115 or 230?

[DonL]

I removed the formula because you are confused enough already.

Yes the formula was correct. http://en.wikipedia.org/wiki/Ohm's_law

Sorry about the smoke comment, but you are playing around, So I needed to too.

http://en.wikipedia.org/wiki/Magnetic_circuit

 

[LLigetfa]

Cutting the voltage in half only results in roughly doubling the current when a motor is rewired for 110, not when a motor wired for 220 is fed 110.

I think there is already too much smoke being inhaled by some folk.

 

[Sponsor]

 

[DonL]

Cutting the voltage in half only results in roughly doubling the current when a motor is rewired for 110, not when a motor wired for 220 is fed 110.

I think there is already too much smoke being inhaled by some folk.

That may be true for a resistive load, but this is a motor.

I have smoked a few things, but did not inhale.


What is the make and model number of this pump motor that we are smoking ?

 

[FrustratedinNJ]

Cutting the voltage in half only results in roughly doubling the current when a motor is rewired for 110, not when a motor wired for 220 is fed 110.

I think there is already too much smoke being inhaled by some folk.

I think you're right about that. DonL said in an earlier post that "When you cut the voltage in half, You double the current." I don't think so. I did some research and fiddling around with the formulas that apply.

See Question 60 here and the answers to it:

http://www.physicsclassroom.com/reviews/circuits/circuitsans4.cfm

Specifically, question and answer e, which says that if the voltage is halved and the resistance is doubled, the new current will be 1/4 of what it was.

DonL said that in my case, where I'm running a motor wired for 230 volts but powering it with 115 volts (halving the voltage) I'd be doubling the current. But I think I'm correct in saying that the windings when wired for 230 volts have more resistance than the windings when wired for 115 volts (4 times the resistance, I think) so according to question and answer e, not only wouldn't the current be doubled, it would be less than 1/4 of what it was (if I'm right that the resistance is higher by a factor of 4 as opposed to being merely doubled).

Because, according to question and answer e, when the voltage is halved and the resistance is [more than doubled], the new current will [less] than 1/4 of what it was (because doubling the resistance results in current that's 1/4 of what it was, so increasing the resistance above doubling will result in even lower current).

My pump is a 1 HP pump regardless of how it's wired (as long as it's matched to the power supply).

Which means it's 1 HP when wired for 115 and powered by 115, and it's 1 HP when wired for 230 and powered by 230.

So since it has the same horsepower either way, it makes sense to me that when you run it at higher voltage (230) the resistance in the windings has to be higher if you're going to maintain the same horsepower. Seems to me that if the resistance was the same (or less) the motor would produce more power, not the same power. So to maintain the same horsepower, the resistance has to be higher the higher the voltage.

And if the resistance is higher, the current certainly wouldn't double, as DonL claims. Even if the resistance wasn't higher the current wouldn't double if the voltage was halved. They don't give that exact scenario in question 60, but the implication pretty clearly is that halving the voltage results in less current since halving the voltage and doubling the resistance results in 1/4 the current.

So I think that's what's going on when I'm running the motor wired for 230 volts and powering it with 115 volts is that I'm pushing 115 volts through higher resistance at slower motor speed (thus there's more heat produced) coupled with slower cooling fan speed (dissipation of less heat resulting in more overall heat). And there's probably other factors that also are goin' on (valveman opined that "slip" was a factor here, saying that: "The more slip, the more the rotor bars are passing through lines of magnetic flux, and the more heat it produces."

And in any event, DonL seems to be too much of a non-helpful smart-ass for my tastes. I frankly wonder if he knows what he's talking about at all.

 

[DonL]

LOL. I have been known to have my head up my ass.

You post on a thread that was started in 2008, and tell me I am "too much of a non-helpful smart-ass for my tastes.". You crack me up.

Most motors are made to operate within 10-20% of their Voltage and Frequency design range. Some require 5%. Depends on the motor.

You are right, duty cycle is the Key. The motor will last a long time at 0%.

Seems like fixing your original well problem would be a better solution. Or control flow with a valve, as was suggested.

I would like to know how you are going from 120v to 240v to wash clothes ?


Have a Great Weekend.

 

[Jadnashua]

What you don't understand is that the amount of power sent to a motor is Volts * amps. A motor needs a certain amount of power to run. If you halve the voltage, to get the same power to it, you must then DOUBLE the current. One big reason why people tend to use the higher voltage option on a motor is that you can run smaller wires to it because the amount of current going through them is half of what is required at 120vac. Especially if the supply wires are too small, this makes their resistance higher at load, and they will heat up as well as the motor.

 

[LLigetfa]

Jim, you need to understand the difference between internally strapping a motor to run on 110 or 220 versus feeding a motor 110 when it is strapped for 220.

A motor is an inductive load, not a simple resistive load so one cannot apply basic Ohm's Law here. The motor is designed to run 3600 RPM and lowering the RPM by reducing the voltage is not advised.

 

[FrustratedinNJ]

What you don't understand is that the amount of power sent to a motor is Volts * amps. A motor needs a certain amount of power to run. If you halve the voltage, to get the same power to it, you must then DOUBLE the current. One big reason why people tend to use the higher voltage option on a motor is that you can run smaller wires to it because the amount of current going through them is half of what is required at 120vac. Especially if the supply wires are too small, this makes their resistance higher at load, and they will heat up as well as the motor.

First of all, thanks for the 3 replies since my last long-winded post.

Jim -

Not sure if you were talking to me or to DonL, but I think it must be me since it seems you agree with his claim that I'm doubling the current. If you were talking to me, I did know that P = IE (and I think that might have been the "formula" (I should have said "equation") that DonL had intended to put in his prior post but then removed from the post because (he claims) I was "confused enough already." Assuming he had the equation correct, that logic escapes me. But most of what he's said here doesn't make a whole lot of sense to me. Of course, he acknowledges that he's "just a old fart" (shouldn't that be an old fart?) who has been know to have his head up his ass, so perhaps that's why he seems a tad confused himself, not to mention a tad too confrontational and unhelpful for my tastes. But perhaps he has "issues."

I'm thinking he removed the equation because he didn't get it quite right.

I'm also familiar with Ohm's law, E = IR (or V = IR), where E is energy and V is Voltage and they are interchangeable (or so I'm led to believe)

But, as I said, I'm no electrician (or electrical engineer). I'm sure I had this stuff in college and probably also in high school, but there's been a lotta water under the bridge since then.

Long story short (tho' likely not short enough) I think I agree with LLigetfa in that because my situation involves an electrical motor, basic Ohm's law isn't really applicable here. Perhaps ironically so, DonL also made reference to that fact in what appears to be his opposition to LLigetfa's contention that he (DonL) was incorrect in his assertion that I'm doubling the current. See: Yesterday at 9:15 PM

I've already learned a lot from this discussion (and probably also have a lot of misconceptions as well, but that's part of the learning process, after all) but the bottom line for me is that everyone seems to agree that running the motor powered by 115 volts (or 110, if you prefer) when wired for 230 volts (or 220 if you prefer) is not a good idea and could burn the motor out.

So I'm thinking that what I should do -- or try -- is run the motor with 115 volts, wired for 115 volts, but do so via a good old-fashioned rheostat such that I'm running it on less voltage and thus slowing it down that way.

No one has given me feedback yet on that idea but I'm thinking that this approach -- if it works at all -- would have less likelihood of damaging the motor. The way I'm doing it now appears to have the very real probability that I'll burn the windings in the 230 volt circuit and/or fry the motor.

To DonL: In answer to your question (I would like to know how you are going from 120v to 240v to wash clothes ?) ...

I simply rewired the motor back to 115 volts and ran it normally. As I said previously, the motor can be wired for either 115 or 230 and it just involves switching each of 2 wires from one terminal to another. I think this is a fairly common thing with these motors.

I could add a great big "DUH!" but I'll cut you some slack since you're just an old fart with your head -- at least at times -- up your ass.

Also, your suggestion that I just fix my well problem is, of course, the best solution. I already knew that (duh!) but my reasons for doing what I'm doing are irrelevant to this discussion. So let's try to keep the discussion on topic.

Why can't we all just get along?

If anyone does have a suggestion as to how I might solve the basic problem by some means other than pumping the water more gently by slowing my pump down, please chime in. I don't think controlling the flow with a valve, as has been suggested, is going to work. I considered a centering gizmo, which is actually commercially available, but I think that would just move / vibrate / stir things up as is apparently going on now. I think the commercially available (stainless steel) gizmo is intended for use with an in-well pump, so adapting it for use with my jet pump could be difficult.

The only real solution(s) I can come up with is to get casing in the well, or try to center the foot valve in the well by dumb luck (which seems unlikely without straightening the poly pipes), or go to an in-well pump (which could have the same problem I'm having now and probably would), or fine tune the venturi/nozzle combo to produce much slower, gentler pumping when running the pump normally. That last one would involve pulling the pipes out of the well and putting them back in, probably many times, before I got the combo right -- if ever.

So I'm not liking that idea much.

 

[DonL]

Maybe this will help you some.

http://en.wikipedia.org/wiki/Electrical_impedance


You are fun to play with... Stop Throwing sand in my eye. lol

 

[LLigetfa]

A rheostat is just a voltage divider and not much different than what you are doing now except that it is variable. None the less you would be running a motor under low voltage and pump motors are not designed to run that way.

 

[FrustratedinNJ]

A rheostat is just a voltage divider and not much different than what you are doing now except that it is variable. None the less you would be running a motor under low voltage and pump motors are not designed to run that way.

My thought was that since I'd be running it wired for 115 volts, albeit at lower voltage, I'd at least be using the windings that were designed for 115 volts as opposed to running it the other way, where I'd be running it at 115 volts but through windings designed for 230 volts.

Jim said: One big reason why people tend to use the higher voltage option on a motor is that you can run smaller wires to it because the amount of current going through them is half of what is required at 120vac.

This suggests to me that the windings utilized when wired for 230 volts are not as heavy as those when wired for 115 volts and/or the 230 volt windings have more resistance than the 115 volt windings. Either way, it seems more likely that the 230 volt windings could be more susceptible to burning out than the 115 volt windings, even when those windings are run at lower voltage.

Even tho' Ohm's law doesn't quite fit an electric motor situation, I think the 230 volt windings would necessarily have to have higher resistance since P = IE. If E=V (that is, "energy" = "voltage") then P = IV

Since 230 volts is higher than 115 volts (by a factor of 2) then I in the equation (current) would have to have a lower value than when the pump is wired for and run by 115 volts. That is, if the motor is to have the same power (P) as when run on 115 volts. Which is the case with my motor because it's a 1 HP motor no which way it's wired (as long as it's powered accordingly)

So if I (the current) is lower re: the 230 volt windings, those winding must have more resistance than the 115 volt windings. Because those windings are designed for 230 volts which is twice the voltage of 115 volts and the only way to keep the power the same is to lower the current when powered by 230 volts. That is, the only other variable in the equation that can be changed is I (current). And it has to be lower since P remains the same and V (or E) is higher (twice as high in fact).

That's a long-winded way of telling you and others what you already know, unless I'm totally off base.

Anyway, that's my thinking. I know and expect that running the pump with a rheostat will produce more heat than when it's run without a rheostat (at full power), but I think that slowing it down that way is a safer way of doing it than the way I'm doing it now. That is, it's more likely to do less harm -- hopefully no harm -- to the motor.

Or maybe it'll just be the same? If so, perhaps someone could 'splain to me where I'm going wrong?

P.S.: I know that P = IE is not Ohm's law.

 

[DonL]

You should get a Amp Probe and a Volt Meter and do some testing. Make sure you have your PPE while testing.

Something that can measure PF would be helpful also. http://en.wikipedia.org/wiki/Power_factor

You seem to think you are playing with a DC motor, and pure resistance.

You never did say the model of your pump motor, or I missed it. Do you have a link to your old thread ?

Let the motor run awhile and make a Utube video. It will get a lot of views.

You crack my ass up, But I like it. lol

 

[Jadnashua]

P=v*a. Power=volts*amps (often written P=I*E). The pump wants a certain amount of power, so if you drop the voltage by half, the current must double to keep the overall power the same. Basic electronics and algebra. Not rocket science. To carry more current requires bigger wires, or they begin to act like resistors or heating elements. Think of a light bulb...the filaments are quite small in diameter compared to the power cord going into the lamp for a reason - so the wires don't get too hot. You don't want to make your pump into a light bulb filament! Wire it as it was intended, or you're likely to induce an early demise.

 

[FrustratedinNJ]

P=v*a. Power=volts*amps (often written P=I*E). The pump wants a certain amount of power, so if you drop the voltage by half, the current must double to keep the overall power the same. Basic electronics and algebra. Not rocket science. To carry more current requires bigger wires, or they begin to act like resistors or heating elements. Think of a light bulb...the filaments are quite small in diameter compared to the power cord going into the lamp for a reason - so the wires don't get too hot. You don't want to make your pump into a light bulb filament! Wire it as it was intended, or you're likely to induce an early demise.

Uhhhhhhh, P = IE is the same as what you wrote (P=v*a. Power=volts*amps (often written P=I*E)) so I'm not sure what your point is in repeating it unless you didn't think that "IE" is just another way of writing "I*E"

The pump wants a certain amount of power, so if you drop the voltage by half, the current must double to keep the overall power the same. Basic electronics and algebra. Not rocket science.

This is the same point you made in your prior post with which LLigetfa disagreed (or so it seemed to me) ( Today at 5:09 AM )

The rest of your post is just what I already said and your conclusion is what's already been recommended by just about everyone and doesn't address my current proposal to "wire it as it was intended" but power it at some lower voltage via a rheostat.

BTW, not that it matters, but my motor runs at 3450 RPM and draws 14.8 / 7.4 amps max load (115/230). It's also a 3/4 HP motor, so says the plate on it. I thought it was 1 HP, but it's not.

 

[LLigetfa]

I give up.

unsubscribing

 

[DonL]

I give up.

unsubscribing

Don't be a party pooper, You should stay and play.

 

[Reach4]

Not in the way you envision it. It is not like the head of a vacuum cleaner where proximity matters.

I think it could be like the head of a vacuum cleaner hose. I could imagine some type of material near the intake that could be loosened or eroded by turbulence.

A shroud should make it easy to ensure that the higher-velocity intake is not right next to the bore wall. I don't see why it would take "many times" to get the intake away from the well sides. Yes, the picture is for a submersible pump, but it seems to me that the same general construction could be adapted. Maybe extend the sleeve a foot below the jet assembly intake to let the flow smooth out.

 

[DonL]

A easy fix may just be to shorten the drop pipe a few feet, to get the foot valve out of the mud.

 

[LLigetfa]

Don't be a party pooper, You should stay and play.

Well... it would appear that the board won't let me unsubscribe.

 

[LLigetfa]

I think it could be like the head of a vacuum cleaner hose.

I addressed that in post #38. A long shroud would act as a sand separator if the sand is entering from above the pump.