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*Electric motor load* means the current drawn by an electric motor. It depends upon the load on the shaft of the motor. For a given electric motor, it can estimated using input power, amperage or speed of the motor. Most electric motors are designed to run at 50 % to 100 % of rated load. Maximum efficiency is usually near 75 % of rated load.

A motor is considered under-loaded when it is in the range where efficiency drops significantly with decreasing load. Overloaded motors can overheat and lose efficiency.Many electric motors are designed with a service factor that allows occasional overloading. The service factor is a multiplier that indicates how much a motor can be overloaded under ideal ambient conditions.

When direct-read power measurements are available, we can use then to determine electric motor load. By measuring the parameters of the motor using a hand-held device, equation (1) can be used to determine the 3-phase input power to the loaded motor. Then, we can quantify the motor load by comparing the measured input power under-load to the power required when the motor operates at rated capacity.

The input power of a 3-phase electric motor being,

$$\mathrm{P_{i}=\frac{\sqrt{3}\times\:V\times\:I\times\:\cos\varphi}{1000}\:\:\:...(1)}$$

Where,

**P**_{i}is 3-phase input power in kW,**V**is RMS line voltage,**I**is RMS line current, and**cos φ**is the power factor of the motor.

The rated input power of the 3-phase motor will be,

$$\mathrm{P_{i_{rated}}=hp\times\:\frac{0.746}{\eta\:f_{1}}\:\:\:...(2)}$$

Where,

**P**is 3-phase input power at rated-load in kW,_{irated}**hp**is rated horse-power from nameplate.**ηf**_{i }is full-load efficiency.

Therefore, the motor load will be

$$\mathrm{Electric\:Motor\:Load=\frac{P_{i}}{P_{i_{rated}}}\times\:1000\%\:\:\:(3)}$$

Where,

**P**_{i}is the measured 3-phase power in kW,**P**_{irated}is the input power at rated-load in kW.

Using the equation (3), the electric motor load can be determined.

The current load estimation method is recommended when only amperage measurement is available. The current drawn by a motor varies approximately linearly with respect to load, down to about 50% of full load. Below the 50% of rated-load, the current becomes nonlinear, thus in the low load region, current measurement method is not used for load determination. Therefore, the electric motor load can be calculated using the below equation.

$$\mathrm{Electric\:Motor\:Load=\frac{I}{I_{r}}\times\:\frac{V}{V_{r}}\times\:100\%\:\:\:...(4)}$$

Where,

**I**is the RMS current,**I**_{r}is the rated motor current,**V**is RMS line voltage and**V**_{r}is rated voltage of the motor.

When the operating speed measurements are available, the slip method can be used to estimate the motor load. The actual speed of the motor is less than the synchronous speed.Therefore, the difference of synchronous speed and actual speed is called slip. The slip is directly proportional to the load imposed upon the motor by the driven equipment.

By using a tachometer to measure actual motor speed, it is possible to calculate motor loads. Using the below equation, the electric motor load can be determined in terms of slip.

$$\mathrm{Electric\:Motor\:Load=\frac{Slip}{N_{s}-N_{r}}\times\:100\%\:\:\:...(5)}$$

Where,

**Slip**= (Synchronous speed – Measured speed),**N**_{S}= Synchronous Speed in RPM,**N**_{r}=Rated motor speed.

An electric motor is of 40 hp, 1500 RPM. The motor is 10 years old and has been rewound.Determine the load of the motor, if the electrician makes the measurements as −

Line Voltage | Line Current | Power Factor |
---|---|---|

V_{RY} = 465 V | I_{R} = 35 A | Cos φ_{R} = 0.78 |

V_{YB} = 475 V | I_{Y} = 32 A | Cos φ_{Y} = 0.76 |

V_{BR} = 470 V | I_{B} = 30 A | Cos φ_{B} = 0.75 |

**Solution** −

Here,

$$\mathrm{Voltage,V=\frac{V_{RY}+V_{YB}+V_{BY}}{3}=\frac{465+475+470}{3}=470\:V}$$

$$\mathrm{Current,I=\frac{35+32+30}{3}=32.33A}$$

$$\mathrm{Power\:Factor,\cos\varphi=\frac{0.78+0.76+0.75}{3}=0.763}$$

Therefore, the load taken by the motor will be,

$$\mathrm{P_{i}=\frac{\sqrt{3}\times\:V\times\:I\times\:\cos\varphi}{1000}=\frac{\sqrt{3}\times\:470\times\:32.33\times\:0.763}{1000}=20.08\:kW}$$

Hence, the electrical load imposed by the motor on the supply system is 20.08 kW.

An electrical motor has synchronous speed equal to 1500 RPM, nameplate full-load speed is 1450 RPM and the measured speed is 1470 RPM. The rated horsepower of the motor is 20hp. Determine the actual output horsepower of the motor.

**Solution**−

$$\mathrm{Electric\:Motor\:Load=\frac{1500-1470}{1500-1450}\times\:100\%=60\%}$$

Therefore, the actual output horsepower would be,

$$\mathrm{Actual\:output\:hp = 20 \times 60\% = 12\:hp}$$

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