For some imported equipments with motors rated at 60Hz (480V, 460V, 440V, 400V, 380V) which are different with the power conditions of 50Hz countries, but, can such appliances be used in the 50Hz countries? In addition, for exporting appliances, whether the motor rated at 50Hz frequency can be used in the power supply with 60Hz?

**Analysis of the motor rated at 60Hz frequency is used in 50Hz power supply when the voltage is the same (e.g. 380V)**

The magnetic flux, Φ, for per pole of the motor can be expressed in the formula below:

**Φ = k _{e}U / 4.44fWk_{dp1}**

In the formula:

k_{e} - depressurization coefficient

U - extra supply voltage (phase voltage) (V)

f - power frequency (Hz)

W - turns for every series coil of stator winding

k_{dp1} - stator winding coefficient

For induction motors, k_{dp1}, W, k_{e} are all fixed values. When the voltage, *U*, is fixed, the frequency, *f*, is proportional to the flux, Φ, for per pole, i.e.

**Φ _{2} = f_{1}/f_{2} * Φ_{1} = 60/50 *Φ_{1} = 1.2Φ_{1}**

In the form: the "1" in the lower corner stands for all variables of 60Hz, and "2" stands for the variables of 50Hz.

In addition, it is necessary to note that the magnetization curve of the motor core has nonlinear characteristics, and the flux margin is very small during design.

*(1) No-load current*: When the power frequency drops from 60Hz to 50Hz, **Φ _{2} = 1.2Φ**

_{, }that is, the flux for per pole increases by 20% accordingly, the magnetic flux density of each part of the motor will increase by 20%.

As the magnetic flux margin during the design is very small, the no-load current will increase by more than 20%.

If the no-load current approaches or exceeds the original rated current, the motor cannot be used. If there is a big gap between the no-load current and the original rated current, it can be used. But generally speaking, the motor capacity is less than before at least 20%.

*(2) Speed*: As the speed is determined by the form below

**n _{1} = 60f / P**

the pole log, P, is constant, when the power frequency is changed from 60Hz to 50Hz, the speeds is decreased by

**∆n _{1} = (f_{1} - f_{2}) / f_{1} = (60 -50 ) / 60 = 16.6% ≈ 17%**

Therefore, the motor speed is decreased by about 17%.

*(3) Starting current:* The motor is an inductive load, the reactance value, *X*, of which is proportional to the power frequency. The lower the power frequency is, the lower the *X* is. While starting current is inversely proportional to the reactance value, *X*, so the starting current of the motor will be increases by about 20% correspondingly.

*(4) Torque*: The magnitude of the torque is inversely proportional to the square of the power frequency, that is M∝1/f^{2}. As a result, the torque will increase when the power frequency turns from 60Hz to 50Hz:

**M _{1} = f_{1}^{2} / f_{2}^{2} * M_{2} = 60^{2} / 50^{2} * M_{2} = 1.44M_{2}**

Means an increase of about 44%. Similarly, the maximum torque and minimum torque of the motor will increase accordingly.

*(5) Efficiency*: In the main consumption of an electric motor;

- Iron consumption: It is approximately proportional to the square of the flux density and the 1.3 power of the frequency, so the iron consumption, P
_{Fe}, is about 14% higher than before. - Stator copper consumption: If the load current is same, the stator copper consumption, P
_{cu}, is unchanged. - Rotor copper consumption: As the flux density increases by 20%, the rotor current will be reduced by 16.6% in order to maintain the same torque. Therefore, the rotor copper consumption, P
_{cu2}, will decline. - Additional consumption: Wind friction loss, P
_{F}, decreased as the speed decreased, which is 60% of the original approximately. Additional consumption will fall considerably.

But the output power of motor is reduced greatly, so the efficiency will decrease generally.

*(6) Power factor*. As the no-load current increases, although the reactance value of the motor drops, it is still not enough to compensate for it. Therefore, the power factor will also drop.

*(7) Temperature rising*. The core flux density will be saturate because the magnetic flux density is 20% higher than the original. In addition, the ventilation effect deteriorates with the decrease of the speed, so, the temperature rising of the motor is much higher than the original.

**Analysis of 60Hz motor being used in 50Hz power supply with voltage reduction**

(1) The use of reduction voltage and the voltage determination of the reduction voltage: In order to make the motor with the frequency of 60Hz to be used on 50Hz power grids to generate heat without through the current, the magnetic flux of the motor must be maintained. It is known from

**Φ = k _{e}U / 4.44fWk_{dp1}**

That the only variable is the supply voltage.

Now, take the motor with 60Hz and 380V as example. Making Φ_{1}=Φ_{2}, keep Φ unchanged in the 50Hz power supply, then

**U' = f _{2}/f_{1} * U = 50/60 * 380 = 317 (V)**

(2) The changing of motor speed and power after voltage reduction

*1) Speed*: Because the pole log, *P*, is invariant, and asynchronous motor speed is only proportional to the power frequency *f*. So, as mentioned earlier, the speed dropped by about 17%.

*2) Power*: Since setting Φ is constant, the current, *I*, is constant after voltage reduction. According to the formula of output power:

**P = √3*UI*cosψ**

put U' = f_{2}/f_{1} * U = 50/60 * U = 0.83U to above formula, then the motor power of the reduction voltage motor is about 83% of the original motor power. If the stator winding of the original motor is ∆ connection method and converted it to Y Connection method for voltage reduction, its phase voltage will drop to 1/√3 of the original, that is U' = 380 / √3 = 220 (V). At this time, the flux, Φ', is also decreasing, while the current I∝Φ. Then I' = (220/317)I = 0.694I, the current is 0.694 times than the original. Therefore, the output power is

**P _{2}' = √3*U'I'*cosψ = √3 * ((U/√3)*0.694I) * cosψ = 0.4 * (√3*UI*cosψ) = 0.4P**

That is the 40% of original power.

(3) Methods and possibilities to achieve voltage reduction; There are two ways to solve the problem of reducing voltage by 17%:

- As for areas with lower power supply voltage, it can be realized by adjusting the tap changer of the power supply transformer. Normally, the voltage regulating range of the power transformer is ±10%.
- Adding a voltage regulator. The cost of adding a regulator is about 30%~60% of the cost of replacing an electric motor, so it is economically worthwhile.

However, it must be paid attention that motor speed and power will decline to 83% of the original after voltage reduction. If the normal use of mechanical equipment is not prevented, the method of reducing voltage can be adopted. Generally, when mechanical equipment is equipped with an electric motor, they all have the margin of 20%~30%, or even greater, to prevent the motor from blocking when the voltage is reducing. In addition, many mechanical equipments have the load characteristics of the decline of speed and the decrease of torque, so the method of voltage reduction can be used. No more than, the decline in speed will have a slight effect on output (processing machinery) or air volume (blower).

(4) The requirements of motors with 60Hz 480V, 460V, 440V, 420V and 380V that are used in power grids with 50Hz, 380V and voltage reduction applications are shown in Table 1.

Table 1. Requirements for 60Hz motors being used in 50Hz power supplies

Original motor with 60Hz (V) |
480v |
460v |
420v |
440v |
380v |

Be used in 50Hz power supply with voltage reduction (V) |
400v |
383v |
367v |
350v |
317v |

The output power is 83% of the original power |
|||||

Be used in 50Hz 380v power supply directly |
It can be used with reducing output power. |
It can not be used for a long time, or even can not be used. |

**Analysis of 50Hz motors being used in 60Hz power supplies **

The method of analysis is the same as before. Now, the motor with 50Hz 380V is used in power supplies with 60Hz, 380V, 60Hz, 420V and 60Hz, 440V. The parameters are listed in table 2.

Table 2. The situation of the 50Hz 380v motor being used in 60Hz power supply

Parameters |
Be used in power supply with 60Hz, 380V |
Be used in power supply with 60Hz, 420V |
Be used in power supply with 60Hz, 440V |

Percentile value of the original motor with 50Hz (%) |
|||

Output power P |
100% |
110% |
116% |

Rated torque M |
83% |
91% |
96% |

Max torque M Rated torque M |
85% |
94% |
98% |

Starting torque M Rated torque M |
70% |
85% |
95% |

In addition, the efficiency, power factor, temperature rising and starting current of the motor are all better than those of the original when it was 50Hz.

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