Transformers...Electrical Machines Unit-1 No -99, T.P.K. Road, Madurai – 01 Cell: 7373 0077 31,...

Electrical Machines Unit-1 No -99, T.P.K. Road, Madurai – 01 Cell: 7373 0077 31, 7373 0077 34

Website: www.maduracoaching.com Email: [email protected]

Transformers

Transformation ratio:

Emf equation



For an ideal transformer:

Winding resistance = 0

Magnetizing current ( ) is equal to

No Magnetic leakage

Transformer’s rating is specified in KVA.

Problems

1. The emf per turn of a single phase, 2200/220V, 50Hz transformer is 10V. Calculate

i) no.of Primary and secondary turns ii) The net cross sectional area of core for max

magnetic flux density of 1.5 Tesla

Solution:

(1)

(2)



From (1);

2) A 10KVA, 500 / 250V, 50Hz, Single phase transformer has a net area of cross section

90cm2 and maximum flux density is 1.2 Tesla. Calculate the no.of turns on both primary

and secondary.

A =



3) A single Phase transformer 10 : 1 turns ratio and rated at 50KVA, 2400/ 240V, 50Hz is

used to step down the voltage of a distribution sys. A 10 w tension is to be kept constant

find the value of load impedance of the LT side so that transformer will be loaded fully

find also the value of max flux inside the core if LT. side has 23 turns.

Effect of voltage variation and frequency variation

Variation in voltage and frequency affects iron losses

generally take 1.6



If a transformer is operated with frequency and voltage changed in same proportion, the flux

density will remain unchanged and remains apparently the no load current will be unaffected.

( )

If ratio is constant transformer core will not be saturated.

Phasor diagram of transformer when no load:

1. Iron loss reduces

2. Iron loss Increases

3. V is constant



No load power

is called exiting current

Note:

No load primary current is very small (2 to 3% of full load current) no load power factor

is in the range of 0.1 – 0.3 lag-, no load primary copper loss is very small and can be neglected.

The wave form of magnetizing current is not sinusoidal.

Problem:

1. At 2200 / 250V transformer takes 0.5A at a power factor of 0.3 on 0.1 find the

Magnetizing and active components of no load primary current.



I = 0.5A,

ϕ =

2) A no load current of is 5A at 0.25 pf when supplied at 235V, 50Hz, the no.of turns on

the primary winding is 200. calculate A) the max value of flux in the core b) core loss c) the

magnetizing component.

A.

B.

=



Phasor Diagram of a Transformer:



Problem:

A Single phase Tr. with a ratio of takes a no load current of 5A at 0.2 pf lagging if

secondary supplies a current of 120 A at. pf of 0.8 lagging estimate the current drawn by

primary.



= 0.97







Voltage Regulation

The secondary voltage varies with load depends on the load current, internal impedance, load

power factor.

(+ → lag; – → load)

Voltage regulation should be as minimum as possible.

Zero voltage regulation



Zero voltage Regulation occurs at leading power factor.

Condition for maximum voltage regulation

Maximum voltage Regulation occurs at lagging pf.

% Resistance and % Reactance

%

Problems:

1. The Primary and secondary windings of a 440KVA, 6600/250V single phase have

resistance of 10Ω and 0.02 Ω respectively the total leakage reactance is 35 Ω as referred

to primary winding. Find the full load regulation at 0.8pf lagging

KVA = 40,



2. A 10 on full load has impedance drop of 20V and resistance drop of 10V. Calculate value

of pf when its voltage Regulation is zero.

2) The full load copper loss on the H.V side of a 100KVA, 11000/317V, single phase is

0.62KW on the L.V side is 0.48KW (i) calculate secondary resistance referred to primary (ii)

The toal reactance is 4% find secondary reactance referred to primary if the reactance is divided

in the same proportion as resistance

KVA = 100





Note: %

2) Copper loss or ohmic loss

Testing of Transformer:

Open circuit test (no load test)

The purpose of the test is to determine the core loss, no load current I0 and theory shunt branch

parameter R0 and X0 of the equivalent Ckt .

Iron power on no load = p0 (watts) No load current

= I0(A)

Applied voltage to the primary = V1 (Voltas)



Note:

Since no load current I0 is very small the pressure coils of wattmeter and voltmeter

should be connected such that the current drawn by the do not flow through current coils

of wattmeter is ammeter; since the p.f at no load is low & low pf wattmeter must be used.

Separation of Hysteresis & Eddy current losses:

Short circuit Test:

The purpose of this test is to determine fall load copper loss and equivalent reactance

referred to metering side.



Problems:

1. In a 400V, 50Hz, the total iron loss, is 2500 watts when the supply voltage is frequency is

reduced to 200V and 25Hz respectively, the corresponding loss is 850W calculate the

eddy current loss at normal voltage & frequency.

Case (i)

Case (ii)



2) At 10KVA 200/400V, 50Hz 1ϕ gave the following test result

220V 1.3A 120W

22V 30A 200W

find the parameter of cq. Ckt referred to L.V. side



Star Delta

1)

2)

Transformer Efficiency



Condition for maximum efficiency

A 10KVA, 5000/440V, 25Hz 1ϕ tr. has copper, Eddy current and hysteresis loss of 1.5, 0.5, and

0.6% of O/P on full load what will be the % losses if the transformer used on 10KV, 50Hz

system keeping full load current constant. Assume unit p.f operation compare the full load

efficiencies for the two cases.

KVA = KW cosϕ

10KVA; pf = 1

O/P = 10KW

Case (ii) When f = 50 Hz.



in both cases same

Since is constant

2) If P1 and P2 be the iron loss and cu loss of on full load find the ration of P1 and P2 such that

max. efficiency occur at 75% of full load.



3) A 200KVA has an efficiency of 98% at full load if the max efficiency occurs at 3/4 of full

load. Calculate the efficiency at half load, arrume negligible magnetizing current & pf of 0.8 at

all load.

Gn cosϕ = 0.8, = 98%

…(1)

…(2)

All day efficiency

All day efficiency =

Power transformers are designed to supply maximum efficiency at full load.

Distribution transformer are designed to supply maximum efficiency of 70-80% of full

load.

Insulating Oil

Insulating oil has & functions it provides additional insulation, protects the insulation

from dirt and moisture and it carries away heat generated in cor and oil

Properties :

Low viscosity purity, high flash point

free from sludging under normal operating conditions



Buchholz’s Relay:

Is used for transformer internal faults. It is used to give alarm in case of minor fault and

to disconnect transformer from supply in case of severe internal faults.

Methods of Reducing leakage flux

By increasing the height of window

By arranging primary and secondary winding concentrically

By sandwiching primary & Secondary Winding

By Adopting shell type construction,

Transformer:

Transformer is cheapest (about 15%) than bank of transformer. It is easy to transport

also efficiency of 3ϕ is tr. is always greater than 1ϕ transformer

Transformer Connections

Group 1 Yy0, Dd0, Dz0

Group 2 Yy6, Dd6, Dz0

Group 3 lags Dy1, Yd1, Yz1

Group 4 lead Dy11, Yd11, Yz11

1) A 3ϕ step down is connected to 6.6KV mains and takes 10A calculate the secondary line

voltage, line voltage, line current O/P for the following cond.

1) 2) Y-Y 3) 4) Y

Ration turns per phase is 12.

1)

,



V-V connection:

One of the Transformers in bank is failed, so that service may be maintained at a

reduced capacity until a faculty transformer is repaired with new one.

If the transformers rating is S1 the V – V transformer rating is,

of S1

The power rating of the and rated current is reduced to 57% of earlier value.

For V-V bank if the two ‘s are delivering power to a load one is operating at a

of cos tr1 and other is operating at of cos tr2 load pf

angle.

1) Load pf = Unity; angle =

2) Load pf angle = 30 (Pf = 0.866)



3) Load pf =0.5, load pf angle = 60

Two ‘s are connected in open delta and supply a balanced 3ϕ load of 240 KW at 400V and a

power factor of 0.866

1) Determine Secondary line current

2) The KVA load on each Transformer

3) Power delivered by individual

tr1= 0.5 tr2 = 1



A 3 single phase 1100 / 110V connected in suppling a lighting load of 100KW

one of there is damaged and hence removed for repairs what currents will be flowing

in each when 3 ’s wile in service.

Case (ii) when two ‘s are in service

Gn data : Case (i)

P =

Case (ii)

Secondary

Primary

Transformer Notice:

Under the no load condition in the one caused by energized power originates

in the core when the laminations tend to vibrate by Magnetic forces, the permeability of

ferromagnetic material changes when they are subjected to varying magnetization as B.H

chalateristie curves of term magnetic material are non – linear. As a result, dimensional

changes occurs both parallel to + to the direction of magnetization. This phenomenon is

called Magnetostriction.



Necessary condition for parallel open:

Polarity of the ‘s are same

The Voltage rating of both secondary & primary are identical, this implies that

turns ratio are the same.

ratio should be same

Phase displacement b/w both line voltage primary & Secondary of the ‘s are

the same (Same group ref number)

Phase Sequence of are same

1) 4) 5) are essential condition

2) & 3) are should be satisfied for closure delivered

If point no.3 is satisfied, the better will be load sharing.

Two 1ϕ tr’s with equal turns have impedence of (0.35 + j 3) Ω and (0.6 + j 10) Ω

with respect to secondary if they operate in Determine how they will share a total

load of 100KW at 0.8 lagging.



Power conducted directly =

of ‘auto transformer’

Power conducted interactively = (1-K) KVA

Saving of copper = K weight of copper in 2 winding

% full load loss = 1- K (Full load loss in 2 winding )

Per unit impedance of 2 winding = 1-K



% Voltage Regulation = (1-K) (2 winding Regulation)

KVA conducted = New rating – old rating

KVA inducted = Old rating

Scaling of transformer:

11500/2300V. is rated at 100 KVA as 2 winding transformer if the 2 winding are

connected in series to from an autotransformer, what will be the voltage ratio and O/P

Case (i)

Rated O/P of tr. =



Case (ii)

Rated O/P of tr. =

Date post:	01-May-2021
Category:	Documents
Upload:	others
View:	1 times
Download:	0 times

Transformers...Electrical Machines Unit-1 No -99, T.P.K. Road, Madurai – 01 Cell: 7373 0077 31,...

Documents