# How Do You Calculate Armature Current and Efficiency in a DC Shunt Motor?

• Engineering
• zsolt2
In summary, the conversation discusses a problem involving a 20 kW, 500 V d.c. shunt wound motor and its efficiency at half torque. The armature current, speed, and efficiency of the motor are calculated using the given data, and the missing 2.7A is accounted for. The conversation also clarifies the calculation for the fixed loss and the efficiency at half torque.
zsolt2

## Homework Statement

Hello I have a bit of a problem solving this
A 20 kW, 500 V d.c. shunt wound motor draws a current of 45 A when running at full load with a speed of 600 rev min–1. On no load, the current drawn from the supply is 5 A. If the armature resistance is 0.3 Ω and the shunt field resistance is 220 Ω, calculate the armature current, speed and efficiency of the motor when the torque falls to half of its full load value, assuming that the flux per pole does not change.

## Homework Equations

Armature current: Ia=20000/500=40A
Field current= 500/220=2.3A

## The Attempt at a Solution

Now first i don't understand where is the missing 2.7A or is the field current=5A?
If the torque falls half of its value the armature current falls to half as well Ia=20A
Speed:
On full load E=V-Ia x Ra=500- 40 x 0.3=488V
When Ia falls to half= E=500-20 x 0.3=494V
if the back e.m.f. increases the speed increases: 494/488=1.012 new speed= 600 x 1.012=607 rev/min
Efficiency:
V x Ia=Ia^2 x Ra + E x Ia
500 x 20=20^2 x 0.3 + 494 x 20=10000W

Can anyone tell me if I am right so far
Thank you

zsolt2 said:
Now first i don't understand where is the missing 2.7A
zsolt2 said:
Armature current: Ia=20000/500=40A
That's not correct.
5kW is the mechanical power output of the motor and not electrical power input.
zsolt2 said:
Field current= 500/220=2.3A
That's 2.2728 A(≈2.3 A).
So the armarure current would be simply supply current minus field current.

zsolt2

So the armature current Ia=45-2.3=42.7
Then the power=V x Ia=500 x 42.7=21350W
And that 1350W is a fixed lost?
Ia on half torque = 42.7/2=21.35A
Speed:
E=V-ia x Ra=500-42.7 x 0.3=487V
Half torque
E=500-21.35 x 0.3=494V
The speed increases if the back e.m.f. grows (Ia decreases) 494/487=1.014
n2=n1 x 1.014=600 x 1.014=608 rev/min

zsolt2 said:

So the armature current Ia=45-2.3=42.7 A
Ia on half torque = 42.7/2=21.35A
Speed:
E=V-ia x Ra=500-42.7 x 0.3=487V
Half torque
E=500-21.35 x 0.3=494V
The speed increases if the back e.m.f. grows (Ia decreases) 494/487=1.014
n2=n1 x 1.014=600 x 1.014=608 rev/min
Looks good!

zsolt2
Thank you!
And the efficiency :
Total power used: V x Ia = 500 x 21.35= 10675W
Losses:
Armature loss: Ia^2 x Ra =21.35^2 x 0.3=136.7W
Field loss: If^2 x Rf= 2.3^2 x 220=1163.8W
Brush losses: 2 x Ia= 2 x 21.35=42.7W
Fixed loss: 1350W
All losses:136.7+1163.8+42.7+1350=2693.2W
Is that looks right?

zsolt2 said:
And that 1350W is a fixed lost?
I didn't qoute this part in my previous post. You should use the no load data to find the fixed loss.
zsolt2 said:
Brush losses: 2 x Ia= 2 x 21.35=42.7W
You need not take brush drop into account if it is not mentioned in the problem.

zsolt2
Ok
So the current drawn on no load is 5A the power consumed is 500 x 5=2500W
Field loss is 2.3^2 x 220=1163.8~ 1164W
Then the fixed loss: 2500-1164=1336W
Efficiency at half torque :
10675/10675+1336+1164+137=0.8
Efficiency =80%

zsolt2 said:
Then the fixed loss: 2500-1164=1336W
To be accurate, you should also subtract armature copper loss from this. Usually, because of low armature resistance, armature copper loss is neglected on no load but here, since nothing is mentioned as such, you should calculate it.

zsolt2
Ok thank you very much for the help!

cnh1995

## 1. What is a DC motor?

A DC motor is a type of electric motor that converts direct current (DC) electrical energy into mechanical energy. It is commonly used in various applications such as electric vehicles, industrial machinery, and household appliances.

## 2. What is armature current?

Armature current is the electrical current that flows through the conductors of the armature of a DC motor. It is responsible for creating the magnetic field that interacts with the stator field to produce the rotational motion of the motor.

## 3. How is armature current controlled in a DC motor?

Armature current can be controlled by adjusting the voltage applied to the motor. Increasing the voltage will increase the armature current, while decreasing the voltage will decrease the armature current. Additionally, the number of turns in the armature winding and the strength of the magnetic field can also affect the armature current.

## 4. What factors can affect the armature current in a DC motor?

The armature current in a DC motor can be affected by several factors, such as the voltage applied to the motor, the number of turns in the armature winding, the strength of the magnetic field, and the resistance of the armature winding. Other factors include the load on the motor, the speed of the motor, and the overall efficiency of the motor.

## 5. How does armature current affect the performance of a DC motor?

Armature current plays a crucial role in determining the performance of a DC motor. It affects the speed, torque, and power output of the motor. Higher armature current results in higher torque and power output, while lower armature current results in lower speed and power output. However, excessive armature current can lead to overheating and damage to the motor.

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