Question

A permenant magnet DC generator has a k value of 2 Wb and an RA of 5 . The prime mover rotates the armature at a speed of 1200 rpm. Determine the no-load voltage, the full-load voltage, and the torque required by the prime mover at full load if the full-load power required by the load is 120 W (assumming most of the power is delivered to the load) neglecting rotational losses.

Answer 1

The no-load voltage of the DC generator is 40 V, the full-load voltage is 25 V, and the torque required by the prime mover at full load is 0.955 N·m.

Step-by-step explanation:

To find the no-load voltage of a permanent magnet DC generator, we use the formula for electromotive force (emf), which is emf = k × n, where k is the constant (2 Wb in this case) and n is the rotational speed in revolutions per second. Since the generator operates at 1200 rpm, we need to convert this to revolutions per second by dividing by 60. Therefore, n = 1200 / 60 = 20 rps. The no-load voltage will then be emf = 2 Wb × 20 rps = 40 V.

For the full-load voltage, we need to account for the voltage drop across the armature resistance RA. The full-load current I can be found using the formula I = P / V, where P is the power (120 W) and V is the no-load voltage (40 V). Thus, I = 120 W / 40 V = 3 A. The voltage drop is then RA × I = 5 Ω × 3 A = 15 V. So, the full-load voltage is 40 V - 15 V = 25 V.

To calculate the torque τ required by the prime mover at full load, we first find the power delivered to the load (120 W) and use the relation P = τ × ω, where ω is the angular velocity in rad/s. The angular velocity ω = 2π × n = 2π × 20 rps. Therefore, ω = 125.66 rad/s and τ = P / ω = 120 W / 125.66 rad/s = 0.955 N·m.