ELECTRIC MOTOR AND OVERCURRENT PROTECTION

Parts of a Motor Branch Circuit

A - Motor circuit conductor 125 % of current full load

B' - Motor disconnecting means

B - Motor Branch Circuit; over current protection usually 300% of full load current for safety switch (SS) or 250% of full load current for air circuit breaker (ACB)

NOTE: These values are for general purposes motor only, values of B will depend on type and class of motor. The range is from 150% to 300% of the full load current, and shall in no case exceed 400% of full load current.

C - Motor Controller - rated according to motor horsepower

D - Motor Running Over current and overload protection: setting is from 115% to 125% of full load current. Value of 1.15 nd 1.25 is called service factor (SF)

1. Circuit Conductor - for the size of conductor having a maximum ampacity greater than the calculated value to protect the conductor from burning due to overheating. Faulty wiring is caused by the undersized conductor wire.

2. Branch Circuit Over current Protection - for protection against short circuit ground fault which cause over current flow. It should be capable of carrying the starting current of the motor.

3. Motor - controller ( magnetic contactor ) - used to start and stop the motor. It includes any switch or device capable of interrupting the stalled rotor current of motor.

4. Running Over current and Overload Protection - used to protect motor controller and motor against excessive heating due to motor overload and failure to start.

CIRCUIT FOR MOTOR LOAD

Name plate of the motor

25 hp,220 volts, 3 phase; 3 wires
60 Hz, 0.84 power factor 90.5% efficiency

SOLUTION:

1. Solve for the current load:

1 horsepower = 746 watts

I = (load in hp x 746 w)/ (k x E x pf x n)

where:

k - 1.0 for 2 wire single phase DC
1.73 for 3 wire, 3 phase AC
2.0 for 3 wire single phase AC or DC
3.0 for 4 wires, 3 phase AC

E - voltage between the neutral and live wire or between two live wires if no neutral line exists

I - Current in any live wire except Neutral Line

pf - power factor

N - efficiency

2. Applying the formula

I = (25 hp x 746)/(1.73 x 220 volts x .84 x 90.5%)

= 64.45 A

3. Determine the Motor feeder. The current load of a motor multiplied by 125% (Code requirements)

4. Find the size of the conductor wire. Refer to Table, for 80.56 A, use any of the following:

3 - 38 mm² THW or RHW copper wire
3 - 50 mm² TW copper wire
3 - 80 mm² TW aluminum or copper clad alum
3 - 50 mm² THW or RHW aluminum or copper clad aluminum

The allowable ampacities of the above wires in Table was derated by 80% to carry the 80.56 A current load.

5. Solve for the size of conduit pipe. Refer to Table, use 50 mm diameter pipe.

6. Determine the size or rating of the over current protection. The Code provides "The maximum over current protection for a single motor or a combination of motors should be, 250% of the ampacity of the largest motor plus the sum of the full load current of the other motors.

Therefore:
64.45 A x 250% = 161.15 A minimum

7. Refer to Table. Use 150 A fuse or trip breaker. It is the nearest standard rating which does not exceed the 161.12 A current load as computed.



MATERIALS FOR MOTOR INSTALLATION:

1. 25 hp Induction motor 230 volts, 3 phase, 1800 rpm, 60 Hz at 40 degree temperature rise

2. Magnetic thermal overload control with contractors.

3. Service entrance cap 50 mm with locknut

4. 38 mm² THW or RHW copper wire.

5. 50 mm² diameter IMT or RSC conduit pipe

6. Conduit clamp with screw, 50 mm conduit pipe

7. TPST safety switch or circuit breaker 150 or 250 volts

The quantity of materials depends upon the area and choice of the designing Engineer


CALCULATION PROCEDURES IN FINDING THE SIZE OF FEEDER AND THE OVERLOAD CURRENT PROTECTION FOR A GROUP OF MOTORS

Four 3 - phase motor 220 volts squirrel cage induction motor designed for 40 degree Celsius temperature rise at 1800 rpm, 60 Hz


SOLUTION:

1. Determine the main feeder of the motors. Apply 25% of the biggest motor current load plus the sum of the other motors.

(45 x 1.25) + 39 + 29 +21
= 145.25 A

2. Refer to Table. For the 145.25 A current load use any of the following conductor wires:

3 - 80 mm² THW or RHW copper wire
3 - 100 mm² TW copper wire
3- 125 mm² THW or RHW aluminum or copper clad aluminum
3 - 150 mm² TW clad aluminum

3. Determine the main over current protection. The National Electrical Code provides that:

" The protection rating or setting of a motor shall be 250 % percent (maximum) of the full load current of the biggest motor being served plus the sum of the full load current of the other motors."

(45 x 125%) x (250% +39 + 29 + 21)
140.625 + 89 = 229.625 A (maximum)

4. Refer to Table. Select a fuse or trip breaker that is nearest to standard rating that will not exceed 229.62 A. Use 200 A.

OVERLOAD PROTECTION
Overcurrents

An overcurrent exists when the normal load current for a circuit is exceeded. It can be in the form of an overload or short-circuit. When applied to motor circuits an overload is any current, flowing within the normal circuit path, that is higher than the motor’s normal full load amperes (F.L.A.). A short-circuit is an overcurrent which greatly exceeds the normal full load current of the circuit. Also, as its name infers, a short-circuit leaves the normal current carrying path of the circuit and takes a “short cut” around the load and back to the power source. Motors can be damaged by both types of currents.Single-phasing, overworking and locked rotor conditions are
just a few of the situations that can be protected against with the careful choice of protective devices. If left unprotected, motors will continue to operate even under abnormal conditions. The excessive current causes the motor to overheat, which in turn causes the motor winding insulation to deteriorate and ultimately fail. Good motor overload protection can greatly extend the useful life of a motor. Because of a motor’s characteristics, many common overcurrent devices actually offer limited or no protection.


Motor Starting Currents

When an AC motor is energized, a high inrush current occurs. Typically, during the initial half cycle, the inrush current is often higher than 20 times the normal full load current. After the first halfcycle the motor begins to rotate and the starting current subsides to 4 to 8 times the normal current for several seconds. As a motor reaches running speed, the current subsides to its normal running level. Typical motor starting characteristics are shown in Curve 1.




Because of this inrush, motors require special overload protective devices that can withstand the temporary overloads associated with starting currents and yet protect the motor from sustained overloads. There are four major types. Each offers varying degrees of protection.

Fast Acting Fuses
To offer overload protection, a protective device, depending on its application and the motor’s service factor (S.F.), should be sized at 115% or less of motor F.L.A. for 1.0 S.F. or 125% or less of motor F.L.A. for 1.15 or greater S.F. However, as shown in Curve 2, when fast-acting, non-time-delay fuses are sized to the recommended level the motors inrush will cause nuisance openings.

A fast-acting, non-time-delay fuse sized at 300% will allow the motor to start but sacrifices the overload protection of the motor. As shown by Curve 3 below, a sustained overload will damage the motor before the fuse can open.