Does a 5.5kW 3-phase motor use 5.5kWh or more per hour at normal load?

Question

I would like to find out whether my 5.5 kW three-phase motor, operating at normal load for an hour, will consume 5.5 kWh of electricity or three times more. Thank you.

kindlar · Answer

Show the nameplate and see the cosφ (its efficiency). The engine will draw enough energy from the network to compensate for the load, which may result in its heating. If you load it more, the rotor`s slip in relation to the rotating magnetic field will be greater, and it will draw more current to keep up with the stator field rotating at a constant speed. (Rotor slippage is normal, so don`t worry).

hrabia24 · Answer

The question is: will it use 5.5 kWh when running without load for an hour, or 3 times 5.5 kWh?

mczapski · Answer

If First of all, it`s best to check it practically. Power is power regardless of the number of power phases. Going further, power is work over time. So if you`re not putting any load on the engine, you`re basically not doing any work. Something doesn`t seem right.

jurek84 · Answer

It will consume 5.5kWh.

Krzysztof Reszka · Answer

Is it exactly that much or should something be added to this bill, e.g. some losses or resistances.

czesiu · Answer

You would have to add it, because engines are determined by shaft power. The power consumed from the network is the product of voltage, current, cosφ and √3. Cosφ is not efficiency.

jurek84 · Answer

Right, I forgot to add that for the perfect engine. Divide the power by the efficiency and you will have the power drawn from the network during continuous operation under the maximum allowable load.

Krzysztof Reszka · Answer

These are heat losses, bearing resistance, and engine cooling fan. Engine efficiencies are listed in engine catalogues.

retrofood · Answer

Apart from these kW/h, which please correct, only a fortune teller knows how much the engine will consume under normal load, because apart from the fortune teller, no one knows how much is normal. Especially since you are about to write about working without any burden. However, the technical term is load rated . And for the rated load, all values are given on the nameplate.

Kris555 · Answer

The energy consumed by the receiver is the product of its power and time unit. In other words, your engine will consume approximately 5.5kWh per hour at rated load, of course with a constant and unchanging load.

Anonymous · Answer

At rated conditions it will consume 5.5 kWh/energy efficiency. The rated conditions are continuous mechanical power received from the shaft at a level of 5.5 kW, rated speed, etc. The efficiency given in the catalogs also applies only to rated conditions. Over the full range of engine load is different at each point.
In summary, an upper limit can be estimated. In practice, when working with a receiver that probably works with variable loads, you need to measure it yourself.

PS. The rated energy efficiency of the motor is given on each plate, which contains information about its power, rated current and power factor. How to calculate?
N/P = η
N - power from the plate on the motor shaft
P - electrical power drawn from the supply network
P = √3 * U * I * cosφ

mczapski · Answer

Well, well. And which of the distinguished informants can provide the engine/system with a constant, nominal load for a long period of time? Of course, outside the laboratory or measurement station. And the questioner insisted on knowing for rated parameters or maybe without load. Does this not bother you?
As I have already testified, power is power regardless of how it is obtained. You consume and pay for work (energy consumption). So, in real conditions, it is unlikely that you will load the engine with just 5.5 kW (never mind the little things) and you will pay for the number of work units multiplied by the unit rate plus additional fees. The measuring system reads the amount of work.

Anonymous · Answer

The question is about normal conditions, i.e. variable loads can be assumed to an indefinite degree, from a few percent to overloads. Upper limit - it won`t be more expensive. Isn`t this comprehensive news?

Aleksander_01 · Answer

Oh, gentlemen, but these 5.5 kW are shaft power, not absorbed power.
At rated load, the power of such a device drawn from the network is greater by the efficiency, cos φ (to put it briefly).
All data is listed on the nameplate, including power consumption.

jurek84 · Answer

This is not a problem with fans or pumps, and these are just 2 examples. I think there is no point in continuing this thread any longer because the author`s question has already been answered several times.

mczapski · Answer

There`s really no point in continuing, but when experts write something like this, it`s hard not to react. Did your friend know the laws of thermodynamics? For example, a change in the medium temperature by 1°C causes changes in the engine`s operating conditions. So be careful with exposing your messages.

hrabia24 · Answer

I have capitulated and am closing the issue.

retrofood · Answer

In technology, you should use strict terms, if you call something normal, everyone can understand it in their own way. Hence, there is no chance to obtain a clear answer.

Does a 5.5kW 3-phase motor use 5.5kWh or more per hour at normal load?

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