Question

4. A 4 pole, 250 V DC series motor has a wave connected armature with 1200

conductors. The flux per pole is 20x10-3 wb, when the motor is drawing 45 A.

Armature and field resistances are 0.25 Ω and 0.15 Ω respectively. Calculate:

(i) Speed (ii) Torque​

Answer 1

Explanation:

Voltage (V)= 250V

Poles (P)= 4

Armature (Z)= 1200

Current (I) = 45A

Flux (∅) = 0.02 Wb

Armature resistance (Ra) = 0.25Ω

Series resistance (Rs) = 0.15Ω

Back emf

Eb = V - I (Ra + Rs)

     = 250 - 45(0.25 + 0.15)

      = 232

Eb = $\frac{(P)(∅)(Z)(N)}{60A}$

N = speed of armature

N = $\frac{(Eb)(60)(A)}{(P)(∅)(Z)}$

N = $\frac{(232)(60)(1)}{(4)(0.02)(1200)}$

N = 145 RPM

Torque  = 0.02 x 0.25

             = 0.005

Answer 2

Answer:

1374 RPM

Explanation:

Given Data

Supply Voltage

$V = 250 V$

Number of Poles

$P = 4$

Total number of armature conductors

$Z = 1200v$

Number of parallel paths in the armature winding A = 2 (for wave winding number of parallel path is always 2)

Line current

$I_{l} = 45 A$

Armature current Ia = 45 A ( In DC series motor the line current is equal to the armature current i.e IL = Ia)

$Flux Φ = 20 \times 10-3$

Armature Resistance

$R_{a} =0.25Ω$

field Resistance

$R _{se} = 0.15Ω$

The Back EMF of DC series Motor is given by

$Eb = V − Ia(Ra + Rse)$

$= 250 − 45 (0.25+ 0.15)$

$= 250 − 14.05</p><p></p><p> = Eb = 235.95$

Back EMF in DC Motor Eb.

$E_{b} = \frac{ P ϕ Z N   }{60 A}$

Where

P – Number of poles of the machine

ϕ – Flux per pole in Weber.

Z – Total number of armature conductors.

N – Speed of armature in revolution per minute (r.p.m).

A – Number of parallel paths in the armature winding.

$N = \frac{E_{b} \times 60 A }{ P ϕ Z}$

$\frac{235.95 \times 45 \times 2}{4 \times 25 \times {10}^{ - 3} \times 1200}$

$N = 159266.25\: r.p.m$

$</p><p>T = F × r.$

Let in a DC Motor

r = average radius of armature in m

= effective length of each conductor in m

Z = total number of armature conductors

A = number of parallel paths

i = current in each conductor = Ia/A

B = average flux density in Wb/m2

ϕ = flux per pole in Wb

P = number of poles

Force on each conductor, (F) = newtons

Torque due to one conductor = F × r newton- meter

$Total armature torque, (Ta) = Z F r newton- meter$

∴ newton- meter ... (1)

Now,

$i = Ia/A, (current in each conductor)$

$B = ϕ/a; where a is the x-sectional area of flux path per pole a radius r.$

And,

Substitute these values in equation (1),

TORQUE=

$= 0.25 \times 0.15 \\ = 0.0375$

The project code is #SPJ3