Author	[Modern View]
joe fabregas 08-12-2012 08:06:36 12.75.112.169 Report to Moderator	I'm working on digesting George's experiment. I have a 1/2 hp, 120v, cap start well pump motor. If I add a 25mf, 600v cap in line I should see a drop in amps? If I add another 25mf cap in line with the first I should see another drop in amps? An so on until the amps begin to rise again. Once I achieve the optimum capacitance I should be running efficiently and my genny should have an easier life when the power goes off? Have I got it right? Thanks for the inspiration! Joe Fabregas [Reply]   [No Email]
David G 08-14-2012 07:10:21 205.215.206.18 Report to Moderator	Re: George Marsh experiment? in reply to joe fabregas, 08-12-2012 08:06:36   It is IMPORTANT not to have capacitors hooked up when generator is starting. They will be a dead short when it is coming on line. The capacitors should be installed after the contactor. [Reply]  [No Email]
JMOR 08-14-2012 05:35:02 72.181.173.171 Report to Moderator	Re: George Marsh experiment? in reply to Roger in Iowa, 08-12-2012 08:06:36   Quoting Removed, click Modern View to seeNot to put words in Mark's mouth, but I don't think he is saying, " you're saying the genny will have to work the same either way?" P=IE is the fallacy here. P=I x E x cosine of phase angle VA =I x E. Generator is needing to provide VA and if you correct the power factor it will only need to provide P (the smaller of P and VA). [Reply]  [No Email]
David G 08-14-2012 05:54:15 205.215.206.18 Report to Moderator	Re: George Marsh experiment? in reply to JMOR, 08-14-2012 05:35:02   JMOR, you are correct. The generator must provide both real and apparent power. [Reply]  [No Email]
buickanddeere 08-13-2012 15:39:49 209.240.125.254 Report to Moderator	Re: George Marsh experiment? in reply to showcrop, 08-12-2012 08:06:36   Quoting Removed, click Modern View to see Only place a cap will benefit if it high enough mf and high enough voltage and placed in series with the start winding circuit. More phase shift = higher starting torque. [Reply]  [No Email]
John T 08-12-2012 16:40:27 216.249.82.117 Report to Moderator	Re: George Marsh experiment? in reply to joe fabregas, 08-12-2012 08:06:36   You only add enough capacitance (in paralell with motor NOT series) until a unity (1) Power factor is achieved such that the current comes back in phase with the voltage AND NO MORE. The existing starting capacitor is only in the start winding and only for start up to improve torque. John T [Reply]  [No Email]
joe fabregas 08-13-2012 09:43:06 12.75.111.53 Report to Moderator	Re: George Marsh experiment? in reply to John T, 08-12-2012 16:40:27   Thanks John T. Guess I forgot all about my series/parallel lessons. Probably be better if I let this dog sleep. Good thing you don't live closer I'd be following you around like a puppy dog. Thanks for all your interesting posts! joe- [Reply]  [No Email]
John T 08-13-2012 10:27:47 216.249.82.117 Report to Moderator	Youre Welcome in reply to joe fabregas, 08-13-2012 09:43:06   Thanks for the kind words, youre welcome John T [Reply]  [No Email]
George Marsh 08-12-2012 17:13:22 50.104.223.0 Report to Moderator	Re: George Marsh experiment? in reply to John T, 08-12-2012 16:40:27   JohnT, You are our expert EE here, so tell me if I got this about right. When the voltage and current are in phase, the inductive reactance is canceled out by the capacitive reactance. Therefore there are two major things using the applied power. One is the power lost because wire has resistance, I squared R. The other power is the actual work being done by the motor. SOOOOOO, When the capacitors cancel out the coils, XL = XC, the power applied to a motor will be close to 746 watts per horse plus more watts because the motor's wire are getting hot. If you designed the most efficient motor, power factor it, you can't use super conductors, just how close could you come to 746 watts/hp when dealing with fractional hp induction start motors? If I don't have it about right, Please explain it to me, like I'm a first grader, what I'm missing. Thanks, George [Reply]  [No Email]
MarkB_MI 08-12-2012 18:44:55 75.198.63.201 Report to Moderator	Re: George Marsh experiment? in reply to George Marsh, 08-12-2012 17:13:22   The efficiency of a motor is essentially unchanged as its power factor is adjusted. It takes 746 watts to make a horsepower, whether that 746 watts is applied in the form of 746 volt-amps (unity power factor) or 1492 volt-amps with of .5 power factor. Sure the heating losses go up at the higher power factor, but the difference isn't that great. Remember, your power factor correction isn't changing the current through the MOTOR, it's changing the current requirement seen by the GENERATOR. [Reply]  [No Email]
John T 08-12-2012 18:17:34 216.249.82.117 Report to Moderator	Re: George Marsh experiment? in reply to George Marsh, 08-12-2012 17:13:22   George, thanks for the kind words and while YES Im an EE and YES Century Electric Motor Co was my first EE job (1970), I'm rusty on BOTH AC Motor Theory and the latest NEC (retired from power distribution design in 1991), Im far better versed on AC Power Distribution then motors nowadays. That being said: Youre pretty well on track as I understand it: Power factor is the ratio of the real power flowing to the load to the apparent power in the circuit. Real power is the capacity to perform work (spin the motor shaft) while apparent power is the product of the current and voltage of the circuit. When XL = XC the power factor is unity one same as a pure resistive load since voltage and current are in phase with each other (no lagging no leading current) Its my understanding the figure of 1 HP = 746 watts is a pure energy conversion (HP versus watts) figure, its NOT the real actual energy a motor must consume to produce 1 HP outout since as you noted the motor is NOT a 100% efficient machine due to heat losses i.e. If a 110 volt 1 HP motor draws 15 amps thats like 1650 watts not only 746. Things such as power factor, density of motor windings, mutual inductance factor, resistivity of windings, magnetic material properties, eddy currents, friction etc etc all play a part in motor efficiency, but the bottom line is non perfect mutual inductance, friction and heat losses is why it takes more then the ideal 746 watts in to yield 1 HP out. I guess a 100% efficent motor would consume 746 watts for 1 HP out. Power factor is ONLY ONE compoment to consider when improving a motors efficiency although YES it does help An induction motor is a type of motor NOT the definition of how it starts. A single phase induction motor is not self starting as it will just sit there and hum if voltage is applied, so yes it can be started inductively using an out of phase seperate start winding or repulsively by a commutator and brushes. There are some good electricians on here who are more current and still practicing who may explain this better or correct me if Im wrong, Im probalby as good as them on distribution theory (did that for most of my career) but for now this is my best short answer to your question (without dusting off my old books and studying, but Im too tired n lazy to do all that lol) Fun chat for me at least but I bet the others are bored to tears, sorryyyyyyyyy John T [Reply]  [No Email]
George Marsh 08-12-2012 19:10:57 50.104.223.0 Report to Moderator	Re: George Marsh experiment? in reply to John T, 08-12-2012 18:17:34   JohnT It sounds to me that the other losses you mentioned will show up in the form of heat energy, right? So the electrical energy applied to a motor is converted in to mechanical work and the rest heat energy. So the energy converted to mechanical energy is 1 hp = 746 watts = 550 ft-lb/seconds. BYW, James Watt came up with the 550 mumber when he wanted to sell steam engines to replace horses, Right? Take a wild educated guess, what percentage is converted to mechanical energy and what percentage makes heat energy? Who cares if they anyone is bored to tears. I have dry eye syndrome and need more tears:) Thanks John, George [Reply]  [No Email]
John T 08-12-2012 19:41:32 216.249.82.117 Report to Moderator	Re: George Marsh experiment? in reply to George Marsh, 08-12-2012 19:10:57   Looks to me like if a 1 HP single phase split phase AC Induction motor draws around 15 amps at 110 volts, thats 1650 watts, and since the pure mechanical energy conversion is 1 HP to 746 watts, then theres about (little more actually) as much heat losses as there are mechanical conversions. Thats my best wild guess as you asked for but NO WARRANTY. Many modern electronic devices may be like 80% efficient so figuring induction losses and all that wire and reistance and heat generated, less then 80% efficiency dont really come as a surprise to me... John T [Reply]  [No Email]
George Marsh 08-12-2012 19:55:56 50.104.223.0 Report to Moderator	Re: George Marsh experiment? in reply to John T, 08-12-2012 19:41:32   JohnT, So. if about 25% of a motor applied watts, by your best educated guess, is in the form of heat loss, then the best I may hope for is getting the input watts for my 3/4 hp pump is around 746. After I get my power factor meter, I let you know how low I get the input power. After adding 170 uf, I'm putting in 120 volts from public service and using a little over 9 amps. 120 x 9 = 1080 w. Looks like I still have room to add more capacitors. OK, time to get off. My eyes are better. Sorry to hog the post. Thanks, George [Reply]  [No Email]
George Marsh 08-12-2012 16:02:11 50.104.223.0 Report to Moderator	Re: George Marsh experiment? in reply to joe fabregas, 08-12-2012 08:06:36   Put the capacitors in parallel, not series. Use only AC capacitors. After installing, 2 55 mic, a 35 mic and 25 mic a total of 170 mics to my 3/4 hp pump, my total amps dropped about 3 amps. Each capacitor caused the total current to drop. I stopped there for now. I then went on ebay and purchased a cheap power factor meter, $18. It's only good for 120 v and 15 amps. Just ordered it yesterday. Going to see how close I came to getting it right using an Amprobe. My pic shows how I measured my current and voltage. I plugged the pump into the cord and used another cord to install the caps by plugging them into my short extension cord. Remember when you unplug the capacitor cord, the caps are charged up. It would be a good idea to short out before touching. If you live in Indiana, I have 5 buckets full of old caps I removed from Air conditioners. They are filled with PCB's, so you know they are old. If you want to stop by Terre Haute, I'll be glad to give you what you need. I really don't think power factoring a well pump, used by a single family, may pay for the cost of new caps. In my case, the old caps are FREE. The main reason I did this experiment was to make it easier for my generator to power my well, which it does run the pump every easily. Glad some people find my posts more interesting and not aggravating:) Life is good in my neighbor hood. George [Reply]  [No Email]
36 coupe 08-13-2012 15:20:33 66.186.169.176 Report to Moderator	Re: George Marsh experiment? in reply to George Marsh, 08-12-2012 16:02:11   George,you have spent more monry on toys than you can save on electricity.There aint no free lunch. [Reply]  [No Email]
George Marsh 08-13-2012 17:15:46 50.104.223.0 Report to Moderator	Re: George Marsh experiment? in reply to 36 coupe, 08-13-2012 15:20:33   36 coupe Every $3 I spend on toys keeps me from sending the IRS $1. I love my toys. I don't have a boss that tells me what I can spend my money on either. Feel sorry for those who do. When it comes to toys, this is just the tip of the iceburg. I even mentioned that the savings one can get by adding capacitors to a pump that is only used a litte, won't pay for the caps. Did you also read, my caps were FREE? I did this to reduce the starting amps for my generator. Which it has. Ordered a cheap power factor meter, tax deductable too. I think I should be able to shave 2 or more amps off. Got an order in to my sister for more free used caps. My sister and her husband have a HVAC business. I also read the owner's manual for my genny and it said I shouldn't use it on saws. Guess it doesn't like motors with brushes. So my goal to solve a problem, that some people said didn't exist, was a real problem. I solved that problem 2 ways. One I got an RV adapter. The RV out uses both winding of the generator, 30 amps instead of the 15 amps I was using. The second way was using a 50 ft 14g wire, which to my best estimate, drops the starting voltage 10% or more, starting current 10% and the power from the genny 81% or more. Problem solved, the problem that some self proclaimed experts said didn't exist, RIGHT. What else does a retired person, who has solved problems all his life, have to do? George [Reply]  [No Email]
buickanddeere 08-14-2012 07:17:08 184.151.63.220 Report to Moderator	Re: George Marsh experiment? in reply to George Marsh, 08-13-2012 17:15:46   Only $$$ power factor correction saves is reduced I2R losses in the gen windings and supply conductors. Of course using the 30amp receptacle with both windings will start and operate a 120V load better than using 1/2 of the generator. Is your 50ft 14 gauge cable more magical than the 12,14 or 16 gauge cable anybody else uses.......no. Still as far as inventing anything? Most people's time is worth more than following a 110 year old trail . Or duplicating a high school electrical shop class theory lab. I''ll challenge my list of certificates and qualification .A long with the installed and repaired equipment to your's and find out who the pro is. [Reply]  [No Email]
Samuel LB 08-14-2012 08:47:40 72.181.173.171 Report to Moderator	Re: George Marsh experiment? in reply to buickanddeere, 08-14-2012 07:17:08   OK, buick&d, you win....you are the most experienced, highest credentialed bull-thrower here, hands down! A cranky one, too. [Reply]  [No Email]
36 coupe 08-13-2012 18:47:27 66.186.169.176 Report to Moderator	Re: George Marsh experiment? in reply to George Marsh, 08-13-2012 17:15:46   Go fishing. [Reply]  [No Email]
George Marsh 08-13-2012 21:23:13 50.104.223.0 Report to Moderator	Re: George Marsh experiment? in reply to 36 coupe, 08-13-2012 18:47:27   No fishing, lot of fish kill, too dry and hot. My lake is almost dried up. Lowest it has ever been. Besides, I'm having too much fun inventing something to do:) [Reply]  [No Email]
joe fabregas 08-13-2012 09:39:06 12.75.111.53 Report to Moderator	Re: George Marsh experiment? in reply to George Marsh, 08-12-2012 16:02:11   Thanks for the offer George. I'm in NY so I guess I won't take you up on it. Sounds like I better let this dog sleep. I knew about the series/parallel thing but seem to have forgotten more than I ever learned, eli the ice man, pie, and all that stuff. At least you got the juices flowing. Thanks again! joe- [Reply]  [No Email]
George Marsh 08-13-2012 10:52:16 205.188.116.142 Report to Moderator	Re: George Marsh experiment? in reply to joe fabregas, 08-13-2012 09:39:06   Joe, If you are friends with a heating air company, they will usually scrap out units they replace and give you the capacitors. Most places will never install a used capacitor. I may call my sister and see if she won't save me some newer non PCB capacitors. George [Reply]  [No Email]
MarkB_MI 08-12-2012 12:20:16 75.198.63.201 Report to Moderator	Re: George Marsh experiment? in reply to joe fabregas, 08-12-2012 08:06:36   Power factor correction capacitors only help if you have a lagging power factor. Many new pumps already have built-in PF correction. Capacitance is additive when capacitors are wired in parallel. Two capacitors of the same size in series have half the capacitance of a single cap. I would suggest adding PF correction capacitors in parallel with the motor rather than in series. That way the full motor current doesn't pass through the capacitor. Power factor correction doesn't actually improve motor efficiency. Your generator still has to produce 1/2 hp to turn a 1/2 hp motor. The difference is the current is less with a corrected power factor, which reduces I2R losses. But those are typically fairly small, unless you're going down George's other rabbit hole, which is to ADD resistance to the circuit. (In that case you would be trying to simultaneously increase and decrease I2R losses, and I don't know why you would want to do that.) [Reply]  [No Email]
joe fabregas 08-14-2012 04:29:07 12.75.110.54 Report to Moderator	Re: George Marsh experiment? in reply to MarkB_MI, 08-12-2012 12:20:16   Thanks Mark. Re-reading your last paragraph. Say my motor needs 8 amps and with caps I can get it to run on 6 amps. Using PIE I would have 960 watts vs 720 watts, you're saying the genny will have to work the same either way? Seems the 720 would be less to push? I must be over-simplifying the process. Probably why me & sparkies never hit it off. joe- [Reply]  [No Email]
MarkB_MI 08-14-2012 11:17:08 198.208.251.23 Report to Moderator	Re: George Marsh experiment? in reply to joe fabregas, 08-14-2012 04:29:07   "Using PIE I would have 960 watts vs 720 watts" Joe, You're assuming watts and volt-amps are the same. That's true in a DC circuit. It's also true in a resistive AC circuit. It's not true in a reactive AC circuit. When you do a power factor adjustment, you are reducing the current, but you are not reducing the power delivered to the load. By adjusting power factor, you time-shift the current to be in phase with the voltage, so you don't need as much current to deliver the same power. So what is this extra current and what is it doing? It's stored energy, bouncing back and forth between the source and load at 120 times per second. It isn't adding to the load of the generator, nor is it delivering any net power to the load. It creates some I2R losses, but those are relatively small compared to the actual power delivered to the load. Adding the power factor correction capacitor just causes energy to be stored in the capacitor rather than the generator. The generator doesn't have to supply as much CURRENT, but it still has to supply the same POWER as before. Imagine this: You take the shock absorbers off your car and go for a drive. You run over a pothole and the car starts bouncing. Do you need to accelerate to keep the car bouncing? No. The springs sapped a little energy from the car when you went over the hole, but after that the stored energy in the springs and the vehicle's inertia keeps it bouncing (until friction in the suspension soaks up the stored energy). [Reply]  [No Email]
joe fabregas 08-15-2012 05:24:57 12.75.112.34 Report to Moderator	Re: George Marsh experiment? in reply to MarkB_MI, 08-14-2012 11:17:08   Mark, thanks again. So what I learned in the old days was over-simplified. I can believe that. Seems my Father tried to explain that to me a long time ago. Funny about the shocks. I had a chevy wagon that the rear shocks both broke. I drove it like that for nearly a year until inspection time. No safety issues but what a blast, especially to make little kids laugh when you got it bouncing with a hard brake. Guess I better stick to nuts & bolts and let the sparkies sleep. joe- [Reply]  [No Email]
JMOR 08-12-2012 08:25:10 72.181.173.171 Report to Moderator	Re: George Marsh experiment? in reply to MasseyHarrisGuy, 08-12-2012 08:06:36   Quoting Removed, click Modern View to seeno. not "in line" [Reply]  [No Email]