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A 250 V, dc shunt motor has an armature current of 20 A when running at 1000 rpm against some load torque. The armature resistance is 0.5 and brush contact drop is one volt per brush. By how much must the main flux be reduced to raise the speed by 50% if the developed torque is constant? Ignore effects of armature reaction and magnetic saturation.?
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A 250 V, dc shunt motor has an armature current of 20 A when running a...
Problem Statement:
A 250 V, dc shunt motor has an armature current of 20 A when running at 1000 rpm against some load torque. The armature resistance is 0.5 and brush contact drop is one volt per brush. By how much must the main flux be reduced to raise the speed by 50% if the developed torque is constant? Ignore effects of armature reaction and magnetic saturation.
Solution:
Step 1: Calculation of Developed Torque
Developed torque (Td) is given by the formula:
Td = (Ia * V - Ia * Ra - 2 * Vb) / ω
where:
Ia = armature current = 20 A
V = applied voltage = 250 V
Ra = armature resistance = 0.5 Ω
Vb = brush contact drop = 1 V per brush (2 brushes in total)
ω = angular velocity = 2πn/60
n = speed = 1000 rpm
Substituting the given values in the formula, we get:
Td = (20 * 250 - 20 * 0.5 - 2 * 1) / (2π * 1000/60)
Td = 9.39 Nm
Step 2: Calculation of Flux
Flux (Φ) is given by the formula:
Φ = (V - Ia * Ra - 2 * Vb) / (K * ω)
where:
K = constant
Substituting the given values in the formula, we get:
Φ = (250 - 20 * 0.5 - 2 * 1) / (K * 2π * 1000/60)
Φ = 0.057 Wb / K
Step 3: Calculation of New Speed
New speed (n1) is given by the formula:
n1 = (Td / (K * Φ)) * 60 / (2π)
Substituting the values of Td and Φ, we get:
n1 = (9.39 / (K * 0.057)) * 60 / (2π)
n1 = 2910 / K
Step 4: Calculation of Reduced Flux
Using the condition that new speed is 50% greater than the original speed, we get:
n1 = 1.5 * n
Substituting the values of n and n1, we get:
2910 / K = 1.5 * 1000
K = 2.94
Substituting the value of K in the formula of flux, we get:
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A 250 V, dc shunt motor has an armature current of 20 A when running at 1000 rpm against some load torque. The armature resistance is 0.5 and brush contact drop is one volt per brush. By how much must the main flux be reduced to raise the speed by 50% if the developed torque is constant? Ignore effects of armature reaction and magnetic saturation.?
Question Description
A 250 V, dc shunt motor has an armature current of 20 A when running at 1000 rpm against some load torque. The armature resistance is 0.5 and brush contact drop is one volt per brush. By how much must the main flux be reduced to raise the speed by 50% if the developed torque is constant? Ignore effects of armature reaction and magnetic saturation.? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A 250 V, dc shunt motor has an armature current of 20 A when running at 1000 rpm against some load torque. The armature resistance is 0.5 and brush contact drop is one volt per brush. By how much must the main flux be reduced to raise the speed by 50% if the developed torque is constant? Ignore effects of armature reaction and magnetic saturation.? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 250 V, dc shunt motor has an armature current of 20 A when running at 1000 rpm against some load torque. The armature resistance is 0.5 and brush contact drop is one volt per brush. By how much must the main flux be reduced to raise the speed by 50% if the developed torque is constant? Ignore effects of armature reaction and magnetic saturation.?.
A 250 V, dc shunt motor has an armature current of 20 A when running at 1000 rpm against some load torque. The armature resistance is 0.5 and brush contact drop is one volt per brush. By how much must the main flux be reduced to raise the speed by 50% if the developed torque is constant? Ignore effects of armature reaction and magnetic saturation.? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A 250 V, dc shunt motor has an armature current of 20 A when running at 1000 rpm against some load torque. The armature resistance is 0.5 and brush contact drop is one volt per brush. By how much must the main flux be reduced to raise the speed by 50% if the developed torque is constant? Ignore effects of armature reaction and magnetic saturation.? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 250 V, dc shunt motor has an armature current of 20 A when running at 1000 rpm against some load torque. The armature resistance is 0.5 and brush contact drop is one volt per brush. By how much must the main flux be reduced to raise the speed by 50% if the developed torque is constant? Ignore effects of armature reaction and magnetic saturation.?.
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