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I have a 2 hp, 2 speed pump on a 24'x54" AG pool. I know, I don't need 2 hp, but I didn't find this great site until after the fact! We've had the pool for a couple of weeks and have been running the pump on low 24x7. We just got a timer yesterday, and know I should run the pump early morning and at night. What speed should I use, and what times do you run your pump?
Anyone have any tips on this? The last 2 years I've run the pump 24/7. Jacks up the electricity but made me feel better about the water. We put our pool up at the new house this year and I ran the pump 24/7 for about a month. I now run it 8pm-1am & 11am-1pm.
I know it's probably one of those personnal preference things. Knowing what others do would help to determine what I want/need to do.
Pump run time should be based upon turnover (i.e. how long it takes to pump the pool volume). Turnover can be difficult to determine accurately but you can use a few rules of thumb to determine it.
First you need to determine head loss. In most pools that is between 45-65 ft on high speed and 11-16 on low speed. Or you can calculated it from the filter PSI reading. Head loss is usually 2.5 to 3.5 times the the PSI reading. I would use 3x.
Next look up the flow (GPM) on the pump head curve for both high and low speeds. Low speed should be 1/2 of high speed GPM. Low speed head loss should be 1/4 of high speed.
Then determine your turnover rate.
Turnover (hours) = Pool Volume / GPM / 60
For low usage pools, 1 turnover is usually enough. For high usage pools, two turnovers may be required. Most of the swim season, I run mine 1.5 turnovers.
Given your high HP pump and low volume pool (15000 g?), my guess is a turnover < 3 hours on high speed and 6 hours on low speed. So you probably don't need more than 6 hours on high speed or 12 hours on low speed.
I thought that the conversion was 2.3 feet of water per PSI so that you would multiply PSI by 2.3 to get head loss in feet. Am I missing something?Quote:
Originally Posted by mas985
Richard
2.31 times the filter PSI will give you only the return head loss from the filter to the pool. Head loss consists of both return and suction losses. The most accurate way to measure suction loss is with a suction guage at the pump. Unfortunately, most people do not have one nor willing to buy one so I usually skip that suggestion. So without a suction guage, a pretty good estimate for suction loss is 30% above the return head loss which results in a factor of about 3 (2.31*1.3 ~ 3). It gives an answer with a little more accuracy than 45-65 feet of head.Quote:
Originally Posted by chem geek
BTW if you do have a suction guage, then multiply inches of mercury by 1.13 to get the suction head.
Thanks Mark. I figured it was something simple I was overlooking. I really appreciate your clear explanation.
Richard
I would add that you might want to consider breaking up your run time into 2 or more increments, especially if you are running for less than 12 hours a day. Of course, you need a pump timer to do this.
I just find that my pool seems to stay cleaner and the SWG seems to work better if I add a 2 to 3 hour night time run to the 10 hours or so I run during the day (we run during the day due to solar heating). Stuff has less opportunity to sink and it just makes me feel better to know that I am not leaving the pool completely stagnant for 14 hours straight.
Could someone put this formula into layman's terms? I don't see where head loss fits into the formula..
>>First you need to determine head loss. In most pools that is between 45-65 ft on high speed ...
Then determine your turnover rate. Turnover (hours) = Pool Volume / GPM / 60<<
Where is head loss used in the equation?
Obviously I'm missing something (brains?). :D
TIA,
Steve
Here is what I had posted regarding converting head loss to GPM:Quote:
Originally Posted by ePoolUser
Next look up the flow (GPM) on the pump head curve for both high and low speeds. Low speed should be 1/2 of high speed GPM. Low speed head loss should be 1/4 of high speed.
Usually, the pump head curve is in your pump manual. The vertical axis is head loss and the horizontal is GPM.
It takes three complete turnovers to remove 98% of the suspended solids. It's the Gage and Bidwell Law of Dilution.Quote:
Originally Posted by mas985
Don't think I'm smart or anything, I was just coincidently reading up on it yesterday.
And I'll probably forget it tomorrow.
While true, it still does not mean you need to do three turnovers. One turnover will hit about 70% of the water which is usually sufficient for proper sanitation and removing most of the solids. It is not necessary to remove all of them in one day since as you are filtering, more solids are being dropped so you will never have water without solids. That is the main reason for a chlorine residual.Quote:
Originally Posted by giroup01
As far as gage and biwell goes you get 67% at the first turnover and
89% at the second. This also assumes no new dirty stuff goes in.
I run 1 turn most days. With balanced water I have never had a water
problem. I have A 2 HP @ 100gpm measured with Triton TR-100 sand filter and
1 cup of DE.