voltage drop

[iaov]

Motors are not "different" . I still equals E over R or in this case over Z. Impedence is a vector sum of resistence and reactence in an AC cicuit. Inductive reactence depends on frequency not votage.

 

[don_resqcapt19]

If the work is also less to do,then the motor should be content with less current.

Why would it have less work to do? It is still getting the same frequency and trying to run at the same speed.

 

[T.M.Haja Sahib]

Motors are not "different" . I still equals E over R or in this case over Z. Impedence is a vector sum of resistence and reactence in an AC cicuit. Inductive reactence depends on frequency not votage.

Impedance of a motor is a fascinating thing:it can represent external load on the motor as part of its resistive component.Due to this resistive component,the impedance of the motor depends on the applied voltage.

 

[T.M.Haja Sahib]

Why would it have less work to do? It is still getting the same frequency and trying to run at the same speed.

Observe a ceiling fan provided with a speed regulator;observe current taken by the ceiling fan at minimum and maximum fan speed.Current will be less at minimum speed.Why?The fan load and hence the work done by the fan motor is less.

 

[don_resqcapt19]

Observe a ceiling fan provided with a speed regulator;observe current taken by the ceiling fan at minimum and maximum fan speed.Current will be less at minimum speed.Why?The fan load and hence the work done by the fan motor is less.

It does less work as the fan slows down. When you reduce the voltage to a normal blower motor it tries to run at the same speed. If it is running at the same speed the work is the same and the current has to go up. At some point the motor will no longer be able to run at the same speed as the "slip" will increase. When this happens the current will go down as the work is going down. I think that the work changes with the cube of the change in speed so a small reduction in speed will result in a large decrease in the work.

 

[ActionDave]

Motors are not "different" . I still equals E over R or in this case over Z. Impedence is a vector sum of resistence and reactence in an AC cicuit. Inductive reactence depends on frequency not votage.

Vector sum of the angle of the bifocal times Z plus an inductive load vs. a resistive load = Motors are different.

Yes, reluctance, resistance, impedance all equal R; but powering an incandescent bulb or resistive heating element vs. trying to start a motor on too low of a voltage is not the same.

 

[infinity]

I would guess that your typical residential A/C unit can run easily even in a brown out condition. Here's the nameplate of my unit. It will run on as little as 187 volts supplied by a 240 volt system. I would think that many residential units would run the same way.

 

[iaov]

Vector sum of the angle of the bifocal times Z plus an inductive load vs. a resistive load = Motors are different.

Yes, reluctance, resistance, impedance all equal R; but powering an incandescent bulb or resistive heating element vs. trying to start a motor on too low of a voltage is not the same.

Agreeded/ But until that motor starts turning it is basicaly a short circuit. Thus the high inrush currents. When you load a motor down you are changing the impedence of that motor because you are changing the interaction between the magnetic fields of the windings and the armature. They are a moving transformer and Tesla originally patented them as such. There are no exceptions to Ohms Law for motors. Motors are not different!!

 

[CFL]

I was looking for information on voltage drop formulas and stumbled on this discussion about motors. This is a concept I've been trying completey understand for years. I have heard so many people say "less volts=more current to a motor". I know that it's sometimes true, but they usually claim it is always the case. Obviously, reduced voltage starters are designed to reduce the startup current. Once the motor is at running speed, any reduction in voltage will reduce the amount of counter emf.......and so on. Sorry, just looked at the clock, late for work. It'll take me way too long to get this outta my head right now. Can someone steer me in the right direction here, will check back after work.

 

[kwired]

I was looking for information on voltage drop formulas and stumbled on this discussion about motors. This is a concept I've been trying completey understand for years. I have heard so many people say "less volts=more current to a motor". I know that it's sometimes true, but they usually claim it is always the case. Obviously, reduced voltage starters are designed to reduce the startup current. Once the motor is at running speed, any reduction in voltage will reduce the amount of counter emf.......and so on. Sorry, just looked at the clock, late for work. It'll take me way too long to get this outta my head right now. Can someone steer me in the right direction here, will check back after work.

Without getting too technical (which would take a little reasearch from me just to make sure I am providing accurate info) when you use reduced voltage starting, part winding starting, wye-delta starting, or other methods that are not across the line starting the main goal is usually to reduce the amount of inrush current. As has been said a motor that is not turning is a short circuit. Providing only part of the voltage, or part of the windings will lower the amount of current provided to the non turning motor. This is usually only for a couple of seconds or so at the most, by then the motor is turning and the full voltage / full windings is needed to develop enough torque to continue to accelerate the load more efficiently. Like I said this is a short description and I believe accurate enough to get the general idea.

 

[Jraef]

... he said i need a single phase "PHASE LOSE MONITER" .....

Think that one through a little bit.

You have SINGLE phase.

You lose ONE phase.

What's left for the monitor to monitor?

Look, I know your customer wants you to find a way for him to say "Screw everyone else, I want MINE!". But the harsh reality is, this just isn't practical. I would tell him it is possible, and it's going to cost him $13,458.62. If that doesn't shut him up, then you have enough in it to keep wasting time on solutions, knowing that your ultimate fall back is the generator and transfer switch. If he thinks it's too much, then you are wasting your time because whatever you come up with isn't going to work perfectly and he won't pay you.

Think about what he is asking for and what kind of person that makes him out to be. Do younthink he will be likely to treat you fairly if something doesn't please him?

 

[dkidd]

I was looking for information on voltage drop formulas and stumbled on this discussion about motors. This is a concept I've been trying completey understand for years. I have heard so many people say "less volts=more current to a motor". I know that it's sometimes true, but they usually claim it is always the case. Obviously, reduced voltage starters are designed to reduce the startup current. Once the motor is at running speed, any reduction in voltage will reduce the amount of counter emf.......and so on. Sorry, just looked at the clock, late for work. It'll take me way too long to get this outta my head right now. Can someone steer me in the right direction here, will check back after work.

There was an article in EC&M that discusses the fact that when voltage goes down current goes up.

http://ecmweb.com/design_engineering/electric_highs_lows_motor/

 

[dkidd]

These graphs show the relationship between voltage and frequency and motor parameters.

http://s1187.photobucket.com/albums/z390/dkiddpe/