Engineering Just starting out on DC motors and hit a snag

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To solve the problem of calculating torque and operating efficiency for a series DC motor, start by determining the input power using the supply voltage and current. The output power is given as 3.8kW, and the motor runs at 1320 rev/min. Use the equation Power = torque * angular velocity to find torque, ensuring unit consistency. For efficiency, divide the output power by the input power and express it as a percentage. Understanding the relationship between power, torque, and angular velocity is crucial for solving these types of problems.
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Hi Guys
I'm hoping someone can help me out with a problem??
I have just started learning about DC motors and I have hit a wall on how to solve the following problem-

Q.The output of a series d.c. motor when connected to a 230V supply and running at 1320rev/min is 3.8kW. The supply current for this loading is 20.5A.
Calculate the torque developed and the opperating efficiency??

Can anyone explain the equations i need to be looking at, as I thought you needed the armature current and field current to be able to solve this??

Any advice or help is much appreciated

Cas
 
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You can solve this with conservation of energy. Find the input power (easy for DC), find a way to express output power with angular frequency and torque.
 
carolinern said:
Q.The output of a series d.c. motor when connected to a 230V supply and running at 1320rev/min is 3.8kW. The supply current for this loading is 20.5A.

Calculate the torque developed

Power = torque * angular velocity......(1)

The problem gives you the output power and enough info to calculate the angular velocity. Just take care with the units.

If you have difficulty remembering that equation perhaps remind yourself of the similar equations for linear motion..

Work = force * displacement
Power = work/time
so
power = force * displacement/time
but
displacement/time = velocity
so
Power = force * velocity.........(2)

see the similarity between (1) and (2)?

the operating efficiency?

The problem statement gives enough info to calculate the input power (electrical) and the output power is stated. Efficiency is one divided by the other expressed as a percentage.
 

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