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marysya [2.9K]
4 years ago
12

A permanent magnet DC motor has an armature resistance of 0.5 Ω and when a voltage of 120 V is applied to the motor it reaches a

steady‐state speed of rotation of 20rev/s and draws 40A. What will be:
(a) the power input to the motor?
(b) the power loss in the armature?
(c) the torque generated at that speed?
Engineering
1 answer:
BaLLatris [955]4 years ago
3 0

Answer:

12 N-m

Explanation:

The dc motor is operating at 24 V that is its terminal voltage V =24 V

Armature resistance  = 0.2 ohm

No load speed = 240 radian /sec

For motor we know that  as the motor is on no load so  so

Power developed in the motor

Now we know that power = torque× angular speed

So

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Answer:

A. 288,030.91 cy

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Explanation:

The natural material in the barrow properties are;

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The water content, w₁ = 5.5 %

The net section volume, V_T = 245,000 cy = 6,615,000 ft³

A.  \gamma _d = W_s/V_T

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V = 288,030.91 cy

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W_{w1} = (w₁/100) × W_s = (5.5/100) × 807030000 lbs = 44386650 lbs

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Therefore, 4,035,150 lbs of water must be removed

The density of water, ρ = 8.345 lbs/gal

Therefore, V = 4,035,150 lbs/(8.345 lbs/gal) = 483541.04254 gal  of water must be removed from the natural material

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