What is the Boosting Percentage in Power Rating When an Elevator Will Start to Move

Here is the formula that you use.

I_d+10+t

AP,

This is a better question than your earlier one. The answer from AP2 is meaningless, however. Assuming you are dealing with a traction motor that is starting from a stopped condition, its inrush current is what you are dealing with. This can be a small multiplier of its normal running current (such as 1.2x or 1.5x) all the way up to 6x or even 8x. This multiplier depends on the type of control for the motor. If it is a DC motor then the multiplier is usually near the small size of this range. I'm not an elevator guy, so you probably should find someone from that background to help you.

I suspect that the manufacturers will list the required size of electrical service for an elevator machine in their literature.

--JMM

simply, we just need to add 20% from the desired load, taking in concideration the motor that need to restart more than once..

I'm not clear on what is meant by "boosting percentage", especially in relation to "power rating". The power RATING of an electric motor does not change. In fact one could argue that the POWER consumption of an electric motor is not much different at starting than at running either. Only the starting CURRENT is different, but current is not the same as power. Current is high at starting mainly because the power factor is very very low.

So to the OP, what is the purpose of your question? We, as a group, tend to wax off into obscure data without a stated need, so you may end up with a lot of confusing and distracting information to try to sort through if you make us keep guessing.

AP1,

I am assuming you are fairly new to the electrical trade and trying to learn. I also assume that English is not your first language; if so, I congratulate you on your efforts.

Basic electrical theory:

• Voltage is what moves the electricity through the circuit, measured in volts, symbol is E.
• Current is the quantity moving at any one instant, measured in amps (amperes), symbol is I.
• Power is the work this electricity can do, measured in watts, symbol is P.
• Every conductor has resistance to the flow of electricity, measured in ohms, symbol is R and uses Greek letter omega Ω.
• In DC circuits these are related by the formula P = I*E (or P=IE), and E = I*R (or E=IR). With algebra you can solve for any one of these in given any two others.
• In AC circuits, the voltage and current are oscillating in a sinusoidal waveform (ideally) from zero through a positive peak back through zero to a negative peak and back to zero; the frequency for each complete cycle is typically either 50 or 60 cycles per second, measured using units of Hertz (Hz).
• In AC circuits inductors charge/discharge a magnetic field and capacitors charge/discharge an electrostatic field. These fields cause the current to be out of phase with the voltage--leading it for capacitors and lagging it for inductors.
• In real-world AC circuits, this phase "out of phase" condition is called the phase angle, using the Greek letter theta (Θ).
• Therefore in real-world AC circuits, P = IE(cosΘ), because the real work is only possible for the portion of the cycle where volts and amps overlap. However, equipment such as transformers must still supply both volts and amps to the circuit, so they are given ratings of VA or kVA or MVA (depending on size)--note the cosΘ portion of the equation is ignored.

Back to your question in this post (and in the other post also):

• You cannot figure the power requirements if you don't have (typically) the current AND the voltage. Tornado was repeatedly saying this to you!
• JRaef has pointed out that the term "boosting percentage" is not defined to most electricians. Perhaps this is a poor translation into English of a term better translated as "inrush". When a motor starts, a large amount of current must flow into it to get it to turn (called the inrush current or starting current). As a motor begins to turn, the magnetic field that develops inside the motor generates a voltage that opposes the power line's voltage--the faster it turns the greater this voltage becomes. Thus, the motor windings will effectively work with much less than the imposed circuit voltage when the motor is at its operating speed. At this time, the motor is drawing its running current. The inrush current can be as high as 6x or even 10x the running current, but different methods for starting the motor can reduce this inrush current to as little as 1.2x the starting current (and sometimes it can be lower).
• This was what I was attempting to say in an earlier post.
• When Suz said 20%, for the "boosting percentage", I have to assume this was an inrush of 1.2x the running current, which is very low.
• The motor's supply circuit must be capable of providing the required current and voltage to run a motor. However, because the inrush is very brief (in most cases), transformers and wires normally are not sized for the inrush but for the running load. To provide enough power to start a motor, the transformer must be sized with an extra capacity above the running load--I believe this is the question that your original post is asking.

Therefore--

1. Look at transformer manufacturers' literature to see if they give multipliers of motor loads for sizing a transformer.
2. Look at elevator manufacturers' literature to see if they give minimum required transformer sizes for a particular type of elevator controller/motor.
3. Then, if your question has not been answered, come back to CR4 with a question based on what you have learned and using terms that are familiar to most electricians and engineers.

--John M.

The motor current on start up might also depend on whether you have a lift full of chip eating monsters - a couple of fat lads - of whether there is a size 0 young lady on her own.

PC - never!