Just buy a smaller pulley for the motor. Or a variable pulley that you can set for the speed that gives the air flow you want and then leave alone. Electronics are great but fancy electronic gadgets aren’t necessarily the solution to all problems.
Thanks for all the replies. As far as it being induction motor or not…it looks http://www.electricmotorwarehouse.com/Furnace_Motors.htmlike this], if that helps.
So from what I gather, a variac would be a safe, albeit, expensive option, and a potentiometer would be an unsafe, inexpensive option? I was really looking for a cheap option, along the lines of adding a dimmer switch or the like, but if is unsafe, it is not worth it (I sometimes leave the fan on for long periods while I’m away when trying to dry things (paint, drywall, flooded basements…)).
So what’s the cheapest, safe way of doing what I would like? (I think a Variac is too expensive for this purpose…)
The pulley on the current motor seems to be permanently attached to the shaft- I’ve never worked on a motor, but it seems that it would be too much of a hassle to try to replace it.
What’s the amperage on your blower? A quick google indicates that it’s not uncommon for dimmers to be rated for 600W, so a dimmer might be OK if your blower pulls 5A or less.
i’ll check the amperage on the motor when I get to the job site later today, and I’ll post all the relvant info…thanks again!
This is true of an ideal transformer, but in the real world, ideal transformers do not exist. If you examine an equivalent circuit schematic model for a real-world transformer such as this, you’ll note a capacitance in parallel with the secondary (among other loss paths), which is due to various intrawinding stray capacitances. A small leakage current flows through this capacitance. At 60 Hz, the capacitive reactance of this shunt capacitance is large, and usually negligible, but at higher frequencies it becomes more significant.
To reiterate the best advice you’ve received on this subject - just add a damper to the inlet or outlet of the fan to restrict the flow. It really is that simple.
And since we were discussing, in this thread, devices plugged into 60 Hz AC supplies, why would I include a detailed explanation of the negligible leakage current?
Really? A 600W potentiometer won’t handle a voltage of 115V and a current of 5 Amps? Last time I did the calculations, that meant a 575 W, which is below the pot’s rating. And it’s ‘dangerous’ to fiddle around with the wiring in a device that’s unplugged or that the circuit breaker is turned off? Since when?
Someone with knowledge enough to handle a schematic and the ability to read a data sheet should be able to handle such a thing.
Because what you said was wrong, that’s why.
It’s not that it’s dangerous to work on de-energized circuits–it isn’t, unless there are capacitors present. It’s what might happen after you plug it back in and restore power. If you don’t know what you’re doing the result can be anything from nothing to you or someone else getting electrocuted and your house burning down. Given this, being dismissive of warnings to that effect is highly irresponsible.
None of what I said was wrong.
Anyone with the ability to read and follow a schematic and a data sheet should be perfectly capable of adding a component to a circuit without burning down the house.
Oh, really?
…is wrong for the reasons I stated above.
You do not know that the OP has those skills. He may, he may not. To assume that he does is, again, irresponsible.
Nope, it’s 100% technically correct. Current cannot flow through an incomplete circuit, or are you attempting to say that it can?
I’ve specifically said that it is a reasonable thing for someone who can read a schematic and a data sheet. It’s up to the OP to decide whether or not he can read a schematic and a data sheet.
By the way, current does not flow through capacitors because electrons do not travel through them, because the dielectric does not conduct current. When one plate is charged negatively, electrons in that plate cause electrons in the dielectric to move toward the other plate, which forces electrons in that plate to move and a current will then flow on the other side of the capacitor. But to say that current ‘flows through a capacitor’ is blatantly wrong.
I doubt it. It it’s stuck on and hard to get off Dag Otto’s reiteration of casdave’s suggestion about the shutter is probably the way to go.
That’s not correct. While it’s true that individual electrons do not generally cross the dielectric (there is a leakage current, since no dielectric is perfect, but we’ll ignore that), for all practical purposes, the current can be said to be flowing through the capacitor. There are several equations for calculating this current, which you’re free to look up for yourself. However, even if we stipulate there there is zero electron current flow through a capacitor, you said “a current will not exist in the secondary unless it is placed under load.”, which is still wrong. Period.
While we’re arguing theoretical fine points, don’t forget the secondary and primary current that flows as a result of radiation from the transformer.
And by the way, caymus28, if to can borrow a puller to get that pulley off, there are matched sets of stepped pulleys so that you can have a range of blower speeds.
Of course you can also have various sizes of shutters.
Actually, it is correct. Current does not flow through a capacitor, because electrons do not pass through the dielectric, aside from the fact that no dielectric is perfect. You chastise me for not stating what is technically correct, and then you say ‘for all practical purposes’? You’re wrong on this point. While current does appear on the “other side” of a capacitor, it’s not the same current that was applied, and it definitely does not pass through. So, you’d be getting that one ticked against you on my circuits exams.
Convenient then that you ignored the rest of that sentence? but a current will not exist in the secondary unless it is placed under load (there must be a complete path for current to flow).
Now that I have highlighted it for you, can you accept that I specifically stated a closed path must exist for current flow? Or are you going to haggle over the fact that you can push electrons to or away from the ends of a conductor not in a closed loop, which is not considered to be ‘current flow’?
You left off a very important part of what I said so that you could call me wrong. Of course, I’ve had students who do this also, so perhaps you would fit in with them.
So, if I connect a large capacitor across the secondary terminls, you’re going to say there is still zero current flow? Mind telling what my ammeter is reading when I connect it in series with this capacitive load?
How did we get from “speed control of an AC motor” to a theoretical discussion on capacitors?
For the record, you’re both correct, but Q.E.D. is (at least) looking at it from a practical and useful perspective.
Yes, it is true an electron can’t tunnel through the dielectric of a perfect capacitor. So from the physical perspective, catsix is correct. But for circuit analysis purposes, the physical perspective of a component should rarely (if ever) be used. Instead, we must use lumped & distributive parameter models based on h-parameters, I-V relationships, transfer functions, etc. So from a circuit analysis perspective, current does flow trough a capacitor, and the magnitude & phase of the current on the capacitor’s two leads are always identical (just like a resistor or inductor). The time-domain I-V relationship for an ideal capacitor, as I’m sure you’re all aware of, is i = C dv/dt.
Because of capacitive reactance, which by the way is measured in Ohms, you are placing a load condition on the circuit at that point in time and you will get a current out of it.
If you’d like me to explain how capacitive reactance leads to apparent power but not true power because there is not a true resistance, sit in on my circuits class next week.
Which is why every one of my college circuits students is taught that the current they measure on the two leads of a capacitor are equal in magnitude, direction, and phase, it is still not technically correct to say that a current ‘flows through’ a capacitor.
Yup. First semester circuits course. Not something currently covered in my circuits courses because my students are not learning calculus based circuits and would have a heart attack if they saw dv/dt in any formula, but the books on my reference shelf definitely cover it, and I tend to give them problems like that when they think the approximations aren’t good enough for them.