EMF and Torque Equation of DC Machine

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EMF and Torque Equation of DC Machine
EMF and Torque Equation of DC Machine

EMF and Torque Equation of DC Machine

EMF EQUATION OF DC MACHINE

When the armature of the DC machine rotates under the influence of magnetic field, then an EMF is induced (or a voltage is generated) in the armature winding. In case of DC Generator, we call it as Generated EMF. While in case of DC motor, it is known as Back EMF.

In both the cases expression for induced EMF is same, which is

E = (ⲫ*N*P*Z) / 60*A

DERIVATION

Let,  = Useful Flux per pole

       P = Total Number of Poles.

       N = Speed of Rotation of armature per minute

       n = Speed of Rotation of armature per second = N/60

       A = Number of Parallel Paths between brushes of opposite polarity.

       Z = Total Number of conductors in armature

Since flux per pole is Ⲫ. So each conductor will cut a flux P.Ⲫ in one revolution.

EMF Induced per conductor = Flux Cut per revolution / Time taken for 1 revolution in seconds

Since, n revolutions are made in one second. So one revolution will take 1/n second.

EMF Induced per conductor in second = P.Ⲫ/(1/n) = n.P.Ⲫ

Now, number of Armature conductors for each parallel path = Z/A

Total EMF generated in Armature is determined by the number of conductors in any path between two brushes. Therefore,

Total EMF Induced = (Avg. EMF per conductor)*(Number of conductors per path)

E = n.P.Ⲫ.Z/A,  or

E = (N*ⲫ* P*Z) / 60*A

TORQUE EQUATION OF DC MACHINE

When armature conductor carrying current is placed in a magnetic field, then it experiences a force which exerts a torque on the armature.

Let, Ⲫ = Useful Flux per pole

        P = Total Number of Poles.

       A = Number of Parallel Paths between brushes of opposite polarity.

       Z = Total Number of conductors in armature

       r = Radius of armature

       a = Cross sectional area of flux path

       B = Flux Density = Ⲫ/a

       l = length of armature conductors

       I = Armature current

       i = Current flowing in each armature current = I/A

So, the Torque equation of DC Machine is,

T = (0.159*Z*Ⲫ*P*I)/A   Newton-metre

DERIVATION

Average force on one conductor, F = B*i*l

Torque produced due to one conductor = Force * Distance

So, Torque = (B*i*l) * r

Since there are total Z conductors in the armature, so total torque developed in the armature will be

Torque = B*i*l* r*Z……..eq1

Since, i is the current in each conductor and total number of parallel paths is A. So,

i = I/A

Now, eq 1  will become

Torque = (B*I*l* r*Z)/A

Torque = (Ⲫ*I*l* r*Z)/(A*a) …….eq2             [B = Ⲫ/A]

As we know ‘a’ is the cross sectional area of flux path at a radius of r meters. So,

a = (2*π*r*l)/P

Putting the value of ‘a’ in eq2, we get

T =(Z*Ⲫ*P*I*l*r)/2*π*r*l*A

T = (Z*Ⲫ*P*I)/2*π*A  Newton-metre

T = (0.159*Z*Ⲫ*P*I)/A   Newton-metre     [1/2Π=0.159] 

So, the above equation is the Torque Equation of DC Machine.

As we can see from above equation that,

T∝ Ⲫ*I

Torque developed by armature is directly proportional to the product of flux and armature current.

In case of series DC machine, Ⲫ is proportional to the armature current. So

T∝ I²

In case of shunt DC Machine,Ⲫ is practically constant

T∝ I

READ HERE  Transfer Function of Field Controlled DC Motor

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