Accumulator settings and calculations

[chris w]

Accumulator Calculations: to prove that the best setting for an accumulator is a couple of psi BELOW the water pump's cut-in pressure.

 

 

Where:

 

Ppc = pre-charge pressure

Pco = water pump cut-out pressure

Pci = water pump cut-in pressure

V1 = volume of accumulator

 

The above pressures are gauge pressures (psig) ie: as measured on a normal gauge. I have also assumed a Non-Return Valve (NRV) exists just before the calorifier, so that any changes in the calorifier volume due to heating do not affect the accumulator

 

All pressures need to be converted to absolute”pressures (psia), by adding 15psi to them, to account for atmospheric pressure and for Boyle's law to apply.

 

From Boyle's law P1V1 = P2V2, ie: the product of Pressure and Volume remains constant.

 

If the accumulator is pre-charged and we assume the air bag inside fills the whole accumulator and then we switch on the water pump, the pump pressure will compress the bag until the pressure of the air in the bag is equal to the water cut-out pressure.

 

Therefore:

 

Ppc x V1 = Pco x V2 where V2 is the new (compressed bag) air volume

 

Thus V2 = Ppc/Pco x V1

 

 

 

When a tap is turned on, the pressure in the bag pushes the water out of the accumulator until the pressure falls to the pump's cut-in pressure.

 

In other words:

 

Pco x V2 = Pci x V3 where V3 = volume of accumulator air when pump cuts in

 

So V3 = Pco/Pci x V2

 

But V2 = Ppc/Pco x V1 from above

 

Therefore V3 = Pco/Pci x Ppc/Pco x V1 = Ppc/Pci x V1

 

 

 

The amount of water discharged (D) by the accumulator is the difference in the two volumes = V3 - V2

 

Which is: (Ppc/Pci x V1) - (Ppc/Pco x V1) = (Ppc/Pci - Ppc/Pco) x V1

 

 

Although the primary reason for using an accumulator is to prevent the pump's rapid cycling, it is also useful to maximise the water discharged before the pump switches on, allowing people to run off a small amount of water, at night say, without waking other people for example should the pump cut in.

 

 

For a typical pump of 30psig cut-out pressure and 20psig cut-in pressure and a typical 2 litre accumulator, we can plot D against various settings of Ppc (the pre-charge pressure) by using the formula above, viz: D = (Ppc/Pci - Ppc/Pco) x V1

 

The results are as follows:

 

pre charge........Discharge (D)

(psig)...............(litres)

0.........................0.19

1.........................0.20

2.........................0.22

3.........................0.23

4.........................0.24

5.........................0.25

6.........................0.27

7.........................0.28

8.........................0.29

9.........................0.30

10.......................0.32

11.......................0.33

12.......................0.34

13.......................0.36

14.......................0.37

15.......................0.38

16.......................0.39

17.......................0.41

18.......................0.42

19.......................0.43

20.......................0.44

 

 

It can be seen that D increases right up to where the pre-charge pressure is equal to the pump's cut-in pressure and, in this example, is 440ml.

 

But what happens if we increase the pre-charge pressure further above the pump's cut-in pressure?

 

The results are as follows:

 

pre charge..........Discharge (D)

(psig).................(litres)

21.......................0.46

22.......................0.47

23.......................0.48

24.......................0.50

25.......................0.51

 

 

It can be seen that D increases even further, so would this not be desirable?

 

Actually not, if we think about it. If we increase the pre-charge pressure to be above the pump's cut-in pressure then, when the bag is empty of water, the water flow will cease (because the bag is empty) but the line pressure is still above the pump's cut-in pressure, so pressure will have to fall another few psi, due to the tap's being open, before water will flow again. Although this will only be momentary, it does mean that the flow is not truly constant but has a "blip" in the middle.

 

Therefore it is more desirable to keep the pre-charge pressure just below the pump's cut-in pressure to avoid this and to ensure a steady flow. The loss of additional discharge, by doing this, is pretty minimal and amounts to only 70ml in this example even if the pre-charge pressure were increased to 25 psig. So the "best" setting is 2-3psi below the pump's cut-in pressure, or 18psig in the example above.

 

Chris

 

PS: Interestingly, I measured my pump's cut-in and cut-out pressures which are labelled as being 20psi and 30 psi respectively. In reality, they are 16psi and 27psi respectively.

Edited November 23, 2008 by chris w

[RLWP]

Hi Chris,

 

I see that your pressure gauges have now arrived! Did you measure the 440ml discharge or is this a theoretical figure? I can imagine charging up the system then turning off the pump and measuring how much water is discharged vs pressure.

 

Thanks for this contribution.

 

Richard

[wonderdust]

Have you confirmed your readings with Terry

[blackrose]

Although the primary reason for using an accumulator is to prevent the pump’s rapid cycling, it is also useful to maximise the water discharged before the pump switches on, allowing people to run off a small amount of water, at night say, without waking other people for example should the pump cut in.

 

I've never really understood what rapid cycling meant in this respect? I've got an isolator valve on my accumulator (if the pump cycles unexpectedly and there are any concerns about leaks in the system, then isolating the accumulator really helps with the diagnosis). Anyway, my waterpump seems to cycle at the same speed with or without the accumulator. I've always thought that the main function of an accumulator was as a kind of 'buffer', smoothing out an otherwise pulsing and erratic delivery from the waterpump.

[Gibbo]

Accumulator Calculations: to prove that the best setting for an accumulator is a couple of psi BELOW the water pump's cut-in pressure.

 

 

Where:

 

Ppc = pre-charge pressure

Pco = water pump cut-out pressure

Pci = water pump cut-in pressure

V1 = volume of accumulator

 

The above pressures are “gauge pressures” (psig) ie: as measured on a normal gauge. I have also assumed a Non-Retrn Valve (NRV) exists just before the calorifier, so that any changes in the calorifier volume due to heating do not affect the accumulator

 

All pressures need to be converted to “absolute” pressures (psia), by adding 15psi to them, to account for atmospheric pressure and for Boyle’s law to apply.

 

From Boyle’s law P1V1 = P2V2, ie: the product of Pressure and Volume remains constant.

 

If the accumulator is pre-charged and we assume the air bag inside fills the whole accumulator and then we switch on the water pump, the pump pressure will compress the bag until the pressure of the air in the bag is equal to the water cut-out pressure.

 

Therefore:

 

Ppc x V1 = Pco x V2 where V2 is the new (compressed bag) air volume

 

Thus V2 = Ppc/Pco x V1

 

 

 

When a tap is turned on, the pressure in the bag pushes the water out of the accumulator until the pressure falls to the pump's cut-in pressure.

 

In other words:

 

Pco x V2 = Pci x V3 where V3 = volume of accumulator air when pump cuts in

 

So V3 = Pco/Pci x V2

 

But V2 = Ppc/Pco x V1 from above

 

Therefore V3 = Pco/Pci x Ppc/Pco x V1 = Ppc/Pci x V1

 

 

 

The amount of water discharged (D) by the accumulator is the difference in the two volumes = V3 –V2

 

Which is: Ppc/Pci x V1 – Ppc/Pco x V1 = (Ppc/Pci – Ppc/Pco) x V1

 

 

Although the primary reason for using an accumulator is to prevent the pump’s rapid cycling, it is also useful to maximise the water discharged before the pump switches on, allowing people to run off a small amount of water, at night say, without waking other people for example should the pump cut in.

 

 

For a typical pump of 30psig cut-out pressure and 20psig cut-in pressure and a typical 2 litre accumulator, we can plot D against various settings of Ppc the pre-charge pressure by using the formula above, viz: D = (Ppc/Pci – Ppc/Pco) x V1

 

The results are as follows:

 

pre charge........Discharge (D)

(psig)...............(litres)

0.........................0.19

1.........................0.20

2.........................0.22

3.........................0.23

4.........................0.24

5.........................0.25

6.........................0.27

7.........................0.28

8.........................0.29

9.........................0.30

10.......................0.32

11.......................0.33

12.......................0.34

13.......................0.36

14.......................0.37

15.......................0.38

16.......................0.39

17.......................0.41

18.......................0.42

19.......................0.43

20.......................0.44

 

 

It can be seen that D increases right up to where the pre-charge pressure is equal to the pump’s cut-in pressure and, in this example, is 440ml.

 

But what happens if we increase the pre-charge pressure further above the pump’s cut-in pressure?

 

The results are as follows:

 

pre charge..........Discharge (D)

(psig).................(litres)

21.......................0.46

22.......................0.47

23.......................0.48

24.......................0.50

25.......................0.51

 

 

It can be seen that D increases even further, so would this not be desirable?

 

Actually not, if we think about it. If we increase the pre-charge pressure to be above the pump’s cut-in pressure then, when the bag is empty of water, the water flow will cease (because the bag is empty) but the line pressure is still above the pump’s cut-in pressure, so pressure will have to fall another few psi, due to the tap’s being open, before water will flow again. Although this will only be momentary, it does mean that the flow is not truly constant but has a “blip” in the middle.

 

Therefore it is more desirable to keep the pre-charge pressure just below the pump’s cut-in pressure to avoid this and to ensure a steady flow. The loss of additional discharge, by doing this, is pretty minimal and amounts to only 70ml in this example even if the pre-charge pressure were increased to 25 psig. So the "best" setting is 2-3psi below the pump's cut-in pressure, or 18psig in the example above.

 

Chris

 

PS: Interestingly, I measured my pump's cut-in and cut-out pressures which are labelled as being 20psi and 30 psi respectively. In reality, they are 16psi and 27psi respectively.

 

What about Peukert?

 

 

Gibbo

[Guest]

What about Peukert?

 

 

Gibbo

What has wooden flooring got to do with it?

[chris w]

Hi Chris,

 

I see that your pressure gauges have now arrived! Did you measure the 440ml discharge or is this a theoretical figure? I can imagine charging up the system then turning off the pump and measuring how much water is discharged vs pressure.

 

Thanks for this contribution.

 

Richard

Actually my own exact water pump values are 27psig cut-out and 16psig cut-in and my accumulator is set at 15psig. The predicted discharge value for these figures is 0.51 litres and the actual was 0.5 litres, measured exactly as you suggested.

 

Chris

[kerching]

What about Peukert?

 

 

Gibbo

 

I got half way through reading it and i said Peukert as well.

[chris w]

I got half way through reading it and i said Peukert as well.

I have to admit to being disappointed that there were no exponential coefficients in the equations.

[RLWP]

Actually my own exact water pump values are 27psig cut-out and 16psig cut-in and my accumulator is set at 15psig. The predicted discharge value for these figures is 0.51 litres and the actual was 0.5 litres, measured exactly as you suggested.

 

Chris

 

IIRC there is a hanging question from the first time through this subject. I guess that the water pressure and the air pressure in the accumulator are the same?

 

Richard

 

It is really useful to get real life measurements of what is going on so that we can prove the mathematical model is correct. Thanks Chris.

[kerching]

I have to admit to being disappointed that there were no exponential coefficients in the equations.

 

[chris w]

I guess that the water pressure and the air pressure in the accumulator are the same?

 

Richard

Yes they are.

 

Chris

[Guest]

Hope you have P.M.'d TerryHell with your findings

[RLWP]

Yes they are.

 

Chris

 

So, bottom line, based on your theory and practical experience, set you accumulator charge pressure to just below you actual pump cut-in pressure. Right?

 

Richard

 

Excellent work. Thank you Chris

Edited November 23, 2008 by RLWP

[chris w]

So, bottom line, based on your theory and practical experience, set you accumulator charge pressure to just below you actual pump cut-in pressure. Right?

 

Richard

Absolutely so.

 

Chris

[Batavia]

I have to admit to being disappointed that there were no exponential coefficients in the equations.

You could have postulated that the compression of the gas might have been partially adiabatic and got a gamma in - and then demonstrated that this was not in fact applicable!

 

Chris G

[Timleech]

What about Peukert?

 

 

Gibbo

 

I must admit that when I saw the title I assumed it would be another battery charging thread

 

Tim

[alan_fincher]

I must admit that when I saw the title I assumed it would be another battery charging thread

I've read it all, and still thought it was!

 

Seriously though, very useful, as I'm busy completely reworking our plumbing, and haven't rechecked the accumulator pre-charge since I set it up about 3 years ago.

 

A good prompt to include that in what I'm doing.

 

Incidentally, I will be fitting an expansion vessel as well as the accumulator.... (Where's the licensed plumber!?).

[chris w]

Seriously though, very useful, as I'm busy completely reworking our plumbing, and haven't rechecked the accumulator pre-charge since I set it up about 3 years ago.

With non-diaphragm type accumulators, the air at the top will eventually dissolve in the water and the unit will need refreshing with air every few weeks. Ergo, these types are not recommended.

 

However, even with diaphragm type accumulators, there will be a very slow decrease in the air pressure due to the butyl rubber diaphragm being ever-so-slightly porous to air. We're probably talking about 3psi per year. The valve may also leak at the 1 or 2 psi per year rate.

 

Incidentally, I will be fitting an expansion vessel as well as the accumulator.... (Where's the licensed plumber!?).

Don't worry, I'm working on the full maths for the EV too!!

 

Chris

[alan_fincher]

There were some doubts about the state of health of our diaphragn based accumulator when refitted about 3 years ago, (probably 13 years old now). It will be interesting after 3 years of no attention to see what state it's in, before any recharging takes place.

 

It seems to do the business on pump cycling, but I'm not sure we can draw off a lot of water before the pump first kicks in - I could measure that too....

[chris w]

There were some doubts about the state of health of our diaphragn based accumulator when refitted about 3 years ago, (probably 13 years old now). It will be interesting after 3 years of no attention to see what state it's in, before any recharging takes place.

 

It seems to do the business on pump cycling, but I'm not sure we can draw off a lot of water before the pump first kicks in - I could measure that too....

To measure the pressure in the accumulator, you MUST switch off the pump and open a tap to allow all the accumulator water to run out. Leave the tap open and then measure the accumulator pressure.

 

Knowing the accumulator pressure, the accumulator size and the pump's cut-in and cut-out pressures, you can use my formula to calculate the amount of water the accumulator should be holding and compare that to what you actually get. If they correlate, the accumulator is OK. If there is a noticeable discrepancy, the diaphragm has probably got a pin hole in it. Or send me the data and I'll compute it for you.

 

To measure the pump's cut-out pressure, keeping the taps closed, simply measure the accumulator pressure when the pump cuts out. The two will be equal.

 

To measure the pump's cut-in pressure is slightly less easy. Keeping the taps closed, I found the easiest way is to have a helper with their finger on the water pump switch. Then you slowly depress the air valve on the accumulator and allow the air to escape gradually. At the INSTANT the pump comes on immediately switch off the pump at its switch. Measure the accumulator pressure and that will now be equal to the pump's cut-in pressure. Don't forget to pump the accumulator up again!

 

As I said above, my actual pump pressures were 3-4psi below the stated values on the label (Shureflow Aquaking).

 

Chris

[Keeping Up]

With non-diaphragm type accumulators, the air at the top will eventually dissolve in the water and the unit will need refreshing with air every few weeks. Ergo, these types are not recommended.

That is true but the timescale is much longer than that. I find that once every 3 months is perfectly adequate. For that minor inconvenience they are massively cheaper. Ergo those types are recommended.

 

Mine has no diaphragm, and no pre-pressurisation facility. There is a screw at the top, and all that is needed to recharge the air is to turn off the pump and undo the screw, then open a tap somewhere as low as possible. This empties all the water out of the accumulator and fills it with air. Then all that's left to do is to remember to re-tighten the screw in the top before turning the pump back on. Total time taken, about 2 minutes.

[smileypete]

Mine has no diaphragm, and no pre-pressurisation facility. There is a screw at the top, and all that is needed to recharge the air is to turn off the pump and undo the screw, then open a tap somewhere as low as possible. This empties all the water out of the accumulator and fills it with air. Then all that's left to do is to remember to re-tighten the screw in the top before turning the pump back on. Total time taken, about 2 minutes.

 

You could rig up a schrader valve and one of those little 12v air compressors

 

cheers,

Pete.

[Keeping Up]

You could rig up a schrader valve and one of those little 12v air compressors

 

cheers,

Pete.

Indeed I could. It would probably take me only a couple of hours; about the same time as I would spend, at 2 minutes per 3 months, in 15 years. I seriously doubt that I'll still be boating in 15 years time.

[chris w]

That is true but the timescale is much longer than that. I find that once every 3 months is perfectly adequate. For that minor inconvenience they are massively cheaper. Ergo those types are recommended.

The other inconvenience with a non-diaphragm type is the fact that they must be installed vertically, ie: with the air pocket always at the top. Diaphragm types, on the other hand, can be installed at any angle.

 

Chris