Electrical Machines Test 8

Fundamental voltage (E₁) = 230 V

Fifth harmonic voltage (E₅) = 230 × 0.06 = 13.8 V

In star connector, the third harmonic components of the three phases cancel out at the line terminals.

Hence, the line emf is composed of the fundamental and the fifth harmonic only.

Phase voltage (E_ph) \(= \sqrt {E_1^2 + E_5^2} = \sqrt {{{\left( {230} \right)}^2} + {{\left( {13.8} \right)}^2}} = 230.4\)

RMS value of line emf \(= \sqrt 3 \times 230.4 = 399\;V\)

From the emf equation of an alternator,

E ∝ ϕf

And ϕ ∝ I_f

⇒ E ∝ I_f f

\(\Rightarrow \frac{{{E_1}}}{{{E_2}}} = \frac{{{I_{f1}}}}{{{I_{f2}}}} \times \frac{{{f_1}}}{{{f_2}}}\)

E₁ = 360 V, I_f1 = 3.6, f₁ = 60 Hz

I_f2 = 2.4, f₂ = 40 Hz

\(\Rightarrow \frac{{360}}{{{E_2}}} = \frac{{3.6}}{{2.4}} \times \frac{{60}}{{40}}\)

⇒ E₂ = 160 V

Unsaturated synchronous reactance,

\({X_{s\left( {unsat} \right)}} = \frac{V}{{{I_{SC}}}} = \frac{{\frac{{30 \times {{10}^3}}}{{\sqrt 3 }}}}{{10 \times {{10}^3}}} = 1.732\;{\rm{\Omega }}\)

\({X_{s\left( {sat} \right)}} = \frac{{{V_{OC}}}}{{{I_{SC}}}} = \frac{{\frac{{25 \times {{10}^3}}}{{\sqrt 3 }}}}{{10 \times {{10}^3}}} = 1.44\;{\rm{\Omega }}\)

Base voltage \(\left( {{V_b}} \right) = \frac{{25}}{{\sqrt 3 }}kV\)

Base current \(\left( {{I_b}} \right) = \frac{{750 \times {{10}^6}}}{{\sqrt 3 \times 25 \times {{10}^3}}} = 17.32\;kA\)

Base impedance \(\left( {{Z_b}} \right) = \frac{{25}}{{\sqrt 3 \times 17.32}} = 0.833\)

\({X_{s\left( {unsat} \right)}} = \frac{{1.732}}{{0.833}} = 2.078\;pu\)

\({X_{c\left( {sat} \right)}} = \frac{{1.4434}}{{0.833}} = 1.732\;pu\)

Since the generator is star connected, so the direct current flows through two windings. Thus, the armature resistance is

\(2{R_a} = \frac{{{V_{dc}}}}{{{I_{dc}}}}\)

\( \Rightarrow {R_a} = \frac{{{V_{dc}}}}{{2{I_{dc}}}} = \frac{{10}}{{2\left( {25} \right)}} = 0.2\;{\rm{\Omega }}\)

Synchronous impedance \(= \frac{{{V_{OC}}}}{{{I_{SC}}}}\)

\(\Rightarrow {Z_s} = \frac{{\frac{{8000}}{{\sqrt 3 }}}}{{800}} = 5.77\;{\rm{\Omega }}\)

Voltage/phase \(\left( {{V_{ph}}} \right) = \frac{{6600}}{{\sqrt 3 }} = 3810.5\;V\)

Resistive drop = 3% of 3810.5 = 114.315 V

Full load current, \({I_a} = \frac{{\frac{{6000 \times {{10}^3}}}{{\sqrt 3 }}}}{{6600}} = 524.86\;A\)

Resistive drop (I_a R_a) = 114.315 V

\( \Rightarrow {R_a} = \frac{{114.315}}{{524.86}} = 0.218\;{\rm{\Omega }}\)

\({X_s} = \sqrt {Z_s^2 - R_a^2} = \sqrt {{{\left( {5.77} \right)}^2} - {{\left( {0.218} \right)}^2}} = 5.76\;{\rm{\Omega }}\)

\({E_0} = \sqrt {{{\left( {V\cos \phi + {I_a}{R_a}} \right)}^2} + {{\left( {V\sin \phi + {I_a} \times s} \right)}^2}} \)

\( = \sqrt {{{\left( {3810.5 \times 0.8 + 114.3} \right)}^2} + {{\left( {3810.6 \times 0.6 + 525 \times 5.76} \right)}^2}} \)

= 6180.77 V

% Regulation, \(\frac{{E - V}}{V} \times 100\)

\(= \frac{{6180.77 - 3810.5}}{{3810.5}} \times 100 = 62.2\)

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

\(\Rightarrow 375 = \frac{{120 \times f}}{{16}}\)

⇒ f = 50 Hz

Flux per pole (ϕ) = 0.03 Wb

Slots per pole \(= \frac{{144}}{{16}} = 9\)

Slot angle \(\left( \beta \right) = \frac{{180}}{9} = 20^\circ \)

Slots per pole per phase \(\left( m \right) = \frac{9}{3} = 3\)

\({K_d} = \frac{{\sin \frac{{m\beta }}{2}}}{{m\sin \frac{\beta }{2}}} = \frac{{\sin \left( {\frac{{3 \times 20}}{2}} \right)}}{{3\sin \left( {\frac{{20}}{2}} \right)}} = 0.96\)

Total number of conductors \(\left( z \right) = \frac{{144 \times 10}}{3} = 480\)

Number of turns per phase \(\left( T \right) = \frac{{480}}{2} = 240\)

E_ph = 4.44 × 1 × 0.96 × 50 × 0.03 × 240 = 1534 V

Line emf. \({E_L} = \sqrt 3 \;\;{E_{ph}} = 2657.76\;V\)

Per phase voltage, \({V_t} = \frac{{6600}}{{\sqrt 3 }} = 3810.5\;V\)

Per phase armature current,

\({I_a} = \frac{{\left( {3000} \right)\left( {{{10}^3}} \right)}}{{\sqrt 3 \left( {6600} \right)}} = 262.43\;A\)

Percentage reactance, \({X_s} = \frac{{{X_s}\;in\;ohms}}{{{V_t}/{I_a}}} \times 100\)

\(= \frac{{20}}{{100}} \times \frac{{3810.5}}{{262.43}} = 2.9\)

At no load, synchronizing power per mechanical degree,

\({P_s} = m\frac{{dp}}{{d\delta }} \cdot \frac{{\pi p}}{{360}} = \frac{{3\;{V_t}{E_f}}}{{{X_s}}}\cos \delta \cdot \frac{{\pi p}}{{360}}\)

\(= \frac{{3 \times {{\left( {3810} \right)}^2}}}{{2.90}} \times \frac{{\pi \left( 8 \right)}}{{360}} = 1048.36\;kW\)

Synchronizing torque,

\({T_s} = \frac{{60}}{{2\pi \times {N_s}}} \cdot {P_s}\)

\(= \frac{8}{{2\pi \left( {100} \right)}}\;\left( {1048.36 \times {{10}^3}} \right)\)

= 13, 348.13 N-m

Active power of total load is,

\({P_L} = 300 + 150 = 450\;kW\) 

Reactive power of total load is,

\({Q_L} = 300 \times \tan \left( {{{\cos }^{ - 1}}\left( {0.8} \right)} \right) + 150\tan \left( {{{\cos }^{- 1}}\left( {0.6} \right)} \right)\) 

= 300 × 0.75 + 150 × 1.33 = 424.5 kVAR

Rating of machine 1, 250 kW at 0.85 pf lag:

P₁ = 250 kW,

Q₁ = 250 × tan (cos^-1 (0.85)) = 154.93 kVAR

P₂ = P_L - P₁ = 450 - 250 = 200 kW

Q₂ = Q_L - Q₁ = 424.5 - 154.93 = 269.57 kVAR

Q₂ = P₂ tan ϕ

⇒ tan ϕ = 1.34785

Z_s = 10 Ω, r_a = 1 Ω, E_f = V_t = 500 V

The maximum power output

\({P_{og\left( {{\rm{max}}} \right)}} = \frac{{{E_f}{V_t}}}{{{Z_s}}} - \frac{{V_t^2}}{{Z_s^2}}.{r_a} = \frac{{500 \times 500}}{{10}} - \frac{{{{500}^2}}}{{{{10}^2}}} \times 1 = 22.5\;kW\)