# A DC motor has the following specifications: 10 hp, 37.5 A, 230 V; flux / pole = 0.01 Wb, number of poles = 4, number of conductors = 666, number of p

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A DC motor has the following specifications: 10 hp, 37.5 A, 230 V; flux / pole = 0.01 Wb, number of poles = 4, number of conductors = 666, number of parallel paths = 2. Armature resistance = 0.267Ω. The armature reaction is negligible and rotational losses are 600W. The motor operates from a 230V DC supply. If the motor runs at 1000 rpm, the output torque produced in (in Nm)  is_______________

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Concept:

Back emf:

• In dc motor also generator action takes place.
• Because of this generator action the rotating conductor's cuts the flux and emf induced in the conductors.
• This induced emf is called back emf. It always opposes the supply voltage.

The back emf induced in the dc motor can be given by,

Eb = (ϕZNP) / (60 A)

Where,

ϕ = flux/pole

Z = total number of conductors

A = number of parallel paths

N = speed in RPM

P = number of poles

Calculation:

$E = \frac{{\phi ZNp}}{{60A}} = \frac{{0.01 \times 666 \times 4 \times 1000}}{{60 \times 2}} = 222 V$

Ia = (V - E) / Ra = (230 - 222) / 0.267

Ia = 29.96 A

Internal power ( Pd)  $= \;EI_a = 222\; \times \;29.96 = 6651.685 Watt$

Shaft Power output,

Pout = P- Rotational losses

$\begin{array}{l} {P_{out}} = 6651.685\;-\;600\; = 6051.685\\ T = \frac{{{P_{out}}}}{\omega } = \frac{{6051.685}}{{2\pi \times \frac{{1000}}{{60}}}} = 57.78\;Nm \end{array}$