Electrical Machines Test 4

Load power = 60 × 40 = 2400 W

Load current \(\left( {{I_L}} \right) = \frac{{2400}}{{100}} = 24\;A\)

Shunt field current \(\left( {{I_{sh}}} \right) = \frac{V}{{{R_{sh}}}} = \frac{{100}}{{50}} = 2\;A\)

Armature current (I_a) = I_L + I_sh = 24 + 2 = 26 A

The current per armature path \(= \frac{{26}}{4} = 6.5\;A\)

Filed current, \({I_{sh}} = \frac{{{V_t}}}{{{R_{sh}}}} = \frac{{250}}{{100}} = 2.5\;A\)

Armature voltage, \({I_a}{R_a} = 7.2\)

Armature current, \({I_a} = \frac{{7.2}}{{0.04}} = 180\;A\)

Load current, \(I = {I_a} - {I_{sh}} = 180 - 2.5 = 177.5\;A\)

Generated emf, \({E_g} = {V_t} + {I_a}{R_a} = 250 + 7.2 = 257.2\;V\)

\(\begin{array}{l}{E_g} = \frac{{\phi NZP}}{{60A}}\\257.2 = \frac{{\phi \times 1000 \times 700}}{{60}}\\ \Rightarrow \phi = 22\;mWb\end{array}\)

Resistance loss in field winding = 400 W

⇒ I²_f R_f = 400 W

Time constant = 0.2 sec

\(\Rightarrow \frac{{{L}_{f}}}{{{R}_{f}}}=0.2\) 

Energy stored in magnetic field \(=\frac{1}{2}{{L}_{f}}I_{f}^{2}\)

\(=\frac{1}{2}\times \frac{{{L}_{f}}}{{{R}_{f}}}\times I_{f}^{2}{{R}_{f}}\) 

\(=\frac{1}{2}\times 0.2\times 400=40~Joules\)

Total mmf required at rated load

= 3 × 1000 = 3000 AT

Total mmf required at no load

= 2 × 1000 = 2000 AT

The mmf to be supplied by series field winding

= 3000 – 2000 = 1000 AT

The line current at full load,

\({I_L} = \frac{{10 \times {{10}^3}}}{{230}} = 43.48\;A\)

Armature current = 43.48 + 2 = 45.48 A

Required series filed turns per pole

\(= \frac{{1000}}{{45.48}} = 21.98\) turns per pole.

Shunt field loss,\(\frac{{V_t^2}}{{{R_{sh}}}} = \frac{{{{125}^2}}}{{62.5}} = 250\;W\)

Load current,\({I_L} = \frac{{{P_{out}}}}{{{V_t}}} = \frac{{12500}}{{125}} = 100\;A\)

Shunt field current,\({I_{sh}} = \frac{{{V_t}}}{{{R_{sh}}}} = \frac{{125}}{{62.5}} = 2\;A\)

Armature current,\({I_a} = {I_L} + {I_{sh}} = 102\;A\)

Generated emf,\({E_g} = {V_t} + {I_a}{R_a} = 125 + \left( {102 \times 0.1} \right) = 135.2\;V\)

Developed power,\({P_d} = {E_g}{I_a} = 135.2 \times 102 = 13790.4\;W\)

Rotational losses,\({P_r} = {P_{in}} - {P_d} = \left( {20 \times 746} \right) - \left( {13790.4} \right) = 1129.6\;W\)

The constant loss is,\({P_{const}} = {P_r} + {V_{sh}}{I_{sh}} = 1129.6 + 250 = 1379.6\;W\)

For maximum efficiency, armature current

\({I_a} = \sqrt {\frac{{{P_{const}}}}{{{R_a}}}} = \sqrt {\frac{{1379.6}}{{0.1}}} = 117.4\;A\)

Load current I₁ for the load 500 kW \(= \frac{{500\; \times \;{{10}^3}}}{{500}} = 1000\;A\)

Load current I₂ for the load 250 kW \(= \frac{{250\; \times \;{{10}^3}}}{{500}} = 500\;A\)

E_g1 = V + I_a1 R_a = 500 (1000) (0.015) = 515 V

E_g2 = V + I_a2 R_a = 500 + (500) (0.015) = 507.5 V

E_g ∝ N

\(\Rightarrow \frac{{{E_{g1}}}}{{{E_{g2}}}} = \frac{{{N_1}}}{{{N_2}}}\)

\(\Rightarrow \frac{{{N_2}}}{{{N_1}}} = \frac{{{E_{g2}}}}{{{E_{g1}}}} = \frac{{507.5}}{{515}} = 0.09854\)

Percentage reduction in speed \(= \frac{{{N_1} - {N_2}}}{{{N_1}}} \times 100\)

\(= \left( {1 - \frac{{{N_2}}}{{{N_1}}}} \right) \times 100\)

= 1.45%