wrv@ihlpm.ATT.COM (Vogel) (04/04/89)
This really belongs in sci.electricity, but since there isn't one, here goes: I have a swimming pool, which has a filter pump that runs all summer. Over the weekend I had a conversation with the local pool supply company service person. We got on the subject of pumps, and he asked me if my pump was ok, and how it was connected. I told him it was connected to 110V (by previous owner), but it could also run at 220. He said I should convert it, that this would give a great savings in the power bill. Well, I went home wondering how this could be true. I examined the pump motor, its rated: 9 amps @ 220v 18 amps @ 110v So, in my mild knowledge of electronics and Ohm's law, Power = E x I 1980 watts in both cases. So somebody please tell me, is the pool guy whacko, or am I missing something? -Bill [waiting for summer so I can swim......] Vogel, ihlpm!wrv
vermilye@penelope.oswego.edu (Jon R. Vermilye) (04/04/89)
In article <3272@ihlpm.ATT.COM> wrv@ihlpm.ATT.COM (Vogel) writes: >Over the weekend I had a conversation with the local pool supply >company service person. We got on the subject of pumps, and he asked >me if my pump was ok, and how it was connected. I told him it was connected >to 110V (by previous owner), but it could also run at 220. He said I >should convert it, that this would give a great savings in the power bill. >So somebody please tell me, is the pool guy whacko, or am I missing >something? > -Bill [waiting for summer so I can swim......] Vogel, ihlpm!wrv He is probably right. The current will be lower for the same amount of power at the higher voltage. This means that you will have less loss due to the resistance of the wire. This is why power it usually transamitted at high voltage/low current. If your wire length is long, you probably would have some savings both in the wire, and the internal wiring of the motor. It the run is short, it won't make much difference. Generally, it is best to run motors and other appliances at the highest voltage available. JRV Jon R. Vermilye 315 341 2138 Department of Theatre vermilye@oswego.Oswego.EDU SUNY Oswego rutgers!sunybcs!oswego!vermilye Oswego, NY 13126
torkil@psivax.UUCP (Torkil Hammer) (04/05/89)
In article <3272@ihlpm.ATT.COM> wrv@ihlpm.ATT.COM (Vogel) writes:
[Dual Voltage motor]
#
#Well, I went home wondering how this could be true. I examined the
#pump motor, its rated:
#
# 9 amps @ 220v
# 18 amps @ 110v
#
#So, in my mild knowledge of electronics and Ohm's law, Power = E x I
#1980 watts in both cases.
#
#So somebody please tell me, is the pool guy whacko, or am I missing
#something?
He is correct. What you are missing is the loss in the wire from the
outlet to the pool, which can be quite large if the pool is away from
the house.
Connecting to 220V will cut the current in half, and Ohms law applies
to the wiring (P=R x I x I) [but not to the motor] so the wire loss
is cut to 25%.
You can expect about the same wattage, but at 220V the pump is more
efficient due to the samller wire loss, so you can pump for shorter time,
hence the saving on the bill. Also, you heat the wires less, which
might mean longer wire life, depending on its quality, as well as less risk
of fire. If you have 220V available, you should convert.
Have fun and don't forget the sunscreen.
torkil of California.jshelton@deimos.ADS.COM (John L. Shelton) (04/05/89)
>In article <3272@ihlpm.ATT.COM> wrv@ihlpm.ATT.COM (Vogel) writes: >>Over the weekend I had a conversation with the local pool supply >>company service person. We got on the subject of pumps, and he asked >>me if my pump was ok, and how it was connected. I told him it was connected >>to 110V (by previous owner), but it could also run at 220. He said I >>should convert it, that this would give a great savings in the power bill. >>So somebody please tell me, is the pool guy whacko, or am I missing >>something? I suspect your pool person was mistakenly thinking of 208v three-phase power, which is much more efficient for motors than single phase as provided by 110v and 220v. I believe the current losses in wire around the house will not make a large efficiency difference in your case. =John=
rsd@sei.cmu.edu (Richard S D'Ippolito) (04/06/89)
In article <3272@ihlpm.ATT.COM> wrv@ihlpm.ATT.COM (Vogel) writes: >This really belongs in sci.electricity, but since there isn't one, here goes: > >I have a swimming pool, which has a filter pump that runs all summer. >Over the weekend I had a conversation with the local pool supply >company service person. We got on the subject of pumps, and he asked >me if my pump was ok, and how it was connected. I told him it was connected >to 110V (by previous owner), but it could also run at 220. He said I >should convert it, that this would give a great savings in the power bill. > >Well, I went home wondering how this could be true. I examined the >pump motor, its rated: > > 9 amps @ 220v > 18 amps @ 110v > >So, in my mild knowledge of electronics and Ohm's law, Power = E x I >1980 watts in both cases. > >So somebody please tell me, is the pool guy whacko, or am I missing >something? > > -Bill [waiting for summer so I can swim......] Vogel, ihlpm!wrv Like, the guy is out to lunch. First of all, the voltages you get at the meter are 120 and 240 by power company tariff. Second, we're not talking about motor efficiency, but circuit efficiency -- the motor doesn't run any differently at the 240VAC connection. Third, in an AC circuit with reactance (the motor inductance) the voltage and current are out of phase and cannot be multiplied to get watts -- you get volt-amperes. To get watts, you must multiply by the power factor. However, let us assume that your filter pump requires a wire run of 100 feet from the breaker box to the pump motor, for a total conductor length of 200 feet. Assume also that the wire gauge used is #10AWG, which has an approximate unit resistance of 1 Ohm/1000ft. Thus, the circuit resistance will be 0.2 Ohm. (It's possible that #12 wire was used, at 1.26 Ohm/1000ft, but the following calculations will not be significantly changed.) Finally, let's assume your energy costs are 9.5 cents per KWHr, about what they are here, that you leave the filter on continuously for 100 days, and that the motor is actually running at full load (this depends on the water head). Now, for the high-voltage connection, the wire (I^2R) losses are: 9A * 9A * 0.2 Ohm = 16.2 watts. For 100 days at 24Hrs/day, your energy cost is: 16.2W * 24Hr/day * 100 Day / 1000W per KWhr * $0.095/Kwhr = $3.69 Doing the same for the present 18-Amp load brings it up to four times that, or a whopping $14.77 for the season, so you will save $11.08 over the season. Play with the numbers, if you like, and insert your own. This is the kind of superficial analysis that the partially educated (the pool folks, not you) do, if they do it at all. Being a well-trained (I think) electrical engineer, however, I will give you the benefit of a true engineering analysis: What's your time worth? Got the number? Let's proceed... You will have to purchase and install a two-pole circuit breaker (if you can free up two circuits) in the breaker box. If not, you will need an auxiliary box. (Fuse boxes are even worse!) You will then have to replace the switch at the pump (you DO have one there for safety, don't you?) with a 240-Volt unit. Now, assuming that you do all of the labor yourself, the job will cost around $50.00 plus your labor plus a building permit plus any fee for a certified inspector, if required by your municipality. When you figure out the economic payback period, remember the time-value of money before you attempt simple division. Today's dollars are not tomorrow's. The above was rather cute, I realize, but the analysis and numbers are correct. NOW, LET'S GET REALLY SERIOUS... Finally, the other issues -- Do you really want 240 volts running around your pumphouse and pool? Have you provided a ground-fault interrupter on ALL of the outdoor circuits? SAFETY FIRST! PLEASE, IF YOU HAVE ANY DOUBTS, DO NOT ATTEMPT THE CONVERSION YOURSELF. MILLIAMPS KILL! Rich -- --------------------------------------------------------------------------- Ideas have consequences. RSD@sei.cmu.edu Richard Weaver ---------------------------------------------------------------------------
tomb@hplsla.HP.COM (Tom Bruhns) (04/06/89)
>In article <3272@ihlpm.ATT.COM> wrv@ihlpm.ATT.COM (Vogel) writes: >>Over the weekend I had a conversation with the local pool supply >>company service person. We got on the subject of pumps, and he asked >>me if my pump was ok, and how it was connected. I told him it was connected >>to 110V (by previous owner), but it could also run at 220. He said I >>should convert it, that this would give a great savings in the power bill. ^^^^^^^^^^^^^ I doubt it... >>So somebody please tell me, is the pool guy whacko, or am I missing >>something? >> -Bill [waiting for summer so I can swim......] Vogel, ihlpm!wrv >He is probably right. The current will be lower for the same amount ^^^^^^^^^^^^^^^^^^^^ Again, I doubt it... You will have about 1/4 the I^2*R loss in the wire leading to the motor if you run it at 240 instead of 120. But I'd be surprised if you have more than .3 ohms (about 100 feet of 2-conductor 12 gauge), and that means around 100 watts max. Maybe 1/2 cent per hour? It adds up, but a "great savings"?? And rewiring the motor for a higher voltage most likely will put two windings in series that used to be in parallel, for the same net I^2*R loss in the motor. I submit that the most noticable change will be faster starting of the motor; it may draw a rather large starting current and the lower drop in the line would be noticable at that time. >of power at the higher voltage. This means that you will have less >loss due to the resistance of the wire. This is why power it usually >transamitted at high voltage/low current. If your wire length is long, >you probably would have some savings both in the wire, and the internal >wiring of the motor. It the run is short, it won't make much difference. >Generally, it is best to run motors and other appliances at the highest >voltage available. > >JRV > >Jon R. Vermilye 315 341 2138 >Department of Theatre vermilye@oswego.Oswego.EDU >SUNY Oswego rutgers!sunybcs!oswego!vermilye >Oswego, NY 13126 >----------
rlf@mtgzy.att.com (r.l.fletcher) (04/06/89)
In article <3146@ae.sei.cmu.edu>, rsd@sei.cmu.edu (Richard S D'Ippolito) writes: > > Like, the guy is out to lunch. First of all, the voltages you get at the > meter are 120 and 240 by power company tariff. Well they try to anyway. [Savings analysis (much thanks for saving me the trouble) deleted.] [ arrives at $14.77 savings for the season due to line losses] [estimates conversion from 110 to 220] > will cost around $50.00 plus your labor plus a building permit plus any fee > for a certified inspector, if required by your municipality. When you I suspect it will be more like $100 to $150. I would guess that the existing circuit is on 12/2 WG and some local codes forbid using the bare ground as a neutral return in a 220V circuit. So you need to add 100 ft of 12/3 WG and the labor to re-install the run. Wonder what the pool companies electricians would charge for the job. > > The above was rather cute, I realize, but the analysis and numbers are > correct. > > Rich If one really wants to save some $$ I suggest you would save more by experimenting with the number of hours you run the pump per day. The manual for my pool filter claims I should run it for 10-12hrs/day, so I did. Until I had the "Free Energy Audit" by the local power company which suggested I experiment. I am down to 7hrs/day with no effect on water clarity. This season I will try to get that down to 5-6 hrs per day. This has made a big difference in my electric bill. 120V * 18A * 1hr/day * 100 Day / 1000W per KWhr * $0.095/Kwhr = $20.52 Thats $20.52 savings over a 100 day season for each hour per day the pump is cut back. Ron Fletcher Followups are directed to misc.consumers.