Understanding Flux Reduction for Speed Increase in 240V Shunt Motors

Introduction to Shunt Motors

Shunt motors are a type of direct current electric motor with a field winding connected in parallel with the armature. They are commonly used in applications where a constant speed is required under varying load conditions. A fundamental aspect of shunt motors is the interplay between armature current, flux, and speed.

Given Parameters and Initial Setup

A 240V shunt motor has the following given parameters:

Voltage ( V 240 ) V Armature resistance ( R_a 0.2 ) Ω Armature current ( I_a 20 ) A

We are tasked with finding how much the flux must be reduced to increase the speed by 40%. Let's break down the process step-by-step.

Step 1: Calculate Initial Back EMF (E_b)

The back EMF is defined by the formula: ( E_b V - I_a cdot R_a ). Substituting the given values:

( E_b 240 ) V - ( 20 ) A cdot 0.2 ) Ω 240 V - 4 V 236 V

Step 2: Relate Speed, Back EMF, and Flux

For a shunt motor, the relationship between the speed ( N ), back EMF ( E_b ), and flux ( Phi ) is given by:

( E_b k cdot Phi cdot N )

Where ( k ) is a constant. With the new speed being ( N_2 1.4 cdot N_1 ), the equation becomes:

( E_b k cdot Phi_1 cdot N_1 )

( E_b k cdot Phi_2 cdot 1.4 cdot N_1 )

Step 3: Maintain Constant Torque and Armature Current

Since the electromagnetic torque is constant, the product ( Phi cdot I_a ) remains constant. Therefore:

( Phi_1 cdot I_a Phi_2 cdot I_a )

( Phi_1 Phi_2 )

Step 4: Calculate New Flux (( Phi_2 ))

From the relationships above, the new back EMF can be expressed as:

( E_b k cdot Phi_2 cdot 1.4 cdot N_1 )

From ( E_b k cdot Phi_1 cdot N_1 ), we can express ( k ):

( k frac{E_b}{Phi_1 cdot N_1} )

Substituting ( k ) into the equation for ( E_b ):

( E_b frac{E_b cdot 1.4 cdot Phi_2}{Phi_1} )

Cancelling ( N_1 ):

( E_b frac{E_b cdot 1.4 cdot Phi_2}{Phi_1} )

Cancelling ( E_b ):

( 1 1.4 cdot frac{Phi_2}{Phi_1} )

Rearranging:

( frac{Phi_2}{Phi_1} frac{1}{1.4} )

Therefore:

( Phi_2 frac{Phi_1}{1.4} )

Step 5: Calculate the Reduction in Flux

The reduction in flux is given as:

( DeltaPhi Phi_1 - Phi_2 Phi_1 - frac{Phi_1}{1.4} Phi_1 left(1 - frac{1}{1.4}right) Phi_1 left(frac{0.4}{1.4}right) )

The reduction in flux is approximately ( frac{0.4}{1.4} ), which is about 28.57% of the original flux.

Conclusion

To increase the speed of a 240V shunt motor by 40%, the flux must be reduced by approximately 28.57% of the original flux. This process involves understanding the relationship between the back EMF, flux, and speed in shunt motors while maintaining constant torque and armature current.

Key Takeaways

Understanding the relationship between back EMF, flux, and speed in shunt motors. Calculating the necessary reduction in flux for a desired increase in speed. Maintaining constant torque and armature current to ensure the motor operates effectively under load.

Further Reading

For further exploration on the principles of shunt motors and their applications, consider reading more about theshunt motor, armature resistance, and electromagnetic torque. These topics will provide a deeper insight into the workings of electrical machinery and energy conversion systems.