Electrical Machines Test 6

General equation for the frequency of an induction frequency changer is

\(f = {f_s}\left( {1 + \frac{{{N_r}}}{{{N_s}}}} \right)\)

+ve for opposite direction of rotation

-ve for same direction of rotation

\(f = {f_s}\left( {1 + \frac{{{N_r}}}{{{N_s}}}} \right)\)

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

\(f = 50\;\left( {1 + \frac{{1500}}{{1000}}} \right) = 125\;Hz\)

We know that,

\(\begin{array}{l} \frac{{{T_{st}}}}{{{T_{fl}}}} = {\left( {\frac{{{I_{st}}}}{{{I_{fl}}}}} \right)^2} \times {s_{fl}} = {\left( {\frac{{x\;{I_{sc}}}}{{{I_{fl}}}}} \right)^2} \times {s_{fl}}\\ \Rightarrow \frac{{1.5\;{T_{fl}}}}{{{T_{fl}}}} = {x^2}{\left( {\frac{{8{I_{fl}}}}{{{I_{fl}}}}} \right)^2} \times 0.04 \end{array}\)  

⇒ x = 0.7654

One complete oscillation of galvanometer corresponds to one cycle of rotor frequency.

Rotor frequency \({f_r} = s{f_1} = \frac{{144}}{{60}} = 2.4\;Hz\)

Slip, \(s = \frac{{{f_r}}}{{{f_1}}} = \frac{{2.4}}{{60}} = 0.04\)

Rotor speed = N_r = N_s (1 – s)

\({N_s} = \frac{{120 \times f}}{P}\)

\({N_r} = \frac{{120 \times 60}}{4}\left( {1 - 0.04} \right)\)

= 120 × 15 × 0.96

Concept:

The ratio of maximum torque to full load torque is

\(\frac{{{T_m}}}{{{T_{fL}}}} = \frac{{s_m^2 + s_{fL}^2}}{{2\;{s_m}{s_{fL}}}}\)

Where s_m is the slip at maximum torque.

s_fl is the slip at full load torque.

Slip at maximum torque, \({s_m} = \frac{{{R_2}}}{{{X_2}}}\)

R₂ is the rotor resistance

X₂ is the rotor reactance

Calculation:

Given that, Number of poles (P) = 24

Frequency (f) = 50 Hz

Synchronous speed \(\left( {{N_s}} \right) = \frac{{120\;f}}{P}\)

\( \Rightarrow {N_s} = \frac{{120\; \times \;50}}{{24}} = 250\;rpm\)

Rotor resistance (R₂) = 0.016 Ω

Rotor reactance (X₂) = 0.265 Ω

Slip at maximum torque, \({s_m} = \frac{{0.016}}{{0.265}} = 0.06\)

Speed corresponding to full load torque (N_r) = 247 rpm

Slip at full load torque \(\left( {{s_{fL}}} \right) = \frac{{250 - 247}}{{250}} = 0.012\)

Synchronous speed, \({N_S} = \frac{{120\;f}}{P}\)

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

N_r = (1 - s) N_s = (1 – 0.04) 1000 = 960 rpm

Output power \(= {T_{sh}} \times \frac{{2\pi N}}{{60}}\)

\(= 2\pi \times \frac{{960}}{{60}} \times 149.3 = 15\;kW\)

Friction and windage losses = 200 W

Rotor gross output = 15,200 W

Rotor input \(= 15,200 \times \frac{{1000}}{{960}} = 15,833\;W\)

Stator cu and iron losses = 1620 W

Stator input = 15,833 + 1620 = 17,453 W

Efficiency \(= \frac{{Output}}{{input}} = \frac{{15000}}{{17453}} \times 100\)

= 85.94%

X₂ = 2πfL

X₂ = 2π × 50 × 3.61 × 10^-3 = 1.13 Ω

Transformation ratio,\(k = \frac{1}{{3.6}}\)

\(X_2' = \frac{{{X_2}}}{{{k^2}}} = \left( {1.13} \right){\left( {3.6} \right)^2} = 14.7\;{\rm{\Omega }}\)

R₂ = 0.1 Ω/phase

\(R_2' = \frac{{{R_2}}}{{{k^2}}} = {\left( {3.6} \right)^2} \times 0.1 = 1.3\;{\rm{\Omega }}\)

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

\({T_{st}} = \frac{{180}}{{2\pi {N_s}}} \cdot \frac{{{V^2}R_2'}}{{{{\left( {R_2'} \right)}^2} + {{\left( {X_2'} \right)}^2}}}\)

\(= \frac{{180}}{{2\pi \times 1000}} \times \;\frac{{{{\left( {\frac{{3000}}{{\sqrt 3 }}} \right)}^2} \times \;1.3}}{{\left( {{{1.3}^2}\; + \;{{14.7}^2}} \right)}} = 513\;Nm\)

\(Resistance\;per\;phase = \frac{{0.04}}{2} = 0.02{\rm{\Omega }},\;{S_{fl}} = 0.02\) 

Torque ∝ N²

\(T \propto \frac{{S{V^2}}}{{{R_2}}}\left( {0\;to\;{S_{mT}}} \right)\) 

\(T \propto \frac{{{R_2}}}{S}\left( {{S_{mT}}\;to\;1} \right)\) 

\({N_s} = \frac{{120 \times 50}}{{12}} = 500\;rpm\) 

\(S_{Fl}' = \frac{{500 - 350}}{{350}} = 0.3\) 

\(\frac{S}{{{R_2}}} \propto {\omega ^2}\) if voltage is not changing

N₁ = 500 (1 – 0.02) = 490 rpm

\(\frac{{{S_1}}}{{{R_1}}} \times \frac{{{R_2}}}{{{S_2}}} = {\left( {\frac{{{N_1}}}{{{N_2}}}} \right)^2}\) 

\(\frac{{0.02}}{{0.02}} \times \frac{{0.02 + {R_{ext}}}}{{0.3}} = {\left( {\frac{{490}}{{350}}} \right)^2}\) 

R_ext = 0.568 Ω

Given that, T_m = 2.5 T_fl

T_st = 1.5 T_fl

\(\Rightarrow \frac{{{T_m}}}{{{T_{st}}}} = \frac{{2.5}}{{1.5}}\)

\(\Rightarrow \frac{{1\; +\; {a^2}}}{{2a}} = \frac{5}{3}\)

⇒ 3a²  - 10 a + 3 = 0

\(\Rightarrow a = \frac{1}{3}\)

\(\Rightarrow \frac{{{R_2}}}{{{X_2}}} = \frac{1}{3} \Rightarrow {R_2} = \frac{{{X_2}}}{3}\)

When full load slip is 0.03

\(\frac{{{T_{fL}}}}{{{T_{in}}}} = \frac{{2{s_m}{s_{fL}}}}{{s_m^2\; + \;s_{fL}^2}}\)

\(\Rightarrow \frac{1}{{2.5}} = \frac{{2{s_m}\left( {0.03} \right)}}{{s_m^2\; + \;{{\left( {0.03} \right)}^2}}}\)

\(\Rightarrow s_m^2 - 0.15\;{s_m} + 0.0009\) = 0

⇒ s_m = 0.1437

\(\Rightarrow \frac{{R_2'}}{{{X_2}}} = 0.1437\)

\(\Rightarrow R_2' = 0.1437\;\;{X_2}\)

% reduction in rotor resistance is