Unit 3

Transformers

Q1) Explain in detail the working of 3-Φ transformer.

A1)

As the supply is in 3-phase so we need this transformer. Its alternating voltage and currents differ in phase by 120 degrees.

                                         Fig: Three phase Transformer

Fig: Line and phase voltage

Construction:

a)     Core Type:

The construction of a core type three phase transformer is as shown in the figure. The core consists of three legs or limbs. The core is made up of thin laminated sheets to reduce eddy current losses. Each limb has primary and secondary windings in cylindrical shape (former wound) arranged concentrically. The construction is shown in the figure.

                                     Fig: Core type 3-phase transformer

b)    Shell Type:

In a shell type three phase transformer, three phases are more independent than they are in core type. Each phase has its individual magnetic circuit. The construction of shell type three phase transformer is illustrated in the figure at right. The construction is similar to that of three single phase shell type transformers kept on the top of each other.

                                         Fig: Shell Type 3-phase Transformer

Q2) List the advantages and uses of open delta connection.

A2)

The advantages are listed below

1)     It can provide power to the three-wire ungrounded service by using only two single phase transformers.

2)     It can provide power to the four-wire grounded service by having the midtap grounded on one leg of the transformer.

3)     Three single phase transformers connected in closed delta can be rewired to open delta in case one transformer is taken out of service for maintenance.

It is not very efficient and hence is used for lower loads.

Q3) what is polarity test of transformer

A3)

The polarity test of a transformer can be done by measuring only voltage at no load. The basic connections are shown in below circuit.

Fig: polarity Test

From above circuit input is applied for one winding. There is another voltage ‘V’ measured between one winding from each terminal. If voltage V’ > V, polarity is additive. If voltage V’ < V, polarity is subtractive.

Q4) Explain parallel operation of 3-Φ transformer

A4)

Parallel operation of 3-Ø Transformer

The conditions are same as in case of 1-Ø transformer.

1. The voltage ratio must refer to the terminal voltage of primary and secondary.
2. The phase displacement between primary and secondary voltages must be same for all transformers.
3. The phase sequence must be same.
4. All three transformers in 3-Ø transformer bank will be of same construction.

Parallel operation of single phase transformer

If the present transformers are not capable of supplying any amount of additional load which is required, then the parallel connections are used to fulfil the requirement. The basic circuit for parallel transformer is shown below.

The condition required for parallel connections: -

1. The primary windings of the transformer should be suitable for supply voltage and frequency.
2. The connections w.r.t. Polarity of transformer should be proper.
3. The voltage ratings of both primaries and secondaries should be identical.
4. In order to avoid circulating current the ratio X/R and percentage impendences should be equal in magnitude.
5. The transformer that have different KVA ratings should have equivalent impendences inversely proportional to individual KVA rating.

Fig. Parallel Connection of 1-ØTransformer

1)     Ideal Case :-

Here both the transformers are considered to have same voltage ratio and impedance voltages triangles identical in size and shape.

Fig. Phasor for ideal Case

Fig: Ideal Case

I = IA + IB

V2 = E - IAZA

IAZA = IBZB

IA =

IB = 

E – No load secondary voltage

V2 – Terminal voltage

IA , IB – Currents of each Transformer

2)     Equal Voltage Ratio :-

Let No-load voltage of each secondary be same E = EA = EB. The change in the secondary is shown below.

Fig: Equal Voltage Ratio

All values are considered referred to the secondary.

ZA , ZB = impedances of transformer

IA , IB = currents of transformer

V2 = common terminal voltage

IAZA = IBZB = IZAB

ZAB = ZA || ZB

ZA = IZABZA ,

IA =

V2IA =

V2IB =

Let V2I × 10-3 = S (Combined Load KVA)

SA =

SB =

3)     Unequal Voltage Ratios :-

Here the no load secondary voltages are unequal.

Fig.: Unequal Voltage Ratios

EA , EB = no-load secondary emp

Z1 = load impedance across secondary.

Here circulating current exist as the voltage are unequal (IC).

IC =

EA = IAZA + V2

EB = IBZB + V2

V2 = IZL = (IA + IB)ZL                     -----------------(1)

EA = IAZA + (IA + IB)ZL                   -----------------(2)

EB = IBZB + (IA + IB)ZL                   -----------------(3)

EA – EB = IAZA – IBZB

IA =

If ZA and ZB are smaller than ZL.

Substitute IA in equation (3) and solve

IA =

IB =

Neglecting ZA ZB in comparison with ZL(ZA + ZB)

We get above equations.

I = IA + IB

I =

V2 = EA - IAZA         and   V2 = EB – IBZB

Q5) Two transformer connected in open delta supply a 430kVA balanced load operating at 0.9 p.f lagging. The load voltage is 460v. What is the kVA supplied by each transformer?

A5)

KVA of each transformer = (430/2)/0.9 = 238.9kVA

The p.f of two transformers will be cos (30-φ) and cos(30+φ)

Cosφ=0.9

φ = 25.8

P1= kVA cos (30-φ) = 238.9 x cos (30-25.8) = 238.27kW

P2= kVA cos (30+φ) = 238.9 x cos (30+ 25.8) =134.28kW

Q6) Two 1-Ø transformers, one of 800 kVA and 400 kVA are connected in parallel to the same bus bars on primary side. Their no load secondary voltage 500 V and 510 V respectively. The impedance voltage of first is 8% and that of second is 5%. Assume ratio of resistance to reactance is same and equal to 0.4 in each. Calculate the cross current when secondary’s are connected in parallel?

A6)

The X/R is same for both; hence they do not have any phase difference.

Let secondary be 480 V

For full load IA = 800 × A

IB = 400 × A

ZA =

ZB =

The cross current is given as

IC = A

Q7) A delta-delta bank consisting of three 10kVA,2200/220v transformer supplies a load of 40kVA. If one transformer is removed, find the resulting V-V connection. a) kVA load carried by each transformer b)% of rated load carried by each transformer.

A7)

KVA rating of V-V bank = 2x10x0.866 = 17.32kVA

= = 0.577 = 57.7%

As we already know  = 3

KVA load supplied by each transformer = 40/3 = 23.1kVA

% of rated load = = 23.1/10 = 231 %

Q8) Two transformer connected in open delta supply a 430kVA balanced load operating at 0.9 p.f lagging. The load voltage is 460v. What is the kVA supplied by each transformer.

A8)

KVA of each transformer = (430/2)/0.9 = 238.9kVA

The p.f of two transformers will be cos (30-φ) and cos(30+φ)

Cosφ=0.9

φ = 25.8

P1= kVA cos (30-φ) = 238.9 x cos (30-25.8) = 238.27kW

P2= kVA cos (30+φ) = 238.9 x cos (30+ 25.8) =134.28kW

Q9) Two, 1-Φ furnaces working at 110v are connected to 3000v,3-Φ mains through Scott connected transformer. Calculate the line current in  3-Φ mains when the power taken by each furnace is 400 kW at a p.f of 0.8 lagging. Transformer has no losses.

A9)

K= 110/3000 = 0.0367

I2=400000/0.8x110 = 4545.5A

The 3-phase side is balanced as the load is balanced.

The primary phase current = 1.15kI2 = 1.15 x 0.0367 x 4545.5=191.84A

For star connection Iph=IL

IL= 191.84A

Q10) Explain Scott connection of transformer

A10)

For making 3-Ø to 3-Ø transformation from two transformers we use this connection.

Fig. : Scott Connection

• The above figure shows that the centres of one transformer are tap, and has 0.866 tap and is known as Leaser Transformer.
• One ends of both the primary and secondary of Leaser transformer are joined to centre tap on both primary and secondary of main transformer.
• There is one transformer having its centre taps on both primary and secondary and is known as main transformer.

The voltage diagram is shown below

Fig. : Voltage and Phase diagram

• EDC and EDB are each 50 V and 180˚ out of phase because coils DB & DC are on same circuit but in opposition.
• EDA is equal to 86.6 V and lags behind the voltage across main by 90˚.
• Idb lags behind voltage Edb by 30˚ and Idc leads Edc by 30˚.
• The teaser transformer and each half of the main transformer all operate at different power factors.
• Full rating of transformer is not utilized. The teaser transformer operate at 0.866 of its rated value and main transformer coil operates at cos 30˚ = 0.866 p.f

Q11) Explain tap changing of transformer.

A11)

On-Load Tap changing:

For uninterrupted supply we use on-load tap changing transformer. Such a transformer is known as a tap-changing under load transformer. While tapping, two essential conditions are to be fulfilled.

• The load circuit should not be broken to avoid arcing and prevent the damage of contacts.
• No parts of the windings should be short–circuited while adjusting the tap.

The tap changing employing a centre tapped reactor R show in the figure.

Fig: On load Tap changing

Here S is the diverter switch, and 1, 2, 3 are selector switch. The transformer is in operation with switches 1 and S closed. To change to tap 2, switch S is opened, and 2 is closed. Switch 1 is then opened, and S closed to complete the tap change. It is to be noted that the diverter switch operates on load, and no current flows in the selector switches during tap changing. It is to be noted that the diverter switch operates on load, and no current flows in the selector switches during tap changing. During the tap change, only half of the reactance which limits the current is connected in the circuit.

No-Load Tap Changing:

The cheapest method of changing the turn ratio of a transformer is the use of Off Load Tap Changer. As the name indicates, it is required to deenergize the transformer before changing the tap. A simple Off Load Tap Changer is shown in Figure below

                                            Fig: No-Load Tap Changing

It has eight studs marked one to eight. The winding is tapped at eight points. The face plate carrying the suitable studs can be mounted at a convenient place on the transformer such as upper yoke or located near the tapped positions on the windings. The movable contact arm A may be rotated by handwheel mounted externally on the tank.

If the winding is tapped at 2% intervals, then as the rotatable arm A is moved over to studs 1, 2; 2, 3; . . . 6, 7; 7, 8 the winding in circuit reduces progressively by it from 100% with arm at studs (1, 2) to 88% at studs (7, 8).The stop F which fixes the final position of the arm A prevents further anticlockwise rotation so that stud 1 and 8 cannot be bridged by the arm.

Q12) Explain autotransformer?

A12)

It is a transformer in which only one winding is common it both primary and secondary because it has one windy it uses loss copper and is cheaper.

It can be step -up and step down auto-transformer. The circuit for both is shown below.

Fig a: Step up auto transformer

Fig b: Step Down Autotransformer

CB-Primary winding

BA- Secondary windings

Negating iron loss and no-load current-

===K

The current in AB() is vector difference (-) >

     An auto transformer when compared to as ordinary 2- winding transformer having same output, given higher efficiency and is smaller in size with superior voltage regulation.

Q13) Explain the methods used for cooling of transformer.

A13)

The main source of heat generation in transformer is its copper loss or I2R loss. If this heat is not dissipated properly, the temperature of the transformer will rise continually which may cause damages in paper insulation and liquid insulation medium of transformer. So, it is essential to control the temperature with in permissible limit to ensure the long life of transformer by reducing thermal degradation of its insulation system.

1)     ONAN (Oil Natural Air Natural):

This is the simplest transformer cooling system. In this the natural convectional flow of hot oil is utilized for cooling. In convectional circulation of oil, the hot oil flows to the upper portion of the transformer tank and the vacant place is occupied by cold oil. This hot oil which comes to upper side, will dissipate heat in the atmosphere by natural conduction, convection and radiation in air and will become cold. In this way the oil in the transformer tank continually circulate when the transformer put into load.

As the rate of dissipation of heat in air depends upon dissipating surface of the oil tank, it is essential to increase the effective surface area of the tank. So additional dissipating surface in the form of tubes or radiators connected to the transformer tank. This is known as radiator bank of transformer.

2)     ONAF(Oil Natural Air Forced):

The Heat dissipation can be increased, if dissipating surface is increased but it can be made further faster by applying forced air flow on that dissipating surface. Fans blowing air on cooling surface is employed. Forced air takes away the heat from the surface of radiator and provides better cooling than natural air. As the heat dissipation rate is faster and more in ONAF transformer cooling method than ONAN cooling system.

3)     OFAF(Oil Forced Air Natural):

In OFAF cooling system the oil is forced to circulate within the closed loop of transformer tank by means of oil pumps. The main advantage of this system is that it is compact system and for same cooling capacity OFAF occupies much less space than farmer two systems of transformer cooling. Actually, in oil natural cooling system, the heat comes out from conducting part of the transformer is displaced from its position, in slower rate due to convectional flow of oil but in forced oil cooling system the heat is displaced from its origin as soon as it comes out in the oil, hence rate of cooling becomes faster.

4)     OFWF (Oil Forced Water Forced):

In OFWF cooling system of transformer, the hot oil is sent to water heat exchanger by means of oil pump and there the oil is cooled by applying sowers of cold water on the heat exchanger’s oil pipes.

5)     ODAF (Oil Directed Air Forced):

The forced circulation of oil directed to flow through predetermined paths in transformer winding. The cool oil entering the transformer tank from cooler or radiator is passed through the winding where gaps for oil flow or pre-decided oil flowing paths between insulated conductor are provided for ensuring faster rate of heat transfer. This cooling method is generally used in very high rating transformer.

6)     ODWF(Oil Directed Water Forced):

This is just like ODAF only difference is that here the hot oil is cooled in cooler by means of forced water instead of air. Both of these transformer cooling methods are called forced directed oil cooling of transformer.

Q14) Explain equivalent circuit, phasor diagram, voltage regulation.

A14)

The basic transformer and its equivalent circuit both are shown below,

  Fig.   Equivalent   Transformer   Circuit

Iµ - magnetising component of current

Iw = working component

R0 – Non- inductive resistance

I0 – No load current

X0 = E1/I0.  R0 = E1/Iw

E2/E1 = N2/N1 = K

E’2 = E2/K = E1

V’2 = V2/K

I’2 = K I2

The total equivalent circuit is again given as,

But the above circuit is exact equivalent but harder to solve so, it can be further simplified as,

Z = Z1 + Zm || ( Z’2 + Z’L )

= Z1 + Zm(Z’2 + Z’L)/Zm + (Z’2 + Z’L)

Z’2 = R’2 + jX’2

Zm = impedance of exciting circuit

V1 = I1[ Z1 + Zm(Z’2 + Z’L) / Zm + (Z’2 + Z’L) ]

Q15) A 20kVA, 1-Φ transformer,50Hz,500/250v gave following results

OC test (LV side): 230v,2.8A,200W

SC test (LV side): 15v,30A,300W

Calculate the efficiency at full load 0.8 p.f lagging.

A15)

Full load current = 20000/230 = 86.96A

SC current is measured till 30A on low voltage side.

I2R losses 40A n low voltage side = (40/30)2 x 300 = 533.3W

Iron loss for OC test = 200W

Full load output at p.f 0.8 = 20000 x 0.8 =16kW

Efficiency = {(16x1000)/[16000+200+533.33]} = 0.956 = 95.6%