Motors and Controls

Question 1

A 1ph electric motor draws 28A and is suplied by a size 10 awg stranded uncoated copper conductor. The length of the circuit from the panel to the motor is 200ft. The approximate voltage drop caused by the circuit conductors when the motor is running is:

Answer 1

Vd = I X R
Vd = 28A X .496 ohm (#10awg)= 12.9V

Question 2

3 electric devices are connected in parallel on a circuit with the first device having a restance of 6 ohm, the second having a resistance of 10 ohms, and the third having a resistance of 15 ohms. What is the total resistance?

Answer 2

Rt = R1 x R2 / R1 + R2
Rt = 6 x 10 / 6 + 10
Rt = 60 / 16 = 3.75
Rt = 3.75 x 15 / 3.75 + 15
Rt = 56.25 / 18.75 = 3
Rt = 3 ohms

Question 3

A 3 phase, 460V, 40 horsepower continuous duty wound-rotor induction motor has a nameplate primary full-load current of 45.5 A. A nameplate secondary full-load current of 82A, and a temperature rise of 40*C. The resistor bank classification is “medium intermittent duty” and is located separate from the controller. What is the minimum PRIMARY conductor current rating permitted?

Answer 3

1. Use the FLC listed in Table 430.250 and NOT the current of the nameplate as stated in430.6(A)(1)
2. Multiply that value by 1.25 by 430.22(A)

Table 430.250- 40hp @ 460V = 52A
continuous - 52A x 1.25 = 65A

Question 4

A 3 phase, 460V, 40 horsepower continuous duty wound-rotor induction motor has a nameplate primary full-load current of 45.5 A. A nameplate secondary full-load current of 82A, and a temperature rise of 40*C. The resistor bank classification is “medium intermittent duty” and is located separate from the controller. What is the minimum SECONDARY conductor current rating permitted?

Answer 4

The wound-rotor has windings on the rotor that connect to a resistor bank. The rotor circuit is often called a secondary circuit.

The full load secondary current must be given on the name plate. The conductor ampacity to the resistor bank is determined according to 430.23(A). The secondry current is multiplied by 1.25 (125%).

82A x 1.25 = 103A

Question 5

A 3 phase, 460V, 40 horsepower continuous duty wound-rotor induction motor has a nameplate primary full-load current of 45.5 A. A nameplate secondary full-load current of 82A, and a temperature rise of 40*C. The resistor bank classification is “medium intermittent duty” and is located separate from the controller. The circuit is protected from short-circuits and ground-faults with time-delay fuses. What is the MAX time-delay fuse rating?

Answer 5

1. Go to Table 430.52 multiplier factor too used with 430.250
2. Table 430.250. Full load current
3. 52A x 1.5 (multiplier) = 78A
4. 240.6- round up to next standard size = 80A

Question 6

A feeder supplies two 25hp 3 phase, 460V, design B induction motors with a service factor of 1.15, and a 3 phase 460V, 40hp wound-rotor induction motor with a primary full load current of 45.5A, and a temp rise of 40*C. *All motors operate independent of each other.The minimum feeder conductor current rating permitted to supply this load is:

Answer 6

1. The feeder conductor minimum current rating- determined by 430.24
2. Takes the sum of all motor FLCs and adds to that 0.25 the FLC of the largest motor
3. 34A + 34A + (1.25 x 52A) = 133A

Question 7

A feeder supplies two 25hp 3 phase, 460V, design B induction motors with a service factor of 1.15, and a 3 phase 460V, 40hp wound-rotor induction motor with a primary full load current of 45.5A, and a temp rise of 40*C. *All motors operate independent of each other and each is protected by time-delay fuses. What is the maximum time-delay fuse rating permitted for this feeder?

Answer 7

1. Not permitted to be greater than 430.62A
2. Determine the Maximum permitted rating for each motor - 430.52
3. Wound-rotor- 52A x 1.5 = 78A (80)A
4. two 25hp motors- FLC(430.150) x factor from (430.52 = 1.75). 34A x 1.75 = 59.5A(60A)
5. 430.62(A) gives the maximum overcurrent rating as not more than the maximum rating for any motor circuit protection rating plus the full load current of all other motors served by that feeder.
6. the 40hp is protected by 80A- the 25hp motors are protected by 60A fuses
7. The maximum feeder OC rating is determined by adding the highest rated motor circuit overcurrent device and the full-load currents of all other motors served by that feeder.
8. 240.6– 80A + 34A + 34A = 148A (round down 125A)

Question 7

A 3 phase 460V, design Bmotor is drawing 21A with apower factor of 0.79 and is producing 15hp. The motor is operating with an efficiency of:

Answer 7

Eff = hp x 746 / 1.73 x V x I x pf
Eff = 15 x 746 / 1.73 x 460 x 21 x 0.79 =
Eff = 11,190 / 13,202W x 100 =** 85%**