Electrical Machines Test 2

V_t = 230 V

I_a = 36 A, R_a = 0.278 Ω

DC generator:

E_g = V_t + I_a R_a

= 230 + (36) (0.278) = 240 V

\({\omega _m} = \frac{E}{{k\;{\phi _p}}} = \frac{{240}}{{\left( {212.21} \right)\left( {0.091} \right)}}\) 

= 113.09 rad/s = 1080 rpm

DC Motor:

E_b = V_t - I_aR_a

= 230 – (36) (0.278) = 220 V

\({\omega _m} = \frac{E}{{k\;{\phi _p}}} = \frac{{220}}{{\left( {212.21} \right)\left( {0.01} \right)}}\) 

= 103.67 = 990 rpm

At maximum power condition, δ = 90°,

\({I_a} = \frac{{{{\bar E}_f} - {{\bar V}_{an}}}}{{j{X_S}}}\)

\( = \frac{{11,500\angle 90^\circ - \frac{{13,800}}{{\sqrt 3 }}\angle 0^\circ }}{{20\angle 90^\circ }}\)

Developed torque \(\left( T \right)=\left( \frac{PZ}{2\pi A} \right)\phi {{I}_{a}}\)

P is number of poles

Z is number of conductors

ϕ is flux per pole

I_a is armature current

A is number of parallel paths

Total flux = 0.14 wb

⇒ Pϕ = 0.14 wb

Current in each conductor \(=\frac{{{I}_{a}}}{A}\)

Armature ampere conductors \(=Z\left( \frac{{{I}_{a}}}{A} \right)\)

= 4500

\(\Rightarrow T=\frac{0.14\times 4500}{2\pi }=100.26~N-m/r\)

Copper loss = I²_a(R_a + R_s) = 50² × (0.1 + 0.3) = 1000 W

Power loss in brush contacts = 2 × I_a = 2 × 50 = 100 W

Friction and windage losses = 200 W

Total power loss = 1000 + 100 + 200 = 1300 W

\(efficiency = \frac{{10000}}{{10000 + 1300}} = 88.49\%\)

\(syn.\;speed\;{N_S} = \frac{{120 \times 50}}{4} = 1500\;rpm\)

Full load speed given as 1440 rpm

\(\therefore \% slip = \frac{{{N_S} - N}}{{{N_s}}} \times 100 = \left( {\frac{{1500 - 1440}}{{1500}}} \right) \times 100 = 4\% \)

Rotor voltage frequency = s × f = 0.04 × 50 = 5%

2 cycles/sec

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

\(= \frac{{120 \times 50}}{6}\)

= 1000 rpm

Slip, \(s = \frac{{{N_S} - N}}{{{N_S}}} = \frac{{1000 - 960}}{{1000}}\)

= 0.04

Resistance of backward field \(= \frac{{\frac{{10}}{2}}}{{2 - s}}\)

\(= \frac{5}{{2 - s}} = 2.55{\rm{\Omega }}\)

Concept:

Active power is directly proportional to the load angle i.e. P ∝ δ

Reactive power is directly proportional to the voltage ∝ V

Application:

Machine A is supplying an active power to Machine B i.e.

⇒ δ_A > δ_B

Machine B is supplying reactive power to Machine A

1) The speed is constant, hence the load does not know that the line voltage has dropped. Therefore, the mechanical power will remain unchanged.

2) \(P = \frac{{{E_f}{V_t}}}{{{X_S}}}\sin \delta \)

P, E_f, and X_S are the same but V_t has fallen consequently sin δ must increase, which means δ increases.

3) As δ increases, the poles fall slightly behind their former position.

4) The terminal voltage is smaller than before, the motor internal voltage is bigger than the terminal voltage and as a result, the power factor will be less than unity and leading.

S_P = 2.4 ∠cos^-1 0.707

= P_P + jQ_P = 1.697 + j 1.697 MVA

The reactive power rating of synchronous motor.

Q_p = 1.697 MVAR

The motor is used to drive a new load of 0.405 MW and the motor is 91% efficient.

\({S_m} = \frac{{{P_m}}}{{0.91}} + j{Q_m}\) 

\(= \frac{{0.45}}{{0.91}} + j{Q_m}\) 

= 0.5 – j 1.697

MVA rating = 1.769 MVA

Concept:

The efficiency of the Transformer \((\eta ) = \frac{{XS\cos \phi }}{{XS\cos \phi + {P_i} + {X^2}{P_e}}}\)

Where, X = Fraction of load

S = Apparent power in kVA

Pi = Iron losses

Pcu = Copper losses

Maximum efficiency of transformer occurred at a fraction of load, \(X = \sqrt {\frac{{{P_i}}}{{{P_{cu}}}}}\)

In a transformer, a short circuit test is used to find copper losses and the open circuit is used to find core losses.

Calculation:

From the given table, iron losses (W_i) = 120 W

Copper losses (W_cu) = 600 W

kVA rating of the transformer at maximum efficiency \( = \sqrt {\frac{{{W_i}}}{{{W_{cu}}}}} \times kVA = \sqrt {\frac{1}{5}} \times 48 = 21.467\;kVA\)

Maximum efficiency, \(\eta = \frac{{21.467 \times {{10}^3} \times 0.8}}{{21.467 \times {{10}^3} \times 0.8 + 120 + 600{{\left( {\frac{1}{{\sqrt 5 }}} \right)}^2}}} \times 100 = 98.621\;\% \)

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%

Per phase power \(=\frac{30}{3}=10~kW\)

\({{I}^{2}}\left( {{r}_{1}}+r_{2}^{'} \right)=10\times {{10}^{3}}\)

\(\Rightarrow {{\left( 100 \right)}^{2}}\left( {{r}_{1}}+r_{2}^{'} \right)={{10}^{4}}\)

\(\Rightarrow {{r}_{1}}+r_{2}^{'}=1\)

Given, \({{r}_{1}}=r_{2}^{'}\)

\({{r}_{1}}=r_{2}^{'}=0.5~\text{ }\!\!\Omega\!\!\text{ }\)

\(Starting~torque~\left( {{T}_{s}} \right)=\frac{180}{2\pi {{N}_{s}}}\times {{I}^{2}}\times \frac{r_{2}^{'}}{s}\) 

At starting, slip(s) = 1

Synchronous speed (N_s) \(=\frac{120f}{P}=\frac{120\times 50}{4}=1500~rpm\)

\({{T}_{s}}=\frac{180}{2\pi \times 1500}\times {{\left( 100 \right)}^{2}}\times \frac{0.5}{1}\) 

starting capacitor is added to produce 90° phase difference b/w the I_a, I_m

\(\begin{array}{l} {{\rm{\Phi }}_{\rm{m}}} + {ϕ _{\rm{a}}} = 90\\ ta{n^{ - 1}}\left( {\frac{{{X_a} - {X_c}}}{{{R_a}}}} \right) + ta{n^{ - 1}}\left( {\frac{{{X_m}}}{{{R_m}}}} \right) = 90\\ ta{n^{ - 1}}\left( {\frac{{5 - {X_c}}}{7}} \right) + ta{n^{ - 1}}\left( {\frac{4}{{2.3}}} \right) = 90\\ {X_c} = 0.975 \end{array}\)

\({\tan ^ - }^1\left( {\frac{{5 - {X_c}}}{7}} \right) = 90^\circ - 60.10^\circ \)

\(\frac{{5 - {X_c}}}{7} = tan 29.89^\circ \)

Solving this we get

∴ C = 3.264 mf

Note:

To obtain the maximum torque at starting itself. Then

ϕ_m+ 2ϕ_a = 90⁰