old pump and estimation of the turnover time

Diver

0
May 5, 2011
480
South of Boston
I’ve been trying to figure out my turnover time for my pool and I can’t really come up with a number. I know the rule of thumb on this forum is to set the pump time higher and then gradually decrease the time window as long as the water stays clear and sparkling. The problem for me is that ”clear and sparkling” is somewhat subjective and this being my first full season I’m not sure I can trust my judgment on this.

I have a 27k in-ground pool I have a Hayward Superpump with the following on the nameplate:

C48K2N143B1 SP-1607-Z1-M series 8D00 1HP, SF 1.1, 3450 RPM, 0.75KWt (see pic below)

It appears that this model is discontinued and looking at the current superpump lineup on Hayward site the closest current model is SP2607X1.

I have two intake lines (skimmer and MD) merging into on 1-1/2” pipe before pipe, DE filter (model in my signature) and then one 1-1/2” return pipe splitting into three 1-1/2” pipes going to the return jets. Doing some repiping I discovered some water in one of the 3 return pipes. The equipment pad is on the level with the pool and is 5 feet away. I would estimate skimmer and MD lines to be 25-30 feet each. The return lines are about 50, 50 and 60. No water features.

I looked at the hydraulics 101 post and pump basics post. The GPM estimate XLS from the first post doesn’t have my pump model and I tried to plug-in my numbers as much as it made sense, but it seemed that I was lacking too much input data.

The pump basics post has a simple table for turnover time per give volume and HP:


HP SFHP BEP GPM BEP Head Pool Volume (8hr Turnover)
0.5 1.05 62 37 29,616
0.75 1.39 67 51 32,168
1 1.85 85 58 40,850
1.5 2.40 102 59 48,965
2 2.70 121 61 57,893
3 4.80 148 76 70,804


My pump being 1HP and my pool being 27k, it seems that a good estimate for my turnover time would be slightly more than 5 hours. Which seems a little low. I ran the pump for 6 hours a day last Septmber before I closed the pool, but too be honest I had no idea was I was doing back then. The water seemed to be ok though.

Old owner had the pump running for 12 hours a day, but then he had calcium hardness increaser in the pool shed for the vinyl pool.

Can someone who had a similar size pool and 1HP pump with somewhat similar piping situation tell me their turnover times? Or maybe someone who has a lot of experience to simply estimate the time for my situation?

Thank you a lot, and sorry for the long post 
 

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What's your filter pressure?

My turnover on a 13,000 gal pool is about 3.5 hours with a 1.5 SFHP Dynamo pump, so 5 hours may not be too far out of range for your turnover. I'd probably say 6 just to be safe.

[edit] Jason and I had the same idea[/edit]
 
Jason, zero at the moment, since the pool is still closed. From your reply and what I read before it sounds that I should have this number on hand to get a turnover time? I will have it after memorial day weekend, right now I'm just doing my homework for the upcoming season.
 
Since intake is 1.5 inch that’s going to be the max you will be able to flow. Most agree that 44 gallons minute is about the max you will get though that. So in theory all things perfect you would need at least 11 hours for your pool for one turn over. In reality I would think 14 or more hours.
 
mikespoolfl said:
Since intake is 1.5 inch that’s going to be the max you will be able to flow
It doesn't work that way. If the pump is large enough you can draw far more water through a 1.5" pipe than that. That max flow rate guideline is the flow rate at which metal pipes start to erode away. With PVC you can easily exceed those flow rates and not have any problems (other than wasting electricity).

Diver, it is much easier to guess at the flow rate when we know the normal/clean filter pressure.
 
Jason, thanks for comforting answer on 14 hour turnover time! :) Mike, don't get me wrong, but I really hope that your 14 is off. My electric bill would be astronomical.

Jason, "wasting electricity" that you're referring to has something to do with efficiency curve of the pump, doesn't it? I understand that it makes more sense to pump slower and longer, however my pump is a single speed. Is there any way to lower the RPMs and reduce the power?

I will post the filter pressure in a week and a half.
 
Based just on the information you gave, I would guess about 65 GPM @ 32' of head and a 7 hour turnover. My calculations put the filter pressure around 8 PSI but an actual measurement would help improve the accuracy. The estimate assumes that you have two suction pipes pool to pad and 3 return pipes pool to pad with the combine/split at the pad and not at the pool.
 
Mark, your assumptions is accurate. The return split is about 4 feet away from the filter, so i think you can call it at the pad.

I will post actual filter pressure number when it becomes available.

Thank you very much for all your answers.
 

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Shutting off the main drain and running with only the skimmer increases the suction side head loss thereby reducing the pump flow rate which in turn reduces the pressure side head loss (but less than the suction increase) and reduces the filter pressure. If possible, I would recommend that you always run the pump with both the main drain and skimmer open. This will be your most efficient setting, highest flow rate and shortest turnover.
 
With the new information on settings and pressure:

With both MD and Skimmer open, flow rate should be around 67 GPM @ 30' of head, 1310 watts.

With only the skimmer, 61 GPM @ 36' of head, 1265 watts.
 
thank you Mark! i hope you don't ask me asking, but you're hydrolics guru here :) i was putting fish eye nozzles on my returns yesterday and could't keep thinking how restrictive they looked! my returns are 1.5" and the eyes are maybe 3/4". i understand the purposes, but my returns a located in a good pattern and i'm wondering if removing the eyes would help the circulation by reducing head loss. if i remove them, should i see filter psi going up?
 
Removing the eyeballs means you can't aim your return jets, a significant disadvantage. If you do remove them, the filter pressure should go down and flow rate go up, as there is less flow restriction. I use 1" eyeballs, which are not as bad, though there is still some flow restriction.
 
Jason,

here is my pool (little squares are the returns, big one is the skimmer). will removing the eye balls in this configuration have a bad impact? right now they are pointing towards the middle. I think the one by the steps (short side of the L) will be bad without an eye ball.

previous owner told me the volume was 27k, but i recalculated and it's at least 30k!

i might just take the eyeballs out and make the hole bigger
 

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You could go to 1" eyeballs but I suspect that you won't see that much of a difference (<1 GPM). Since you have three eyeballs, the flow rate is divided between them so the head loss experienced by any one of the eyeballs is much much less than if you had only one return.

Plus the pressure loss at the eyeball is converted to water exit velocity so the smaller eyeballs will actually have faster water (ft/sec) water coming out of the return (16ft/s vs 9 ft/sec). So it isn't all bad to have the smaller eyeballs.
 
mas985 said:
With the new information on settings and pressure:

With both MD and Skimmer open, flow rate should be around 67 GPM @ 30' of head, 1310 watts.

With only the skimmer, 61 GPM @ 36' of head, 1265 watts.

Mark,

I just noticed the Watts you mentioned. The pump is rated at .75kW. Is this number nominal power consumption and the ones your gave me peak ones?
 
The label rating, 0.75 kw which is equivalent to 1 HP, is the output power rating of the motor (i.e. maximum output power) and does not reflect the input power required to achieve that output power. For that you also need the efficiency of the motor. However, the input power required to drive the motor is also dependent on the operation point (i.e. point on head curve). What I gave you was the electrical input power required for the given operating point (GPM & Head). The input power will always be greater than the output power of the motor determined by the efficiency.
 

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