Question Bank

Unit-4

Question Bank

Unit-4

Q 1) A 10000/ 250 v , 120KVA 1-

Solution : (1) load KVA for max efficiency

= F.L KVA * = 120 * =

= 97.98KVA

At   cu loss = Iron loss

Total  Loss = 1.2+1.2 = 2.4kw

Output       = 97.98*1  = 97.98kw

= =1.59%

2) cu Loss at half full load =1.8*()2=0.45kw

Total loss =1.2+0.45=1.65kw

Half F.L  output at 0.8 pf =*0.8= 48 kw

n= = 0.97

Q2) A 2500/200 V transformer draws a no-load primary current of 0.5 A and absorbs 400 W. Find magnetising and loss currents.

Sol: 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

Q3). 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.

Sol : 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

Q4)  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.

Sol.

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

Q5) 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

Q7) 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

Q8) 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

Q9) 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?

Sol: 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

Q.10) 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?

Sol: 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

Q.11) 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.

Sol: The Turn ratio is given as E1/E2 = N1/N2 = K

K=6

i)                   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

Ii)                 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

Q.12) 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.

Sol: 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

Bmaxα E/f

So, the above equations become Wh = P (E/f)2 f = P E1.6f-0.6

                                                             600= P x 12001.6 x 50-0.6

                                                             P=0.074

      The eddy current loss               We= Q (E/f)2f2 = 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 E1.6f-0.6 = 0.074 x 20001.6 x 100-0.6 = 893.06W

We= QE2 = 0.243 x 10-3 x 20002 = 972 W

Q.13) A short circuit test when performed on HV side of a 12kVA,2000/400v,1-Φ transformer the data is:50v, 3A, 100W. If the LV side is delivering full load current at 0.8 p.f and at400v, find the Voltage applied to HV side.

Sol:   As the test is done from HV side which is primary side

Z01=50/3 = 16.67ohm

R01= 100/32 = 11.11ohm

X01= 16.672 – 11.112 = 12.42ohm

I1= 12x1000/2000 = 6A

Voltage drop referred to primary is E = I1(R01cosφ + X01sinφ)

                                                                       = 6 (11.11 x 0.8 + 12.42 x 0.6)

E = 98.04V

Q.14) A transformer has cu loss of 1.5% and the reactance drop of 3.2% when tested at full load. Calculate its full load regulation at unity power factor.

Sol: Approx. Voltage regulation at unity p.f (full load) = 0.015 cos φ + 0.032 sin φ= 0.015 = 1.5%

Approx. Voltage regulation at 0.08 lagging p.f = 0.015 x 0.8 + 0.032 x 0.6 = 0.032=3.2%

Approx. Voltage regulation at 0.08 leading p.f = 0.015 x 0.8 – 0.032 x 0.6

= 7.2x10-3= -0.72%

Q.15) A 20kVA 440/220V 1-phase transformer 50Hz has iron loss of 289W. The Cu losses found to be 60W and delivers half of full load current. Determine percentage of full load when the efficiency will be maximum?

Sol: =

Full load Cu loss = 22 x 100 = 400W

= = = 0.85

Hence, efficiency will be maximum at 85%

Q16) 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.

Sol: 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%

Question Bank

Unit-4

Question Bank

Unit-4

Q 1) A 10000/ 250 v , 120KVA 1-

Solution : (1) load KVA for max efficiency

= F.L KVA * = 120 * =

= 97.98KVA

At   cu loss = Iron loss

Total  Loss = 1.2+1.2 = 2.4kw

Output       = 97.98*1  = 97.98kw

= =1.59%

2) cu Loss at half full load =1.8*()2=0.45kw

Total loss =1.2+0.45=1.65kw

Half F.L  output at 0.8 pf =*0.8= 48 kw

n= = 0.97

Q2) A 2500/200 V transformer draws a no-load primary current of 0.5 A and absorbs 400 W. Find magnetising and loss currents.

Sol: 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

Q3). 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.

Sol : 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

Q4)  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.

Sol.

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

Q5) 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

Q7) 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

Q8) 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

Q9) 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?

Sol: 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

Q.10) 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?

Sol: 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

Q.11) 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.

Sol: The Turn ratio is given as E1/E2 = N1/N2 = K

K=6

i)                   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

Ii)                 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

Q.12) 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.

Sol: 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

Bmaxα E/f

So, the above equations become Wh = P (E/f)2 f = P E1.6f-0.6

                                                             600= P x 12001.6 x 50-0.6

                                                             P=0.074

      The eddy current loss               We= Q (E/f)2f2 = 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 E1.6f-0.6 = 0.074 x 20001.6 x 100-0.6 = 893.06W

We= QE2 = 0.243 x 10-3 x 20002 = 972 W

Q.13) A short circuit test when performed on HV side of a 12kVA,2000/400v,1-Φ transformer the data is:50v, 3A, 100W. If the LV side is delivering full load current at 0.8 p.f and at400v, find the Voltage applied to HV side.

Sol:   As the test is done from HV side which is primary side

Z01=50/3 = 16.67ohm

R01= 100/32 = 11.11ohm

X01= 16.672 – 11.112 = 12.42ohm

I1= 12x1000/2000 = 6A

Voltage drop referred to primary is E = I1(R01cosφ + X01sinφ)

                                                                       = 6 (11.11 x 0.8 + 12.42 x 0.6)

E = 98.04V

Q.14) A transformer has cu loss of 1.5% and the reactance drop of 3.2% when tested at full load. Calculate its full load regulation at unity power factor.

Sol: Approx. Voltage regulation at unity p.f (full load) = 0.015 cos φ + 0.032 sin φ= 0.015 = 1.5%

Approx. Voltage regulation at 0.08 lagging p.f = 0.015 x 0.8 + 0.032 x 0.6 = 0.032=3.2%

Approx. Voltage regulation at 0.08 leading p.f = 0.015 x 0.8 – 0.032 x 0.6

= 7.2x10-3= -0.72%

Q.15) A 20kVA 440/220V 1-phase transformer 50Hz has iron loss of 289W. The Cu losses found to be 60W and delivers half of full load current. Determine percentage of full load when the efficiency will be maximum?

Sol: =

Full load Cu loss = 22 x 100 = 400W

= = = 0.85

Hence, efficiency will be maximum at 85%

Q16) 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.

Sol: 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%