Unit - 3
Single phase Transformers
Types of transformers
- According to input supply: 1-phase and
phase - According to construction: core and shell type
- According to O/P: step up and step down
Construction of transformers
- Laminated steel core
Material used for core is (silicon steel) it is used for its (high permeability) and (low magnetic reluctance) 
magnetic field produced is very strong
The core is formed of (stacks of laminated thin steel sheets) which are electrically isolated from each other. They are typically (0.35 to 0.5 mm thick)
We can used 2 ‘L’ shaped sheets or 2 
shaped sheets for laminations
Construction and types
There 2 types of winding
- Concentric or cylindrical
- Sandwiched type
- Cylindrical
L.V. = low voltage
H.V. = high voltage are mounted on same limb to minimum leakage.
L.V. Winding placed inside and H.V. Winding placed outside with (proper insulation between the winding as it is easy to insulated L.V. Winding) than H.V. Winding.
2. Sandwiched
The H.V. And L.V. Winding are divided into no. Of small coils and there small windings are interleaved. (the top and bottom winding are L.V. Coils because they are close to core)
- Transformer tank: wholes assembly of winding and core placed inside the Transformer tank (sheet metal tank) which is filled with Transformer oil or insulating oil which acts as an (insulator or coolant)
- Transformer oil: (The function of oil is to remove efficiently the heat generated in core and in winding)
- Moisture should not be allowed which creeps the insulation which achieved by closed Transformer tank. (To increase cooling surface are tubes or fins are provided)
- Conservator tank: above tank T/F tank there is one small tank in which same empty space is always provided above the oil level. (This space is required for oil to expand or contract due to temperature change). However during contraction outside air can have moisture which will deteriorate the insulating properly of oil.
- Breather: the air goes in or out through the breather (To reduce the moisture content of their air same drying agents such as (silica gel or) calcium chloride) is used in the breather (The dust particles present in air are also removed by breather)
- Buccholz Relay: (for incipient (slowly increasing) faults
There is pipe connecting rain tank and conservator. On the pipe a protective device called Buccholz Relay is mounted.
When the Transformer is about to be faulty and draw range current the oil becomes very hot and decompose.
During this process different types of gases are liberated. (The Bucchoz Relay get operated by these gases) and gives an alarm to the operator. ɡȴ the fault continues to persist then there lay will trip off main circuit breaker to protect the Transformer.
- Explosion Vent:
An explosion Vent or relief value is the bent up pipe filled on the main tank.
(The explosion vent consists of aluminium of oil) when the T/F becomes faulty the cooling oil get decomposed and various types of gases are liberated
(ɡȴ the gas pressure exodus certain level then the aluminium of oil (diagram) in explosion vent will burst) to release pressure. They will save main tank from getting damaged.
- Single phase transformer
(Symbol and principal of operation)
It is a static device which can transfer electrical energy from one ac circuit to another ac circuit without change in frequency.
It can increase or decrease the voltage but with corresponding decrease or increase in current.
It works on Principle 
“Mutual Induction”
A major application of transformer is to increase voltage before transmitting electrical energy over a long distance through conductors and to again reduce voltage at place where it is to be used.
Symbol
- Primary winding: (ac supply side)
The winding which is connected to supply is called primary side
2. Secondary winding: (load side)
The winding which is connected to load is called as secondary winding
Principle of operation
Working:
It works on principle of mutual Induction i.e. “when 2 coil are inductively coupled and if current in one coil is changed uniformly then an emf get induced in another coil”
- When alternating voltage V1 is applied to primary winding am alternating current I1 flows in it producing alternating flux in the core.
- As per Faradays Law of Electromagnetic Induction 1 an emf E1 is induced in the primary winding E = N1
- The emf induced in the primary winding is nearly equal and opposite to applied voltage V1
- Assuming leakage flux to be negligible almost whole flux produced in primary winding links with secondary winding thence emf induced in the secondary winding E2 = N2
Emf Equation of Transformer:
N1 = No of primary turns
N2 = No of secondary turns
Φm = max flux in cone (wb)
Φm = Bm x A
f = Frequency of input(ac) Hz
Average rate of change of flux
= Φm / ¼
= 4 ΦmF volts
Average emf per turn = 4F ΦmV
As Φ varies sinusoidally rms value of induced emf is given as
Form factor = r.m.s value / average value = 1.11
r.m.s value of emf/turn = 1.11 x 4f Φm
= 4.44 f Φm volt
r.m.s value of induced emf in primary winding
= (induced emf/turn) x No of primary turns
E1 = 4.44 fN1 Φm
E1 = 4.44 fN1BmA
r.m.s emf induced in secondary
E2 = 4.44 fN2 Φm
E2 = 4.44 fN2BmA
E1/E2 = N1/N2 = K
K – voltage transformation
(i). If N2 > N1 i.e K > 1 STEP UP TRASFORMER
(ii). If N1 > N2 i.e K<1 STEP DOWN TRASFORMER
FOR Ideal Transformer
V1 I1 = V2 I2 = 1/K
Hence, current is inversely proportional to the voltage transformation ratio.
Q1>. A 25 KVA transformer has 500 turns on primary and 50 on secondary. The primary is connected to 2000V, 50 Hz supply. Find full load primary and secondary currents, the secondary emf and max flux in core. Neglect leakage drop and no load primary current.
Sol: K = N2/N1 = 50/500 = 1/10
I1 = 25,000/2000 = 12.5 A
I2 = I1/K = 10 x 12.5 A = 125 A
Emf/turn on primary side = 2000/500 = 4 V
E2 = KE1
E2 = 4 x 50 = 200 v
E1 = 4.44fN1 Φm
2000 = 4.44 x 50 x 500 x Φm
Φm = 18.02 m Wb
An ideal transformer has no losses i.e. its winding have no magnetic leakage and no ohmic resistance. Hence, an ideal transformer has only two purely inductive coils wound on a loss-free core.
Fig. Ideal Transformer
For above transformer when secondary is open and primary is having input sinusoidal voltage V1. An alternating current flows due to difference in potential. As primary coil is purely inductive so, Iµ current is drawn through it. This current is very small and logs V1 by 900.
The current Iµ produces magnetic flux φ and hence are in same phase. The flux is linked with both the windings and hence, self induced emf is produced E1 which is equal and opposite of V1. Similarly E2 is induced in secondary which is mutually induced emf E2 is proportional to rate of change of flux and number of secondary windings.
The phasor is shown below.
Fig. Phasor for Ideal Transformer
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) ]
(1). Core or Iron Loss:
This includes both hysteresis loss and eddy current loss. Iron or core loss is found from open circuit test.
Hysteresis Loss(Wh) = nB1.6max f V Watt
Eddy current Loss(We) = P B2max f2 t2 Watt
(2). Copper Loss:
This is due to ohmic resistance of the transformer windings.
Total cu loss = I21 R1 + I22 R2
= I21 R01 + I22 R02
Cu loss α I2
The value of cu loss is found from short-circuit test.
Efficiency of transformer:
Basically efficiency is defined as
n = output/input
But for transformer there are small amount of losses so the improved way to find efficiency is
n = output/output + losses
n = output/output + cu loss + iron loss
Or n = Input – losses/Input
= 1 – Losses/Input
Condition for maximum efficiency:
For n to be maximum dn/dI1 = 0
(Ww) cu loss = I21 R01 or I22 R02
Iron loss = Hysteresis loss + Eddy current loss
= Wn + We = Wi
n = Input – losses/Input
Primary Input = V1I1 Cosφ1
n = V1 I1 Cosφ1 – losses/V1 I1 cos φ1
n = V1 I1 cos φ1 – I21 R01 – Wi / V1I1 cosφ1
= 1 – I1R01/V1cosφ1 – Wi/V1I1cosφ1
Differentiating w.r.t I1 both sides of above equation we have
Dn/dI1 = 0 – R01/V1cosφ1 + Wi/V1I21 Cosφ1
For max value dn/dI1 = 0
R01/V1cosφ1 = Wi/V1I21 cosφ1
Wi = I21 = I21 R01
Hence,
Wi = Wcu
Iron loss = copper loss
The value of output current for maximum efficiency will be
I2 = √Wi/R02
The maximum efficiency can also be given as,
nmax = full load x √ Iron loss / F.L .cu loss
Or
nmax = R’ x full load KVA x pf / R’ x full load KVA x pf + Wi + Wcu x 100
R’ = ratio of actual to full load KVA
Wi = iron loss (KW)
Wcu = copper loss (KW)
Q1). In a 50 KVA, 2200/200 V, 1-φ transformer, the iron and full-load copper losses are 400 W and 450 W respectively. Calculate n at unity power factor on (i). Full load (ii). Half-full load?
Sol. (i). Total loss = 400 + 450 = 850 W
F.L output at unity power factor = 50 x 1
= 50 KVA
n = 50 / 50 + .850 = 50/50.850 = 0.98 = 98%
(ii). Half full load, unity pf
= 50 KVA/2 = 25 KVA
Cu loss = 400 x (1/2)2 = 100 W
Iron loss is same = 450 W
Total loss = 100 + 450 = 550 W
n = 25/25 + 0.55 = 25/25.55 = 0.978 = 97.8 %
Q>. A 40 KVA 440/220 V, 1- φ, 50 Hz transformer has iron loss of 300 W. The cu loss is found to be 100 W when delivering half full-load current. Determine (i) n when delivering full load current at 0.8 lagging pf (ii) the percentage of full-load when the efficiency will be max.
Sol. Full load efficiency at 0.8 pf
= 40 x 0.8/(40 x 0.8) + losses
Full load cu loss = (440/220)2 x 100
= 400 W
Iron loss = 400 + 300
= 700 W
n = 40 x 0.8/(40 x 0.8) + 0.7 = 97.8 %
(ii). KVA for maximum / F.L KVA = √ iron loss / F.L cu loss
= √300/400 = 0.866
Testing-OC and SC Test
(1). Open circuit test
(2). Short circuit test
(1). OPEN CIRCUIT TEST:
Basically these tests are performed to find the basic parameters of transformers. The open circuit test is performed to determine no-load loss(core loss) and no-load current I0 which is helpful in finding X0 and R0.
Fig. Open Circuit Test
In this test the high voltage winding is left open and the other is connected to supply.
A wattmeter W, voltmeter V and an Ammeter A are connected to L.V. Winding.
Due to applied voltage in primary, flux is set up in core, and hence normal iron loss occurs. This is recorded in Wattmeter. The current I0 is measured by Ammeter A.
Wattmeter reading represents core loss under no-load condition.
The wattmeter reading will be given as
W = V1 I0 cos φ0
Iµ = I0 Sin φ0
Iw = I0 cos φ0
X0 = V1/Iµ, R0 = V1/Iw
At no load I0 ≈ Iµ
I0 = V1 Y0
Y0 – Exciting Admittance
W = V21 G0
G0 – Exciting conductance
B0 = √ Y20 – G20
B0 – Exciting susceptance
(2). SHORT CIRCUIT TEST:
This test is used to determine
(i). Equivalent Impedance (Z01, Z02), leakage reactances (X01 or X02), Total resistance (R01, R02).
(ii). Cu losses are also calculated at full load.
(iii). The regulation of transformer can be determined as Z01, Z02 can be calculated.
Here low voltage is short circuited. And the supply is through high voltage (primary here).
As applied voltage is small, hence the flux φ induced is also small.
As core losses are very small, the wattmeter reading is the cu-loss at full load for whole transformer.
The equivalent circuit is shown below,
Z01 = Vsc/I1
Vsc = voltage to circulate rated load current
W = I21 R01
R01 = W / I21
X01 = √ Z201 – R201
R’2 = R01 – R1
Polarity test
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.
Back to Back Test:
Sumpner's test or back to back test on transformer is another method for determining transformer efficiency, voltage regulation and heating under loaded conditions. Sumpner's test or back to back test is done with two identical transformers. Both transformers are connected to supply such that one transformer is loaded on another. Primaries of the two identical transformers are connected in parallel across a supply. Secondaries are connected in series such that emf's of them are opposite to each other. Another low voltage supply is connected in series with secondaries to get the readings, as shown in the circuit diagram shown below.
Fig: Back to Back Test
In above diagram, T1 and T2 are identical transformers. The secondaries are of opposite polarity the emf's cancel each other, as transformers are identical. In this case, as per superposition theorem, no current flows through secondary. The current drawn from V1 is 2I0, where I0 is equal to no load current of each transformer. Thus, input power measured by wattmeter W1 is equal to iron losses of both transformers. So, we can write iron loss per transformer Pi = W1/2.
Now, a small voltage V2 is injected into secondary with the help of a low voltage transformer. The voltage V2 is adjusted so that, the rated current I2 flows through the secondary. In this case, both primaries and secondaries carry rated current. Thus, short circuit test is simulated and wattmeter W2 shows total full load copper losses of both transformers.i.e. copper loss per transformer PCu = W2/2.
Examples
Q1>. The primary and secondary windings of a 30 KVA, 5000/330 V, 1- φ transformer having resistance of 15 Ω and 0.02 Ω. The reactance referred to primary is 34 Ω. Calculate primary voltage required to circulate full-load current when the secondary is S.C. Also calculate the pf?
Sol. K = 330/5000 = 33/500
X01 = 34 Ω
R01 = R1 + R2/K2 = 15 + 0.02(500/33)2 = 19.59 Ω
Z01 = √ R201 – X201
= √ 19.592 + 342
Z01 = 39.23 Ω
F. L I1 = 30,000/5000 = 6A
Vsc = I1 Z01 = 6 x 39.23 = 235.4 V
S.C. Power factor = R01 / Z01 = 19.59/39.23 = 0.5
Q2). In no-load test of 1- φ transformer, the test data are
Primary voltage = 200 V,
Resistance of primary = 0.6 Ω
Secondary voltage = 100 V
Primary current = 0.6 A
Power input = 30 W
Find (a) Turns ratio (b) Magnetising component of no-load current (c) Working component (d) Iron loss
Sol>. (a). N1/N2 = 200/100 = 1
(b). W = V1 I0 cos φ0
Cos φ0 = 30/200 x 0.6 = 0.25, sin φ0 = 0.97
Iµ = I0 sin φ0 = 0.6 x 0.97 = 0.58 A
©. Iw = I0 Cos φ0 = 0.6 x 0.25 = 0.15 A
(d). Cu loss = I20 R1 = (0.6)2 x 0.6 = 0.216 W
Iron loss = 30 – 0.216 = 29.78 W
Q.3 Obtain the secondary voltage when delivered 10 KW at 0.8 PF.logging . The primary volt be 220v. The 300/600 v 50H2, 1-
Transformer has following test results.
O.C test-:200v, 0.8A 70W-L.V side
S.C test-: 12V,10A 80W-HV side
Sol: O.C TEST
W=
COS 
70=200*0.8*Cos
Cos
= 0.436
Sin
=0.899
=
COS
=0.8*0.436=0.35A
I
=
sin
= 0.8*0.899=0.72A
=
=
=571.4
=
=
=277.8𝛺
S.C TEST:
=
=
=1.2A
K=
=2
Z01=
=
=0.3
I22
=W
RO2=
=0.7
R01=
=
=0.175𝛺
X01=
= 
X01 = 0.244
Output KVA =
=12.5
I2=
= 20.83A
Z02=1.2A R02=0.7𝛺
X02=
=0.975𝛺
Total transformer drop refereed to secondary
= I2 (RO2COS
+ X02Sin
)
= 20.83(0.7*0.8+0.975*0.6)
= 23.85v
= 600 - 23.85=576.15v
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:
- The primary windings of the transformer should be suitable for supply voltage and frequency.
- The connections w.r.t. Polarity of transformer should be proper.
- The voltage ratings of both primaries and secondaries should be identical.
- In order to avoid circulating current the ratio X/R and percentage impendences should be equal in magnitude.
- The transformer that has 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
Parallel operation of 3-Ø Transformer
- The conditions are same as in case of 1-Ø transformer.
- The voltage ratio must refer to the terminal voltage of primary and secondary.
- The phase displacement between primary and secondary voltages must be same for all transformers.
- The phase sequence must be same.
- All three transformers in 3-Ø transformer bank will be of same construction.
Key takeaway
- The primary windings of the transformer should be suitable for supply voltage and frequency.
- The connections w.r.t. Polarity of transformer should be proper.
- The voltage ratings of both primaries and secondaries should be identical.
- In order to avoid circulating current the ratio X/R and percentage impendences should be equal in magnitude
Example
Que 1. What should be the kVA rating of each transformer in V-V bank when 3-Ø balanced load is 50 kVA. If a third transformer is connected for operation find % increase in rating?
→ kVA for each transformer has to be 15%
KVA per transformer = 
= 28.75
Δ-Δ bank rating = 28.75 × 3
= 86.75
Increase =
= 72.5 %
Que 2. A balanced 3-Ø load of 120 kW at 1100V 0.866 lag p.f is supplied from 1500V, 3-Ø main through 1-Ø transformer connected – (i) Δ-Δ (ii) Y-Y
Find current in the windings of each transformer and p.F at which they operate in each case.
→ (i) Δ-Δ
VLILcosØ = 120,000
× 1100 × IL × 0.866 = 120,000
IL = 72.73 A
Secondary line current = 72.73 (= IL)
Secondary phase current = 
= 41.99 A
K = 
Primary phase current = 
= 20.99 A
(ii) Y-Y connection
× 1100 × I × 0.866 = 120,000
I = 72.73 A
Secondary phase current = 72.73 A
Primary phase current = 
= 41.99 A
Transformer p.f = 86.6 % of 0.866
= 0.75 (lag)
Que 3. Two transformers are required for a Scott connection operating from 410 V 3-Ø supply for supplying two 1-Ø furnaces at 200 V on the two-phase side. If total output is 130 kVA, calculate secondary to primary turn ratio and the winding of each transformer?
→ For main Transformer
Primary voltage = 410 V
Secondary voltage = 200 V
Secondary current = 
= 325 A
Primary current = 325 × 
= 153.54 A
For teaser Transformer
Primary volts = 
= 355 V
Secondary volts = 200 V
Load Sharing under various conditions
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
Key takeaway
SA = 
SB = 
I = 
V2 = EA - IAZA and V2 = EB – IBZB
An auto Transformer is a special types of transformer such that a part of the winding is common to both primary as well as secondary
It has only One winding wended on a laminated magnetic core
With the help of auto Transformer the voltage can be stepped up and stepped down at any desired value
Figure. Auto Transformer
Fig A shows auto T/F as step down T/F variable terminal B C is connected to load and it acts as secondary wdg.
The position of point C is called as topping point can be selected as per requirement
Fig. B show auto T/F as step up T/F variable terminal B C is connected to supply side ie ac side and it acts as secondary winding
The operating principle of auto Transformer is same as that of 2 winding Transformer
Advantages
- Weight of copper required in an auto Transformer is always loss than that of the conventional 2 winding Transformer and hence it is chaper
- Com[act in size and loss costly.
- Losses taking place in Transformer is reduced hence efficiency is higher than conventional Transformer.
- Due to reduced resistance, voltage, regulation is better than conventional T/F.
Disadvantage:
As low voltage and high voltage sides are not separate then there is always risk of electric shocks when use for high vtg.
Applications
- Starting squirrel cage induction motor and synchronous motor.
- Auto transformer as dimmer stat
- Used as various ac to vary a. c Voltage
FOR AUTO TRANSFORMER:
Fig: (a) Set-Up Auto-Transformer Fig: (b) Step-down Auto-Transformer
From above fig (a)& (b)
Weight of copper in AC∝(N1-N2)I1
Weight of copper in BC ∝ (N2(I2-I1)
Total weight of copper ∝ (N1-N2) I1+N2(I2-I1) 
auto transformer
For 2- winding transformer:
Weight of 
on primary ∝N1I1
Weight of 
on primary ∝N2I2
Total weight of 
∝ N1I1+N2I2
Weight of 
in auto transformer =(1-k)*(weight of 
in ordinary transformer)
Saving = Wo-Wa
= W0-(1-k)W0
Saving = Kwo
Key takeaway
Power transformer inductively =(1-k) input hence, saving will increase at K approaches unity.
Q. An auto transformer suppliers load of 4KW at 100v at unity pf. IF the applied primary voltage is 220v. Calculating power transferred to load (a) Inductively (b) conductively.
Sol: Power transferred inductively = Input(1-k)
Power transferred conductively = K* Input
K= 
Input= Output =4KW
Inductively transferred power =4(
)
=3.82KW
Conductively transferred power = 
= 0.182 kw
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.
References:
1. A. E. Fitzgerald and C. Kingsley, "Electric Machinery”, McGraw Hill Education,2013.
2. M. G. Say, “Performance and design of AC machines”, CBS Publishers,2002.
3. P. S. Bimbhra, “Electrical Machinery”, Khanna Publishers,2011.
4. I. J. Nagrath and D. P. Kothari, “Electric Machines”, McGraw Hill Education,2010.
5. A. S. Langsdorf, “Alternating current machines”, McGraw Hill Education,1984.
6. P. C. Sen, “Principles of Electric Machines and Power Electronics”, John Wiley & Sons,2007.
