Speed and Torque Equation of DC Motor

SPEED AND TORQUE EQUATION

For a DC motor, the speed equation is obtained as follows.

E_b = V - I_aR_a

For a given machine, Z, A and P are constants

Where K is constant

Speed equation becomes

The speed of the motor is diretly proportional to back emf (E_b) and inversly proportional to flux (ϕ).

The torque equation of a motor is given by

Τ α ϕ Ia

Flux is directly proportional to tue current flowing through the field winding ϕ α I_ƒ For a DC shunt motor, shunt field current I_sh, is constant as long as input voltage is constant. Therefore flux is also constant.

T α l_a

For DC shunt motor torque is directly proportional to the armature current.

For a DC series motor, the series field current is equal to the armature current Ia. The flux ϕ is proportional to the armature current I_a.

ϕ α I_a

Τ α ϕ I_a

T α I_a²

For a DC series motor, the torque is directly proportional to the square of the armature current.

Example: 12

A 4 pole DC motor has a wave wound armature with 594 conductors. The armature current is 40 A and flux per pole is 7.5 mwb. Calculate the torque developed by the motor.

P = 4, Z = 594, I_a = 40 A, ϕ = 7.5 mwb

For wavewound A = 2

Torque developed

T_a = 0.159 ϕ 

= 0.159 x 7.5 × 10-3

Ta = 56.66 N-m

Example: 13

 A DC shunt motor takes as current of 5A at no load with terminal voltage 230 V and run at 1000 r.p.m. The armature and field resistances are 0.2Ω and 230Ω respectively. Under load conditions motor takes a current of 30A. Determine the motor speed under load condition.

I_L1 = 5A, I_L2 = 30 A, V = 230 V, N₁ = 1000 rpm, R_a = 0.2Ω, R_sh = 230Ω

I_sh = V/R_sh = 230/230 =1A

I_a1 = I_L1 - I_sh

= 5 - 1

= 4 A

E_b1 = V - I_al R_a

= 230 - (4 × 0.2)

E_b1 = 229.2 V

I_a2 =   I_L2 - I_sh

=30-1

I_a2 = 29 A

E_b2 = V – I_a2 R_a

=230-(29 × 0.2)

E_b2 = 224.2 V

Flux is constant

ϕ1 = ϕ2

N2/N1 = E_b2/E_bl

N₂ = 978 rpm

Example: 14

A 6 pole lap connected 220 V shunt motor has 400 armature conductors. It (pitain takes 40 A on full load. The flux per pole is 0.06 weber. The armature and field resistances are 0.1Ω and 220Ω respectively contact drop per brush is 1 V. Determine the speed of the motor at full load.

P = 6, V = 220 V, Z = 400, ϕ = 0.06 wb, I = 40 A, R_a = 0.1Ω, R_sh = 220V,

V_brush = 1 V/brush.

I_sh = V / R_sh = 220/220 = 1A

I_a = I - I_sh = 40 - 1 = 39 A

E_b = V- I_aR_a - V_brush

= 220 - (39 × 0.1) - (2 × 1)

E_b = 218 V

[A=P=6 for lap winding]

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N = 545 rpm