The armature circuit resistance of a 25 hp 250 v series moto

The armature circuit resistance of a 25 hp, 250 v series motor is 0.1 ohms the brush volt drop is 3 v and the resistance of the series field is 0.05 ohms When the series motor takes 85 A, the speed is 600 rpm cal- culate (a) The speed when the current is 100 A (b) The speed when the current is 40 A Neglect armature reaction and assume that the machine is operating on the linear portion of its saturation curve at all times (c) Recompute speeds in (a) and (b) using a 0-05 diverter at these speeds.

Solution

voltage=250 volts.

Ra=0.1 ohms

brush voltage drop= 3 volts.

Rf=0.05 ohms

I1=85 amps

N1=600 rpm

in series motors Eb is directly proportional to (field current * speed)

a) (Eb2/Eb1)=(I2/I1)*(N2/N1)

we know v=Eb1+I*Ra+2* brush drop

Eb1=250-85*0.15-2*3=231.25 volts.

I1=85 amps, N1=600 rpm

I2=100 amps

v=Eb2+I2*R+2* drop

where R=Ra+Rf.=0.1+0.05=0.15 ohms.

Eb2=250-100*0.15-2*3=229 volts.

(Eb2/Eb1)=(I2/I1)*(N2/N1)

(229/231.25)=(100/85)*(N2/600)

N2=505.75 rpm.

b) I2=40 amps

v=Eb2+I2*R+2* drop

Eb2=250-40*0.15-6=238 volts.

(238/231.25)=(40/85)*(N2/600)

N2=1312.216 rpm