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Effects of Harmonics on Power Cables

The power cables are used to carry power from supply source to the load point. The cable has DC resistance which depends on the size of the cable. When current flows through the cable the heat loss I2R takes place in the cable. The heat loss in the cable is proportional to the square of the current flowing through the cable and proportional to the cable resistance. The harmonic current cause additional heating in the power cable due to increased I2R losses.

Image result for XLPE cable

The DC resistance of the cable does not change if the cable carries the DC power, however in the case of AC power transmission the resistance gets increased with increase in the frequency and also increase in the current. The AC resistance of the cable is higher than the DC resistance of the cable. Thus the power loss caused by the heating of the cable gets increased if the cable carries high frequency AC current.

Increased I2Losses due to Harmonics

The RMS value of the current increases when the magnitude of the harmonics current increase. In the harmonic rich supply system, the cable losses increase due to two reasons.

  • Increase of the cable resistance with harmonics
  • Increase of RMS current due to Harmonics

How the RMS current increase with Harmonics?

The harmonic current waveform is not sinusoidal. According to Fourier series, the non sinusoidal waveform can be resolved into sinusoidal and the integral multiples of the sinusoidal waveform. The integral multiples of fundamental frequency is known as harmonics. For example, take a non sinusoidal current waveform as given below.

This sort of current waveform contains harmonics. If we synthesize this waveform let the magnitude of the harmonics are as follows.



The harmonic spectrum of the current waveform is as given below.


The RMS value of the non-sinusoidal current waveform cab be calculated by taking the harmonic current into account.


Where,

I= fundamental current
I= 5th harmonic current
I= 7th harmonic current
I1111th harmonic current

The RMS value of the current increase by 5 % of the fundamental RMS current. If the fundamental current is 100 amperes and the load draws harmonic current from the supply source then the total current flowing will be 105 amperes.The increase of  current due to harmonic current cause 10% additional power loss in the cable.The power cable needs to be de-rated according to the harmonic current flowing through the cable.

While selecting cable for a particular load the harmonic current must be taken into consideration.

The other factor that also contribute to the increase of power loss is the AC resistance of the cable. The harmonic current has higher frequency -250 Hz, 350Hz,550 Hz etc.At higher frequency the current tends to flow on the outer surface of the conductor.The phenomenon is called the SKIN EFFECT. The effective resistance gets increased when the power cable carry the high frequency current. The AC resistance of the power cable is always more than the DC resistance because of the skin effect and the proximity effect. The skin effect is proportional to the frequency and the proximity effect is proportional to the square of the frequency. Thus the effective resistance of the conductor carrying harmonics current gets increased.

Over Voltage due to Harmonics

If the capacitor banks installed for power factor correction resonate with the inductance of the transformer, the resonance phenomenon can boost the system voltage. In this condition the power cables experience voltage stress and the cable insulation may get failed.

Cable Derating

The effect of heating due to harmonic current must be taken into consideration while sizing the cable. The usual derating curves as per IEEE519 have been plotted for a number of cable sizes for 6 pulse harmonic distribution.





Tuesday, October 15, 2019

Shell Type Transformer

The transformer can be categorized into core type transformer and shell  type transformer depending on the construction of the magnetic core and arrangement of the winding of the transformer. In the shell type transformer the core encircles the winding. 

Core of Shell Type Transformer

The core of the shell type transformer is made from the E-I and E-E type laminated sheet.


The core of the single phase shell type transformer has three limbs.This arrangement increase the mechanical strength of the core and also the winding remain protected from the mechanical shock.


Winding of Shell Type Transformer

The HV and LV winding is wound around the central limb and the central limb carries the entire flux (Ф ) and the side limbs carries the half (Ф/2 )of the total flux(Ф ).That is why the cross section area of the central limb is two times of the cross section area of the side limbs.The arrangement of the winding in a shell type transformer around the central limb is as given below.




The magnetic flux flows through two closed magnetic path and therefore the losses gets reduced and the efficiency of the transformer improves. The shell type transformer gives more output as compared to similar core type transformer. The core also provides better mechanical support to winding and under short circuit condition the electromagnetic force developed has less effect on the winding.

The HV and the LV winding is wound around the central limb in the sandwich arrangement. The HV winding is sandwiched between LV winding.




The quantity of the conductor required for the shell type transformer is less because both the winding are wound on the central limb. However, the insulation requirement is more as compared to the core type transformer because HV and LV winding is wound alternatively on the same limb. The design of the shell type transformer is complex compared to the core type transformer.

In case of any defect in the inner winding of shell type transformer, all the outer winding need to be removed for repairing or rewinding.


Cooling of Shell Type Transformer

In case of core type transformer natural air cooling may be sufficient. However, the forced air or forced oil cooling is required in a shell type transformer because the winding is surrounded by yoke and limbs.

Applications of Shell Type Transformer


The shell type transformer is used for the low voltage applications and are generally used in low voltage power circuits and electronic circuits.The cost of the shell type transformer for low voltage applications is less because the square or rectangular cross section area core can be used.

Due to complexity of the construction, the shell type transformer requires special fabrication facilities which increase the cost of production. 


Monday, October 14, 2019

Transformer Rating

The maximum current that transformer can deliver to loads as known as the current rating of the transformer. The voltage, maximum current delivery of the transformer and the product of the voltage and maximum current known as VA rating are engraved in the nameplate of the transformer.

Related image


The rating of the transformer is adversely affected with abnormal temperature rise caused by the losses. The no load losses increases with an increase in the frequency and or voltage and the copper loss increase if the load power factor is more lagging which demands more current from the transformer and the copper loss gets increased. The losses can be kept within the limit by maintaining the rated voltage ,frequency and load current. 

The losses cause temperature rise of the transformer and if transformer is operated above its rated safe temperature limit, the winding insulation is apt to fail. The temperature can be held within the safe limit by proper cooling.

The losses depends on the V and I. The rating of the transformer depends on the losses. The losses  are independent of the power factor that is why the rating of the transformer does not depend on the load and it depends on the V x I which is known as VA rating of the transformer. The designer designs the transformer taking into account the maximum VA rating of the transformer. The transformer is specified for apparent power rating- VA rating.

The transformer primary VA is equal to the secondary VA plus the losses in the transformer. If the losses increase, the secondary VA delivery will get affected because of the temperature rise of the transformer. Generally the losses of the transformer is negligible the primary VA is equal to the secondary VA.

The VA rating marked on the nameplate is applicable for both the winding and VA rating of both the winding of transformer is the same. For a 440/220 Volts, 100 KVA transformer, the primary winding VA is 100 kVA and the secondary winding kVA is also 100 kVA.



Friday, October 11, 2019

Voltage Regulation of Transformer


What is Voltage Regulation?

The secondary voltage of the  transformer should not vary with load when the input supply is constant. The transformer can be treated as a voltage source and the variation in the secondary terminal voltage from no load to full load  depends on the  total impedance of the transformer.


When current flows through the transformer the voltage drop takes place due to reactance and resistance of the transformer.The variation in the secondary output voltage of the transformer  from its no load to full load  is called the voltage regulation of the transformer.

Voltage Regulation Formula

The voltage regulation of the transformer shows how well a transformer maintains constant secondary output voltage from no load to full load condition when the primary voltage is constant.



The secondary output voltage decrease as the current flowing through the transformer is increased.This happens  because of the voltage drop in the transformer winding. The secondary output voltage at full load is always less the the output voltage at no load.The lesser the voltage difference of  no load to full load condition, the better the regulation of the transformer. 

Example:

The transformer secondary output voltage is 220 volts at no load. The secondary voltage decrease from 220 to 210 volts when the transformer is loaded up to its full load current capacity.

The drop in the secondary output voltage from no load to full load = 10 Volts













If the secondary voltage decrease from 220 to 215 volts when the transformer is loaded up to its full load current capacity. The regulation of the transformer in this case is 2.32 %. The % regulation of the transformer should be as minimum as possible to have almost constant secondary output voltage. Also the copper loss gets decreased when the regulation of the transformer is improved.

What Factors affects the Voltage Regulation?

The transformer have primary and the secondary winding. The winding has resistance and rectance. The primary resistance and reactance can be referred at the secondary side or vice versa. The  voltage regulation of the transformer depends on the reactance and the resistance of the transformer. The equivalent secondary circuit of the transformer is as given below.



When the transformer is at no load, the secondary current I2 =0 and the transformer draws only no load current. The voltage drop I2Z2  across secondary impedance  takes place with an increase in the secondary current. The voltage drop in the secondary winding is maximum when transformer operates at its full kVA capacity delivering the rated secondary current to the load.

At no load the secondary terminal voltage = E2

At rated secondary current the voltage drop = I2Z2

At rated secondary current the terminal voltage = V2

According KCL,

     E2   =  I2Z2 + V2

    I2Z2  = E2 - V2


The regulation of the transformer  depends on the power factor of the load. Now, we will discuss the regulation of the transformer at lagging, leading and unity power factor. 

Voltage Regulation of Transformer for Lagging Power Factor

Let the angle between the secondary  terminal voltage V2 and secondary current I2 is θ2. The phasor diagram of the no load voltage(E2 ) , full load voltage (V2) and current (I2)is as shown below. 


From above diagram,

                     
              OC  = OA + AB + BC

In triangle ABE,

              AB = AE Cosθ2


In triangle DEF,



              DC = DE Sinθ2


The angle between OC and OD is very small and OC is equal to OD.

       OC = OD

         OC = OA + AB + BC

        E2  = V2  + AE Cosθ2 + DE Sinθ2

        E2  = V2  + I2R2 Cosθ2 + I2X2 Sinθ2

      E2  - V2  =  I2R2 Cosθ2 + I2X2 Sinθ2








Voltage Regulation of Transformer for Leading Power Factor

The phsor diagram of the transformer operating at leading power factor is as shown below.


From above diagram,

                     
              OC  = OA + AB - BC

In triangle ABE,

              AB = AE Cosθ2


In triangle DEF,



             BC = DE Sinθ2


The angle between OC and OD is very small and OC is equal to OD.

       OC = OD

         OC = OA + AB - BC

      Here, OA = V2

        E2  = V2  + AE Cosθ2 - DE Sinθ2

        E2  = V2  + I2R2 Cosθ2 - I2X2 Sinθ2


      E2  - V2  =  I2R2 Cosθ2 - I2X2 Sinθ2






Zero Voltage Regulation of Transformer 

The secondary voltage at no load can't be equal to the secondary voltage at load. The zero voltage regulation of the transformer is the ideal case and practically it is not possible.

Saturday, October 5, 2019

Open Delta Connection of Transformer

When two single phase transformer is connected in open delta connection the arrangement can provide three phase supply to the load. The open delta connection of transformer is also known as V-V connection. 

The closed delta connection of transformer provides the three phase supply when connected to the balanced load with individual sharing of 1/3 of the load. If one of the delta winding gets opened the transformer can still be operated with two phase delta winding. The connection diagram of the open delta connection of the transformer is as shown below. 


The efficiency of the open delta connection is lower than the closed delta connection. However, the transformer can be operated with reduced efficiency in the open delta or V- V connection. Let us discuss how the transformer delivers three phase power in open delta connection when one of the phase is missing. Two of the phases Vab and Vbc is available and Vac is missing. The vector sum of all the three phase voltage is zero. Therefore,  the voltage across open delta terminal is ;



Vca = - ( Vab + Vbc)

The phasor diagram of open delta connection is as shown in below figure.





Vab and Vbc are equal in magnitude, therefore the resultant will be equal to V and 120 degree apart from the reference voltage Vab and Vbc. From this it is clear that the open delta connection can supply the three phase supply to the load. The VA rating of the transformer in open delta connection gets reduced as compared to the  VA rating of the closed delta connection of transformer.

Case 1

When  all the three winding of transformer in the circuit

Let,

Vph - Phase Voltage of each secondary
Iph  - Phase current of each secondary

In closed delta connection the line voltage is equal to the phase voltage.

VL = Vph

VA rating of the closed delta transformer

=  √3 VL IL

=  1.732 Vph  x √3 Iph
= 1.732 Vph x 1.732 Iph
= 3 Vph Iph

Case 2

Open Delta Connection

In a closed delta connection the line current is 1.732 times of the phase current, but in open delta connection the line current is equal to the phase current. The phasor diagram of the open delta transformer is as given below.




VA rating of the open delta transformer ;

=  √3 VL IL

=  1.732 Vph  x  Iph     IL  =Iph ]
= 1.732 Vph Iph

VA rating of Open Delta / close Delta















From above it is clear that the VA delivering capacity of the open delta transformer is 57.7 % of the VA delivering capacity when all the three transformers are connected in delta.

The total VA available when the two transformer is connected in delta connection is 2 Vph Iph. However, the actual VA delivery of  the open delta connection is 1.732 Vph Iph. 

Transformer Utility Factor(TUF)

The ratio of actual VA available to the total VA available in open delta connection is known as the transformer utility factor(TUF).










Examples: 


Suppose the load VA demands is 60 VA. What will be the rating of each transformer if connected ;

  • Closed Delta Connection
  • Open Delta Connection

If all three transformer is connected in closed delta connection the VA delivered by each transformer is 60/3 = 20 VA, therefore the VA rating of each transformer  is 20 VA.

If one of the transformer is removed from the closed delta and two transformer are connected in the open delta connection then the total VA delivery will be  60 x 0.557= 33.42VA and each  transformer will deliver 33.42/2=16.71 VA. The circuit load is 60 VA, hence the 20 VA transformers connected in open delta connection can't deliver 60 VA to the load.The new VA rating of the transformers for open delta connection to deliver 60 VA  will be equal to 60/0.557=107.71 VA.Thus the each transformer VA rating in open delta connection to deliver 60 VA to load will be 107.71/3=35.90 VA.

Core Type Transformer

The transformer has a magnetic core and the primary and the secondary winding around the core. Types of the transformer can be categorized according  to the shape of the magnetic core. The transformer can be categorized into core type and the shell type.In a core type transformer the winding encircles the core whereas in a shell type transformer the core encircles the winding. The winding encircles the core of a transformer. The Single phase transformer has two limbs and three phase transformer has three limbs.



The core of a single phase transformer has a single window. The core of the single phase transformer has two limbs.



Both HV and LV winding is wrapped around  both the  limbs.The LV winding is placed near to the core and then the HV winding is placed after LV on the same limb. This way the  transformer size becomes compact due to less insulation requirement. The transformer cost reduces with this arrangement of placing the LV and HV winding on the same limb.

The main disadvantage of core type transformer is that all the flux produced in the primary does not link to the secondary and some parts of the flux does not link. Thus, the leakage flux in the core type transformer is more as compared to the flux utilization in shell type transformer. However, the core type arrangement is best suited for the large rating transformer because of easier access of the winding during maintenance.If there is fault in the inner winding, the inner winding can be removed after removing the outer winding.

Transformer Core Types and Applications

There are different types of lamination used in the core type transformer.



L-L and U-I stamping lamination when connected together form the requires core shape.

The transformer core shapes is chosen according to the rating of the transformer. In a small rating transformer, the winding is square or rectangular in the shape therefore, the square or rectangular cross section core is used.The small rating transformer has lower current carrying capacity conductors and it is easy to wrap the conductor in a square or rectangular shape. The square or rectangular core is economic for the small rating transformer.

For large rating transformer, the thick winding conductor is used to carry the more current. It is difficult to bend the thick conductor in a square or in a rectangular shape. The round cylindrical shaped winding is the best choice for the large rating transformer in view of optimize use of copper conductor.However, the significant amount of space between the winding and core is unused when the round cylindrical shaped winding on a square cross sectional core limb is used. To reduce this unused space, the stepped cross sectional core is used. The lamination of different shapes are staged to form nearly circular cross sectional core.The cross section of the core may be one stepped, two- stepped or multi stepped.

Tuesday, September 24, 2019

Air Gap Power in Induction Motor

The motor converts electrical energy into the mechanical rotational energy. The input voltage is fed to the stator of an induction motor. The stator coil opposes the applied voltage and this opposition limits the starting current of the motor. The flux is generated in the stator and the flux travels through the air gap and gets linked to the rotor conductors. The air gap power is equal to the rotor input power. 

The input power drawn by the induction motor is expressed by the following formula.

P=√3* VL*IL* cosΦ ----------(1)
Where,
VL-Line voltage, IL- Line current, cosΦ- Power Factor
The electric power drawn is converted into the magnetic power. The air gap between the stator and the rotor is unwanted but necessary for motor operation. Without having an air gap, the rotation of the rotor is not possible. However, the losses in the motor can be minimized by keeping the air gap length as minimum as practically possible. The following losses occurs in the motor at no load.
Core Losses or No Load Losses
The loss comprises the eddy current and hysteresis loss. The loss is more or less constant so it is called the fixed losses.
Stator I2R loss :
The loss occurs due to heat loss in the resistance of the stator because of the current flowing through the stator coils.
Thus the total losses in the stator at no load is given by;
PstL=Ps(h+e)+Ps --------------(2)
Where,
PstL - Total losses in the stator at no load
Ps(h+e)- Core losses-Eddy current and hysteresis loss
Ps- Copper loss in the stator

The air gap power(Pg) can be calculated as;
=Total Input power - Losses in the stator
Pg=√3VL x IL cosΦ-[(Ps(h+e)+Ps)] ------(3)
The air gap power (Pg) is transferred to the input to the rotor.
The air gap Power= Rotor Input Power
The air gap power gets converted into the kinetic rotational power. The following losses occur in the rotor.
1. Rotor copper loss
2. Rotor core loss- Negligible
3. Friction, windage and stray losses
The air gap power(Pg) can also be calculated by following formula.
Pg  = The mechanical power output(Pm) + Rotor Ohmic loss + Friction and Windage loss

Calculation of Air Gap Power

The equivalent rotor circuit of an induction motor is shown below.
The X2 and E2 is the rotor reactance per phase and EMF /Phase induced in the rotor circuit. R2/s is the rotor winding resistance.
The rotor input power  = Air Gap Power








The rotor copper loss = sPg
Mechanical power output(Pm)= (1-s)Pg

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