How we can measure mechanical load in kW to select proper AC Motor in place of existing DC motors.?

Question

This is an old machine in process of textile (SOAPER), here all DC Motors are connected with gere box to rotate cylendrical roll(Metellic) or series of rollers ,I want to change all DC MOtors with AC motors for sake of maintenance and its dc drive which is older and not in market easily available ,and all motors run with syncronised speed,so that cloth process can take place smoothly.

Answer 1

The old DC motors should have power ratings on their name plate. Chose an AC motor with 5% more power output than the DC motors.

Suppose the rated output power of the DC motor is not given on the name plate. Then you will have to do a little experiment to find it out. If you have an electrodynamometer and tachometer, then you can measure different speed of the motor under different mechanical loads. Use the formule P = 2*pi*N*T. N is the speed in revolutions per second and T is the mechanical torque.

If you don't have the proper instruments then there is another wat. Measure the line current the DC motor draws by using an DC ammeter. Multiply it by the line voltage to get the input power. Multiply this by the motor efficiency to get the output power. If the motor efficiency is unknown, then use an eduacated estimate of around 70% efficiency.

One of these methods should solve your problem. Remember to chose AC motors with a higher power rating than the DC motors to be on the safe side.

Hope this was helpful.

Answer 2

All machines are required (by law) to have a nameplate that includes shaft power, base speed, top speed, rated voltage, rated current, service factor, and temperature rise. Most also include insulation class of the windings as well.

If the shaft power is in HP, multiply by 0.746 to get kW. This is the "output power" of the machine at 100 percent rated load conditions (nameplate voltage and nameplate current).

When deciding whether to go with AC motors over DC, there are several considerations. First, it's almost guaranteed that the DC machines are not runnning at a multiple of synchronous speed (i.e. you can't get 350 rpm from a synchronous design). Second, the DC machine can operate in two areas; the "constant torque" region (from 0 to base speed, where speed is varied by armature voltage), and the "constant power" region (from base speed to top speed, where speed is varied by reducing main field current).

To obtain a similar performance from a variable-speed AC machine, the drive pretty much has to be a Pulse Width Modulation (PWM) type. Depending on the power requirements, this can be pretty expensive compared to a simpler DC drive. Primarily, this is because the PWM is effectively TWO AC-DC converters connnected back-to-back, whereas the DC drive is a single AC-DC converter scheme.

With PWM applications, special care needs to be taken regarding grounding - the high frequency of the waveform chopping tends to introduce far more harmonics to the system (both downstream at the motor and upstream on the facility bus). With the higher frequencies, the ground signal is highly attentuated ... and if you're not careful, will cause severe bearing damage from shaft currents.

Also - in a fault condition, the DC machine simply 'flashes over' ... the mechanical contact of brush-to-commutator acts like a circuit breaker. Sure, you have to replace a few brushes and maybe a brush box or two ... and probably clean up the commutator surface. But the actual COST is fairly low - especially compared to the same fault occurring on the PWM system, where the place that exhibits a fault is going to be one of the cards (or, heaven forbid, one of the thyristors!) in the drive - a much more expensive repair!

In general, an AC machine (of the same age) as a DC machine sized for the same constant torque application will be physically larger (either in diameter, or in axial length). You will want to verify the new machines can fit the existing space, I think ... so you don't have the headache of dismantling the entire process line to put new drive motors in.

Oh yes ... for the size of machine you're talking about (probably in the order of 100 HP or more) ... a reasonable efficiency estimate of the DC machine for using the Chapman method described by another answerer is between 90 - 93 percent - not the 70 percent value that was suggested.

Answer 3

the Basic formula foe calculating power == work done/time
More over 1 Hp is equal to.
75kg-mtr/sec ie 75 kg load has to transfer to a distance of 1 mter with in isec
From this u may calucluate.