Final answer:
The question involves calculating the output torque and bhp for a DC shunt motor using the electromagnetic torque formula, adjusting for mechanical losses, and then converting the torque to bhp.
Step-by-step explanation:
The question pertains to the calculation of the output torque and power in brake-horsepower (bhp) for a 6-pole, lap wound DC shunt motor. The given motor parameters include a current of 400 A, a speed of 350 rpm, flux per pole of 80 mWb, and 600 armature turns, with a loss of 3% in torque due to windage, friction, and iron losses.
To calculate the output torque (Tout), we first find the electromagnetic torque (Te) using the following formula:
- Te = (N x I x Φ x Z x P) / (60 x 2π x A)
Where:
- N = speed in revolutions per minute (rpm)
- I = armature current (A)
- Φ = flux per pole (Wb)
- Z = total number of armature conductors
- P = number of poles
- A = number of parallel paths, which is equal to P for a lap wound armature
The output torque would then be:
To calculate the output power in brake-horsepower, we use:
- Power (bhp) = (Tout x N) / 5252
Note: The question does not provide enough details to complete the full calculations, such as the presence of a gearbox or any mechanical losses that are not mentioned. The formulae given are to be used with the given and assumed parameters for an ideal scenario.