Electrical Drives

T_L=2.78 \times \omega
After neglecting rotational losses,
(T_{sh}=T_a)=T_L
At rated load,
\begin{aligned} T_a&=500Nm=2.78 \times \omega \\ \omega _r&=179.85 \\ \omega _r&\simeq 180 \text{ rad/sec} \end{aligned}

\begin{aligned} V_a &=E_b+I_aR_a \\ 150&=E_b+20 \times 1 \\ E_b&= 130V\\ \text{For half} &\text{ the rated torque,} \\ I_a&=10A \\ V_a&=130+10 \times 1=140 \\ D&=\frac{140}{200}=0.7 \end{aligned}

Let torque developed by the motor =T_m
Load torque =T_L
Accelerating torque,T_a=T_m-T_L
At point A,
T_m=T_L
\Rightarrow \;\;T_a=T_m-T_L=0
So, no accelerating , hence the motor rums at speed N_A.
When a smal disturbance decreases speed to N_c
At point C,
T_m \gt T_L
\Rightarrow \;\;T_a \gt 0
So, motor will accelerate and comes to point A. When the disturbance increases speed to N_D.
At point D,
T_m \lt T_L
\Rightarrow \;\;T_a \lt 0
Motor will decelerate and comes at point A
So, motor is stable at point A.
When the motor is operating at point B.,
T_m = T_L
\Rightarrow \;\;T_a= 0
If a small disturbance decreases speed to N_E.
T_m \lt T_L
\Rightarrow \;\;T_a \lt 0
Motor will decelerate and motor speed will keep on decreasing.
If the disturbance increases speed to N_F.
At point F,
T_m \gt T_L
\Rightarrow \;\;T_a \gt 0
Motor will accelerate and motor speed will keep on increasing.
So, motor is unstable at point B.

Average output voltage of the converter,
V_0=\frac{2V_m}{\pi}\cos \alpha
Load current =I_0=10A
Back emf =E_b=150V
Armature resistance=R_a=2\Omega
Applying KVL,
\begin{aligned} &V_0-2I_0+150=0 \\ \Rightarrow \;& V_0=-150+2 \times 10\\ &=-130V\\ \Rightarrow \; & \frac{2V_m}{\pi} \cos \alpha =-130\\ \Rightarrow \; &\frac{2 \times \sqrt{2} \times 230}{\pi}\cos \alpha=-130 \\ \Rightarrow \;&\alpha =129^{\circ} \end{aligned}

For a separately excited dc motor
back emf =E_a=V_0-I_aR_a
Since, losses are neglected R_a can be neglected
So,
\begin{aligned} E_a&=V_0 \\ V_0&=E_a =k_a\phi N \;\;...(i)\\ V_0&\propto N \end{aligned}
At rated voltage V_0=440V and N=1500rpm, so, at half the rated speed \left ( \frac{N}{2} =750 \;rpm\right ) output voltage of the bridge (V_0) is 220V.
if I_a is the average value of armature current rms value of supply current will be
I_s=I_a\sqrt{\frac{2}{3}}
Power delivered to the motor
P_0=V_0I_a
Input VA to the thyristor bridge
S_{in}=\sqrt{3}V_sI_s
Input power factor
=\frac{P_0}{S_{in}}=\frac{V_0I_a}{\sqrt{3}V_sI_s}=\frac{220 \times I_a}{\sqrt{3} \times 440 \times I_a\sqrt{\frac{2}{3}}}=0.345

Solar cell emf = 350V
DC current I_{dc}=20A
Solar cell resistance
\begin{aligned} R_{cell}&=0.5\Omega \\ V_0&=\text{Voltage across inverter} \\ &=-(E-I_{dc}R_{cell}) \\ &=-(350-20 \times 0.5)=-340V \end{aligned}
The bridge acts as inverter,
Output voltage of 3-\phi fully controlled bridge
\begin{aligned} V_0&=\frac{3V_{ml} \cos \alpha }{\pi}\\ -340&=\frac{3V_{ml} \cos \alpha }{\pi}\\ -340&=\frac{3 \times 440\sqrt{2} \cos \alpha }{\pi}\\ \Rightarrow \; \alpha &=125^{\circ} \end{aligned}
Therefore, each thyristor will be reverse biased for a period of 55^{\circ}.

For a 3-\phi induction motor, stator voltage per phase is given by
\begin{aligned} V_1 &=\sqrt{2}\pi f_1 N_{pn_1}\phi K_{\omega 1} \\ \phi &\propto \frac{V_1}{f_1} \end{aligned}
To keep the \phi constant, constant V/f control is applied and voltage (V_1) varies linearly with frequency (f_1). For frequency above 50 Hz, such control is not possible because voltage can not be increased above rated voltage. Therefore, for frequency above 50 Hz, voltage is kept constant, During this control induction motor is said to be working in field weakling mode.

T_s= Starting torque developed by the motor =15 N-m
T_L= Load torque =7 N-m
T_a= Accelerating torque =T_s-T_l= 15-7=8 N-m
\alpha =Acceleration = 2 \; rad/sec^2
T_a=I\alpha
I= Moment of inertia =\frac{T_a}{\alpha }=\frac{8}{2}=4kgm^2

Speed changes from -1500 rpm to 500 rpm in 0.5 sec
So angular acceleration,
\begin{aligned} \alpha &=\frac{500-(-1500)}{0.5}\times \frac{2 \pi}{60}\\ &=418.88 \;rad/sec^2\\ \text{Torque}&=T=20N-m\\ T&=I\alpha \\ &\text{Moment of inertia}\\ I&=\frac{T}{\alpha }=\frac{20}{418.88}=0.048kgm^2 \end{aligned}

\begin{aligned} \text{Back emf }&=E_a=k\phi N\\ E_a&=k_bN \end{aligned}
where,
k_b= Back-emf constant =0.25 V/rpm
Average output voltage of 1-\phi half controlled rectifier =V
\begin{aligned} V &=\frac{V_m}{\pi} (1+ \cos \alpha ) \\ &= \frac{230\sqrt{2}}{2 \pi} (1+ \cos 30^{\circ})\\ \Rightarrow \; V &=96.6V \\ E_a&=V-I_aR_a \\ &= 96.6-5 \times 2=86.6V\\ \text{Speed}&=N=\frac{E_a}{k_b}=\frac{86.6}{0.25}=346.4V \end{aligned}