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The electrical power supplied to an electric motor is converted into mechanical power for powering various loads such as pumps. The power output can be rated in KW (kilowatts) or Horsepower (HP). In the United States, the mechanical output of the electric motor is rated in HP while elsewhere, it is in KW.

Note that:

Electric motor horsepower is dependant on speed and torque. When torque

is expressed in lb-ft and speed is expressed in RPM, the following formula can be

used to calculate horsepower (HP):

Note that:

1HP = 746W = 0.746KW1KW = 1/0.746*HP = 1.34HP |

**Electric Motor Horsepower (HP)**Electric motor horsepower is dependant on speed and torque. When torque

is expressed in lb-ft and speed is expressed in RPM, the following formula can be

used to calculate horsepower (HP):

From the above formula, note that any increase in torque or speed or both increases horsepower.

**How to Determine Output Horsepower Overload of an Electric Motor**
To determine whether an electric motor is overloaded, we require:

(a) The electrical power input

(b) The efficiency of the motor or

(c) The mechanical power output or horse power rating- This can be calculated or more accurately measured.

Suppose we have a 1HP, single phase motor with power factor of 75% and efficiency 90% and is driving a fan. If the voltage and currents are respectively 123V and 10Amps, is this motor overloaded?

Let us first compute Electrical power input to motor as shown above:

Power input = Voltage x Current x P.F

= 123 x 10 x 0.75 = 922.5W

In horsepower we have = 922.5/746 = 1.24HP

Now we already know the electrical power input of the motor which is = 1.24HP

With an efficiency of 90%, power output is calculated thus:

Mechanical power output = Efficiency x Electrical power input (HP)

= 0.90 x 1.24 = 1.12HP

A 10HP, 400V, 3-phase AC motor with power factor of 80% and efficiency 85% takes in an average current of 16.5 Amps and is driving a pump load. Is this motor overloaded?

The electrical power input to the 3-phase AC motor is given by:

Electrical power input = Voltage x Current x P.F x 1.732

= 400 x 16.5 x 0.8 x 1.732 = 9,144.96W

In horsepower is = 9144.96/746 = 12.26HP

Motor power output = Efficiency x Electrical input

= 0.9 x 12.26 = 11.034HP

In general, if:

(a) The electrical power input

(b) The efficiency of the motor or

(c) The mechanical power output or horse power rating- This can be calculated or more accurately measured.

**Electrical Power Input for Single and 3-Phase AC Motors**:For a single phase electric motor, power input is given by:Power(Watts) = Voltage x Current x Power factor (P.F)For a three phase AC motor, power input is given by:Power(Watts) = Voltage(line) x Current(line) x P.F x 1.732 |

**Sample Scenario 1**Suppose we have a 1HP, single phase motor with power factor of 75% and efficiency 90% and is driving a fan. If the voltage and currents are respectively 123V and 10Amps, is this motor overloaded?

Power input = Voltage x Current x P.F

= 123 x 10 x 0.75 = 922.5W

In horsepower we have = 922.5/746 = 1.24HP

Now we already know the electrical power input of the motor which is = 1.24HP

With an efficiency of 90%, power output is calculated thus:

**Efficiency = Mechanical power output (HP)/Electrical power input(HP)**Mechanical power output = Efficiency x Electrical power input (HP)

= 0.90 x 1.24 = 1.12HP

This motor is definitely overloaded! |

**Sample Scenario 2**A 10HP, 400V, 3-phase AC motor with power factor of 80% and efficiency 85% takes in an average current of 16.5 Amps and is driving a pump load. Is this motor overloaded?

Electrical power input = Voltage x Current x P.F x 1.732

= 400 x 16.5 x 0.8 x 1.732 = 9,144.96W

In horsepower is = 9144.96/746 = 12.26HP

Motor power output = Efficiency x Electrical input

= 0.9 x 12.26 = 11.034HP

Again, this AC motor is overloaded! |

In general, if:

The calculated or measured power output of an electric motor in
horsepower(HP) is greater than the nameplate rating, then the motor is
evidently overloaded! |