotating wave. This is known as "plugging". The motor can also be stopped quickly by cutting the AC supply and feeding the stator windings instead with a DC (zero frequency) supply. With both of these methods, energy is not returned to the supply but is dissipated as heat in the motor. These techniques are known as dynamic braking.
Regenerative braking
‡Problem in industry: Most AC inverters cannot return the excess power to the power line.There is a diode bridge that takes AC power and charges a capacitor bank.(There are a very few specialized AC inverters that use a bi-directional bridge of transistors to charge the capacitors or return power to the AC line.)The capacitor bank is connected to the motor via a bridge of transistors or gate turn off SCRs.This bridge is bi- directional.So whenever the motor is a generator, the capacitor bank is charged up and potentially overcharged.Some drives turn off to protect the controller.Some drives switch on a resistor to dump the excess energy.Here we are not really regenerative, ‡ If you were to replace the capacitor bank with batteries, like some electric cars, you now have the ability to absorb a lot of energy from the motor, and you can easily be regenerative over the entire speed range of the motor.

(Also see the DC injection braking mechanism from wikipaedia. It basically is breaking the motor by removing the AC supply and applying DC to stator to any two terminals)

Set 2

What are the Typical applications for delta-star transformers?

- generator step-up transformers - with the generator connected to the delta and the star connected to the system.
- substation transformers - with the delta connected to the grid, and the star connected to the load substation.

Star delta is used at transmission levels for step down


Running a 3ph motor on Single phase line?
The way that I accomplish this is to add capacitors from one of the hot leads to the third lead of the three phase motor. It creates the phantom third hot lead. However, for starting the motor I have a momentary contact switch set up, just like the the before mentioned capacitors, except the starting capacitors are much, much larger. I first turn on the motor. It rotates slowly. Then I hit the momentary contact switch which brings online the starting caps. Then varoom. The motor takes off to it's normal operating speed. The largest motor that I've tried this trick on is a 7 1/2hp. If the motor is small enough then I don't need the start switch.

How to decide the current rating of ISOLATORS?
Need to have current breaking capacities of:

Why do we go for bundled conductors in EHV lines?
We go for bundled conductors because this reduces the voltage gradient in the vicinity of the lines hence corona is reduced and also the interference with neighbouring comm. Lines. Also inductance is reduced leading to better voltage regulation and more power handling capacity.

What is form factor?
Form Factor= rms/avg… (=1.11 for sine & 1 for DC and square waveforms…)

What happens if we give 220 volts dc supply to bulb or tube light?
Bulbs [devices] for AC are designed to operate such that it
offers high impedance to AC supply. Normally they have low
resistance. When DC supply is applied, due to low
resistance, the current through lamp would be so high that
it may damage the bulb element.

How are generators rated?
Generator sets are rated in kVA at 0.8 power factor lagging.
This 0.8 power factor is not the load power factor.
It is a nominal power factor used to calculate the kW output of an engine to supply the
power for a particular alternator kVA output.
Example:
Alternator output (kVA) : 100kVA
Engine power output (kW) : 100kVA x 0.8 = 80kW

What are diversity and demand factors in power system?
An example would be a conveyor belt made up of six sections, each driven by a 2 kW motor. As material is transported along this belt, it is first carried by section 1, then each section in succession until the final section is reached. In this simple example only one section of conveyor is carrying material at any point in time. Therefore five motors are only handling no-load mechanical losses (say .1 kW) keeping the belts moving whilst one motor is handling the load (say 1 kW). The demand presented by each motor when it is carrying its load is 1 kW, the sum of the demand loads is 6 kW but the maximum load presented by the system at any time is only 1.5 kW.
The diversity factor for this system is:
∑Demand Loads 6 kW
-------------------- ----- = 4
Maximum Demand 1.5 kW
The demand factor for this system is:
Maximum Demand 1.5 kW
-------------------- ----- = 0.125
∑Connected Load 12 kW


What are the advantages of EHV Lines?
Advantages:
1. Higher generation at statns requires higher voltage transmission
2. SIL is proportional to V2 (3 V2/Zc) and long transmission lines can be loaded to SIL for stability considerations. So SIL can be increased for higher loading of lines
3. EHV also decreases the right of way requirement
4. Line costs are reduced /MW/Km

What type of conductors are used in EHV lines?
Bundled Conductors to reduce Electic field intensity in the vicinity of lines: So Corona & Radio interference are reduced…Also inductance is decreased
Expanded ACSR can also be used but it has to be handled properly to avoid distortion
400kV: 2 or 4 subconductors /ph
765kV: 8 subconductors /ph

Set 1

Here is the first collection of the most basic concepts you would often encounter in the interviews. Hope it helps...

Plant load factor is often defined as the ratio of average load to capacity or the ratio of average load to peak load in a period of time. A higher load factor is advantageous because a power plant may be less efficient at low load factors, a high load factor means fixed costs are spread over more kWh of output (resulting in a lower price per unit of electricity), and a higher load factor means greater total output. If the power load factor is affected by non-availability of fuel, maintenance shut-down, unplanned break down, or reduced demand (as consumption pattern fluctuate throughout the day), the generation has to be adjusted, since grid energy storage is often prohibitively expensive.


The load, or burden, in a CT metering circuit is the (largely resistive) impedance presented to its secondary winding. Typical burden ratings for IEC CTs are 1.5 VA, 3 VA, 5 VA, 10 VA, 15 VA, 20 VA, 30 VA, 45 VA & 60 VA. As for ANSI/IEEE burden ratings are B-0.1, B-0.2, B-0.5, B-1.0, B-2.0 and B-4.0. This means a CT with a burden rating of B-0.2 can tolerate up to 0.2 Ω of impedance in the metering circuit before its output current is no longer a fixed ratio to the primary current. Items that contribute to the burden of a current measurement circuit are switch-blocks, meters and intermediate conductors. The most common source of excess burden in a current measurement circuit is the conductor between the meter and the CT. The rated secondary current is commonly standardized at 1 or 5 amperes.


Why go for sinusoids in power system?
Electrical power system is usually voltage-driven. Voltage drops across various elements thus modify the load voltage with respect to source voltage. These voltage drops are decided by a scaling of current by resistance value in the case of a resistor. It is decided by derivative of current in the case of an inductor and by integral of current in the case of a capacitor. A time-function retains its waveshape when multiplied by a constant. But, in general, it does not maintain its waveshape on differentiation and integration. Therefore, it follows that, in general, voltages and currents at various locations in an interconnected electrical network will have different waveshape even if all sources in the network have same waveshape.

Thus, sinusoidal signals have gained their pre-eminent position in Electrical Power engineering since they belong to a special class of time functions that preserve their waveshape on differentiation and integration. A linear network excited by sinusoidal sources of a particular frequency will have sinusoidal voltages and currents at the same frequency everywhere in the system in the long run.


Why not DC instead of Sinusoids?
If the customer had to be given 220V at his premises, the generators had to generate 220V and all the interconnection system had to work at that voltage level. As the load level increases, the current flow everywhere become excessive and generation/transmission become inefficient due to resistive losses everywhere in the system. It would have been very convenient if generation could be done at a voltage level economical from the point of view of electrical machine design and operation. Similarly, it would have been convenient if the transmission of power through transmission lines could be done at high voltage level so that the current level and consequently losses in lines would decrease. But this calls for generation at low voltage level, transmission at high voltage level and consumption at low voltage level. An efficient ‘voltage level conversion unit’ is needed for this. Such voltage level conversion equipment for dc at high power levels was simply not available at the initial stages of evolution of power systems in late 19th century and early 20th century.


IMPORTANT: Need for Governors in Turbines

When load on alternator increases, the load angle also increases. Hence the electromagnetic torque acting in opposition to prime mover would increase (being proportional to sin(δ) ) , which causes subsequent decrease in speed! However, speed deviations would cause frequency variations and so have to be controlled fixed. This brings in governors (Turbine speed controls)


When we increase load , current drawn from source also increses why ?
The electrical loads are normally connected in parallel, whenever the load(impedance) is increased means the additional load (or you can say impedance) is connected in parallel to the existing load. The parallel combination always reduces the equivelent impedance. Hence the current is increases.


What is the Difference between MCB & MCCB?
MCB is MINIATURE CIRCUIT BREAKER
MCCB is MOULDED CASE CIRCUIT BREAKER
MCB-A circuit breaker is an automatically-operated electrical switch designed to protect an electrical circuit from damage caused by overload or short circuit. Unlike a fuse, which operates once and then has to be replaced, a circuit breaker can be reset (either manually or automatically) to resume normal operation. Circuit breakers are made in varying sizes, from small devices that protect an individual household appliance up to large switchgear designed to protect high voltage circuits feeding an entire city. Rated current not more than 100 A. Trip characteristics normally not adjustable. Thermal or thermal-magnetic operation. Breakers illustrated above are in this category.
MCCB (Moulded Case Circuit Breaker)—rated current up to 1000 A. Thermal or thermal-magnetic operation. Trip current may be adjustable in larger ratings.

Let me help you...

Hello friends, I have created this blog for those who are electrical engineering students to the core and are preparing for interviews. I have myself given a few electrical tech interviews and have been exposed to the most generic type of questions asked. I would be sharing them with you in sets to make sure you get the best and the most relevant of the web here.
All the best!