Question

A 230 V shunt motor has a nominal armature current of 60 A. If the armature resistance is 0.15 ohm, calculate the following: a. The counter-emf [V]. b. The power supplied to the armature [W]. c. The mechanical power developed by the motor, [kW] and [hp]. d. The initial starting current if the motor is directly connected across the 230 V line, and the value of the starting resistor needed to limit the initial current to 115 A.

Answer 1

a)
`E_(b) = 221 V`

b) P = 13,800 W

c)
`P_(mech) = 13260 W`

di)
`I_(initial) = 1533.33 A`

dii)
`R_(start) = 1.85 ohms`

Step-by-step explanation:

Voltage, Vt = 230 V

Armature current,
`I_(a) = 60 A`

Armature Resistance,
`R_(a) = 0.15 ohms`

a) The back emf is calculated as follows:

`E_(b) = V_(t) - I_(a) R_(a) \\E_(b) = 230 - (60 * 0.15)\\E_(b) = 230 - 9\\E_(b) = 221 V`

b) The power supplied to the armature (W)

`P =V_(t) I_(a)`

P = 230 * 60

P = 13,800 W

c) Mechanical power developed by the motor

`P_(mech)` = Power supplied to the armature - Power lost in the armature

Power lost in the armature,
`P_(a) = I_(a) ^(2) R_(a)`

`P_(a) = 60^(2) *0.15\\P_(a) = 540 W`

`P_(mech) = 13800 - 540\\P_(mech) = 13260 W`

d)( i)The initial starting current if the motor is directly connected across the 230 V line

At starting, there is no back emf,
`E_(b) = 0`

`V_(t) = I_(initial) R_(a)`

`230 = I_(initial) * 0.15\\ I_(initial) = 230/0.15\\ I_(initial) = 1533.33 A`

ii) Value of the starting resistor needed to limit the initial current to 115 A

`V_(t) = I_(initial) (R_(a) + R_(start))`

`230= 115 (0.15 + R_(start))\\230/115 = 0.15 + R_(start)\\2 - 0.15 = R_(start)\\ R_(start) = 1.85 ohms`