Water pump electricity

[Papote]

Where I live, the water pressure from the water utility is very low (About 8 PSI average). I am considering getting a jet pump with a 33 gallon pressure tank and a 400 gallon water tank for when there is no water at all. Electricity is a major concern because it is very expensive in my area, about $0.25 a KW. I use on average about 100 gallons of water a day.

Which type of jet pump or full setup in terms of HP and Watts would you guys recommend for my setup that would save me on electricity?

Most people around here just use a 1/2 HP well jet pump with the pressure tank and water storage tank. I've been reading much on Cycle Stop Valves which use a small pressure tank and Variable Frequency Drives, but that seems a little overkill.

 

[bimr]

A jet pump can be used as a pressure booster, with a couple of limitations in mind. First, the overall pressure that the pump will see must be considered. This is the total of the incoming pressure plus the pressure the pump will add to that. For most thermoplastic jet pumps that total pressure cannot exceed about 70 psi. Thermoplastic jet pumps are not really recommended for boost applications. For most cast iron pumps, the total pressure cannot exceed about 100 psi. These are maximums. You don’t want to constantly run a pump at its maximum so you should include some buffer if considering this. The second thing to consider is if the pump cycling will be too much. Pumps like to run, and they like to be off. They do not like to be on/off/on/off, etc. If used in a booster application where the water itself can be turned on and off a lot, it is best to include a pressure tank in the application.

There are minimum opportunities to save electricity in a system like this.

https://www.youtube.com/watch?v=VJS4eO4GF40

https://www.youtube.com/watch?v=s3dNzYjtOok

You may consider some type of disinfectant if you have a large storage tank in a warm weather climate. Legionella may grow and multiply in a building water system.

 

[LittleInch]

This is such a common issue that package supplies are normally freely available.

Your electricity demand will only really vary by the amount of water you use and the pressure you want.

I'm really not sure what you mean by a "jet pump" and think "water booster" pump might be a bit more of a useful description.

what you need to avoid is using electricity inefficiently, either with a poorly designed pump (often difficult to get efficiency numbers for these small units) or by pressurizing the tank too much and then simply leaking all that energy way across your partly open tap.

As ever there is a balance in these things. if you can afford / have space for a large pressurizing tank, then that's a good start, but then you need to look carefully at what pressure does your pump start / stop, how much water do you in one go and how often does the pump start / stop. Anything you save on a large pump being more efficient, you will lose if the number of starts burns the motor out a lot faster.

For you pressurising tank, you probably should aim for operation between 1/3 and 2/3 full.

Set the start pressure switch for your pump as low as you can go and still get good flow and then work out when the turn the pump off.

Also consider how much flow you can actually get from the mains without sucking it dry and / or having your neighbours beat you up for "stealing" their water....

So it's up to you and your circumstances using the outline above. If you really want to avoid electricity charges, you run the risk that your pump isn't big enough on the odd occasion when everything is running, but so long as that's acceptable go for a 1/2 to 3/4 HP pump, try and find the most efficient then run it in the most efficient manner you can.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[electricpete]

It might be instructive to see what kind of energy costs we're talking.
Assume a very low system efficiency Eff (combine pipe losses, pump losses, motor losses). Let's say 20%.
Then assume a dp you want to target. Let's say you want 13-23 psi in the tank or dp 5-15 psid, average 10 psid during pumping
You've told us Q = 100 gallons per day and Rate = 25 cents per kw*hr.

It's an excercize in unit coversion to turn that into money per day.
Cost per time = Q * DP* Efficiency * Rate

My calculations come out on less than 1 cent per day.
Please feel free to do your own calculation..I've been known to make a mistake.

Assuming that's in the ballpark, it suggests that spending anything significant to bump the efficiency is probably not going to be worthwhile (will take a long time to get your money back). I'd think that suggests to focus primarily installation costs/effort along with reliability and maintainability. Efficiency would be a minor piece of the equation.

=====================================
(2B)+(2B)' ?

 

[Artisi]

Mountains out of mole hills:
Easy solution is to head for your local domestic water pump supplier and ask for a house booster pump capable of 4 / 6 gpm at 25/30psi. Should be off the shelf supply for any reasonable stockiest.
Power consumption isn't even worth calculating, see earlier post from electricpete.

You are correct re overkill trying to save a cent or less a day.

For interest, your computer is probably consuming more power per day than what the pump will.

It is a capital mistake to theorise before one has data. Insensibly one begins to twist facts to suit theories, instead of theories to suit facts. (Sherlock Holmes - A Scandal in Bohemia.)

 

[electricpete]

I have to post a correction to the formula (but not the numerical result)

Cost per time = Q * DP* Rate * Efficiency

should've been

Cost per time = Q * DP * Rate / Efficiency

CostPerTime=Q*DP*Rate/Efficiency = (100 gallons / day)*(10psi)*($0.25/(kw*hr))/(0.2) ~ 0.9 cents per day.
So I posted the equation wrong but had the right numerical result for my stated assumptions

=====================================
(2B)+(2B)' ?

 

[LittleInch]

I get something similar.

Even if you do it really badly (pressurizing to higher level than you need, you're not going to exceed 4-5c/day.

Getting the right size is key to match to flow demand. too big and it start and stops too many times and could heat up / burn out. too small and at peak times you don't get enough flow for what you want.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[Papote]

I have found a 800 Watt with a factory set 28/50 PSI pressure switch and a 4 gallon pressure tank. It's the most economical one I could find and easily available at my local Sam's Club for $200 bucks.

 

[EdStainless]

Are you still going to install the 400g holding tank?
Just make sure that there is a check valve so that this water can't flow back into the main.
You may want to restrict the flow into this tank, after all if you only use 100g/day you can afford for this tank to take 3-4 days to fill.
Then just draw off near the bottom of the larger tank for the inlet to your pump/pressure tank.

= = = = = = = = = = = = = = = = = = = =
P.E. Metallurgy, Plymouth Tube

 

[bimr]

That is a good value for a pump. The cost of electricity for the booster pump will be approximately $7 per year.

 

[Papote]

So I got the booster pump and connected to the hot side for now, because it was the most accessible at the moment. The pressure tank holds about 2 gallons of water and between the cut in and cut out, the pump runs for about 25 seconds at about 800 watts.
So, assuming I use 100 gallons of water a day, the pump should run about 50 times a day, at 1,200 seconds a day or 20 mins. (.33 hrs). I get that it would consume 0.264 kWh a day ( 800 Watts x (0.33 hrs / 1000) ). In a 31 day cycle it would be 8.184 kWh at $2.05 (8.184 kWh x $0.25).

Please correct me if I'm wrong.
Thanks.

 

[electricpete]

How are you determining 800 watt?
Current, voltage and power factor angle? (That would be most accurate.)
Or something else?

Do you know what are the pressures when the pump turns on and off (so we can estimate average dp)


=====================================
(2B)+(2B)' ?

 

[Papote]

I have a P3 kill-a-watt meter. It averages about 770 - 790 watts. I rounded off to the rated 800 watts. The pump is set at the factory to cut in at 25 PSI and cut off at 55 PSI. The city line is at 7 PSI, so once the pressure tank is empty the pump will always fire.

 

[LittleInch]

Looks about right to me. As said before, you might be spending money pressurising water only to lose that energy across your tap, but if it works for you then leave it as is.

If you're not pressurizing the hot and cold together then you won't get much combined flow, but maybe that's ok as well?

At $25 per year max, you won't get it to less than $10-15 without spending lots of money.



Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[Artisi]

You need to check with your water authority if it is allowed to pump directly off their main supply, I doubt that it is as in most places this is illegal for many legitimate reasons.

It is a capital mistake to theorise before one has data. Insensibly one begins to twist facts to suit theories, instead of theories to suit facts. (Sherlock Holmes - A Scandal in Bohemia.)

 

[Papote]

Yup, it is legal. I pay for the water utility that I use, no matter if I get a thin stream or none at all. My local water authority isn't obligated to provide a minimum or maximum PSI to thier homes. I know people that get 165 PSI.

 

[electricpete]

Kw meter is good approach, obviously.

You went from 50 times day at 25 sec each to 1200 sec, should've been 1000 sec. So maybe your result should be 1.70 per month or 5.5 cents per day.

25-55 psi pump discharge with 8 psi suction, so dp goes from 17-47. average dp is 32 which is 3.2 times higher than my assumption of 10psid. So to account for higher dp, my result 0.9 cents per day should be multiplied by 3.2 to give around 2.9 cents per day. Still lower than your 5.5 c per day. I assumed 0.2 efficiency. To match your results, I think system efficiency during pumping must actually be 0.2 *2.9 /5.5 ~ 0.1. (10% efficiency)


=====================================
(2B)+(2B)' ?

 

[LittleInch]

Yes but you still need to get their approval - last sentence of your attachment.

A number of them don't like it if it brings the pressure below a certain value as you can then suck in water from the outside if there's any sort of hole or leak in your water connection system.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[Compositepro]

If you want significant water storage you might want to consider the systems used for recreational vehicles that use a diaphragm pump, controlled by a pressure switch, and feeding from an unpressurized storage tank. Most of these are 12 volt, so with a battery and charger you would still have water pressure in a power failure. Afloat valve would keep the tank full.

With such low water pressure it would be wise to have a back-flow preventer valve (BFPV) between you and the water main. With a BFPV installed the water company has no interest in what is on the down stream side.

 

[ScottyUK]

Low voltage systems are generally quite inefficient because a relatively large amount of the available power is lost as I²R losses in the conductors. A couple of volts lost at 110V is trivial, but at 12V a couple of volts lost is significant. You can combat this to some extent with larger conductors, but some volt-drops are inherent in the motor itself.