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Hi Edison123. We have recently blasted a ball mill synchronous motor 3000 HP 32 poles. the rotor was partially removed to reach the windings. Dry ice blasting cleaned nicely the areas reached by the blasting but did not well at the hidden areas between top and bottom conductors. The extreme low...

You will go to the motor and pump speed-torque curves to calculate the starting time required to reach full speed under the nominal or the reduced voltage applied to the motor. But you need the total inertia (motor+pump+coupling) to be accelerated (WK2) (Lb*Ftsq) And yes, your motor torque...

For "High Voltage Motors" the manufacturer designs the number of starts that the motor can endure before thermal damage is developed. They comply the standard of manufacture or the customer requests. The number of starts is close linked to the load inertia since the starting time and the heat...

http://i35.tinypic.com/2vrzpeb.jpg An open delta or V connection could work with 2 contactors per step . The currents are nor balanced. .

Standard NEMA MG1 Part 14 defines USUAL SERVICE CONDITIONS. Environmental Conditions Machines shall be designed for the following operating site conditions, unless other conditions are specified by the purchaser. a. Exposure to an ambient temperature in the range of –15°C to 40 °C or, when...

Only (as synchronous machine) the residual magnetism (phi)in the rotor core could generate very reduced voltage since e= -N d(phi)/dt, and phi---> 0

The voltage reduction (57.73%) due to the wye reconnection will drop the torque similar to the sample lower curve. The load torque will produce high speed SLIP. If the load torque exceeds the new reduced breakdown torque, the motor shaft will stall and the current goes to the locked rotor...

By definition an electric motor converts electric power into mechanical power. The motor losses is the part of electric energy that is converted to heat and it is proportional to the motor efficiency, with exception of the mechanical energy required by the self propelled motor air cooling fans.

The temperature shall be measured in the windings, not the external motor surface. Taking the resistance of one phase with the motor at room temperature and then comparing it to the same phase resistance after working the motor under load is a way to measure the winding temperature. See IEEE Std...

Hello joebems: The winding seems good for 2 poles. The magnetic load shows average flux densities. 59 kL/sqin in the air gap, 116 kL/sqin in the back iron and 104 kL/sqin in the teeth. The Circular mils per ampere are around 267 for the wires posted.However if you drop a single turn the...

Consult NEMA Std Publication No. MW 1000, Magnet Wire.

As WARoss stated, accuracy of temperature and water flow rate measurements is very important. The water temperature rise is pretty straight to measure the energy it removes from the heat source. It takes 1 calorie (4.184 joules) to increase 1 gram of water 1 °C.

kWloss = (liters/sec) x ( delta T) / 0.24 for water. Since 9.4 cum/hr = 2.611 liters/sec, and delta T = 39-33 °C = 6 C kWloss = 2.611 x 6 / 0.24 = 65.275 kilowatts.

The phase angle displacement is dependent on the number of poles (P).In general G° = E°/P/2 For 2 poles the geometric degrees (G°) are the same as the electrical degrees (E°). For 8 poles to displace 90°E you move 90/8/2 = 22.5 ° geometric. WRIM = Wound Rotor Induction Motor.

At 75% load assuming the power factor will be reduced around 10 points or PF = 86.5 % at full voltage and due to the flux reduction it keeps the 87.5 with reduced voltage and 75% load and no change in efficiency ( very optimistic). At full voltage, the current I1= (0.746* 400*.75) / (1.73...

In real life applications of WRIM as voltage regulator, you should take care of: 1- Torque developed by the rotor when the load and rotor current increase. How will you keep the rotor locked and how it will be moved to adjust the desired voltage? 2- If the stator voltage is fed into the rotor...

Typical figures for an induction motor 400 HP, 4 poles are; Eff = 94.5% and PF = 87.5%. This means the input power is 400/. 945 = 423.28 HP and the total losses = 423.28 – 400 = 23.28 HP (17.367 kW) If the efficiency should be increased to 100% (zero losses, ideal motor) the saving is 17.367 kW...

As almost everybody stated above you can not increase the voltage 1.73 times which is equivalent to reconnect the motor in delta and apply 400 Volts. It probably last a few minutes. However the variable frequency driver could increase the accelerating torque from 60% to 100% and only with...

A pair of low noise blowers with independent motor-fan could be adapted to provide at least the same airflow through the cooling pipes. Then the original fan could be removed. Probably you need to measure the flow and pressure of air. That is frequently made for DC machines and variable speed...

The motor torque drops with the squared ratio of voltage reduction and for 1000 HP 8 poles the starting torque at full voltage is only 60% of the full load torque. Based on that the torque at 80 volts is only 2% of the full load torque and 2.5% with 90 Volts. More Voltage is required after the...