A 4 pole DC motor runs at 600 rpm on full load taking 25A at 450 V. The armature is lap- wound with 500 conductors and flux per pole is expressed by the relation. fie = 1.7×1/100 where I is the motor current.If supply voltage and torque are both halved, calculate the speed at which the motor will run. ignore stray losses
Answers
Answer:
The back emf in a DC motor is given by
ϕ is the flux per pole
Z is the total number of conductors
N is the speed
P is the number of poles
A is the number of parallel paths
For lap wound, A = P
For wave wound, A= 2
Mechanical power developed (P) = Eb Ia
Calculation:
Pole P = 4, Z = 400, A = 4, l = 20 cm, D= 30 cm, N = 1500 rpm, Ia= 25 A and B = 0.4 T
As the motor is Lap Connected, therefore A = P i.e. a number of parallel paths is equal to the number of poles.
Flux(ϕ) = Flux density(B) × Area(A)
Area/pole = πDL/P
Area =
Therefore, Flux = 0.4 × 0.0471 = 0.01884 Wb
Answer:
372 rpm
Explanation:
Let us first find R
Now
N = E(60 /P )r.p.m.
600 = Eb /1.7x102 x 25^0.5 *60*4/500 x 4
Eb = 10 x 1.7 x 10 x 5 x 500
=425 v
IaRa=450-425=25 V
: R=25/25 = 1.0 ohm
Now in the 1st Case
T₁ √I1
T1 directly proportional to .7₁ 1.7 x 10^-2 x √25 x 25
T₂ directly proportional to 1.7 x 10^-2 x √1xI
Now T₁ = 2T2
Similarly
1.7 x 10^-2 x 125 =1.7 x 10^-2xI^3/2x2
I=(125/2)^2/3
= 15.75 A
Eb1 425 V:
Eb2=225-(15.75 x 1)
=209.3 V
N2/N₁=Eb2/Eb1*pi1/pi2
Using the relation
N₂/ 600 =209.3/425 *1.7×10² x 5 /1.7 x 10² x√5.75 N₂=372 r.p.m.