How to Get Flow Rate for Cut-out Pressure of Vessels

It looks as though the maximum operating pressure of the vessel is 3 Bar. The cut-out point is to stop vessel overpressure.

The flowrate/pressure numbers in 3 and 4 above cannot come from the pump performance curve.

Thank you PWSlack for your answer.

If maximum operating pressure of vessel is 3 bar, why they are using to 2.84bar without using to exactly 3 bar. Because 3 bar is easily more than 2.84 bar when the calculation.

Moreover, our building needfull flow-rate is 5.96. If we using this pressure and relative flow-rate, will be different require amount of water. I can't clear in this portion.

Ask <...they...>!

If you have to, go all the way back to the architects, who apparently were in charge of the engineers setting up physical plant for the building.

The vessel requirement of 3 bar might be reasonable, but at any rate, there should be a safety relief valve on it?

If you are sure about the "needful" flow rate of 5.96 L/s, is that the minimum rate required, or the maximum requirement?

There is actually no need for a pressure relief facility if there is no prospect of the supply pressure exceeding the rated pressure of the vessel. Process Engineers do this stuff in their sleep; not so Mechanical Engineers.

It actually matters in code within the U.S. what the intended use of the vessel is. If for just cold water supply, or for fire water, I tend to agree that no safety relief valve is required. Strangely enough, I have seen safety relief valves on condensate polishers running on warm to hot condensate in a cogeneration facility. Apparently, this is the case when the possibility of condensate reaching the vessel when a temperature could raise the steam pressure to values higher than the pressure rating of the vessel (as in a steam explosion). Also, any boiler feed water deaeration vessel must have a safety relief valve, as would any other water heating device.

Ask GRUNDFOS to send you a user's manual, or go on-line and download it.

Then read it.

The terms in the OP are altogether uncorrelated and confused. Perhaps this is an ESL (English as Secondary Language) problem.

This whole business seems to be about determining the size of an expansion tank.

a quick Yahoo search on your "vessel size" returned a few street addresses in various towns, but not much related to a tank. When I included that entire line in the search term I get a range of things from container ships to oil tankers to blood vessels to stuff about neuroscience for kids.

Did you mean to type this: ANSI code 2719.53? Unfortunately, no search results obtain on that either. You, sir, have apparently "stumped the band", could you hum a few bars?

Some posters seem to have an idea what you're asking, but yours doesn't make any sense to me. Perhaps I'm too tired!

English probably isn't your 1st language, but can you explain better just what your problem is? Then you might get a more useful answer.

The OP is not an engineer.

You failed to take the variable flow, specified by the OP (cut-in) 5.96L/s...... (cut-out) 8.94L/s into account in your discourse. That might have some impact on duty cycle.

I'll let you use my broom.

I never claimed the OP was an engineer. What i claimed is whom ever is engineering this use very ambiguous although technically workable values for everything.

Cut in and cut out references also use pressure as given in Bars.

The flow rates are as I take it the anticipated maximum flow the system (~95 GPM out flow. 142 GPM in flow @ cut out pressure of ~ 42 PSI) could be requiring and the recommended pump capacity to achieve and surpass that flow requirement. Also given the known anticipated input and output flows and the 20 cycles per hour limit the actual maximum duty cycles per run event or per hour is easily calculated, No BS there.

As far as the remaining math what do you come up with for a suggested surge tank capacity given the 95 GPM out and 142 GPM in with no more than 20 recharge cycles per hour and the estimated run time per cycle and total run time per hour based on those two flow rates?

If I explained what we have and what we are looking for my 14-year-old niece could work this math problem out!

To me, this is basic fluid flow related calculations. The OP just has them grouped oddly by normal expectations of how most of us would be familiar with.

Which as most of us would be more familiar with it would look more like this.

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Required input flow( minimum pump capacity) 142 GPM @ 42 PSI

Anticipated maximum output flow. 95 GPM @ >30 PSI

Pump cycling events. No more than 20 per hour.

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BTW, Those specs show that a pump with a flow rate of ~50% more than the anticipated maximum is being speced. Anyone else catch that speced in engineered reserve capacity ratio?

Now given that what sized surge tank would you recommend?

Doesn't matter.

You could size the vessel to give volume between pump stop/start to ensure max 20 starts/hour, but if your niece is going to work it out from first principles she will need a touch of calculus. No reason why she can't handle that! There is also some estimating of average flow involved, because as lyn says in #11, the pump output varies as the pressure changes.

But most systems use a smaller vessel and limit the start frequency by a minimum run timer, set to 3 minutes for 20/hour. If pump stop pressure is reached before the timer expires the pump runs on, possibly dead-headed, but for the short time that's OK.

Really? Three minutes? We have pumps in our generation plants that would totally overheat (the water) in that period of time if dead-headed, resulting in a steam explosion the like of which has not been seen.

Yes, really! That may be, but I guess those pumps aren't on hydropneumatic systems. Companies selling these usually propose timer control as I described. They do this side of the pond anyway.

I did an estimate of temperature rise. Assuming 6 l/s, pump likely to be 80mm suction, 50mm discharge, volute say 300mm dia x 100mm wide, volume ~ 7 litre. 3bar pressure rise, hydraulic power 1.8kW, efficiency 60%, motor rating 3kW. If all that goes into the water (closed valve power probably lower) and ignoring heat losses, in 3 mins ΔT = 18K, which is acceptable. And that's worst case, pump start and then demand immediately closed off. In practice some (or all) of the 3 mins is likely to have expired, and the flow may be low but not zero,

Effectively 18 L-bar/second of work being done on the water, except for the mechanical loss of the pump, as you stated. 1 bar = 0.98692 atm.

1 L-atm/sec = 101.33 W, according to an engineering site that offers conversion factors, and that number seems familiar to me.

Thus we have 18 x 0.98692 x 101.33 = 1.8Kw, so work done in one second is 1800 J, but energy expended on the water to arrive at the pressure is higher due to efficiency, so we are saying 3 kW, 3000 J/s.

Now, if the volute of said pump is effectively one liter, and the timer is on for a full three minutes we have: 540,000 J applied to heating 1000 g of water (or thereabouts). The heat capacity of water is 4.1855 J/g°C (the mechanical equivalent of heat), thus we now may calculate the rise as: 540,000 J/4185.2 J/°C-Kg = 129°C rise, and if the water was initially at 20 °C, 149 °C is the result, with a steam pressure of ~460 kPa (66.7 psig, 4.6 bar.) Alright, alright, it might not explode in the worst way, but the pump will be hot, and you could wipe the seals and bearings depending on allowable temperatures.

Now go back and do the calculations with a pump having even more confined volute.

Say 500g, then we get a 258 °C (rise), a final temperature of 278 °C, having a pressure of ~62 bar, and then it all depends on what breaks first, especially if the metal is not rated for that pressure, but I would not want to be there when it lets go.

You do give yourself a lot of work with units! Why not say 1 L-bar/sec = 100 W, so 6 l/s at 3 bar = 1800 watt? Saves a load of decimal points!

Without disputing your analysis point by point (though I could), how do account for the fact that the method I described works fine in practice? It may not always be appropriate, but if the customer hasn't checked it's OK in his particular case, you'd expect the supplier to do it.

Obviously, it all depends on the rate of conversion of mechanical power to heat, and the constrained/contained volume of water, obviously, in a power plant it just isn't done.

http://engineersedge.com/pumps/pump_protection.htm

Just one final pass-by on this subject, then I will let you "geniuses" figure it out.

In a high powered, high pressure boiler feed water pump (certainly not what the OP was referring to), If the pump goes to dead-head for more than a few seconds, seals will wipe out, the pump will overheat, and possibly the casing will rupture (perhaps injuring nearby personnel). I suspect we are referring to two completely different situations, although I encountered once in an industrial setting, a centrifugal pump that was supposedly pumping cold water that was neither pumping (moving water) nor cold, as the most horrible clattering sound was taking place, and the pump would produce a second-degree burn if touched.

Fine, I'm sure you're right, and you know far more about that situation than I do. But I stand by what I said about "ordinary" hydropneumatic sets, and the subject of this thread.

Incidentally, I don't see where nt gets vessel volume 2720 litre (he doesn't say litre, but it seems reasonable). If he means that volume between pump start/stop, that's enough for pump flow about 60 l/s, working it out properly .

Actually, I think that is a pretty small vessel? Only 718.6 gallons? What in heck for? He doesn't say. This is one of those posts where it should be hands off, until the OP clears up a few things, but as usual, he is nowhere in sight.

Not only that, is he talking about a hard tank with no air space? A bladder tank? An ion exchange tank? A condensate tank? A deaerating feed-water tank? A bump tank?

Hell, for all we know about this "building" or "homework", it could be an elevated open to atmosphere tank. It might be a Shark Tank.

<unsub>

"You could size the vessel to give volume between pump stop/start to ensure max 20 starts/hour, but if your niece is going to work it out from first principles she will need a touch of calculus."

Calculus? I worked it out with basic math, you know, + - X / =. That sort of stuff.

So did you work out the demand flow which gives maximum start frequency? i.e. worst case for sizing the vessel. I'm not saying it can't be done, just that calculus is probably the easiest way

Taking pump output Q_p ~ 9 l/s ~ 8520 gph and demand Q 6 l/s ~ 5700 gph as you say (I didn't think demand and pump duty were all that clear in the original post), your vessel sizing calculation does not give worst case.

For a given vessel volume V (between pump start/stop) maximum start frequency f occurs when demand = ½ pump output, and f = Q_p/V/4. Demand = ½ pump output is clearly a possibility. That can be shown using calculus, though of course I don't do the calculus on every job, the formula is well known.

So V = Q_p/f/4 and putting in the figures V = 403 litre = 106.5 gal. If you do your calc with demand 4260 gph that's what you get. You might say there's not much difference, but if you used demand further from Q_p/2 it gets worse.

I did some experiments with about 1 Kg water in a small vessel on a "balance beam" scale, a micro-switch to a small hydroponic pump (low head centrifugal), and a siphon tube draining off water at a near constant rate from the vessel.

I have found that my set-up (which has a metal ball in a track on the counterweight arm, and a stabilizing ballast weight below the scale pan) that trips on and off based upon total weight of vessel, will deliver smaller increments of water (shots) when the outflow is a mere stream of droplets (~1 mL/s) compared to a full stream (up to 6 mL/s).

The full stream was about 6.88 mL/s, whereas the pump maximum output was found to be 6.25 mL/s. This obviously resulted in one test at full outflow where the pump made only one cycle, starting about 3 seconds into the run, and ending about 17 seconds after the outflow stopped (when the scales were still indicating low weight, and the pump was still running.

At low drip, each pump cycle was about 13 seconds, when the outflow was ~0.5 mL/s. In this case, pump duty was about 15% at most, and the middle range of outflow with a small stream gave about 35% cycle duty. These are only preliminary tests (mainly to see how well my scales are working, and I learned I have to install magnetic damping), as I plan to get an agricultural spot sprayer pump (DC powered) next, rewire, and test again, at exceedingly low duty factor.

Interesting, but I'm not sure what the object is. Please advise.

If you're saying that when demand is much lower than pump output (or just below pump output) the pump start frequency is reduced, that's true. If you know the flow demand won't vary much, can calculate actual start frequency based on that, and possibly reduce the vessel volume. But in general you don't know that, and it's safer to assume demand can = Q_p/2. And that gives a much simpler calculation for vessel volume than working out the actual pump running and pump stopped times .

This has more to do with my present LER-1 vessel level control, and less to do with topic.

The chart above represents vessel contents, outflow, inflow, and pump status for the following bespoke parameters:

OK.

BTW, I assume by "The chart above" you mean the small thingy at top left, but it hasn't come out. I have same problem trying to paste from eg Mathcad. It seems OK, but when I hit Preview I get what you got.

Very inconvenient, sorry. Cannot get this work any way I slice upon it. Server will not accept if I use camera browse.

I have tried pasting from several applications, but the data goes {poof} each time. No soup for me, no soup for you., sorry.

I keep trying, but it keeps wiping out the chart. I have not figured out how to save the chart as a picture.

Me neither. I can do photos (jpg) OK. Some kind soul once explained how to insert other docs, and I think it worked, but it was a while ago and quite complicated. I do it so rarely I've forgotten how. I tried saving as jpg but it didn't work.

You didn't answer my question in #23 - did you work out the demand flow which gives maximum start frequency? using (elementary) calculus or otherwise.

As I said in #27, your volume gives more than 20 starts/h worst case. Also my calculation for vessel volume is much simpler.

One way to look at it is - at zero demand, the pump never starts, so start frequency is zero. When demand = pump output, the pump never stops, so start frequency is zero. So it's a reasonable guess that max start frequency is when demand is 1/2 way between, Q_p/2, which is correct though it's not a rigorous proof.

Actually not quite half, since the pump must start once on 100% demand or tank runs empty. Oops.

I don't understand the point you're making. Please explain. In my scenario the demand is constant at Q_p/2, the pump is start/stop at Q_p.

I don't know Qp from Shinola, but I do know that if the freakin' pump don't start, the tank goes empty.

I think I explain my reasoning with basic math well enough in post 26.

The OPs output demand is known to be at worst case 5700 GPH and the pumps minimum working capacity is at least 8520 GPH. At any flow rate less than that and the given stored take volume of ~95 gallons per cycle on a 30/40 PSI pressure change there is no way that the pump will need to start more than 20 times an hour.

I don't know where your 1/2 or Qp/2 any other numbers come from though but I do know that those who believe in using calculus where basic math is more than sufficient tend to over think the simple to the point of coming up with wrong answers.

OK, I'll try once more to explain mine.

The OP's output demand is maximum 5700 GPH, but that is not the worst case for pump start frequency. OP didn't say demand is constant at 5700 GPH, and there's no reason why it would be, users will be opening and closing taps (sorry faucets ) at random.

At 5700 GPH demand, the pump is stopped for 95/5700 = 0.0167h. The pump is running for 95/(8250 - 5700) = 0.0337h. Total start-start time = 0.0167 + 0.0337 = 0.0504h, so start frequency = 1/0.0504 = 19.86 per hour.

At worst case demand 8250/2 = 4125 GPH the pump is stopped for 95/4125 = 0.0223h. The pump is on for 95/(8250 - 4125) = 0.0223h. Total start-start time = 0.0446h, so start frequency = 1/0.0446 = 22.4 per hour.

I would say you get the right answer by thinking the problem through and using appropriate tools, including eg calculus if it helps. The formula V = Q_p/f/4 can readily be found using calculus (and in other ways), but as I said before, there's no need for that in a specific case, just use the formula. If you want to give me your email address I'll send you the derivation, can't seem to insert it here.

I'm not sure how you are coming up with a demand of 4125 GPH (50% of pump capacity)being a higher hourly demand and thus equating to a longer runtime or number of cycles than 5700 GPH (66.67%of pump capacity) given the maximum Liters/Second spec the OP gave.

At any outgoing flow rate of less than 5700 GPH the pump cycles per hour will be less than 20 given the fixed pressure change and thus fixed tank volume available per cycle.

In theory anything could happen. In reality no one cares unless the water stops flowing completely which is impossible given the peak demand Vs peak supply numbers of the system.

Personally, I set my home pump system at a 45/55 PSI cycle range with the pressure tank precharged to ~30 PSI so that even if my demand exceeds my pump flow the system has to drop below that ~30 PSI precharge pressure before everything drops down to what my pumps working capacity is.

I've never calculated the exact reserve but on my 40 gallon pressure tank that extra ~15 PSI gives pretty close to 5 extra gallons of reserve capacity over being pre-charged to 45 PSI.

What part of my post didn't you understand? I've shown clearly enough for most people that demand 4125 GPH results in start frequency 22.4 per hour. I thought you were happy using basic arithmetic. As I said, it's entirely possible for the demand flow to be anything up to the stated maximum 5700 GPH, including 4125 GPH.

Using the approach in my #35 (to avoid calculus!) so accepting that for vessel volume V and pump output Q_p, maximum start frequency occurs when demand = Q_p/2. When the pump is stopped outflow = Q_p/2, so stopped time = V/(Q_p/2) = 2*V/Q_p. When the pump is running net inflow = Q_p - Q_p/2 = Q_p/2. So running time also = 2*V/Q_p. Therefore total start-start time = 2*V/Q_p + 2*V/Q_p = 4*V/Q_p. And for maximum start frequency f volume required = Q_p/f/4. This isn't rocket science, the formula is well known (or should be) among designers of eg pump sumps. In this particular case (f = 20 per hour), Q_p = 8250 GPH, V = 8250/20/4 = 103 gal.

The pressures have to be set to give the required vessel in/out volume, with a given air volume, but otherwise aren't directly relevant.

Can anybody else on this thread confirm (or otherwise) what I'm saying?

Alright you two, pistols at 20 paces. No one turns until the signal is given, or the seconds may fire upon whomever turns illegally.

In setting up your basic calculations, TCMtech, are you providing any consideration to the outflow of water from tank while the pump is running, or just considering that the pump will run until a given pressure cutout is reached? Codemaster, are you including the outflow during pump run?

I have witnessed directly using a weighed vessel that rate of outflow (even if constant) makes the "shot" the pump delivers larger and larger as outflow approaches pump flow maximum value, such that at a high enough out flow rate, the pump will never cycle, and the tank may or may not reach a steady state level.

Codemaster, are you including the outflow during pump run? Yes of course. I thought it was clear in my #42. Net inflow = pump flow minus outflow (demand). Worked out start frequency as function of a given demand, but that demand can vary (can't think of a better way of putting that, but I'm sure you know what I mean).

In practice there are also things like delay in pump flow starting when motor start signal given etc. Also the pump output won't be constant as the flow most likely falls as the pressure rises, so in practice the average would be estimated. But neither affects the basic principles we're discussing here.

As you say, at a high enough out flow rate, the pump will never cycle. I said in #35 - when demand = pump output, the pump never stops, so start frequency is zero. In the case here though maximum demand is by definition less than pump output.

Inflow minus outflow until shut off pressure is reached.

At an outgoing flow rate of 71 GPM (50% minimum pump capacity and a ~95 gallon capacity in the tank) I get a cycle time of ~ 1 minute 20. seconds to drain the tank and have the pump start.

From there I get ~ 1 minute 50 seconds to fill for a combined total of 3 minutes 10 second per complete cycle or roughly 19 start/stop cycles per hour.

95/71 = 1.34 minutes drain out time.

71 x 1.34 = 95 gallons more went out during the refill half of the cycle.

142 x 1.34 = 190 gallons came in during that 1.34 minutes on cycle.

71 x 2.68 = 190 gallons out for the complete 2.68 minute cycle.

2.68 x 60 = ~ 2 minutes 41 seconds or roughly 22.37 full cycles per hour.

Sure. He's right. Close enough to 22.4 cycles per hour @ 50% flow. My bad. I forgot that pumping cycle ratios follow a slight bell curve that's highest at half flow.

Still I didn't need anything more than 5 lines of basic math I did in my head (double checked with calculator though) for the calculation and proof to get there.

OK, took some doing but I'm glad we sorted that out!

I must admit to a couple of typos. In #40 I said pump flow 8250 GPH so worst case demand 4125, instead of 8520 and 4260. The times 0.0223h and start frequency 22.4 per hour are correct. Vessel volume in #42 should be 106.6 gal.

Plotting start frequency vs demand flow gives a parabola (strictly speaking a bell curve is from the normal distribution, y = exp(-x²)). It's easy enough to plot on XL, and it shows max frequency at ½ pump flow (as an alternative to calculus). But still need to try various vessel volumes to get specified max frequency. Easier to use my formula!

Doing the basic math(s) in your head or otherwise is one thing, but it doesn't hurt to apply it to the right equations if you want to get the right answer.

Unfortunately, I failed as many higher mathematics classes as I failed English classes.

In my brain none of that stuff translates to anything I can understand let alone work with. It may mean something to you but fro me it's as relevant as if it had been written in ancient Eskimo text.

It does nothing but make for confusion and to be honest I have yet to find any real life scenarios where I couldn't reduce a problem down to basic math to find an answer that gave me a close enough approximation to know whether something would work or not.

In my day to day life I don't deal with mathematical ideals or absolutes as in the case of this topic.

Given a 500 gallon pressure tank the only difference between it holding 95 gallons and 105 gallons per cycle is a bit of pressure manipulation on the air charge side. Sure it might make the pressure differential an extra PSI or two more but when dealing with nothing more than common mechanical pressure gauges that can have over a +10% variance in as the needle reads Vs true pressure.

Add in in realistic flow and pressure loss through the system to the end points and being off by a 1- 2 PSI or so would be unnoticeable to the end user who just wants water come out of their faucet when the turn the knob.

Lastly as the guy who is paying for the system would see things being able to buy a standard sized prebuilt 500 - 600 gallon pressure tank for say $1500 - $2000 that can be on site the next day Vs a custom built one designed to some highly specific volume for $3000+ that may take a week or longer to arrive it's a no brainer.

When it comes to complex theoretical math I stink at it always have always will, It just never proved its worth the effort to learn and use work what little added anything it ever presented itself for, and to be honest when you on a work site and you have to relate something to the average person who only knows basic math being able to tear apart a complex problem and present it to them in basic numbers and ratios they can relate to does way more good than baffling them with BS and calling it math to justify a nitpick that in reality makes very little difference in how the end system works Vs what it will cost to implement.

To be honest, all my life I have just been able to see things in basic numbers and that it. I can;t recall how many math teachers and professoreI have went up against that were trying to solve some supposedly 2, 3 or 4 variable word problem that describes a real life situation only to have me reinterpret the problem with the given knowns to show that there were not 2, 3 or 4 variables at all, but one key one that ultimately defines the others,, and that the whole problem could be easily solved with basic math that everyone in class could follow and relate to.

Same end answers but done in 5 - 10 lines of basic math rather than several chalkboards worth of scribbling and unnecessary BS that only 'the chosen few' could follow.

I don't do complex theory worth crap but I do simple reality very well.

OK, thanks for that, no problem!

Just a minor point about varying the pressure a psi or 2 - my calculation is mainly to do with sump sizing (though it works for pressurised tanks) and if the minimum level for pump suction, and maximum near overflow are determined, there's not much scope for increasing the volume to reduce start frequency.

I agree in practice you'll usually pick an off the shelf tank, but it's worth getting the calculation about right or there's a chance if it's borderline you'll go for one too small.

Good point about how to squeeze the bladder tank. Yes, it really matters.

The pressure ratio matters. The volume of the vessel itself, and the volume taken up by the bladder when the water is completely pushed out matters. Is that 95% of the tank? Maybe. There are tank internals also to consider, so 85-90% is more realistic, depending on tank size and vertical or horizontal.

The same calculation I did that was based on 30.5% recovery between the two pressures, might could have been a lot conservative. The actual volume ratio was more like 0.665, so one could conceivably recover about 33-34%, then there is the matter of how far can the pressure ratio be pushed.

You guys are essentially both wrong and right at the same time, right for the wrong reasons...

The tank mentioned is 720 gallons, of which at the pressures stated will deliver about 220 gallons (30.6% of nominal tank), at half the maximum output of the pump

9L/s -> 1/2 is 4.5 L/s that is 16,200 L/hr demand, so pump has to cycle 220 gallons (832.8 L) each for 19.45 cycles. Amazing.

No! It's the start-start time which needs to = 3 minutes for 20 starts/h. If the vessel is 832 litre, at 4.5 l/s demand it takes 3 mins to fill, then pump stops and it takes another 3 mins to empty and pump restart. That's 10 starts/hr. As I've said a few times V = Q_p/f/4 ~ 106 gal. I imagine you're OK with calculus, if you don't want the trouble of working it out I'll send it, if you let me have your email.

BTW in #24 I was assuming the 2720 L was in/out working volume, it makes a lot more sense if it's working volume. I've been using working volume throughout the thread. I haven't completely digested your #50, you seem to know what nt is looking for, but I admit it's not clear to me, and he hasn't been back to clarify, surprise, surprise.

I say we drop this. If the pump is right, yes, 20 starts/hr is maximum, if the maximum withdraw rate is somewhere near the maximum capacity of the pump. Otherwise it really depends on the size of the bladder tank and the pressure ratio (which leads to volume cycle amount, due to gas compressibility. Also, if the pump cannot keep up, it will burn up eventually. If the pump spends all of its time on the top end of the pressure, attempting to cut out, but can't quite get there, that is a pure burn out situation, since the pump is not designed for (1) too many start/stops, nor (2) continuously running at the pressure set on the tank.

I really don't see the problem. Doing this from scratch I would look at pump curves and select pump start and stop pressures based on what's suitable for the water users, and available from the pump, making sure the pump curve goes well above the stop pressure. Also estimate the actual volume from the pump, taking account of flow variation with pressure, this might mean adjusting the design vessel working volume.

As an example, take start pressure P₁ = 3 bara, stop pressure P₂ = 4 bara. At P₁, volume of air (actual) V₁ = k/P₁, at P₂, volume of air V₂ = k/P₂. So working ΔV = k*(1/P₁ - 1/P₂) = P₁*V₁*(1/P₁ - 1/P₂) = V₁*3*(1/3 - 1/4) = V₁/4. If ΔV = say 100 L, V₁ 400 L, V₂ = 300 L. Similar ratio to your example. Vessel total volume a tad over 400 L so it's not quite empty of water at pump start. Need something to stop the membrane being forced out if system pressure falls to atmospheric, eg during upset or maintenance, some sort of poppet valve is usual. Precharge to ~ 3 bara.

?? Tank seems way too small.

I disagree.

If the pump manufacturer sized their motor correctly to match the pump head pressure and flow curves correctly burning out the motor should not be possible.

Maximum flow at no head pressure, other than inherent piping restrictions, and maximum head pressure at minimum, but not dead head, should not burn out a properly matched pump motor.

OK, I surrender. It won't burn up the pump motor, the pump will not cavitate (since NPSH criterion is met by default.

The tank seems way small, and over the course of time will put a lot more starts on the motor than should be if 400L is the answer. A pump that can do 140 gpm putting 100L in a bladder tank will probably run for start-stops as short as 11-12 seconds, but if this pump maxes out at 9 L/sec (that is not zero pressure, but still more or less meets bespoke cut in pressure), and the draw is 9.01 L/sec, then the pump must lose ground and never shut off, once started, at least until high demand drops back to nominal.

We already know that maximum starts/hr condition is met for the motor. Good.

The reason for having a larger tank is two-fold IMHO: (1) steadier demand pressure for service, and (2) pump runs a bit longer each cycle (the number of cycles remains a constant of the flow conditions), while draw down ratio of the tank can be made less drastic.

I'm not sure where the 400 liter number came from. It's not mine. I say 500 gallons give or take just to meet the ~100 gallon storage capacity on a ~10 PSI pressure differential.

Which BTW on a full 40 PSI to ~0 drop would be a reserve of ~300+ gallons assuming a rather low ~10 - 15 PSI precharge.

I totally agree, and get what you are saying. Nicely, succinctly done.

Sorry to run on about it, but I still don't think you've got this right.

First, I don't get your comment about cavitation. Unless I'm missing something, the pressure vessel is on the pump discharge. The pump suction conditions are unknown to us.

When you say A pump that can do 140 gpm putting 100L in a bladder tank will probably run for start-stops as short as 11-12 seconds, but if this pump maxes out at 9 L/sec (that is not zero pressure, but still more or less meets bespoke cut in pressure), and the draw is 9.01 L/sec, then the pump must lose ground and never shut off, once started, I think you're mixing together 2 scenarios. If demand is zero the pump starts (eventually) and stops after about 11 sec, but unlikely to start again for hours (until the pressure has decayed) so start frequency is very low. Start-stop is ~ 11 s, but start-start is what matters to the motor.

If demand is 9.01 l/s, that's outside the parameters given, but it's not a problem. The pump will be capable of 9.01 l/s at a little lower than design minimum (pump start) pressure. Users will prefer that to having no water! The bladder will expand a bit more, if there is room in the vessel, or until it hits the stop-blowout valve, and the system can work like that as long as the demand exists. It will behave as if solidly piped, with no vessel. During that time the pump doesn't stop (as you say) so start frequency is zero.

If demand is say 1% of pump output, 0.09 l/s, the pump still takes about 11s to input 100 L (and stop on high pressure) but it takes 100/0.09 ~ 1110s to restart, so start-start time is ~ 1120s and frequency ~ 3 per hour. If demand is 99% of pump output, frequency is also ~ 3 per hour, but pump is then running most of the time instead of stopped.

I agree there are benefits in having a larger vessel, but that's changing the basis of the question, 20 starts/h.

BTW in my #55 I should have said "......it makes a lot more sense if it's working total volume".

The comment about NPSH is the typically ignored item when peeps put in pumps, unless they know their poops.

I think we are more or less on same page now, but might as well use the vessel size OP gave us at the start. 400 L too small IMHO, although it would be a good pulsation damper.

I just meant that the fact that the pump delivery pressure being minimum 2-3 barg has nothing to do with the pump NPSHa.

As long as your page doesn't still say working volume 220 gal gives ~ max 20 starts/h. Because mine says it gives 10 starts/h.

I've agreed the bigger the tank the better, but that's not the question posed .

To come back to this - if the vessel is sized for 20 starts/hr when the withdraw rate is somewhere near the maximum capacity of the pump, starts/hr will be much too high when demand = ½ Q_p. The vessel should be sized for that condition. The low and high demand situations can take care of themselves!

As demand goes higher than 1/2 Qp (pump capacity), the pump will obviously put more water into the system each hour, but will run longer trying to catch up, thus less starts per hour. At the highest flow the pump only makes single digit start numbers, I think.

At the highest demand flow 5700 GPH, and tcmtech's vessel volume 95 gal, start frequency is 19.86 per hour. See #40.

START OVER:

OP states:

Tank spec: 2719.53 (L vol, I suspect) --> 718.5 gallons

Qp (pumping rate) (Singapore Std. CP48) 5.96 L/s (94.468 gpm) The pump specified can certainly provide that flow rate within the scope of the optimal portion of the pump curve, and with high efficiency. The pump is to start when draw down reduces pressure to 1.89 bar (19.3 m head, 27.4 psi), and switch off at the cut out pressure of 2.84 bar. Thus the tank operates between 27.4 psi, and 41.2 psi.

CRNE 15-2 GRUNDFOS pump: At the maximum pressure of 2.84 bar (28.965 m head, 41.2 psi), this pump can only do just barely 32.184 m³/h (8.94 L/s), and this is way off the end of the preferred pump curve at 60 Hz (better than 50 Hz), and requires NPSH of over 10 m head, and requires about 2.8 kW power, since Eta% is dropping off in this region.

Poor choice selection of pump, it is undersized, although it might function, and overheat the motor.

OP has his pressure units off by some amount: 10 m water head is NOT 1 Bar, close as in horseshoes, but not exact.

If this is a pressure tank for potable water service in a building (and that appears to be the assumption), the tank should be able to accommodate a wide range of volumes without exceptionally high changes in pressure, this limits draw-down in most 30-50 psi tanks to 30.5% of nominal tank capacity. A 720 gallon nominal tank with that draw down ratio will deliver 220 gallons per cycle.

The pump flow rate is not completely constant during bladder tank refilling, but for the pump in question will be maximized limit as ~9 L/s (142.7 gpm), then as pressure nears cutoff this will be nearer the 8.94 L/s mark. That is practically insignificant in this consideration of pump cycles per hour.

IF the withdrawal rate is ~9 L/s, then the pressure will drop until the lower pressure switch is reached, the pump will switch on once and remain on the entire hour.

IF the withdrawal rate is 0 L/s, then no change will take place and the pump will not switch on.

IF the withdrawal rate is Qp/2 (as stated by Codemaster, smart fellow), at 4.5 L/s, then what? Total withdrawal per hour is 16,200 L (4,280 gallons). At 220 gallons per cycle, the pump needs to switch on (and off) 19.45 times precisely (must round down to 19 or up to 20, depending on tank status at beginning of hour, and the pump will still be incomplete on a cycle.

information from OP:

In one calculation of Pressure vessel,

1. Pump feed area need to 5.96L/s (calculated by CP-48).

2. Choice to GRUNDFOS pump model CRNE 15-2.

3. (cut-in) 5.96L/s >>> 18.94m >>> 1.89 bar

4. (cut-out) 8.94L/s >>> 28.41m >>> 2.84 bar

5. start/stop 20 s/s per hour

6.ANS >>> vessel size is 2719.53

Highest demand flow was 9 L/s, 8559 GPH. I think you missed step #4.

at 9L/s, demand is for 8559 GPH, 142.7 gpm. That is what the pump puts out at about 20+ meters head. At Qp=142.7, then 1/2 Qp = 71.35 gpm. During one hour the pump will take 30 seconds start-stop time. While the pump is on, tank flow is about 35.7 gallons (30 s). The tank determines the stop-start time.

Tank is requiring the off time to be 2.5 min (2m 30s), so that total start-stop-start time is exactly 3 minutes, then the tank will be filled exactly 20 times in that hour for V=0.5Qp

Tank will flow 71.35 gpm for 2.5 min per stop to start = 178.4 gallons, so that must be the draw down on this tank of 720 gallons for a used capacity of 24.8% only. According to my expansion ratio for the bladder, this requires 100.4% of tank capacity to be used, so it will not work as bespoke.

total flow out of tank is (178.4 + 35.7)*20= 4282 GPH which is 50.0% of maximum draw down rate.

My suggestion to the OP is #1 he needs a bigger tank, OR #2 operate the tank with a smaller expansion ratio, i.e.- reset the top pressure to a slighter lower value, or raise the bottom pressure.

To keep things simple, I've assumed constant pump flow, constant demand (for each start-start time calc) and fixed vessel working volume. Apart from #57, I haven't done anything about vessel total volume, as it's not relevant to the points I'm making.

You say "at 9L/s, demand is for 8559 GPH, 142.7 gpm. That is what the pump puts out at about 20+ meters head. At Qp=142.7, then 1/2 Qp = 71.35 gpm. During one hour the pump will take 30 seconds start-stop time. While the pump is on, tank flow is about 35.7 gallons (30 s). The tank determines the stop-start time".

You don't give vessel volume, but for start stop time 30 s it must = your 35.7 gal. I'm saying (I think tcmtech now agrees) it needs to be ~ 107 gal, 406 L. I thought you were pressing for bigger volume . Also I don't understand "During one hour…". In 1 hour there are several pump on and off periods, but looking at it over an hour doesn't help the calculation, confuses things IMHO.

At demand 1/2 Q_p, stop-start time = start-stop time, so if volume = 35.7 gal off time (stop-start) cannot = 2.5 mins, it = 30s. If you try it with volume 107 gal you get both times 107/71.35 = 1.5 mins, total 3 mins and 20 starts/h.

The pump has output that is consistent with the pump curve and the pressure it is pumping into, no one said anything about a VFD or a bypass line on the pump output.

"...,then1/2Qp = 71.35 gpm During one hour cycle"... the pump is pumping at its nominal rate at the system pressure given, and that rate is Qp = 142.7 the entire time it is energized. With 1/2 Qp outlet flow, and 1 Qp inlet flow, the time to pump the difference (1/2 Qp ) is still precisely 30 seconds for the pump to reach cut out from cut in, if there were no out flow, but there is. Tank will outflow another volume x while the pump is on. Cut out volume on tank (V2) - Cut in volume on tank (V1 = time running x( Qp-1/2Qp). And V1-V2 = -time off x (outflow rate 1/2 Qp). We already know there has to be 20 cycles per hour, 0.33333/min.

Thus we have:

time off x (1/2Qp) = time on x (1/2Qp) ; time off + time on = 3 minutes/cycle Clearly time off = time on = 1.5 minutes. I misspoke, and you are totally correct. The cycling (working) volume is then V2-V1 = 1.5 min x 1/2 Qp = 3/4 (142.7) = 107.025 gallons.

Tank size was GIVEN BY THE OP at the beginning of the thread. Tank used capacity is 107.025/720 = 14.9%, so this requires a smaller change in system pressure than that stated by the OP at the beginning of the thread. OR there could be a smaller tank with more % working volume that will work well between the bespoke parameters of pressure.

Do we more or less agree this time?

The pump has output that is consistent with the pump curve and the pressure it is pumping into, no one said anything about a VFD or a bypass line on the pump output. That's right, I was just trying to make sure we were all on the same wavelength.

Yes, I think we do agree now. Phew ! But you do use an elaborate way of getting to the 107 gal!

As you say, even taking OP's tank size as total volume, it's still generous, in practice with that total volume probably better to use a bigger % to get bigger working volume, and so lower the pump starts/h.

It's a pleasure doing business with you . (I've just had most of a bottle of fine wine, St. Emillion 2009, finish it off shortly)

I would have been drinking a cheap beer while doing my calculations but I don't have any.

Common man. Common math. Common drink. No need to over complicate it any further.

I finally apologize about my "roundabout" ways, and "over complicated" way of calculation. Unfortunately, my wife has cut me off from the bar, so that might explain why I thunk the way I thunk.

That's all folks! Last one out turn out the lights.

thunk?

James Stewart, tcmtech, Codemaster,

Thank you for your answers & discussion by actively. This post is very ambiguous and not clear for us. I don't know how singapore engineerings calculated and find to this result. I have report other result for your reference . . .

Cut-in >>> 4.95, 2bar (don't know source)

Cut-out>>>3.3, 3bar (Building unit requirement)

------------------------------

Cut-in >>> 4.44, 2bar (don't know source)

Cut-out>>>2.96, 3bar (Building unit requirement)

------------------------------

Cut-in >>> 4.9, 2bar (don't know source)

Cut-out>>>3.34, 3bar (Building unit requirement)

------------------------------

Cut-in >>> 7.2, 2bar (don't know source)

Cut-out>>>4.8, 3bar (Building unit requirement)

------------------------------

In upper calculation, Using for Cut-out value to Building requirement flow reate.

Black texts is I don't know, how they will get its.

***********************************************

Cut-in>>>3.3, 2bar (Building unit requirement)

Cut-out >>> 4.95, 3bar (don't know source)

------------------------------

Cut-in>>>2.96, 2bar (Building unit requirement)

Cut-out >>> 4.44, 3bar (don't know source)

------------------------------

Cut-in>>>3.34, 2bar (Building unit requirement)

Cut-out >>> 4.9, 3bar (don't know source)

------------------------------

Cut-in>>>4.57, 2bar (Building unit requirement)

Cut-out >>> 6.8, 3bar (don't know source)

------------------------------

Cut-in>>>5.96, 1.89bar (Building unit requirement)

Cut-out >>> 8.94, 2.84bar (don't know source)

In upper calculation, Using for Cut-in value to Building requirement flow reate.

Black texts is I don't know, how they will get its.

***********************************************

My country is not singapore, so i can't ask they.

If this post bored to your, please stop & close.

I'm still not clear what your actual problem is.

You appear to know specified minimum supply pressure, 2 bar (gauge?) so that fixes the pump start pressure. (If minimum supply pressure is at the taps rather than local to the vessel you might need to add something for pipe friction, and don't forget levels might be different).

You give maximum demand flow, so your pump must supply this (at least) at pump start pressure.

Pump stop pressure is arbitrary, but your figure 3 bar seems reasonable. It's normal to size the pump for output rather higher than max demand, so it's running for about 50% of the time, but that depends on how long max demand is likely to go on for eg might be very low all night. In any case, you seem to have settled on ~ 9 l/s at start pressure, which looks OK.

You can then, from the pump curve, estimate average pump output between start and stop pressures. Use that in the formula I put in several posts to calculate the vessel working volume, ie the volume between start and stop pressures, to give not more than 20 starts/h. Or to be safe, use pump output at minimum (pump start) pressure. Following my #57, you can calculate total vessel volume. Note that you need to use absolute pressures. I suspect yours are gauge, in which case your 2 and 3 bar are just what I used in #57.

Let us know how you get on!

Building area is very big including with many towers. These are shorten data of 7 buildings. All have used to CRNE 15-2. But I can understand by generally, I showed these data for your refer only.

Under calculation is one of many towers.

1. Total Loading Unit of building is 827LU and equal with 5.76L/s.

2. Pump discharge head is 18.94m. (CRNE 15-2)

3. Q (cut-in) 5.96 L/s, 357.6 L/m

4. Q (cut-out) 8.94 L/s, 536.4 L/m

5. Head (cut-in) 18.94 m, 1.89 bar

6. Head (cut-out) 28.41 m, 2.84 bar

7. Start/Stop hour 20 s/s per hour

8. Acceptance Factore

(Cut-out - Cut-in)/cut-out = [(2.84 + 1)-(1.89 + 1)]/(2.81+1)=(3.84 - 2.89) / 3.84 = 0.25

9. Mean flowrate

(Cut-in flowrate + Cut-out flowrate) / 2 = (358 + 536) / 2 = 447.0 L/m

10. Mean Time Share by No. of Pump = 30 min

11. Effective Capacity of the pressure tank

(Mean Time Share by No. of pump(s) x Mean flowrate)/(No. of Start/Stop per Hour)

= (30 x 447.0) / 20 = 670.50 Litres

12. Therefore the total capacity of the pressure vessel

Effecitive Capacity / Acceptance Factor = 670.5 / 0.25 = 2719.53 Litres

I understand all statuses except #4 & #6.

Engineering used 8.94 L/s for (cut-out) flow rate & 2.84 bar for (cut-out) pressure.

If I calculate this design, I will use 3 bar for (cut-out) pressure. Why he used 2.84 bar for this & where he get the value of (cut-out) flowrate?

I want to know only this effect. How to determine for (cut-out) flow rate ?

very profound to decide for (cut-out) pressure, So I will use clearly 3 bar. (2.84 bar is very disappointed)

Otherwise, if i use any value of (cut-out) pressure, Is pressure tank show to match flowrate for user pressure? ( I mean that, some equipments are show important datas by square sheet on these body)

But my wish to know is showing above as (How to determine for (cut-out) flow rate ?)

The term "cut-out flow rate" can only be determined from the pump curve. When the pump cuts in at ~2barg, it will have a certain flow rate, such as ~9L/s. As the tank pressurizes to ~3barg, the pump flow rate will decrease to something like ~8L/s, according to the pump curve. For simplicity, you can just average the two flow rates (which will be slightly conservative). Your 2700L tank selection is okay.

this pump curve?

Are you having a laugh? 9 l/s = 32.4 m3/h, this only goes up to < 23.

very try & translate for understand to your writing. Busy with one dictionary for not expert english.

How can I laugh?

I wasn't criticising your English, I'm sure it's much better than any of my attempts at foreign languages But you don't need good English to compare 9 l/s with 23 m³/h.

In your #76

9. Mean flowrate - formula is approximate as the head/flow curve will not be linear.

10. Mean Time Share by No. of Pump = 30 min appears to be arbitrarily chosen. You can't use that and starts/h to get vessel working volume. If you know (max) starts/h allowed, that's all you need to get volume (as I said in several posts). If you use volume 670.5 L and your mean flow 447 l/min, MAX starts/h (at demand ½ pump flow) = Q/V/4 = 447*60/670.5/4 = 10. If you play safe and use pump flow at minimum (pump start) pressure 9 l/s, it's still only about 12 starts/h max. Long-term average starts/h likely to be a lot lower.

I don't know why you're worrying about cut-out flowrate, it's irrelevant, but you've already given a figure in 9.

Thank you Codemaster,

You correct,

I have showed to (CRNE-15-2 Pump curve) JPG according to Engineering choosen. I also confuse this curve, because flow-rate point is outsite of curve.

If this I m conclusion as Engineering choosen pump size is not match with requiring capacity. This calculation is not correct or need anything amendment.

Thank CR4

Bye

I think I now understand why the cut-out flow rate is specified. If the system develops a giant leak, failed pipe, failed vessel, we do not want the pump to continue to feed that. Thus when the pump reaches this flow rate, the entire system shuts down, and should be summoning an operator to begin troubleshooting (alarm, alarm).

Really, there is no other logical reason for flow rate cut-out. The minimum flow rate cut-in is to start the pump no matter what at some flow rate exceeding the minimum cut-in. If one (engineer) were smart, he would have provided for a minimum flow demand pump cut-out as well. I think this minimum flow rate cut-in is largely superfluous (not necessary for the system to basically function).

NT: You are correct, #4 makes no sense. IF they are wanting nearly 9 L/s this is at the limits of the pump.

The pump will do what it does at the range of pressures. I will post other reply on that matter.