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Consider a 350 V DC shunt motor whose nameplate power rating is 25 kW. The motor is taking 2.7 A of current from the source while running at no load. The armature resistance is 0.8 ohm, and field resistance is 500 ohm. If the total brush drop is 4 V, the total constant loss (Pconst​, in W) of the shunt motor is ______

User RasikaSam
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Final answer:

The total constant loss (Pconst) of a 350 V DC shunt motor, with a no-load current of 2.7 A, armature resistance of 0.8 ohm, and field resistance of 500 ohm with a total brush drop of 4 V, is calculated to be 261.632 watts.

Step-by-step explanation:

To calculate the total constant loss (Pconst) of the shunt motor, we need to consider the power consumed by the armature, the field, and the brushes when the motor is running at no load. The armature power loss (Pa) can be calculated using the formula Pa = Ia2Ra, where Ia is the armature current and Ra is the armature resistance. However, since the motor is running at no load, the armature current is approximately equal to the total current drawn from the source, which is given as 2.7 A. The armature resistance Ra is given as 0.8 ohms.

The field power loss (Pf) is calculated using Pf = V2/Rf, where V is the supply voltage and Rf is the field resistance. The power loss due to the brush drop can be included in the armature loss or separately considered as Pb = Ia * Brush drop.

Combining these losses, the total constant loss (Pconst) is given by:

Pconst = Pa + Pf + Pb

Pconst = (2.7 A)2 * 0.8 Ω + (350 V)2 / 500 Ω + 2.7 A * 4 V

After calculating the respective values:

Pa = 2.7 A * 2.7 A * 0.8 Ω = 5.832 W

Pf = (350 V * 350 V) / 500 Ω = 245 W

Pb = 2.7 A * 4 V = 10.8 W

Therefore, the total constant loss is:

Pconst = 5.832 W + 245 W + 10.8 W = 261.632 W

User ZoogieZork
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