Motors and Controls Flashcards
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:
Vd = I X R
Vd = 28A X .496 ohm (#10awg)= 12.9V
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?
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
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?
- Use the FLC listed in Table 430.250 and NOT the current of the nameplate as stated in430.6(A)(1)
- Multiply that value by 1.25 by 430.22(A)
Table 430.250- 40hp @ 460V = 52A
continuous - 52A x 1.25 = 65A
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?
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
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?
- Go to Table 430.52 multiplier factor too used with 430.250
- Table 430.250. Full load current
- 52A x 1.5 (multiplier) = 78A
- 240.6- round up to next standard size = 80A
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:
- The feeder conductor minimum current rating- determined by 430.24
- Takes the sum of all motor FLCs and adds to that 0.25 the FLC of the largest motor
- 34A + 34A + (1.25 x 52A) = 133A
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?
- Not permitted to be greater than 430.62A
- Determine the Maximum permitted rating for each motor - 430.52
- Wound-rotor- 52A x 1.5 = 78A (80)A
- two 25hp motors- FLC(430.150) x factor from (430.52 = 1.75). 34A x 1.75 = 59.5A(60A)
- 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.
- the 40hp is protected by 80A- the 25hp motors are protected by 60A fuses
- 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.
- 240.6– 80A + 34A + 34A = 148A (round down 125A)
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:
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%**
If the temperature of the conductor and circuit breaker is not specified
and the wire size is size 1 AWG and smaller, assume the 60EC column of Table 310.16. Look
up the conductor size directly from the table. For example, if the motor draws 15.2 amperes, what is the minimum size conductor?
Multiply by 1.25 to get 19 amperes. This is a size 14 AWG copper conductor.
When sizing thebranch-circuit short-circuit and ground fault device, (fuse or circuit breaker), look up the multiplier from Table 430.52. Then look up the next standard rating overcurrent device (240.6)
that is higher than the number determined. For example, assume time-delay fuses are used
and the motor draws 15.2 amperes. Find the multiplier of 1.75 from Table 430.52. The
maximum rating overcurrent device permitted for a normal starting motor is 15.2 amperes times 1.75 equals 26.6 amperes. Round up to a 30 ampere overcurrent device. Know the rules forsizing running overcurrent protection from 430.32.
Determine the size of conductors for a 50 kVAR, 460 volt, 3-phase capacitor
bank.
3 AWG
First determine the full-load current of the capacitor bank. In this case it is a 3-
phase bank and the 1.73 must be in the denominator as shown in the above formula.
FLC = 50,000VAR / 1.73 x 460V = 62.8
Next, size the conductor multiplying the full-load current by 1.35 and look up the conductor
size in Table 310.16. The minimum permitted conductor ampacity is 1.35 times 63
amperes which is 85 amperes.** If conductor insulation and termination temperature are not specified in the problem, then use the 60EC column* and find size 3 AWG copper. There is no specific rule for determining the minimum or maximum rating of the overcurrent device.
The minimum branch circuit conductor size, THHN, copper with 75C terminations permitted to supply a 7½ horsepower 3-phase, 240 volt electric motor is:
A. 14 AWG. B. 12 AWG. C. 10 AWG. D. 8 AWG. E. 6 AWG.
10 AWG
430.6(A)(1) requires the motor full-load current of 22 amperes to be looked up in Table 430.150 not taken from the nameplate.
430.22(A) required full-load current to be multiplied by 1.25 and then look up the minimum wire size in Table 310.16.
(22 A × 1.25 = 27.5 A)
A design B, 3-phase, 15 horsepower, 480 volt electric motor has a service factor of 1.15, and a nameplate full-load current of 19 amperes. The maximum rating time-delay fuses permitted to provide branch-circuit short-circuit and ground-fault protection assuming the motor does not start with difficulty is:
A. 20 amperes. B. 25 amperes. C. 30 amperes. D. 35 amperes. E. 40 amperes.
40 amperes
431.430.6(A)(1) requires the motor full-load current of 21 amperes to be looked up in Table 430.150 not taken from the nameplate. 430.52 sets the maximum rating timedelay fuse at a value determined by multiplying the full-load current of 21 amperes by a multiplier found in Table 430.52. Find the multiplier of 1.75 by using the motor design letter B. Next multiply 21 amperes by 1.75 to get 36.75 amperes. Exception 1 of 430.52 permits rounding up to a standard size 40 ampere overcurrent device. The standard sizes of overcurrent devices are listed in 240.6(A). Exception 2 does not apply because it was stated in the question that the motor did not start with difficulty.
In regard to a 7½ hp, 480-volt, 3-phase ac motor with an 80 percent power factor and a full-load ampere rating of 19 amperes indicated on the nameplate, and a service factor of 1.15; when the initial setting of the overload device you have selected is not sufficient to carry the load, what is the MAXIMUM setting permitted for the overload protection?
26.6A
Use 19A because of nameplate.
Table 430.52 for multiplier (175%)
19 x 1.75 x .8 = 26.6A
