|
=
Input= Output =4KWInductively transferred power =4(
)=3.82KWConductively transferred power = 
*4= 0.182 kw Q8) The core of a 110 KVA, 10,000/500v, 50 Hz, 1-Φ core type transformer has a cross section of 18 cm x 18 cm. Find the number of HV and LV turns per phase and the emf per turn if the maximum core density does not exceed 1.3 tesla. Assume a stacking factor pf 0.9.A8) Bm= 1.3TArea = (0.18 x 0.18) = 0.032m2Emf induced in primaryE1 = 4.44 fN1BmA10000=4.44 x 50 x N1 x 1.3 x 0.032N1=1082.8emf induced in secondaryE2 = 4.44 fN2BmA500= 4.44 x 50 x N2 x 1.3 x 0.032N2= 54.14
= 360/2000=0.18AI20 = I2w + I2µMagnetising component Iµ = √I20 – I2w= √ (0.6)2 – (0.18)2Iµ = 0.57 A Q11) A 2200/250 V transformer takes 0.7A at a p.f of 0.4 0n open circuit. Find the magnetising and working component of no load primary current?A11) I0= 0.7Acosφ0=0.4Iw= I0 cosφ0Iw=0.7 x 0.4=0.28AMagnetising component Iµ = √I20 – I2w= √ (0.7)2 – (0.280)2Iµ = 0.64 A Q12) A single phase transformer has 400 turns in primary and 1000 turns in secondary. The cross-sectional area is 80 cm2. If primary is connected to 50hz at 500V. Voltage induced in secondary?A12)As we know Emf induced in primaryE1 = 4.44 fN1BmA500= 4.44x50x1000xBmx(80x10-4)Bm= 0.28Wb/m2The voltage induced in secondary is given asE1/E2 = N1/N2 = KE2=1000x500/400=1250V Q13) A 200 KVA, 1200/200v, 50 Hz, 1-Φ transformer has a leakage impedance of (0.1 + 0.30) Ω for the HV winding and (0.005 + 0.015) Ω for the LV winding. Find the equivalent winding resistance, reactance and impedance referred to the HV and LV side.A13) The Turn ratio is given as E1/E2 = N1/N2 = KK=6Referring to High Voltage SideResistance = R1+K2R2= 0.1+ 62 x 0.005 =0.28ohmReactance = X1+K2X2= 0.30+ 62 x0.015 = 0.84ohmImpedance = √0.282 +0.842 =0.880hmReferring to Low voltage sidesResistance= R1/ K2+R2= (0.1/62) + 0.005=0.007ohmReactance = Reactance HV side/ K2=0.84/62=0.013ohmImpedance = Impedance referring HV/ K2= 0.88/62 = 0.024ohmQ14) When a transformer is connected to a 1200v,50Hz supply the core loss is 900 W, of which 600 is hysteresis and 350 W is eddy current loss. If the applied voltage is raised to 2000V and frequency to 100 Hz. Find new core losses.A14) Hysteresis Loss (Wh) α 
f =P 
fEddy current Loss (We) α P
f2 = Q 
fThe emf equation of transformer is given as E= 4.44fNBmaxABmax α E/fSo, the above equations become Wh = P (E/f)2 f = P E1.6 f-0.6600= P x 12001.6 x 50-0.6P=0.074The eddy current loss We= Q (E/f)2 f2 = QE2 350=Q x 12002Q=0.243 x 10-3Now the applied voltage is raised to 2000V so finding new losses with above found P and QWh = P E1.6 f-0.6 = 0.074 x 20001.6 x 100-0.6 = 893.06WWe= QE2 = 0.243 x 10-3 x 20002 = 972 W UNIT – 2
Magnetic Circuit & Single-Phase Transformer
Q1) A 2500/200 V transformer draws a no-load primary current of 0.5 A and absorbs 400 W. Find magnetising and loss currents.
A1)
Iron-loss current = No load input(W) / Primary voltage
= 400/2500 = 0.16 A
I20 = I2w + I2µ
Iµ = √I20 – I2w
= √ (0.5)2 – (0.16)2
Iµ = 0.473 A
Q2) A 1-φ transformer has 1000 turns on primary and 200 on secondary. The no load current is 4 amp at p.f of 0.2 lagging. Find primary current and pf when secondary current is 280 A at pf of 0.6 lagging.
A2)
Cos-1 0.6 = 53.130 (sin φ = 0.8)
I2 = 280/-53.130A
Φ = cos-1 0.2 = 78.50
Sin φ = 0.98
I1 = I0 + I’2
I’2 = (I2/K) ( -53.130
K = N1/N2 = 1000/200 = 5
I’2 = 280/5 (-53.130
I’2 = 56(-53.130
I1 = I0 + I’2
= 4(0.20 – j0.98) + 56(0.6 – j0.8)
= 0.80 – j3.92 + 33.6 – j44.8
I1 = 34.4 – j48.72
I1 = 59.64 ( -54.770
I lags supply voltage by 54.770
Q3) A 1- φ transformer with ratio of 440/110-V takes a no-load current of 6 A at 0.3 pf lagging. If secondary supplies 120 A at pf of 0.8 lagging. Find current taken by primary.
 |
A 3)
Cos φ2 = 0.8
Φ2 = 36.540
Cos φ0 = 0.3
Φ0 = 72.540
K = V2/V1 = 110/440 = ¼
I’2 = KI2 = 120 x ¼ = 30 A
I0 = 6A
Angle between I0 & I’2
= 72.54 – 36.54
= 35.670
From vector diagram,
I1 = √(62 + 302 + 2 x 6 x 30 cos 35.67)
I1 = 35.05 A
Q 4) A – 100 KVA transformer has 500 turns on primary and 80 turns on secondary. The primary and secondary resistances are 0.3 and 0.01 Ω respectively and the corresponding leakage reactances are 1.1 and 0.035 Ω. The supply voltage is 2400 V. Find
(i). Equivalent impedance referred to primary
(ii). Voltage regulation and the secondary terminal voltage for full load having pf 0.8 lagging?
A 4)
Equivalent impedance referred to primary
Z01 = √R201 + X201 = R01 + jX01
R01 = R1 + R2/K2 = 0.3 + 0.01/K2 = 0.69 Ω
K = 80/500 = 4/25
X01 = X1 + X2/K2 = 1.1 + 0.035/(0.16)2 = 2.467 Ω
Z01 = 0.69 + j2.46
(ii). Secondary terminal voltage Z02 = K2 Z01
Z02 = 0.018 + j 0.063
= 0.065 ( 74.050
No-load secondary voltage = KV1
= 0.16 x 2400 = 384 V
I2 = 100 x 103/384 = 260.42 A
Full load voltage drop referred to secondary
= I2 (R02 cosφ – X02 Sinφ)
Cosφ = 0.8
Φ = 36.860
Sinφ = 0.6
= 260.42(0.018 x 0.8 – 0.063 x 0.6)
= - 6.094 V
% regn = -6.094/384 x 100
= -1.587
Secondary terminal voltage on-load
= 384 – (-6.094)
= 390.09 V
Q4) 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?
A5) (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 %
Q5) 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.
A6)
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
Q6) 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.
A7)
Power transferred inductively = Input(1-k)
Power transferred conductively = K* Input
K= 
=
Input= Output =4KW
Inductively transferred power =4(
)
=3.82KW
Conductively transferred power = 
*4
= 0.182 kw
Q7) The core of a 110 KVA, 10,000/500v, 50 Hz, 1-Φ core type transformer has a cross section of 18 cm x 18 cm. Find the number of HV and LV turns per phase and the emf per turn if the maximum core density does not exceed 1.3 tesla. Assume a stacking factor pf 0.9.
A8)
Bm= 1.3T
Area = (0.18 x 0.18) = 0.032m2
Emf induced in primary
E1 = 4.44 fN1BmA
10000=4.44 x 50 x N1 x 1.3 x 0.032
N1=1082.8
Emf induced in secondary
E2 = 4.44 fN2BmA
500= 4.44 x 50 x N2 x 1.3 x 0.032
N2= 54.14
- The number of turns is N1=1082.8 and N2= 54.14
- Emf per turn = E1/E2 = N1/N2 = K
=10000/1082=9.24V or 500/54.14=9.23V
Q8) A 1-Φ transformer has 400 turns in primary and 110 turns in the secondary. The cross-sectional area of the secondary. The cross-sectional area of the core is 80cm2. If the primary winding is connected to the 50 Hz supply at 500V. Calculate peak flux density in core.
A9)
As we know Emf induced in primary
E1 = 4.44 fN1BmA
500 = 4.44 x 50 x 400 x Bm x (80x10‑4)
Bm=0.704Wb/m2
Q9) A 2000/200v transformer draws a no-load primary current of 0.6A and absorbs 360 watts find the magnetising and iron loss currents.
A10)
Iron loss current =
= 360/2000=0.18A
I20 = I2w + I2µ
Magnetising component Iµ = √I20 – I2w
= √ (0.6)2 – (0.18)2
Iµ = 0.57 A
Q10) A 2200/250 V transformer takes 0.7A at a p.f of 0.4 0n open circuit. Find the magnetising and working component of no load primary current?
A11)
I0= 0.7A
Cosφ0=0.4
Iw= I0 cosφ0
Iw=0.7 x 0.4=0.28A
Magnetising component Iµ = √I20 – I2w
= √ (0.7)2 – (0.280)2
Iµ = 0.64 A
Q11) A single phase transformer has 400 turns in primary and 1000 turns in secondary. The cross-sectional area is 80 cm2. If primary is connected to 50hz at 500V. Voltage induced in secondary?
A12)
As we know Emf induced in primary
E1 = 4.44 fN1BmA
500= 4.44x50x1000xBmx(80x10-4)
Bm= 0.28Wb/m2
The voltage induced in secondary is given as
E1/E2 = N1/N2 = K
E2=1000x500/400=1250V
Q12) A 200 KVA, 1200/200v, 50 Hz, 1-Φ transformer has a leakage impedance of (0.1 + 0.30) Ω for the HV winding and (0.005 + 0.015) Ω for the LV winding. Find the equivalent winding resistance, reactance and impedance referred to the HV and LV side.
A13)
The Turn ratio is given as E1/E2 = N1/N2 = K
K=6
Referring to High Voltage Side
Resistance = R1+K2R2= 0.1+ 62 x 0.005 =0.28ohm
Reactance = X1+K2X2= 0.30+ 62 x0.015 = 0.84ohm
Impedance = √0.282 +0.842
=0.880hm
Referring to Low voltage sides
Resistance= R1/ K2+R2= (0.1/62) + 0.005=0.007ohm
Reactance = Reactance HV side/ K2=0.84/62=0.013ohm
Impedance = Impedance referring HV/ K2= 0.88/62 = 0.024ohm
Q13) When a transformer is connected to a 1200v,50Hz supply the core loss is 900 W, of which 600 is hysteresis and 350 W is eddy current loss. If the applied voltage is raised to 2000V and frequency to 100 Hz. Find new core losses.
A14)
Hysteresis Loss (Wh) α 
f =P 
f
Eddy current Loss (We) α P
f2 = Q 
f
The emf equation of transformer is given as E= 4.44fNBmaxA
Bmax α E/f
So, the above equations become Wh = P (E/f)2 f = P E1.6 f-0.6
600= P x 12001.6 x 50-0.6
P=0.074
The eddy current loss We= Q (E/f)2 f2 = QE2
350=Q x 12002
Q=0.243 x 10-3
Now the applied voltage is raised to 2000V so finding new losses with above found P and Q
Wh = P E1.6 f-0.6 = 0.074 x 20001.6 x 100-0.6 = 893.06W
We= QE2 = 0.243 x 10-3 x 20002 = 972 W
