Question

A 12.5-kW, 230-V shunt motor has 2400 shunt-field turns per pole, an armature resistance (including brushes) of 0.18Ω, and a commutating-field resistance of 0.035Ω. The shunt-field resistance (exclusive of rheostat) is 375Ω. When the motor is operated at no load with rated terminal voltage and varying shunt-field resistance, the following data are obtained:

The no-load armature current is negligible. When the motor is operating at full load and rated terminal voltage with a field current of 0.468 A, the armature current is 58.2 A and the speed is 1770 r/min.

a. Calculate the full-load armature reaction in equivalent demagnetizing ampere-turns per pole.
b. Calculate the electromagnetic and load torques and the rotational loss at the given operating condition.
c. What starting torque will the motor produce with a field current of 0.555 A if the starting armature current is limited to 85A? Assume that the armature reaction under these conditions is equal to 175 ampere-turns per pole.
d. Design a series field winding to give a speed of 1575 r/min when the motor is loaded to an armature current of 58.2 A and when the shunt field current is adjusted to give a no-load speed of 1800 r/min. Assume the series field will have a resistance of 0.045 Ω.

Answer 1

Step-by-step explanation:

According to the magnetization curve, the field current corresponding to a speed of 1770 r/min = 0.482 A

That is,
`I_(shunt, nl) = 0.482 A`

When the motor is operated at full load,
`I_(shunt, fl) = 0.468 A`

a) The armature Reaction,
`R = (I_(shunt, nl) - I_(shunt, fl) )N`

Number of turns, N = 2400

R = (0.482 - 0.468)*2400

The armature reaction at full load R = 33.6 A turns/pole

b) Electromagnetic torque,
`\tau = (E_(a)I_(a) )/(w_(m) )`

The armature emf,
`E_(a) = V_(t) - I_(a) (R_(a) + R_(c) )\\`

`V_(t) = 230 V\\I_(a) = 58.2 A\\R_(a) = 0.18 ohms\\R_(c) = 0.035 ohms`

`E_(a) = 230 -58.2(0.18 + 0.035 )\\`

`E_(a) = 217.487 V`

`w_(m) = ((\pi )/(30) ) 1770\\w_(m) = 185.35`

`\tau = (217.487*58.2 )/(185.35 )\\\tau = 68.29 Nm`

c) Let us calculate the effective field current and get the corresponding motor speed

`I_(eff) = (((E_(a) )/(R_(f) ))N_(f) -R )/(N_(f) )`

`I_(eff) = (((230 )/(375 ))2400 -175 )/(2400 ) \\I_(eff) = 0.5404 A`

The motor speed corresponding to a field current of 0.5404 A is 1310 r/min

The generated voltage will be :

`E_(a) =230 *(1770)/(1310) \\E_(a) = 310 .76 V`

Starting torque,
`\tau = (E_(a) I_(a) )/(w_(m) )`

`\tau = (310.76*85 )/(1770 * ((\pi )/(30) )`

Torque = 142.5 Nm