Unit4 | unit 4 electrical machines

BEE

Unit-4

Electrical Machines

Question Bank

Q1) A 4 pole lap wound dc shunt generator has a useful flux per pole of 0.05 Wb. The armature winding consists of 200 turns each of 0.06 u ohm resistance. Calculate the terminal voltage when running at 900 rpm with armature current of 50 A.

Soln. Terminal voltage V =Ea – Ia Ra

Ia Ra = 50×Ra.

Z = 200×2=400 (each turn has two sides).

N = 900 rpm = 0.05 WbP = A = 4

Total resistance of 200 turns = 200 × 0.004 = 0.8 ohm

As there are 4 parallel paths, so resistance of each path = 0.8/4 = 0.2 ohm

There are 4 resistances in parallel of each of 0.2 ohm

Ra = 0.2/4 = 0.05 ohm

Ia Ra = 50×0.05 = 25 V

V = Ea – Ia Ra = 300 – 25 = 275 V

Q2) A 220 V 4 pole wave wound dc series motor has 780 conductor on its armature. It has armature and series field resistance of 0.75 ohm. The motor takes a current of 30A. Find the speed and gross torque developed if flux/pole is 20 mWb?

Soln. Back emf,

Eb = V – IaRa = 220 – (30×0.75) = 197.5 V

197.5 = 20 × 10-3 ×780×N×0.75

N = 16.88 ≈ 17

Ta = 158.97 Nm.

Q3) 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

Q4) 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

Q5) When operated at 120 V in the primary of an iron core transformer, the current in the primary is 5 amps. Find the current in the secondary if the voltage is stepped up to 400 V.

Sol: VP/VS = IS/IP

120/400 = IS/5

IS = 1.5A

Q6) A transformer with 400 turns on the primary and 60 turns on the secondary draws Solution: 0.5 amps from a 120 V line. Find Is.

Sol: NP/NS = IS/IP

400/60 = IS/0.5

IS = 3.34A

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.

Sol: 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

i) The number of turns is N1=1082.8 and N2= 54.14

Ii) 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.

Sol: 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

Q.9) 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

Q10) A rotor of 6-pole 50Hz IM is rotated by some means at 1000 rpm. Find i) Rotor voltage. Ii) rotor frequency iii) rotor slip.

Sol: N=1000rpm

F=50Hz

The synchronous speed Ns=120f/P=120x50/6=1000rpm

i) Rotor voltage Er=sE

S=Ns-N/Ns x100 = {[1000-1000]/1000} x 100 = 0

Since, s=0 Er=0

Ii) Rotor Frequency fr=sfs

Since, s=0

fr=0

Iii) Rotor slip s=0

Q11) The power input to a rotor of a 400V, 50hz,6-pole, 3-phaseinduction motor is 15kW. The slip is 4%. Calculate i) frequency of rotor current. i)rotor speed. Iii)rotor cu loss iv) rotor resistance/phase if rotor current is 50A?

Sol: i) frequency of rotor current=sf=0.04 x 50=2Hz

Ii) Rotor speed N=(1-s) Ns=(1-0.04) x 1000=960rpm

Ns=120f/P=120 x 50/6=1000rpm

Iii) rotor cu loss=s x rotor input=0.04 x 15=0.6kW

Iv)rotor cu loss/phase=0.6 x 1000/3=200W

602R2=200

R2=0.055Ω

Q12) A 230V shunt motor on no load runs at 1000rpm and takes 5A. Armature and shun field resistances are 0.3 and 250ohm. Calculate the speed when load taking a current of 50A. The armature reaction weakens the field by 4%.

Sol: Ish= 230/250=0.92A

Ia1=5-0.92=4.08A

Ia2=50-0.92=49.08A

Eb1=230-(4.08 x 0.3) =228.776V

Eb2=230-(49.08 x 0.3) =215.276V

==940.99rpm

Q13) A 250V shunt motor has an armature resistance of 0.5 ohm and takes an armature current of 35A on a certain load. What amount the main flux will reduce to raise the speed by 50% if torque is constant.

Sol: As we already know