Electrical Machines Test 1

Number of poles, P = 4

A = 2 (for wave wound armature A = 2)

Slots = 50

conductor per slot = 18

∴ Total no. of conductors Z = 18 × 50 = 900

N = 9000 rpm

E = 357 V

We know that, generated emf

\(E = \frac{{P\phi ZN}}{{60\;A}}\)

\(\Rightarrow 357 = \frac{{4\; \times \;\phi\; \times \;900\; \times \;9000}}{{60\; \times \;2}}\)

⇒ ϕ = 1.32 mWb

Inter-poles are small poles located in the inter-polar region. Inter-poles have two functions:

• To neutralize or cancel reactance voltage in the coil undergoing commutation.
• To remove inequality of the flux density under the pole tips.

These are connected in series with the armature through brushes. Since inter-pole windings carry armature current (high current) it must have higher cross section area and few turns to ensure low resistance and low losses.

Concept:

Per unit voltage regulation = R cos θ ± X sin θ

R and x are in pu.

+ for lagging loads

- for leading loads

\(VR = \frac{{{E_2} - {V_2}}}{{{E_2}}}\)

E₂ is no load voltage

V₂ is full load voltage.

Calculation:

p.u. voltage regulation = (0.02) (0.8) + (0.05) (0.6) = 0.046

For a primary voltage of 6600 V, the secondary no load voltage (E₂) is 440 V.

Change in secondary terminal voltage is (E₂ – V₂)

\(\frac{{{E_2} - {V_2}}}{{{E_2}}} = 0.046\)

⇒ E₂ – V₂ = 0.046 × 440 = 20.24 V

⇒ V₂ = 440 – 20.24 = 419.76 V

The synchronous speed, \({N_s} = \frac{{120f}}{P} = \frac{{120 \times 60}}{6} = 1200\;rpm\)

When the rotor speed is 1140 rpm, slip \( = \frac{{1200 - 1140}}{{1200}} = 0.05\)

When the rotor speed is 1000 rpm, slip \( = \frac{{1200 - 1000}}{{1200}} = 0.167\)

From the equivalent circuit of induction motor, if the value of \(\frac{{R_2^{'}}}{s}\) remains the same, the rotor current I­₂ and the stator current I₁ will remains the same and the machine develops the same torque.

If the rotational losses are neglected, the developed torque is the same as the load torque.

Therefore, for unity turns ratio,

\(\frac{{{R_2}}}{{{s_1}}} = \frac{{{R_2} + {R_{ext}}}}{{{s_2}}}\)

\( \Rightarrow \frac{{0.2}}{{0.05}} = \frac{{0.2 + {R_{ext}}}}{{0.167}}\)

Constant torque load requires constant electromagnetic torque T_e.

For constant torque T_e, air-gap power p_g remains constant because synchronous speed is constant.

For constant air gap power, rotor ohmic losses are proportional to slip.

Slip at full load \(= \frac{{1000 - 960}}{{1000}} = 0.04\)

Slip at reduced speed \(= \frac{{1000 - 400}}{{1000}} = 0.60\)

\(\frac{{Rotor\;ohmic\;loss\;at\;400\;rpm}}{{Rotor\;ohmic\;loss\;at\;full\;load}} = \frac{{0.6}}{{0.04}} = 15\)

\({N_s} = \frac{{120 \times f}}{P} = \frac{{120 \times 50}}{6} = 1000\;rpm\)

Full load slip corresponding to

\(S = \frac{{{f_r}}}{f} = \frac{2}{{50}} = 0.04\)

Thus, rotor speed N = (1 – S) N_s

= (1 – 0.04) × 1000

= 960 rpm

\({P_m} = T \times \frac{{2\pi N}}{{60}} = 200 \times \frac{{2\pi \times 960}}{{60}}\)

= 20106.192 + 600 = 20706.192 = 20.7 kW

The total power available to the rotor form stator \( = \frac{{20.7}}{{0.96}} = 21.56\;kW\)

Concept:

In a single-phase resistor split induction motor, the condition to get the maximum starting torque is

θm = 2θa

\( \Rightarrow {\tan ^{ - 1}}\left( {\frac{{{X_m}}}{{{R_m}}}} \right) = 2{\tan ^{ - 1}}\left( {\frac{{{X_a}}}{{{R_a}}}} \right)\)

In a capacitor start induction motor,

The condition to get the maximum starting torque is

θm + 2θa = 90°

\( \Rightarrow {\tan ^{ - 1}}\left( {\frac{{{X_m}}}{{{R_m}}}} \right) + 2{\tan ^{ - 1}}\left( {\frac{{{X_a}}}{{{R_a}}}} \right) = 90^\circ \)

Where θm is the power factor angle of the main winding

θa is the power factor angle of the auxiliary winding

Calculation:

Given that, power factor angle of main winding (θ_m) = 60°

To get maximum starting torque,

θ_m = 2θ_a

⇒ θ_a = 30°

R_a = 1 Ω, X_a = 1.16 Ω

Let the external resistance to be added is R Ω

Now, \({\tan ^{ - 1}}\left( {\frac{{{X_a}}}{{R + {R_a}}}} \right) = 30^\circ \)

\( \Rightarrow {\tan ^{ - 1}}\left( {\frac{{1.16}}{{R + 1}}} \right) = 30^\circ \)

\(P = \sqrt 3 \;{V_L}{I_L}\cos \phi\)

Mechanical power output = 9 kW

Core loss = 0.8 kW

V_L = 400

cos ϕ = 0.8

Electrical input = Mechanical power output + friction loss + core loss

= 9 kW + 2kW + 0.8 kW

⇒ P_in = 11.8 kW

\(\sqrt 3 \times {V_L} \times {I_L} \times \cos \phi = 11.8\;kW\)

\(\Rightarrow 11800 = \sqrt 3 \times 400 \times {I_L} \times 0.8\)

⇒ I_L = 21.29 A

\({I_a} = \frac{{V\angle 0 - E\angle - \delta }}{{{Z_s}\angle \theta }}\)

V = 1 pu

E = 1 pu

X_s = Z₁ = 1 pu

δ = 60°

θ = 90°

\({I_a} = \frac{{V\angle 0 - E\angle - \delta }}{{{Z_s}\angle \theta }} = \frac{{1\angle 0 - 1\angle - 60}}{{1\angle 90}} = 0.866 - 0.5\;j\)

I_a ∠ϕ = 1∠-30°

∴ pf = cos ϕ = 0.866 lead

\(E = \frac{{P\phi ZN}}{{60\;A}} = k\;\phi \omega\)

\(\omega = \frac{{2\pi N}}{{60}},k = \frac{{PZ}}{{2\pi A}}\;\)

\(T = \frac{{P\phi Z{I_a}}}{{2\pi A}} = k\;\phi {I_a}\)

At no load:

\({E_{b0}} = {V_s} = k\;\phi \omega\)

\(k\phi = \frac{V}{\omega } = \frac{{110}}{{2\pi\; \times \;\frac{{9000}}{{60}}}}\)

⇒ kϕ = 0.2626

At load:

τ = 0.6 Nm

⇒ 0.6 = kϕI_a

\({I_a} = \frac{{0.6}}{{k\phi }} = \frac{{0.6}}{{0.2626}} = 2.284\;A\)

At V₁ = 55V,

E₁ = V₁ - I_a R_a = 55 – 2.284 (1.5) = 51.574 V

E₁ = k ϕ ω₂

\({\omega _2} = \frac{{{E_1}}}{{k\phi }} = \frac{{51.574}}{{0.2626}} = 196.393\;rad/sec\)

\(N = \frac{{60\;\omega }}{{2\pi }} = 1875.4\;rpm\)

V = 220 V

R_a = 0.5 Ω

I_a1 = 50 A

Case 1:

E_b1 = 220 – (40) (0.5) = 200 V

Case 2:

E_b2 = 220 – (I_a2) (0.5) = 220 – 0.5 I_a2

N₂ = 1.5 N₁

E_b ∝ Nϕ

\(\Rightarrow \frac{{{E_{b1}}}}{{{E_{b2}}}} = \frac{{{N_1}}}{{{N_2}}} \times \frac{{{\phi _1}}}{{{\phi _2}}}\)

T remains constant

⇒ ϕ I_a = constant

⇒ ϕ₁ I_a1 = ϕ₂ I_a2

\(\Rightarrow {I_{a2}} = \frac{{40\;{\phi _1}}}{{{\phi _2}}}\)

\( \Rightarrow \frac{{200}}{{220 - 0.5\left( {40\frac{{{\phi _1}}}{{{\phi _2}}}} \right)}} = \frac{{{N_1}}}{{1.5\;{N_1}}} \times \frac{{{\phi _1}}}{{{\phi _2}}}\)

\(\Rightarrow 300\;\left( {\frac{{{\phi _2}}}{{{\phi _1}}}} \right) = 200 - 20\left( {\frac{{{\phi _1}}}{{{\phi _2}}}} \right)\)

Let \(\frac{{{\phi _2}}}{{{\phi _1}}} = x\)  

\(\Rightarrow 300\;x = 220 - \frac{{20}}{x}\)

⇒ 30 x² – 22 x+ 2 = 0

⇒ x = 0.627

⇒ ϕ₂ = 0.627 ϕ₁

⇒ ϕ₂ is 37.3% less than ϕ₁

The rotor current at the start is six times the rotor current at full load.

I_st = 6 I_fl

The synchronous speed, \({N_s} = \frac{{120f}}{P} = \frac{{120 \times 60}}{4} = 1800\;rpm\)

When the rotor speed is 1710 rpm, the full load slip \({s_{fl}} = \frac{{1800 - 1710}}{{1800}} = 0.05\)

\(\frac{{{T_{st}}}}{{{T_{fl}}}} = {\left( {\frac{{{I_{st}}}}{{{I_{fl}}}}} \right)^2} \times {s_{fl}}\)

\( \Rightarrow \frac{{{T_{st}}}}{{{T_{fl}}}} = {\left( 6 \right)^2} \times 0.05 = 1.8\)

\(\frac{{{T_{st}}}}{{{T_{max}}}} = \frac{{2{s_m}}}{{1 + s_m^2}}\)

\(\frac{{{T_{fl}}}}{{{T_{max}}}} = \frac{{2{s_{fl}}{s_m}}}{{s_{fl}^2 + s_m^2}}\)

From the above two expressions,

\(\frac{{{T_{st}}}}{{{T_{fl}}}} = \frac{{s_{fl}^2 + s_m^2}}{{{s_{fl}} + {s_{fl}}s_m^2}}\)

\( \Rightarrow 1.8 = \frac{{{{\left( {0.05} \right)}^2} + s_m^2}}{{0.05 + 0.05s_m^2}}\)

\( \Rightarrow s_m^2 + 0.0025 = 0.09 + 0.09s_m^2\)

⇒ s_m = 0.31

\(\frac{{{T_m}}}{{{T_{fl}}}} = \frac{{s_{fl}^2 + s_m^2}}{{2{s_{fl}}{s_m}}} = \frac{{{{\left( {0.05} \right)}^2} + {{\left( {0.31} \right)}^2}}}{{2 \times 0.05 \times 0.31}} = 3.18\)

Output power = 37 kW

Efficiency (η) = 0.9

\(\Rightarrow η= \frac{output}{input}=0.9\)

\(\Rightarrow input=\frac{37}{0.9}=41.111~kW\)

Losses = input – output

= 41.111 – 37 = 4111 W

Stator copper loss = rotor copper losses = iron loss = P

Mechanical losses = P_m

At no load, iron loss + mechanical losses = no load losses

Given that mechanical losses are one third of no-load losses.

\(\Rightarrow {{P}_{m}}=\frac{1}{3}\left( P+{{P}_{m}} \right)\)     → 1)

Total losses = 4111

⇒ P + P + P + P_m = 4111

⇒ 3P + P_m = 4111     → 2)

By solving both the equations, we get

P = 1174.6 W and P_m = 587.3 W

Rotor input = shaft output + rotor copper losses + mechanical losses

= (37 × 10³) + 1174.6 + 587.3

= 38761.9 W

\(Slip\left( s \right)=\frac{Rotor~copper~loss}{Rotor~input}\)

\(=\frac{1174.6}{38761.9}=0.03\)

Concept:

In a single-phase resistor split induction motor, the condition to get the maximum starting torque is

θm = 2θa

\( \Rightarrow {\tan ^{ - 1}}\left( {\frac{{{X_m}}}{{{R_m}}}} \right) = 2{\tan ^{ - 1}}\left( {\frac{{{X_a}}}{{{R_a}}}} \right)\)

In a capacitor start induction motor,

The condition to get the maximum starting torque is

θm + 2θa = 90°

\( \Rightarrow {\tan ^{ - 1}}\left( {\frac{{{X_m}}}{{{R_m}}}} \right) + 2{\tan ^{ - 1}}\left( {\frac{{{X_a}}}{{{R_a}}}} \right) = 90^\circ \)

Where θm is the power factor angle of the main winding

θa is the power factor angle of the auxiliary winding

Calculation:

Given that, R_m = 1 Ω, X_m = √3 Ω

Power factor angle of main winding, \({\theta _m} = {\tan ^{ - 1}}\left( {\frac{{{X_m}}}{{{R_m}}}} \right)\)

\( = {\tan ^{ - 1}}\left( {\frac{{\sqrt 3 }}{1}} \right) = 60^\circ \)

The condition to get the maximum starting torque is

θ_m + 2θ_a = 90°

⇒ 60° + 2θ_a = 90°

⇒ θ_a = 15°

R_a = 1 Ω, X_a = 1.2 Ω

Let the external capacitance to be added is X_C Ω

Now, \({\tan ^{ - 1}}\left( {\frac{{{X_C} - {X_a}}}{{{R_a}}}} \right) = 15^\circ \)

\( \Rightarrow {\tan ^{ - 1}}\left( {\frac{{{X_C} - 1.2}}{1}} \right) = 15^\circ \)

⇒ X_C = 1.468 Ω

\( \Rightarrow \frac{1}{{2\pi fC}} = 1.468\)

\( \Rightarrow C = \frac{1}{{2\pi \times 50 \times 1.468}} = 2.17\;mF\)

The machine draws 3 kW from the supply.

Power factor = cos ϕ = 1

3V_tI_a cos ϕ = 3 × 10³

\( \Rightarrow {I_a} = \frac{{3 \times {{10}^3}}}{{3 \times 120}} = 8.33\;A\)

E_f = V_t – jI_aX_s

= 120∠0° – 8.33 ∠0° 8 ∠90°

= 137.35 ∠-29°

Excitation voltage, E_f = 137.35 V/phase

The synchronous speed, \({N_s} = \frac{{120f}}{P} = \frac{{120 \times 60}}{4} = 1800\;rpm\)

Maximum torque will be developed at δ = 90°.

\({P_{max}} = \frac{{{E_f}{V_t}}}{{{X_s}}} = \frac{{3 \times 137.35 \times \frac{{208}}{{\sqrt 3 }}}}{8} = 6185.32\;W\)

\({T_{max}} = \frac{{{P_{max}}}}{{{\omega _{syn}}}} = \frac{{6185.32}}{{\frac{{2\pi }}{{60}} \times 1800}} = 32.8\;Nm\)

V = 1 pu, I_a = 1 pu

Pf = 1 (upf)

X_s = 1 pu, R_a = 0

Case-1:

E cos δ = V = 1 pu

\( \Rightarrow \frac{{EV}}{{{X_s}}}\sin \delta = \frac{{E\; \times \;1}}{1}\sin \delta \)

= E sin δ = 1 pu

\(\tan \delta = \frac{{E\sin \delta }}{{E\cos \delta \;}} = \frac{1}{1} = 1\)

⇒ δ = tan^-1 1

⇒ δ = 45°

E sin 45° = 1 ⇒ E = 1.414 V

Case-2:

At 0.8 leading reactive power

E’ sin δ’ = 1 pu

Hence, θ [V – E’ cos δ’] = -0.8 pu

E’ cos δ’ = 1.8 pu

\(\tan \delta ' = \frac{1}{{1.8}} = 0.55\)

⇒ δ’ = tan^-1 0.55 = 28.81° 

\(E' = \frac{1}{{\sin \left( {28.81^\circ } \right)}} = 2.075\)

sin ϕ = 0

cos ϕ = 1

I_a = 50 A

X_q = 2.5 Ω

R_a = 0 Ω

\(V = {V_{ph}} = \frac{{480}}{{\sqrt 3 }} = 277.12\;V\)

\(\tan {\rm{\Psi }} = \frac{{V\sin \phi\; +\; {I_a}{X_q}}}{{V\cos \phi\; - \;{I_a}{R_a}}}\)

\(\tan {\rm{\Psi }} = \frac{{277.12\left( 0 \right)\; + \;50\left( {2.5} \right)}}{{277.12\left( 1 \right)\; + \;50\left( 0 \right)}}\)

\(\tan {\rm{\Psi }} = \frac{{125}}{{277.12}} = 0.45\)

Ψ = tan^-1 (0.45) = 24.27°

δ = Ψ + ϕ

δ = 24.27°

For eliminating n^th harmonic coil span required is

\( = 180 \times \left( {\frac{{n - 1}}{n}} \right)\)

To eliminate 5^th harmonic coil span required is

\(= 180 \times \left( {\frac{4}{5}} \right) = 144^\circ\)

To eliminate 7^th harmonic coil span required is

\(= 180 \times \left( {\frac{6}{7}} \right) = 154.285^\circ\)

To eliminate both 5^th and 7^th harmonic, we should take average of both i.e.

\(= \frac{{144^\circ\; + \;154.28^\circ }}{2} = 149.14^\circ \approx 150^\circ\)

150° is \(\frac{5}{6}\) parts of 180° i.e. pole pitch