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Yes there are solutions. You may not want to use a magnetically latched signal as it may require another signal to release.
Some ideas you can consider.
The voltage level with which you are dealing is TTL. If you do a web seach on “TTL driven Relay” it will bring up a number of possibilities. One possibility is the Tecnec TCR102 (elunits.com/relays). It requires an external power supply but this can come from your PLC I/O supply (there are a number of voltage options available) . The load on the driving signal is only uA at the TTL 5V level so should suit your application. This is an industial type unit which you can wire straight in.
Another possibility is to use a reed reed relay. You can get low power coils in these but they are generally designed for printed circuit board applications so are small and fiddly. You would need to devise a mounting and connecting method. Try a websearch on “reed relay” and check out the range that Farnell has.
The lowest cost and crudest solution would be to select a suitable transistor, drive the base with your 5V signal and use the collector, powered by the PLC 24V supply, to switch a relay coil. Depending on the configuration of yout PLC input module you may even be able to rig to switch it directly (without the relay). May require a bit of mucking around. A disadvantage of this is that it would common the OV rails of your TTL level signal and the PLC and you may not want to do this.
As NYOJ says make sure that you account for transient loads such as starting currents, particularly if you have one or more loads which consttitute a significant proportion of the generator capacity.
Another point to consider is harmonic current. Today, installations often include many non linear loads such as Variable Speed Drives, switched mode power supplies etc. These all produce harmonic currents. You will need to measure, or make an assessment of, the harmonic current content in the system and advise the generator supplier of this as he may need to make an allowance for the additional heating these will cause in the generator.
Smithy
You have the real power (W) for each item.
You have the power factor for each item.
Calculate the apparent power (VA) for each item. This is Watts divided by Power factor.
Add the Watts (W) for each item to get the total real power (W).
Add the apparent power for each item to get the total for apparent power (VA).
The overall power factor is total real power (W) divided by total apparent power (VA).
To get power factor angle take acos of the power factor.
Reactive power (VAR) is sin (power factor angle) x apparent power (VA).
Determine the total reactive power( VAR).
Determine the apparent power (VA) at 0.98 power factor by dividing the total real power (W) by 0.98
Determine power factor angle at power factor 0.98 = acos 0.98 = 11.4degrees
Determine total reactive power (VAR) at 0.98 power factor = total apparent power (VA) x sin 11.4 degrees.
Amount of compensation required = total VAR uncompensated – total VAR at 0.98 power factor.
Good luck. Don’t forget to consider what happens of load is only partial rather than full.
By ‘Reswitching” I suspect that you mean the loss of supply to the bus feeding the drive either due to grid failure of switching from one supply to another and your expectation is that the drive will ride through this event then re-establish speed/flow when the power is restored. This is possible and the principle is known as” kinetic buffering”. It involves using the momentum stored in the driven system to keep the drive control alive whilst there is no incoming feed and reducing speed in a controlled manner to avoid overloads but its success depends on a number of variables including length of supply outage and inertia of the driven system. First of all check that your VFD support “Kinetic buffering”, not all do, and that it is correctly programmed to make use of the feature
I hope that this answers your question but, if not, and you want concise answers you need to be specific in your question indicating the situation and your desired outcome so that people do not have to try to guess what you mean.
The motor does not reverse its direction when you go into regeneration but what does reverse is the direction of current flow. Thus you need a second bridge in regeneration to conduct current in the opposite direction to that when you are motoring. You could also consider using an H-bridge topology. Remember also that when you are regenerating the motor(generator) voltage is higher than that of the battery (which is what forces the current back into the battery) so your voltage regulator will need to sense whether it needs to provide motoring energy (Battery volts higher than motor and forward bridge conducting) or absorb regenerated energy (Motor volts higher than battery and reverse bridge conducting).
There will be some energy saving that is why I said “next to no savings” rather than none but lets us keep it in perpective.
If you had a 1MW load at 0.8 power factor and install correction to 0.95.
1MW @ 0.8 = 1.25MVA
1MW @ 0.95 = 1.05MVA
If one assumes ohmic losses in the cabling system of 5% then the power saving due to current reduction would be approx (1.25-1.05)/1.25 or approx 16%
16% of 5% is approx 0.8%
On a kW tarrif your savings would be approx (0.008 x 1.25)/1 =.001 or 1% of the power bill which is purely the loss reduction saving.
If you were on a KVA tarrif the saving would be the Power factor improvement as well as the loss reduction and would exceed 16%
If you were on a kW tarrif your max demand in kW would only reduce in the order of the loss reduction but if you were on a kVA tarrif the MD kVA reduction would be significant.
Understand the tarrif.
@anilsabaji said:No..the energy saving is by reducing the losses in the system also. However, saving from Max demand reduction may be higher than the saving by reduction in losses in case of small installations.
@Smithy said:Just a point to watch. I have seen a number of people install power factor correction equipment in the hope of making savings and see next to no saving in their electricity costs simply because they were on a kW tarrif rather than a kVA tarrif. There should be something in the calculator to alert people to thoroughtly understanding their tarrif structure.
Just a point to watch. I have seen a number of people install power factor correction equipment in the hope of making savings and see next to no saving in their electricity costs simply because they were on a kW tarrif rather than a kVA tarrif. There should be something in the calculator to alert people to thoroughtly understanding their tarrif structure.
