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

Answer:

a) $E_{b} = 221 V$

b) P = 13,800 W

c) $P_{mech} = 13260 W$

di) $I_{initial} = 1533.33 A$

dii) $R_{start} = 1.85 ohms$

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

$V_{t} = I_{initial} (R_{a} + R_{start})$

$230= 115 (0.15 + R_{start})\\230/115 = 0.15 + R_{start}\\2 - 0.15 = R_{start}\\ R_{start} = 1.85 ohms$