Synchronous Machines: Important Solved Examples Problems

 

SOLVED EXAMPLES

 

EXAMPLE 1

A 3‒phase, 50 Hz Star connected alternator has 150 conductors per phase and flux per pole is 0.05 Wb. Find (i) emf generated per phase and (ii) Line voltage. Assume the winding to be full pitched and distribution factor to be 0.96.

Given Data:

z = 150 conductors, ϕ = 0.05 Wb, f = 50 Hz

To find: (i) EMF generated per phase, (ii) Line voltage

Solution

(i) EMF generated per phase

E_ph = 2.22 K_pK_dfϕz

For full pitch coil,

K_p = 1

K_d = 0.96

E_ph = 2.22 K_pK_dfϕz

E_ph = 2.22 × 1 × 0.96 × 50 × 0.05 × 150

E_ph = 799.2 Volts

(ii) Line voltage

In a star connected system,

E_L = √3 E_ph

= √3 × 799.2

E_L=1384.214 Volts

 

EXAMPLE 2

Find the flux per pole of a 3‒phase, 50 Hz, 10‒pole alternator. The number of armature conductors in series per phase is 372. The winding is star connected to give a line voltage of 10 kV. Assume K_p=1 and K_d=0.96.

Given Data:

f = 50 Hz, p = 10, Z=372 conductors, K_p = 1, K_d=0,96,

E_rms/phase = 2.22 K_pK_dfϕz,

Flux per pole, ϕ = E_rms/phase  / 2.22 K_pK_dfz

To find: Generated emf/phase.

Solution

Generated emf/phase,

E_rms/ph = E_L / √3

= 10000 / √3

= 5773.5 volts

ϕ = 5773.5 / (2.22 × 1 × 0.96 × 50 × 372)

 ϕ = 0.1456 Wb

 

EXAMPLE 3

The armature of an 8‒pole, 3‒phase, 50 Hz alternator has 18 slots and 10 conductors/slot. A flux of 0.05 Wb is entering the armature from one pole to other. Calculate the induced emf per phase.

Given Data: p=8, f = 50 Hz, No. of slots = 18, No. of conductors/slots = 10

To find: Induced emf per phase

Solution

∴ Total no. of conductors = No. of conductors/slot × No. of slots

= 10 × 18

= 180 conductors

No. of conductors/phase = 180 / 3 = 60 conductors/phase

Flux, ϕ = 0.05 Wb

Induced emf per phase E_ph = 2.22 K_pK_dfϕz

= 2.22 × 1× 1 × 50 × 0.05 × 60

E_ph=333 volts

 

EXAMPLE 4

A 3‒phase star‒connected, 50 Hz alternator has 96 conductors per phase and a flux/pole 0.1 Wb. The alternator winding has a synchronous reactance of 5Ω/phase and negligible resistance. The distribution factor for the stator winding is 0.96. Calculate the terminal voltage when three non‒inductive resistance of 10Ω/phase, are connected in star connection across the terminals.

Given Data: Z= 96 conductors, ϕ = 0.1 Wb, f = 50 Hz, X_S = 50Ω/phase, K_d=0.96, R_a = 10Ω/phase

To find: Terminal voltage

Solution

Assume that, K_p = 1

Generated emf/phase, E = 2.22 K_pK_dZfϕ

= 2.22 × 1 × 0.96 × 96 × 50 × 0.1

E = 1023 V

Impedance / phase, Z=√ [R_a²+X_S²]

=√ [(10)²+(5)²]

 Z = 11.18 Ω

Current / phase, I_a = E/Z

 = 1023 / 11.18

 I_a = 91.5 A

The above fig. shows the equivalent circuit for one phase of the alternator while the other figure shows its phasor diagram.

Terminal voltage/phase, V = √ [ E² + (I_aX_S)² ]

V= √[(1023)² ‒ (91.5×5)²]

V=915 V

In a star connected system,

Terminal line voltage V_L =√3V

V_L =√3×915

V_L = 1585 V

 

EXAMPLE 5

A 1500 kVA, 6.6 kV, 3‒phase, star ‒ connected alternator has a resistance of 0.50Ω/phase and a synchronous reactance of 50Ω/phase. Find its voltage regulation for (i) Unity p.f (ii) 0.8 lagging p.f and (iii) 0.8 leading p.f.

Given Data: P=1500 kVA, V_L = 6.6 kV = 6600 kV, R_a = 0.5Ω/phase, X_S = 50Ω/phase

To find: Line current, Armature current/phase, Voltage phase

Solution

Line current, I_L = P / √3V_L

= 1500×10³ / √3×6600

I_L=131 A

Armature current/phase, I_a = I_L = 131 A

 V_L = √3 V_ph

Voltage/phase, V_ph = V_L / √3

 = 6600 / √3

= 3810 V

I_aR_a = 131 × 0.5 = 65.5 V

I_aX_S = 131×5=655 V

(i) Unity p.f

For Unity p.f (Ref Fig.).

E = √[ (V+I_aR_a)² + (I_aX_S)² ]

= √[(3810+65.5)² + (655)²]

E = 3930 V

% Regulation = [(E‒V)/V] × 100

= (3930‒3810)/3810 × 100

% Regulation = 3.15%

(ii) 0.8 lagging p.f

For lagging p.f (Ref Fig.) current has been taken as the reference phasor.

cos ϕ = 0.8

sin ϕ  = 0.6

E = √[ ( Vcosϕ + I_aR_a)² + (Vsinϕ + I_aX_S)² ]

= √[ (3810×0.8 + 65.5)² + (3810×0.6 + 655)²]

= 4283 V

% Regulation = [(E‒V)/V] × 100

= (4283‒3810)/3810 × 100

= 12.4%

(iii) 0.8 p.f leading

For leading p.f (Ref Fig.), current has been taken as reference phasor.

E = √[(OB)² + (BC)²]

OB = OA + AB

= Vcosϕ + I_aR_a

BC = BD‒CD

= V sinϕ ‒ I_aX_s

E = √[ ( Vcosϕ + I_aR_a)² + (Vsinϕ ‒ I_aX_S)² ]

= [(3810×0.8 + 65.5)² + (3810×0.6 − 655)²]

= 3515 V

% Regulation = [(E‒V)/V] × 100

= (3515‒3810)/3810 × 100 = ‒7.7%

 

EXAMPLE 6

A 500 kVA, 3.3 kV, 3 phase, star connected alternator is found to give a short circuit current of 290 A at normal field current. Its effective winding resistance per phase is 0.7Ω. Estimate the full load voltage regulation for (i) 0.8 p.f lagging and (ii) UPF.

Given data: Power = 500 kVA, V = 3.3 kV, short circuit current, I = 290 A, R= 0.7Ω

Solution

V_ph = 3.3×10³ / √3

 = 1905.25 V

Z₂ = V_ph / I

= 1905.25 / 290

= 6.57 Ω

X_S = √[Z_S²‒R_a²]

= √[6.57² ‒0.7²]

X_S = 6.57 Ω

I_FL = (500×10³) / (3×1905.25)

I_FL = 87.47 A

0.8 p.f lagging

E = √[ ( Vcosϕ + I_aR_a)² + (Vsinϕ + I_aX_S)² ]

= √ [ (1905.25×0.8 + 87.47×0.7)² + (1905.25×0.6 + 87.47×6.57)² ]

= 2334.99 V

% Regulation = [(E₀‒V)/V] × 100

= ( 2334.99‒1905.25 / 1905.25) × 100

% Regulation = 22.55%

(ii) Unity power factor'

E₀ = √[ (V+I_aR_a)² + (I_aX_S)² ]

= √[ (1905.25+87.47 × 0.7)² + (87.47 × 6.57)² ]

 E₀ = 2047.8

% Regulation = [(E₀‒V)/V] × 100

= ( 2047.8‒1905.25 / 1905.25 ) × 100

% Regulation=7.48%

 

EXAMPLE 7

A 600 V, 600 KVA single phase alternator has R_a=0.3 Ohm. An exciting current of 5A produces an emf of 400 V, On open circuit and an armature current of 200 A on short circuit. Calculate: 1. Synchronous impedance and reactance, 2. The full load regulation with 0.8 p.f lagging

Given Data: The phase voltage V = 600 V; Capacity = 600 kVA;, R_a = 0.3 Ω, Exciting Current = 5 A, Open Circuit voltage Voc= 400 V, Short circuit current Isc=200 A

To find: 1. Zs, Xs, 2. % R (at 0.8 p.f lag)

Solution

1. Zs, Xs

Voc=400 V for I_f=5 A

Isc=200 A for I_f=5 A

Zs = (Voc/Isc)|_{Same If}

Zs = 400/200 = 2Ω

Now, Zs=√[R_a²+X_S²]

Zs = √(0.3)² +X_S²]

4 = (0.3)² + X_S²

X_S²=4-0.09

Xs=1.97≈ 20Ω

2. For lagging pf load (% R at 0.8 P.flag)

E_ph² = (Vcosϕ +IaRa)² + (Vsinϕ + I_aX_a)²

= (600×0.8 + (200)(0.3))² + (600×0.6 + (200) (2))²

E²_ph = 869200

E_ph = 932.309 V

% Reg = [ ( E_ph ‒ V_ph ) / V_ph ] × 100

= [ (932.309 – 600) / 600 ] × 100

% Reg=55.4%

 

EXAMPLE 8

From the following test results, determine the voltage regulation of a 2000 V single phase alternator delivering a load current of 100 A at 0.8 lagging pf.

Test results: An excitation of 2.5 A produces a current of 100 A in the stator winding on short circuit and an emf of 500 V on open circuit. Assume an effective resistance of 0.8.2Ω

Given Data: The phase voltage Vph = 2000 V, Ra=0.8Ω, Load Current I_L= 100 A, Isc=100 A for I_f=2.5 A, Voc=500 V for I_f=2.5 A

To find: % Regulation (at 0.8 p.f lag)

SOLUTION

Zs = (Voc/Isc)|_{Same If}

Zs = 500/100 = 5Ω

Now, Zs=√[R_a²+X_S²]

5 = √(0.8)² +X_S²]

X_S²=25-0.64

X_S = 4.93 Ω ‒ 5Ω

For lagging pf load,

E_ph² = (Vcosϕ +IaRa)² + (Vsinϕ + I_aX_s)²

= (2000×0.8 + (100) (0.8))² + (2000×0.6 + (100) (5))²

E²_ph = 5712400

E_ph = 2390.06 V.

% Reg = [ ( E_ph ‒ V_ph ) / Vph ] × 100

= [ (2390.06‒2000) / 2000 ] × 100

% Reg=19.5%