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# 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.

1
2021-06-10T03:46:50-0400

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 \\%"

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