Solution Electrical Numerical 9

Numerical 9: A 440V shunt motor takes an armature current of 30A at 700 rev/min. The armature resistance is 0.7ohm. If the flux is suddenly reduced by 20 per cent, to what value will the armature current rise momentarily? Assuming unchanged resisting torque to motion, what will be the new steady values of speed and armature current? Sketch graphs showing armature current and speed as functions of time during the transition from initial to final, steady-state conditions.

Solution: Given Voltage = 440V

Armature current, $\displaystyle \mathrm{I_{a}}$ = 30A

Speed, N = 750 RPM

Armature resistance, $\displaystyle \mathrm{R_{a}}$ = 0.7ᘯ

Reduced flux, $\displaystyle \mathrm{ф_{1}}$ = 0.8 ф

Momentarily Raised Armature current = $\displaystyle \mathrm{I_{a1}}$ = ?

Steady state Armature current = $\displaystyle \mathrm{I_{a2}}$ = ?

Steady state Speed $\displaystyle \mathrm{N_{1}}$ = ?

Voltage = Back emf + (Armature current x Armature resistance)

V = $\displaystyle \mathrm{E_{b}+I_aR_a}$ 

440 = $\displaystyle \mathrm{E_{b}+30\times 0.7}$

$\displaystyle \mathrm{E_{b}}$ = 419V

We know,

Back emf is proportional to the product of RPM and Flux

Given The RMP doesn't change during the momentarily change in flux as resisting torque to motion is unchaged.

thus, 

$\displaystyle \mathrm{\frac{E_{b1}}{E_b}=\frac{RPM\, \phi_1}{RPM\, \phi}}$ or $\displaystyle \mathrm{\frac{E_{b1}}{E_b}=\frac{ \phi_1}{ \phi}}$  or $\displaystyle \mathrm{\frac{E_{b1}}{419}=\frac{0.8 \phi}{ \phi}}$ 

thus, $\displaystyle \mathrm{E_{b1}=335.2\, V}$ 

Again using 

V = $\displaystyle \mathrm{E_{b1}+I_{a1}R_a}$

440 = $\displaystyle \mathrm{335.2+I_{a1}0.7}$

thus, $\displaystyle \mathrm{I_{a1}=149.71\, A}$ 

Steady State Condition

Given Torque is constant 

we know that the Torque is proportional to the product of flux and Armature current.

Thus, To find steady state armature current:- $\displaystyle \mathrm{\frac{Torque\ initial}{Torque\ final}=\frac{\phi I_a}{\phi_1\times I_{a2}}}$ or $\displaystyle \mathrm{1=\frac{\phi\times 30}{0.8\phi\times I_{a2}}}$ 

Thus, $\displaystyle \mathrm{I_{a2} = 37.5\, A}$ 

To find steady state Back emf

$\displaystyle \mathrm{V = E_{b2} + I_{a2}R_a }$ 

$\displaystyle \mathrm{440 = E_{b2} + 37.5\times 0.7 }$

$\displaystyle \mathrm{E_{b2} = 413.75\, V }$ 

To find steady state RPM

$\displaystyle \mathrm{\frac{E_b}{E_{b2}}=\frac{N\phi }{N_1\phi _1}}$  or $\displaystyle \mathrm{\frac{419}{413.75}=\frac{700\phi }{N_1\times 0.8\phi }}$ 

Thus, $\displaystyle \mathrm{N_1 =864.04\, RPM}$