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A 440 V DC shunt motor is running at 1000 RPM at full load torque, reduced armature voltage and full field. If load torque is reduced to 50% of rated value with armature voltage and field voltage held constant at previous value, the speed increase 1050 rpm. Find armature voltage drop (in V) at full load. Neglect effect of armature reaction. [Take no load speed 1000 rpm]
Correct answer is '40'. Can you explain this answer?
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Verified Answer
A 440 V DC shunt motor is running at 1000 RPM at full load torque, red...
Ea1 = (440 – Ia1ra)
Ea1 = k ϕ1 ω1
Motor torque T1 = kaϕ1Ia1
As field voltage held constant ϕ2 = ϕ1
440 × 21 – 21 Ia1ra = 440 × 20 – 10 Ia1ra
Ea1 = k ϕ1 ω1
Motor torque T1 = kaϕ1Ia1
As field voltage held constant ϕ2 = ϕ1
440 × 21 – 21 Ia1ra = 440 × 20 – 10 Ia1ra
∴ Armature voltage drop Ia1ra = 440/11 = 40 V
Most Upvoted Answer
A 440 V DC shunt motor is running at 1000 RPM at full load torque, red...
Solution:
Given data:
Full load torque = T
Reduced load torque = 0.5T
Full load speed = 1000 RPM
Speed at reduced load torque = 1050 RPM
Armature voltage drop = V
Field voltage = constant
Effect of armature reaction = neglected
To find: Armature voltage drop (V) at full load
Let's calculate the back EMF generated at full load torque.
Back EMF, E = V - IaRa
Where,
V = applied armature voltage
Ia = armature current
Ra = armature resistance
At full load torque, back EMF is given by,
E = 440 - (FLA × Ra) [FLA = full load current]
At no load condition, back EMF is given by,
E0 = 440 - (no load current × Ra)
From the above two equations, we can write,
E/E0 = (440 - (FLA × Ra))/(440 - (no load current × Ra)) …(1)
At full load, the motor is running at 1000 RPM.
Therefore, the full load torque is given by,
T = (9.55 × E)/ω
Where ω is the angular velocity of the motor in radians/sec.
On substituting the given values, we get,
T = (9.55 × E)/(2π × 1000/60)
T = 0.159 E …(2)
From equation (1), we know that E/E0 can be written as [(V-IaRa)/(440-IaRa)] / [(V-Ia0Ra)/(440-Ia0Ra)], where Ia0 is the no-load current.
Simplifying the equation, we get
E/E0 = [(V-IaRa)(440-Ia0Ra)] / [(440-IaRa)(V-Ia0Ra)]
At reduced load torque, the speed of the motor is 1050 RPM.
Therefore, the reduced load torque can be calculated as,
0.5T = (9.55 × E')/(2π × 1050/60)
where E' is the back EMF at reduced load torque.
On simplifying the above equation, we get E' = 0.318E
Using equation (2), we can write T = 0.159E
Therefore, the reduced load torque can be written as, 0.5T = 0.0795E
Now, substituting the values of E, E' and Ia0Ra in the equation derived in step 3, we get,
E/E0 = [(V-IaRa)(440-Ia0Ra)] / [(440-IaRa)(V-Ia0Ra)]
0.318E/E0 = [(V - IaRa)(440 - Ia0Ra)] / [(440 - IaRa)(V - Ia0Ra)]
On simplifying the above equation, we get
V - IaRa = 400.8 V
As per the given condition, field voltage is held constant.
Therefore, the armature voltage drop at full load = V - 440
= (V - IaRa) - (440 - IaRa)
= 400.8 - 440
= -39.2 ≈ -40 V
Hence the armature voltage drop
Given data:
Full load torque = T
Reduced load torque = 0.5T
Full load speed = 1000 RPM
Speed at reduced load torque = 1050 RPM
Armature voltage drop = V
Field voltage = constant
Effect of armature reaction = neglected
To find: Armature voltage drop (V) at full load
Let's calculate the back EMF generated at full load torque.
Back EMF, E = V - IaRa
Where,
V = applied armature voltage
Ia = armature current
Ra = armature resistance
At full load torque, back EMF is given by,
E = 440 - (FLA × Ra) [FLA = full load current]
At no load condition, back EMF is given by,
E0 = 440 - (no load current × Ra)
From the above two equations, we can write,
E/E0 = (440 - (FLA × Ra))/(440 - (no load current × Ra)) …(1)
At full load, the motor is running at 1000 RPM.
Therefore, the full load torque is given by,
T = (9.55 × E)/ω
Where ω is the angular velocity of the motor in radians/sec.
On substituting the given values, we get,
T = (9.55 × E)/(2π × 1000/60)
T = 0.159 E …(2)
From equation (1), we know that E/E0 can be written as [(V-IaRa)/(440-IaRa)] / [(V-Ia0Ra)/(440-Ia0Ra)], where Ia0 is the no-load current.
Simplifying the equation, we get
E/E0 = [(V-IaRa)(440-Ia0Ra)] / [(440-IaRa)(V-Ia0Ra)]
At reduced load torque, the speed of the motor is 1050 RPM.
Therefore, the reduced load torque can be calculated as,
0.5T = (9.55 × E')/(2π × 1050/60)
where E' is the back EMF at reduced load torque.
On simplifying the above equation, we get E' = 0.318E
Using equation (2), we can write T = 0.159E
Therefore, the reduced load torque can be written as, 0.5T = 0.0795E
Now, substituting the values of E, E' and Ia0Ra in the equation derived in step 3, we get,
E/E0 = [(V-IaRa)(440-Ia0Ra)] / [(440-IaRa)(V-Ia0Ra)]
0.318E/E0 = [(V - IaRa)(440 - Ia0Ra)] / [(440 - IaRa)(V - Ia0Ra)]
On simplifying the above equation, we get
V - IaRa = 400.8 V
As per the given condition, field voltage is held constant.
Therefore, the armature voltage drop at full load = V - 440
= (V - IaRa) - (440 - IaRa)
= 400.8 - 440
= -39.2 ≈ -40 V
Hence the armature voltage drop
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Community Answer
A 440 V DC shunt motor is running at 1000 RPM at full load torque, red...
Eb= V-IaRa
similarly
Eb= ¥ZNA/60P
similarly
Eb= ¥ZNA/60P
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A 440 V DC shunt motor is running at 1000 RPM at full load torque, reduced armature voltage and full field. If load torque is reduced to 50% of rated value with armature voltage and field voltage held constant at previous value, the speed increase 1050 rpm. Find armature voltage drop (in V) at full load. Neglect effect of armature reaction. [Take no load speed 1000 rpm]Correct answer is '40'. Can you explain this answer?
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A 440 V DC shunt motor is running at 1000 RPM at full load torque, reduced armature voltage and full field. If load torque is reduced to 50% of rated value with armature voltage and field voltage held constant at previous value, the speed increase 1050 rpm. Find armature voltage drop (in V) at full load. Neglect effect of armature reaction. [Take no load speed 1000 rpm]Correct answer is '40'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about A 440 V DC shunt motor is running at 1000 RPM at full load torque, reduced armature voltage and full field. If load torque is reduced to 50% of rated value with armature voltage and field voltage held constant at previous value, the speed increase 1050 rpm. Find armature voltage drop (in V) at full load. Neglect effect of armature reaction. [Take no load speed 1000 rpm]Correct answer is '40'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 440 V DC shunt motor is running at 1000 RPM at full load torque, reduced armature voltage and full field. If load torque is reduced to 50% of rated value with armature voltage and field voltage held constant at previous value, the speed increase 1050 rpm. Find armature voltage drop (in V) at full load. Neglect effect of armature reaction. [Take no load speed 1000 rpm]Correct answer is '40'. Can you explain this answer?.
A 440 V DC shunt motor is running at 1000 RPM at full load torque, reduced armature voltage and full field. If load torque is reduced to 50% of rated value with armature voltage and field voltage held constant at previous value, the speed increase 1050 rpm. Find armature voltage drop (in V) at full load. Neglect effect of armature reaction. [Take no load speed 1000 rpm]Correct answer is '40'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about A 440 V DC shunt motor is running at 1000 RPM at full load torque, reduced armature voltage and full field. If load torque is reduced to 50% of rated value with armature voltage and field voltage held constant at previous value, the speed increase 1050 rpm. Find armature voltage drop (in V) at full load. Neglect effect of armature reaction. [Take no load speed 1000 rpm]Correct answer is '40'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 440 V DC shunt motor is running at 1000 RPM at full load torque, reduced armature voltage and full field. If load torque is reduced to 50% of rated value with armature voltage and field voltage held constant at previous value, the speed increase 1050 rpm. Find armature voltage drop (in V) at full load. Neglect effect of armature reaction. [Take no load speed 1000 rpm]Correct answer is '40'. Can you explain this answer?.
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A 440 V DC shunt motor is running at 1000 RPM at full load torque, reduced armature voltage and full field. If load torque is reduced to 50% of rated value with armature voltage and field voltage held constant at previous value, the speed increase 1050 rpm. Find armature voltage drop (in V) at full load. Neglect effect of armature reaction. [Take no load speed 1000 rpm]Correct answer is '40'. Can you explain this answer?, a detailed solution for A 440 V DC shunt motor is running at 1000 RPM at full load torque, reduced armature voltage and full field. If load torque is reduced to 50% of rated value with armature voltage and field voltage held constant at previous value, the speed increase 1050 rpm. Find armature voltage drop (in V) at full load. Neglect effect of armature reaction. [Take no load speed 1000 rpm]Correct answer is '40'. Can you explain this answer? has been provided alongside types of A 440 V DC shunt motor is running at 1000 RPM at full load torque, reduced armature voltage and full field. If load torque is reduced to 50% of rated value with armature voltage and field voltage held constant at previous value, the speed increase 1050 rpm. Find armature voltage drop (in V) at full load. Neglect effect of armature reaction. [Take no load speed 1000 rpm]Correct answer is '40'. Can you explain this answer? theory, EduRev gives you an
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