Three Phase Induction Machines Practice Questions

Given,

Slip at maximum torque, $S_m=\frac{R'_2}{X'_2}=\frac{0.1}{0.92}=0.1087$ We know that,

Therefore, ratio of maximum torque to full-load torque $=\frac{T_{em}}{T_{fl}}= 1.949$

Frequency of voltage across slip ring = Frequency of voltage induced in the rotor winding. Let speed corresponding to stator frequency, $f=N_s$ . and speed corresponding to $f_r=N_r$ Then, frequency of rotor induced emf $=f+f_r$ $=f_1\left ( \text{corresponding to} \; f_1 =\frac{(N_s+N_r)P}{120}\right )$ Also, frequency of induced voltage across commutator = stator frequency $=f.$

$S=\frac{N_s-N_r}{N_s}$ For given system, slip=(2-s)(speed =negative) Below point P, $T_M-T_l \lt 0$ $\Rightarrow$ speed decreases $\therefore$ S increases $\therefore$ slip=(2-s) decreases Above point P, $T_m-T_l \gt 0$ $\Rightarrow$ speed increases $\therefore$ S decreases and (2-s) increasaes. Thus, system is stable near point P. POINT R: Below point R: $T_M-T_l \lt 0\;\Rightarrow \;$ speed decreases Above point R: $T_M-T_l \gt 0\;\Rightarrow \;$ speed increases Thus, system can't regain stability at point R. Therefore, R is unstable point. POINT S: Below point S: $T_M-T_l \gt 0\;\Rightarrow \;$ speed increases Above point S: $T_M-T_l \lt 0\;\Rightarrow \;$ speed decreases Thus, system can regain stability near point S.

Given , motor draws 12 MW at 0.6 pf lag i.e. $\cos \phi =0.6 \;\;\therefore \;\; \sin \phi =0.8$ $P=12MW Q=\frac{P}{\cos \phi }\times \sin \phi$ $\;\;\;=\frac{12}{0.6} \times 0.8 =16MVAR$ For improving the power factor to 0.8 lag, Q should be reduced i.e. $\phi$ reduces $\phi '=(16-Q)MVAR$ Where, Q is the reactive power delivered by the capacitor. Now, power factor, $\cos \phi '=0.8$ or $\phi '=\cos ^{-1}(0.8)=36.86^{\circ}$ from power traingle,

Reactive power delivered by the capacitor, Q=7.07 MVAR

Currently no explanation available

For negative sequence component,

At standstill, the rotor current is $I_2=\frac{E_2}{\sqrt{R^{2+}X^2}}$ As losses are zero

$\text{Slip}=\frac{N_s-N_r}{N_s}$ From the above formula slip depends upon: (1) Synchronous speed $(N_s)$ (2) Rotor speed $(N_r)$ as the shaft torque depends upon rotor speed therefore the slip also depends on shaft torque. And core-losses are independent of slip.

Stator and rotor magnetic field rotates at same speed. So difference in speed is zero. Speed of stator magnetic field is $N_s$ ( synchronous speed) $N_s=\frac{120f}{P}=\frac{120 \times 50}{6}=1000 \; \text{rpm}$ Rotor speed $=N_s(1-s)=950 \; \text{rpm}$ So, speed of rotor with respect to stator magnetic field $= N_r-N_s$ $=950-1000=-50 \; \text{rpm}$

As both the machines are coupled, rotor of both the machines rotate at same speed. Rotor speed of DC machine = 1600 rpm Rotor spedd of induction machine = $N_r$ rotor speed of DC machine $N_r$ =1600 rpm Speed of stator field in induction machine $=N_s$

Slip of rotor wrt stator field in induction machine $s=\frac{N_s-N_r}{N_s}=\frac{1500-1600}{1500}=-0.067$ As slip is negative, induction machine acts as a generator and the dc machine acts as a motor.

Starting current in induction motor can be as large as 5 to 6 times the full load current. As compared to starting current $(I_{st})$ , exciting current $(I_{m})$ is very small $(I_m \lt lt I_{st})$ Therefore, shunt branch can be neglected in the circuit model of the motor. $I_{st}=\frac{V}{(r_s+r_r)+j(X_s+X_r)}$ $|I_{st}|=\frac{V}{\sqrt{(r_s+r_r)^{2+}(X_s+X_r)^2}}A$

$\text{Slip}=\frac{90}{1500}=0.06$ Total power input to induction motor within $=1800-200=1600W$ Power output of induction motor $P_0=(1-s)P_{\in}$ $\;\;=1600(1-0.06)=1504 W$ Since all the losses are neglected. Induction motor power output = d.c. generator input = d.c. generator output = 1504 W $\therefore \;\;I^2R_{ex}=1504$ $\therefore \;\;I=\sqrt{\frac{1504}{10}}=12.26A$

Line to line supply voltage $=V_{l-l}=400V$ Current drawn by the motor, $I_L=50A$ at 0.8 pf Input power,

Air-gap power $(P_g)$ in the power that is transferred from the stator to the rotor via the air-gap magnetic field. $P_g=$ Input power -Core loss -Copper loss in stator $=27.71-1.5-1.2=25.01kW$

For max. torque slip $=s_m=\frac{R'_r}{x_{sm}+s'_{rm}}$ For starting torque, $s_m=1$ $x_{sm}+x'_{rm}=R'_r$ $2 \pi f_mL_s+0.2 \pi f_mL'_r=1$ (frequency at maximum torque) $f_m$ $=\frac{1}{2 \pi (L_s+L_r)}$

In const V/f control method, $\frac{V_1}{f_1}=\frac{400}{50}=8$ $\therefore \;\; V_2/f_2=8$ $\therefore \;\; V_2=f_2 \times 8$ $\;\;\;\;\;=16.7 \times 8=133.3$ volts

In the case of direction line starting,

Per unit starting current $I_{sc}=3.16 pu$

Accelerating torque $=T_a=T_e-T_L$ At point A, C, D $T_L=T_e$ $T_a=T_e-T_L=0$ if due to some disturbance speed decreases $T_e \gt T_L$ $\Rightarrow \;\;T_a=T_e-T_L \gt 0$ So, rotor accelerates and speed increases. If speed increases due to some disturbances $T_e \lt T_L$ $\Rightarrow \;\; T_a=T_e-T_L \lt 0$ So, rotor decelerates and speed decreases. So, point A, C, D are still stable At point B, $T_e=T_L$ $T_a=T_e-T_L=0$ If speed decreases due to some disturbance $T_e \lt T_L$ $T_a=T_e-T_L \lt 0$ So rotor decelerates, and rotor speed keeps on decreasing. If speed increases, $T_e \gt T_L$ $T_a=T_e-T_l \gt 0$ Rotor accelerates, and rotor speed keeps on increasing. So, point B is unstable.

Assuming rated torque $=T_{rated}$ Starting torque $T_s=150\% \; \text{of} \; T_{rated}=1.5T_{rated}$ Maximum torque, $T_m=300\% \; \text{of} \; T_{rated}=3.0T_{rated}$ as we know,

as we know, maximum torque occurs at

$s_m=3.73$ is rejected, as for this value of slip motor works in braking mode. Therefore, for motoring mode $s_m=0.2679$ or $26.79\%$

rated line to line voltage, $V_{l-l}=400V$ Phase to neutral voltage, $V_{p}=\frac{V_{l-l}}{\sqrt{3}}= \frac{400}{\sqrt{3}}V$ Synchronous speed at 50 Hz