Parallel Operation of Transformer

Why Parallel Operation of Transformers is required?

It is economical to installe numbers of smaller rated transformers in parallel than installing a bigger rated electrical power transformer. This has mainly the following advantages,

1. To maximize electrical power system efficiency: Generally electrical power transformer gives the maximum efficiency at full load. If we run numbers of transformers in parallel, we can switch on only those transformers which will give the total demand by running nearer to its full load rating for that time. When load increases, we can switch none by one other transformer connected in parallel to fulfill the total demand. In this way we can run the system with maximum efficiency.
2. To maximize electrical power system availability: If numbers of transformers run in parallel, we can shutdown any one of them for maintenance purpose. Other parallel transformers in system will serve the load without total interruption of power.
3. To maximize power system reliability: if any one of the transformers run in parallel, is tripped due to fault of other parallel transformers is the system will share the load, hence power supply may not be interrupted if the shared loads do not make other transformers over loaded.
4. To maximize electrical power system flexibility: There is always a chance of increasing or decreasing future demand of power system. If it is predicted that power demand will be increased in future, there must be a provision of connecting transformers in system in parallel to fulfill the extra demand because, it is not economical from business point of view to install a bigger rated single transformer by forecasting the increased future demand as it is unnecessary investment of money. Again if future demand is decreased, transformers running in parallel can be removed from system to balance the capital investment and its return.

   Conditions for Parallel Operation of Transformers

   When two or more transformers run in parallel, they must satisfy the following conditions for satisfactory performance. These are the conditions for parallel operation of transformers.
5. Same voltage ratio of transformer .
6. Same percentage impedance.
7. Same polarity.
8. Same phase sequence.

Same Voltage Ratio

If two transformers of different voltage ratio are connected in parallel with same primary supply voltage, there will be a difference in secondary voltages. Now say the secondary of these transformers are connected to same bus, there will be a circulating current between secondaries and therefore between primaries also. As the internal impedance of transformer is small, a small voltage difference  may cause sufficiently high circulating current causing unnecessary extra I2R loss.

Same Percentage Impedance

The current shared by two transformers running in parallel should be proportional to their MVA ratings. Again, current carried by these transformers are inversely proportional to their internal impedance. From these two statements it can be said that,impedance of transformer running in parallel are inversely proportional to their MVA ratings. In other words, percentage impedance or per unit values of impedance should be identical for all the transformers that run in parallel. 

Same Polarity

Polarity of all transformers that run in parallel, should be the same otherwise huge circulating current that flows in the transformer but no load will be fed from these transformers. Polarity of transformer means the instantaneous direction of induced emf in secondary. If the instantaneous directions of induced secondary emf in two transformers are opposite to each other when same input power is fed to both of the transformers, the transformers are said to be in opposite polarity. If the instantaneous directions of induced secondary emf in two transformers are same when same input power is fed to the both of the transformers, the transformers are said to be in same polarity.

Same Phase Sequence

The phase sequence or the order in which the phases reach their maximum positive voltage, must be identical for two parallel transformers. Otherwise, during the cycle, each pair of phases will be short circuited. 
The above said conditions must be strictly followed for parallel operation of transformers but totally identical percentage impedance of two different transformers is difficult to achieve practically, that is why the transformers run in parallel may not have exactly same percentage impedance but the values would be as nearer as possible.

INDUSTRIAL APPPLICATION OF ELECTRIC MOTOR

Textile Mills

—They are classified into following types

—Cotton mills

—Woolen mills

—Dying and finishing plants

—Knitting mills

—Rayon and synthetic fiber mills

Motors Used In Textile Industries

    Normally totally closed fan cooled motors are used in textile mills. Drip proof motors are used in wet locations.

— Loom Motors

    Loom motors are used for picking up of cotton in short time and are designed to have a high torque. These motors should have a starting torque of 230% of rated torque and pull out torque of 270% of the rated value. Recommended number of poles for such motors are 6 to 8. for light fabrics such as cotton, rayon or silk kw requirement of loom may vary from 0.37 to 1.50 and for heavy fabrics from 2.2 to 3.7.

—Card motors

      They are similar to loom motors but possess still higher torque i.e. starting torque of order of 350% and 275%  0f running torque and a pull out torque of 375% and 300% of running torque for 8-pole and 6-pole motors respectively. The card drum is a heavy mass and possess a high moment of inertia. The motor therefore undergoes a prolonged starting time and a starting currant of 2.5 times the rated current for nearly 2 minutes.

—Spinning Motors

      These motors are required to make threads i.e. Final drawing, twisting and winding of cotton. They must possess smooth acceleration to avoid breakage of threads. They are recommended to have a starting torque of 150-200% of running torque and pull out torque of 200-275% of running torque. Usually smooth acceleration is required because faster acceleration may cause more breakage and small acceleration may cause knots in yarn as a result of insufficient tension.

—In paper mills, pulp making and paper making are the two main processes that take place. The drive required for larger grinders involved in making pulp is provided by large motors 3000-4000 kw in size.

— Paper formation involves formation of sheets, removal of water from them and pressing followed by reeling. Both A.C. and D.C. drives are used for paper making machines. In D.C. drives speed of paper machine is controlled by changing the armature the armature voltage of separately excited D.C. motor.

Sugar Mills

— Sugar mills are driven by variable speed induction motors. Eddy current drives are generally used. When the juice is extracted remaining matter is known as bygasse. Variable speed motors are used to  drive conveyors. The juice is then evaporated through boiling stages to get sugar crystals.

 

— Bygasse is used to produce steam in the boilers. The generation of electricity using bygasse bring down the cost of production of sugar.

Speed Control In Sugar Mills

—For separating crystals from syrup centrifugal pumps are used. These centrifuges are run by variable speed induction motor having two distinctly set of stator windings.

— The motor is connected for regenerative braking to return energy to supply. This is done by connecting the motors from high pole to low pole operation.

— The motors used have an insulation which can withstand the humidity of surroundings.

Cement Mills

        Port land cement is a mixture of oxides of calcium 60-66%, silicon 20-24%, aluminum 4-6%, and iron 2-3% in various composition. Raw materials used are lime stone, marble, chalk e.t.c. During manufacture, grinding and proptioning the raw materials which are sent to rotary kilns and brought out white heat. The resulting clinker and retarder like gypsum are pulversied together to form finished port land cement.

    Two methods are used for preparation of cement i.e. Dry and Wet process.

Motors Used In Cement Mills

— Squirrel cage motors are used because of their low cost and ruggedness and where constant speed is required. Slip ring motors are used where high starting torque is required and low starting current is in use for crushers and hammer mills.

— Adjustable speed D.C. motors are used for greater speed range and close speed control in kilns. All these motors are dust proof and place where moisture from precipitation Is present totally enclosed motors are used.

Motors For Mining Industry

—Pumps

    Centrifugal pumps are used for removing water from any depth ground level. Pumps used are driven by squirrel cage motor which possess low starting torque.

— Ventilating Fans

    They are used for proper working in mines and are made of synchronous or induction motor.

— Drilling Motors

     These are generally induction motors with flame proof enclosures. Sometimes high frequency (150 to 200 Hz) motors are used to drive drills. These drills have reduced size and high power.

—Coal Cutting

   Coal cutting is done by coal cutters fitted with squirrel cage rotor provided with flame proof enclosures. In case of D.C. supplies compound motors with interpoles are employed. Coal cutting in a big block of coal facilitate introduction of explosives to break up big blocks of coal.

— Conveyors

   Conveyors are driven by squirrel cage motors which are totally closed with flame proof materials.

LOAD CHARACTERISTICS & ESTIMATION OF LOAD

BASIC DEFINITIONS

LOAD

Any device or circuit that consumes electric power Or

Any device or apparatus that draws electric current from supply system is said to impose a load on the system. The term “LOAD” (in electrical) is used to:

§  To indicate any device that consumes electrical energy.

§  To indicate power required from given supply circuit.

§  To indicate the current or power passing through transmission    line.

The load may be resistive, inductive, capacitive, or some combination of them.

CONNECTED LOAD

It is ‘ the sum of the continuous ratings of all loads connected to the system or any part thereof.

DEMAND

The demand of an installation of a system is “the load that is drawn from the source of supply at the receiving terminals averaged over a certain period of time”. e.g., Daily demand, Weekly, fortnightly, monthly, yearly, etc. 

The load may be given in

Ø  Kilowatts (KW)

Ø  Kilovars (KVAr)

Ø  Kilovoltamperes (KVA)

Ø  Kiloamperes (KA)

Ø  Amperes (A)

MAXIMUM DEMAND OR PEAK LOAD

The greatest load drawn during the specific period of time. For example, the specified demand might be maximum of all demands such as daily, weekly, monthly, or annual.

Knowledge of maximum demand helps in determining the installed capacity of a generating stations. The generating station must be capable of meeting the maximum demand. Hence the cost of plant and equipment increases with the increase in maximum demand.

DEMAND INTERVAL

It is the period over which the load is averaged. There are two types of demands:

1. Instantaneous Demand: Demand at any particular time.

2. Sustained Demand: Demand over a certain period of time.

DEMAND FACTOR (DF)                               

 In practice consumers do not use all the devices at full load simultaneously. The maximum demand of each consumer is, therefore, less than his connected load. The demand factor depends upon the nature of load. Lighting loads have higher demand factors than power loads. The demand factor is usually less than 1.0

AVERAGE LOAD OR AVERAGE DEMAND                               

 LOAD FACTOR                                  

It is the ratio of the average load over a given period of time to the maximum demand (peak load) occurring in that period.

Load factor plays an important role on the cost of generation per unit (KWh). The higher the load factor, the lesser will be the cost of generation per unit for the same maximum demand.

DIVERSITY FACTOR                                  

The maximum demands of the individual consumers of a group are not likely to occur simultaneously. Thus, there is a diversity in the occurrence of the loads. Due to this diverse nature of the load, power is never required to supply all connected loads to their full capacity at the same time.

Diversity factor can be defined as;

 Diversity factor can be defined for loads, substations, feeders, and generating stations.

 The value of diversity factor is generally greater than 1.0. with a high value representing a good diversity and 1.0 represents a poor diversity.

UTILIZATION FACTOR Fu                         

It is the ratio of maximum demand of a system to the rated capacity of the system   

The utilization factor can also be found for a part of the system

PLANT FACTOR OR CAPACITY FACTOR

LOSS  FACTOR

Problem #01

A consumer has the following connected load:

10 Lamps each of 60 W, 2 Heaters each of 1000 W, Maximum demand 1500 W. On the average he uses 8 lamps for 5 hours per day. Each heater 3 hours per day. Find: (a) average load. (b) monthly energy consumption. (c) load factor.

SOLUTION

Problem #02

There are 4 consumers of diversity having different load requirements at different timings.

Consumer #01

Average load = 1 KW

Maximum demand = 5 KW at 8 p.m.

Consumer #02

Maximum demand = 2 KW at 9 p.m.

Demand of 1.6 KW at 8 p.m.

Daily load factor = 0.15

Consumer #03

Maximum demand = 2 KW at 12 noon.

load of 1 KW at 8 p.m.

Average load of 500 W.

Consumer #04

Maximum demand of 10 KW at 5 p.m.

load of 5 KW at 8 p.m.

Daily load factor = 0.25.

The maximum demand of the system occurs at 8 pm. Determine:

1.The diversity factor

2.Average load and load factor of each consumer

3.Average load and load factor of the combined load.

solution

POWER FACTOR FUNDAMENTALS

What we will learn

qMost Industrial loads require both Real power and
      Reactive power to produce useful work
q   You pay for BOTH types of power
q   Capacitors can supply the REACTIVE power thus
      the utility doesn’t need to
q   Capacitors save you money!

WHY POWER FACTOR CORRECTION IS REQUIRED?

»  Reduces Power Bills

»  Reduces I2R losses in conductors

»  Reduces loading on transformers

»  Improves voltage drop

What is PF ?

Most plant loads are Inductive and require a magnetic field to operate:

§– Motors

§– Transformers

§– Florescent lighting

» The magnetic field is necessary, but produces no useful work

» The utility must supply the power to produce the magnetic field

   and the power to produce the useful work: You pay for all of it!

» These two types of currents are the ACTIVE and REACTIVE

    components

The Power Triangle:

q  Similarly, motors require REACTIVE power to set up the magnetic field while the ACTIVE power produces the useful work (shaft horsepower). Total Power is the vector sum of the two & represents what you pay for:

§Power Factor is the ratio of Active Power to Total Power:

§Power Factor is a measure of efficiency (Output/Input)

Why do we Install Capacitors?

Capacitors supply, for free, the reactive energy required by inductive loads.

» You only have to pay for the capacitor !

» Since the utility doesn’t supply it (kVAR), you don’t pay for it!

Released system capacity:

» The effect of PF on current drawn is shown below:

Decreasing size of conductors required to carry the same 100kW load at P.F.

ranging from 70% to 100%

Reduced Power Losses:

» As current flows through conductors, the conductors heat.    This heating is power loss

» Power loss is proportional to current squared (PLoss=I2R)

» Current is proportional to P.F

» Conductor loss can account for as much as 2-5% of total load

Capacitors can reduce losses by 1-2% of the total load

Voltage Improvement:

» When capacitors are added, voltage will increase.

» Typically only a few percent – Not a significant economic or system benefit.

q  Severe over-correction (P.F.>1) will cause a voltage rise that can damage insulation & equipment; or result in utility surcharges!

– Usually a result of large fixed capacitors at mains

Summary of Benefits:

Reduced Power Costs:

» Since Capacitors supply reactive power, you don’t pay the

   utility for it

» You can calculate the savings

 Off-load transformers

» Defer buying a larger transformer when adding loads

 Reduce voltage drop at loads

» Only if capacitors are applied at loads

» (minimal benefit at best)

What we learned..

 Most Industrial loads (i.e. motors) are Inductive and draw REACTIVE power

The Utility supplies this energy therefore you pay for it

 Power Factor Capacitors supply REACTIVE energy thus the utility doesn’t need to

 Power Factor Capacitors save money

There are other benefits to correcting power factor,

  Reduced heating in cables

  Reduced heating in transformer (s)

  Frees up system capacity

Distribution Transformers

ANSI/IEEE defines a transformer as a static electrical device, involving no continuously moving parts, used in electric power systems to transfer power between circuits through the use of electromagnetic induction.

The term power transformer is used to refer to those transformers used between the generator and the distribution circuits, and these are usually rated at 500 kVA and above.

Power systems typically consist of a large number of generation locations, distribution points, and interconnections within the system or with nearby systems, such as a neighboring utility. The complexity of the system leads to a variety of transmission and distribution voltages. Power transformers must be used at each of these points where there is a transition between voltage levels.

Power transformers are selected based on the application, step – up transformer are used at generator site and step – down transformers are used at distribution systems.

 Any transformer that takes voltage from a primary distribution circuit and “steps down” or reduces it to a secondary distribution circuit or a consumer’s service circuit is a distribution transformer.

Although many industry standards tend to limit this definition by KVA rating (e.g., 5 to 500 KVA), distribution transformers can have lower ratings and can have ratings of 5000 KVA or even higher, so the use of KVA ratings to define transformer types is being discouraged (IEEE, 2002b).

What is DISTRIBUTION SUBSTATIONS

DISTRIBUTION SUBSTATIONS

SUBSTATION & FEEDER 

Substations transform the electrical energy to a different voltage and transfer electrical energy from one line to another. Usually planners try to locate a substation near the center of load.
A feeder is an electrical distribution circuit fed from a single source point: through a breaker or a fuse at the substation.
It operates at the primary distribution voltage and delivers power to an assigned area. Together, the feeders emanating from a substation serve all the loads and cover all the areas assigned to that particular substation.
This area should be an approximate circle, polygon or hexagon and the substation should be located approximately at the central point in cases where the load is more or less uniform. Some times there are constraints of geography – river, canal, forest etc. or just poor planning in the past which can cause an exception to this rule. A feeder consists of a single route, leaving the substation which branches with spurs. Feeders are planned by starting from the substation with the main trunk portion of the largest economical conductor and generally follow streets, roads, highways and property boundaries.

DISTRIBUTION SUBSTATION

The distribution substation is the convenient point for the control and protection of the distribution network. A typical substation may have the following equipment: 1. Power Transformer(s)
 2. Circuit Breakers
3. Disconnecting Switches And Isolators
 4. Station Buses
 5. Current Limiting Reactors
 6. Shunt Reactors
 7. Current Transformers
 8. Potential Transformers
 9. Capacitor Voltage Transformers.
10.Coupling Capacitors
11.Series Capacitors
 12.Shunt Capacitors
13.Grounding System
14.Lightning Arresters And/Or Gaps
15.Line Traps
16.Protective Relays
17.Station Batteries
18. And Other Apparatus

SUBSTATION BUS SCHEMES 

The electrical and physical arrangements of the switching and busing at the sub-transmission voltage level are determined by the selected substation scheme (or diagram). On the other hand, the selection of a particular substation scheme is based upon safety, reliability, economy, simplicity, and other considerations. The most commonly used substation bus schemes are:
1. Single bus scheme
 2. Double bus – double breaker (or double main) scheme
3. Main – and – transfer bus scheme
 4. Double bus – single breaker scheme 5. Ring bus scheme
 6. Breaker – and – a half scheme

Single Bus Scheme

Advantage: Lowest cost.
Disadvantages:
Failure of bus or any circuit breaker results in shutdown of entire substation.
 Difficult to do any maintenance.
Bus cannot be extended without completely de energizing substation.
Can be used only where loads can be interrupted or have other supply arrangements.

Double Bus – Double Breaker

Advantages:
Each circuit has two dedicated                              
breakers.
Has flexibility in permitting
feeder circuits to be connected
to either bus.
Any breaker can be taken out
of service for maintenance.
High reliability.
Disadvantages:
Most expensive.
Would lose half the circuits for
breaker failure if circuits are
not connected to both buses.

Main – And – Transfer

Advantages:
Low initial cost and ultimate                                          
cost.
Any breaker can be taken out for
service or maintenance.
Potential devices may be used on
the main bus for relaying.
Disadvantages:
Requires one extra breaker for
the bus tie.
Switching is somewhat
complicated when maintaining a
breaker.
Failure of bus or any circuit
breaker results in shutdown of
entire substation.

Double Bus – Single Breaker                 

Advantages:
Permits some flexibility with two
operating buses.
Either main bus may be isolated for
maintenance.
Circuit can be transferred readily from
one bus to the other by use of bus –tie
breaker and bus selector disconnect
switches.
Disadvantages:
One extra breaker is required for the bus tie.
Four switches are required per circuit.
Bus protection scheme may cause loss of substation when it operates if all
circuits are connected to that bus.
High exposure to bus faults.
Line breaker failure takes all circuits connected to that bus out of service.
Bus – tie breaker failure takes entire substation out of service.

Ring Bus Scheme                                                                       

Advantages:
Low initial and ultimate cost.                                                    

Flexible operation for breaker
maintenance without interrupting
load.
Requires only one breaker per circuit.
Does not use main bus.
Each circuit is fed by two breakers.
All switching is done with breakers.
Disadvantages:
If a fault occurs during a breaker maintenance period, the ring can be separated into
two sections.
Automatic reclosing and protective relaying circuitry rather complex.
If a single set of relays is used, the circuit must be taken out of service to maintain
the relays. (common to all schemes.)
Requires potential device on all circuits since there is no definite potential reference
point. These devices may be required in all cases for synchronizing, live line, or
voltage indication.
Breaker failure during a fault on one of the circuits causes loss of one additional
circuit owing to operation of breaker – failure relaying.

Breaker – And – a Half Scheme

Advantages:                                                                                   

Most flexible operation.
High reliability.
Breaker failure of bus side breaker removes
only one circuit from service.
All switching is done with breakers.
Simple operation; no disconnect switching
required for normal operation.
Either main bus can be taken out of service
at any time for maintenance.
Bus failure does not remove any feeder
circuits from service.
Disadvantages:
1½ breaker per circuit.
Relaying and automatic reclosing are
somewhat involved since the middle breaker
must be responsive to either of its
associated circuits.