Answer to Question #204409 in Electrical Engineering for Mck

Question #204409

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.

Expert's answer

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




"I_a= 50A+333.33A=383.33A"

b) Copper loss


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}"


"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 \\%"

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!


No comments. Be the first!

Leave a comment

New on Blog