Answer to Question #204409 in Electrical Engineering for Mck

Question

A 2000V, 100KW, 9000 rpm series connected DC motor has an armature resistance of 6Ω and a negligible field resistance.

(a) Determine the armature current at the rated load.

(b) Determine the copper loss at the rated load.

(c) Determine the mechanical loss knowing that the no-load armature current is 7 A at a speed of 9000 rpm.

(d) Determine the full load mechanical torque delivered to the load at a speed of 9000 rpm.

(e) Determine the efficiency of the motor.

Answer 1

a) Armature current ,$I_2=I_L+I_{sh}$

$I_2=\frac{100*10^3}{2000}$

$=50A$

$I_{sh}=\frac{2000}6=333.33A$

$I_a= 50A+333.33A=383.33A$

b) Copper loss

$p=I^2R=333.33^2*6=881651.3334W$

C) Mechanical loss

No load current = $7A \space and \space N=9000rpm$

Power output= power development across the armature

copper losses across the armature

mechanical losses

Power output $I_{full}-I_{a(full)}^2 R_a- P\space mechanical$

P mechanical $=VI,_{full}-I_{a(full)}^2 R_a-P _{output}$

$=[(2000-7*50)*383.33]+383.33^2*6-100$

$1414.1458334 W$

d) $\frac{2\pi N}{60}T_{out}= Power \space output$

$100*10^3= \frac{2\pi N}{60}T_{out}$

$T_{out}=100*10^3* \frac{60}{2 \pi *9000}$

$=106 Nm$

$\eta =\frac{P_{out}}{P_{in}}*100=\frac{100*1000}{141414.8334}*100 =70.71 \%$