Electrical Machines Test 9

Power delivered by the synchronous motor

\( \Rightarrow P = \frac{{EV}}{{{X_s}}}\sin \delta \)

From data, other parameters such as P, E and V are constant.

\(\begin{array}{l} \Rightarrow {X_s} \propto \sin \delta \\ \Rightarrow f \propto \sin \delta \\ \Rightarrow \frac{{{f_2}}}{{{f_1}}} = \frac{{\sin {\delta _2}}}{{\sin {\delta _1}}}\\ \Rightarrow \sin {\delta _2} = \sin 20^\circ \times \frac{{110}}{{100}}\\ \Rightarrow {\delta _2} = 22.09^\circ \end{array}\)

The input power

\({P_m} = \sqrt 3 \;{V_L}{I_L}\cos \theta = \sqrt 3 \times 480 \times 40 = 33.26\;kW\)

∴ P_out = P_in = 33.26 kW

\(\begin{array}{l} {N_m} = \frac{{120\;F}}{P} = \frac{{120 \times 60}}{4} = 1800\;rpm\\ T = \frac{{{P_{out}}}}{{{N_m} \times \frac{{2\pi }}{{60}}}} = \frac{{33.26}}{{1800 \times \frac{{2\pi }}{{60}}}} = 176.54\;N - M \end{array}\)

Power drawn by the synchronous motor per phase, P is

\(P=\frac{EV}{{{x}_{d}}}\sin \delta +\frac{{{V}^{2}}}{2}\left( \frac{1}{{{x}_{q}}}-\frac{1}{{{x}_{d}}} \right)\sin 2\delta \) 

If field current is reduced to zero, E = 0.

\(\Rightarrow P=\frac{{{V}^{2}}}{2}\left( \frac{1}{{{x}_{q}}}-\frac{1}{{{x}_{d}}} \right)\sin 2\delta\) 

It is maximum at δ = 45°

Maximum load that the synchronous motor can carry.

\(\Rightarrow {{P}_{max}}=\frac{{{V}^{2}}}{2}\left( \frac{1}{{{x}_{q}}}-\frac{1}{{{x}_{d}}} \right)\) 

\(=\frac{1}{2}\times {{\left( \frac{400}{\sqrt{3}} \right)}^{2}}\times \left( \frac{1}{6}-\frac{1}{8} \right)\) 

= 1111.11 W = 1.11 kW

Concept:

T ∝ sinδ

\(P = \frac{{{V_s}{V_R}}}{X}sin\delta\)

\(N = \frac{{120f}}{P}\)

Calculation:

Given that N₁ = 1200 rpm

f₁ = 60 Hz

T₁ = 398 N-m

At f₂ = 36 Hz

\({N_2} = \frac{{120f}}{P} = \frac{{120 \times 36}}{6} = 720\;rpm\)

T ∝ N²

\(\frac{{{T_1}}}{{{T_2}}} = {\left( {\frac{{{N_1}}}{{{N_2}}}} \right)^2}\)

\(\Rightarrow \frac{{398}}{{{T_2}}} = {\left( {\frac{{1200}}{{720}}} \right)^2}\)

⇒ T₂ = 143.28 N-m

Now,

\({P_1} = \frac{{2\pi {N_1}}}{{60}} \times {T_1} = 50.014\;KW\)

\({P_1} = \frac{{{V_S}{V_R}}}{X}sin{\delta _1} = \frac{{{V^2}}}{X}sin{\delta _1}\)

\(\Rightarrow sin{\delta _1} = \frac{{2.5 \times 50.014 \times {{10}^3}}}{{{{460}^2}}}\)

⇒ sinδ₁ = 0.59

\(\frac{{{T_1}}}{{{T_2}}} = \frac{{sin{\delta _1}\;}}{{sin{\delta _2}}}\)

\(\Rightarrow \frac{{398}}{{143.28}} = \frac{{0.59}}{{sin{\delta _2}}}\)

⇒ sinδ₂ = 0.2127

⇒ δ₂ = 12.5°

Synchronous reactance, (X_s) = 0.8 pu

V_t = 1.0 pu, R_a = 0, E_f = 1.2 pu

As E_f > V_t synchronous motor is working under leading power factor.

\({E_f} = \sqrt {{{\left( {{V_t}\cos \theta + {I_a}{R_a}} \right)}^2} + {{\left( {{V_t}\sin \theta - {I_a}{X_s}} \right)}^2}} \)

\(\Rightarrow 1.2 = \sqrt {{{\cos }^2}\theta + {{\left( {\sin \theta - 0.8} \right)}^2}}\)

⇒ 1.44 = cos² θ + sin² θ + 1.64 – 1.6 sin θ

⇒ sin θ = 0.125

⇒ cos θ = 0.9922 leading

Mechanical power developed by the motor = V_t I_a cos θ

= 1 × 1 × 0.9922 = 0.9922 pu

At maximum power input,

\({I_a}{Z_s} = \sqrt {\left[ {E_f^2 + V_t^2 - 2{E_f}{V_t}\cos \left( \delta \right)} \right]} \)

δ = θ_z + 2 α_z

Given that, V_t = 500, E_f = 600 V

Z_s = 0.4 + j 3 Ω = 3.026 ∠82.4°

θ_z = 82.4°

α_z = 90 - θ_z = 7.6°

δ = 82.4 + 2(7.6) = 97.6°

\({I_a}\left( {3.020} \right) = \sqrt {{{\left( {500} \right)}^2} + {{\left( {600} \right)}^2} - 2\left( {500} \right)\left( {600} \right)\cos \left( {97.6^\circ } \right)} \)

⇒ I_a = 274.38

Line current \(= 274.38 \times \sqrt 3 = 475.24\;A\)

\({V_t} = \frac{{6.6}}{{\sqrt 3 }} = 3.81\;kV\)

I_a = Power factor = 0.8 lead

I̅_a = I_a ∠36.9°

\({I_a} = \frac{{800}}{{\sqrt 3 \times 6.6 \times 0.8}} = 87.47\;A\)

 Z̅_s = 2 + j 20 = 20.1 ∠84.3° Ω

E_f = 3.81 – 20.1 ∠84.3 × 0.0875 ∠36.9

= 4.724 – j 1.504

E_f = 4.96 kV (phase)

Power input increases to 1200 kW and no change in excitation

\({P_e} = 3\;\left[ {\frac{{V_t^2}}{{{Z_z}}}\cos \theta - \frac{{{V_t}{E_f}}}{{{Z_s}}}\cos \left( {\delta + \theta } \right)} \right]\)

\(\Rightarrow \frac{{1200}}{{1000 \times 3}} = \frac{{{{\left( {3.81} \right)}^2} \times 0.1}}{{20.1}} - \frac{{3.81 \times 4.96}}{{20.1}}\cos \left( {\delta + 84.3^\circ } \right)\)

⇒ δ = 26.1°

\({I_a} = \frac{{3.81 - 4.96\angle - 26.1}}{{20.1\;\angle 84.3}} = 113\;\angle 22.1^\circ \)