Just a matter of physics

Alright everyone, I've got a phire ground fysics problem for ya'll, or something like.

Ok, lets say some time ago a small department on an island out on the pacific coast of Canada, lets say, The Halamat Fire Department, bought itself a portable pump that they had mounted and plumbed into their tanker. The system worked fairly well as the pump was able to fill the 3300 gallon tank in 14-16 minutes. The only problem being that this here portable pump, when it was ordered and showed up on the Halamat, came with some spare head gaskets... never a good sign... Anyway, some time later, lets say about a month ago, this pump basically has a core meltdown. Another pump is purchased to replace it, a nice little (large) Honda. It moves 508 gallons a minute and has 4inch plumbing. The old pump had 3inch, so the tanker is going to have to be re-plumbed from scratch.

Ok, here is where we get phyisical:

The pump is mounted lower then the tank, therefore, the plumbing goes up into the bottom of it. Taking into account friction lose and the gravitational force of the water it is pumping, and all that happy number stuff, what would be effective,

1) plumbing the 4inch to the bottom of the tank

2) plumbing the 4inch up through the tank near the top

3) Something we have not considered

See below for details:

Other question:

Concerning #1, would the pump be working harder because it has the entire volume of the tank to pump against, vs #2 has only the volume of the pipes?

Would the friction loss on the longer pipe (#2) be more then the fluidic friction loss in #1?

Would the gravitational force from the water against the pump be more in the pipes (#2) because of it's volume and height VS #1 volume and height of the tank?

So, are we, I mean the Halamat Fire Department, totally out to lunch on all of this? We basically want to set up the best system we can in order to use the full capacity of the pump and, hopefully, reduce the turn around time for our tanker (especially since we have NO hydrants)

I hope everyone will take a stab at this one (Dal90 you always have great answers), and for those number junkies out there, the more evidence the better (gonna show them to a white hat).

Did ya ever think about hydrants? :D

In my non-physics oriented mind, I would say your second picture, or piping to the top of the tank would seem better. You could then use the bottom fitting for a discharge as well, couldn't you? You need to understand that to my department a tank sits on the back of the toilet and you have to lift the lid quick before it overflows......:D

Dave

In at the top, out at the bottom.

If you consider that water weighs roughly 8 lbs. per gallon, that pump has to do a lot of work to overcome the head pressure has the tank fills. You will also likely defrease your fill time.

The difference is (or should be) so small as to be not worth worrying about. If 2 or 3 pounds pressure will make or break your operation, there are a lot of other factors that will have to be dealt with. Stay Safe....

The quick answer: Normally tank fills go in at the top of tank so you don't get wet or back-fill the supply line when you open the handle. We already have tank-to-pump plumbed low to draft and available to backfeed if necessary (like filling suction hose when the primer dies).

Now the raw physics:

The only factor that matters is the height.

Water exerts a force of 0.434 pounds per square inch for every foot of height of the column.

Say your tank is 6' high, that's 6 x 0.434psi = 2.6 psi

Whether you have a 1" inlet at the bottom of the tank, a 4" inlet at the bottom of the tank, or run a 4" pipe up 6' and then into the top of the tank, you need 2.6psi to overcome gravity. That's it.

Take your hand and try to lift 30 pounds of water. Now I just said use your hand -- not a bucket or other container. Kinda tough, isn't it? Water just keeps going around your hand! It's a liquid and pumping it isn't like lifting a solid. If we were lifting a solid the mass would matter. Doesn't when pumping liquids.

Concerning #1, would the pump be working harder because it has the entire volume of the tank to pump against, vs #2 has only the volume of the pipes?
Nope. As explained above, only the height of the water matters. Whether it's a big, wide tank or a narrow column you need the same work to overcome gravity of 0.434psi/foot.

Would the friction loss on the longer pipe (#2) be more then the fluidic friction loss in #1?
Yes, but so small with 4" at 500gpm as to be irrelevant.

Would the gravitational force from the water against the pump be more in the pipes (#2) because of it's volume and height VS #1 volume and height of the tank?
Nope. The same. Only height matters, not volume.

If you have true 3" piping now, that's adequate. If you have 1.5", 2" piping with 3" fittings I'd replace it. In my area we look for dual 3" or single 4" to support 1500gpm tank-to-pump flows.

If you run a 4" pipe to the top of the tank (say 6') you will have to constantly push against 6 feet of water, plus the friction loss. If you pump directly into the bottom of the tank, you only have to overcome the amount of water in the tank. Since the pressure being exerted is 0.434 PSI (pounds per square inch)only the column of water directly over the input matters. For a 4 inch pipe, the area is 12.56 square inches. When the tank is empty you will need to overcome 0 PSI, at 1 foot you will need to overcome 5.45 PSI, at 2 feet 10.9 PSI, up to 32.7 PSI at 6 feet. If you pump to the top of the tank through a pipe you will have to overcome 32.7 PSI the entire time.

Our tankers have both top fills and bottom fills. When shuttling water it is filled from the bottom fill.

As for Hydrants, we only have those in 2% of our entire district.

Your best bet is to pipe 4 inch pipe to the bottom of the tank and fill from there.

No Devoe, your math and physics are wrong.

12.56 square inches x 0.434 pounds per square inch

The "per" is means divided by -- so you get 12.56 x 0.434 pounds, or a column weight of 5.45 pounds per foot, i.e. 32.7 pounds over the 6' column. Not 32.7psi since the "si" cancel each other out.

But this isn't a solid we're lifting, we're pumping fluid and the weight doesn't matter. Only height of the column matters since it determines how many PSI you need to force more fluid in. For a 6' column, that's 2.6 psi. Whether a pipe 4" in diameter or tank 100' in diameter, you only need 2.6psi to put more water into a column 6' high.

Here's an alternate version of the math:

A 4" pipe has a surface area of pi(radius squared) or 12.56 square inches.

A 1' long section of 4" pipe has 12" x 12.56 square inches or 150.72 cubic inches.

There's 231 cubic inches in a gallon (remember, a gallon is measure of volume). So our 1 foot section of 4" pipe has 150.72/231 = 0.65 gallons of water.

Water weighs 8.34 pounds per gallon, so 0.65 gallon x 8.34 pounds per gallon = 5.42 pounds. Where have we seen that number earlier? Within a reasonable rounding error, that's 5.45 pounds!

So what's the deal with 0.434 pounds-per-square-inch?

The area of a 1" square is...one square inch. That's easy math. Let's go up a foot in height -- now we have 12 cubic inches. 12cubic inches/231cubic inches per gallon = 0.052 gallon.
0.052 gallon x 8.34 pounds per gallon = 0.4336, which (drum roll) rounds to 0.434 pounds.

we had mounted a small honda pump to fill our brush truck up...had it plumbed into the top of the tank like you had talked about. Well the cheif got the idea to mount it for good there.....But he kinda forgot about the pump needing to be primed before it could fill the tanks, just want you to think about if your pump needs to be primed each time. should you mount it in the lower position on a tank and put a check valve on it, it should never loose its prime.

Devoe's math is off a bit, but he has the correct idea. If you fill from the bottom, you're average flow will be greater than if you fill from the top. Centrifugal pumps will produce increasing flow at decreasing head. If you will from the top, as devoe states, you will be pumping against basically a constant head during the entire fill. If you fill from the bottom, initially that head will be low (and consequently your centrifugal pump will produce more flow). The head will then increase as you fill the tank reducing flow from the pump until you get to the max fill level where then essentially you will have the same flow and head as the top fill arrangement. If you are going for pure max gpm's, bottom fill will beat a top fill because the avg gpm for a bottom fill will beat the top fill constant gpm flow. More gpm = less time to achieve specified volume in gallons.

And on the 4.33 value, that is 4.33 psi PER FOOT and originates from a cubic foot of water weighing 62.4 lbs and covering 144 sq in = 0.433 psi/ft. On an inch basis, that's 0.036 psi/in. You can talk about any volume of water, but if you are using 0.433 psi/ft, you need to be talking about columns of water in so many feet. So as has been stated, 6 ft x 0.433 psi/ft (ft cancel) and you you get 2.6 psi. In inches 72 in x 0.036 psi/in (in cancel) you get 2.6 psi again. To prove that the head has nothing to do with volume, take any volume with height of 1 ft. (0.087 cu ft. for the 4" pipe cross section 1 ft high), find the weight of water - 0.087 cu ft. x 62.4 lbs/cu ft. (cu ft. cancel) and you get 5.45 lbs per foot height. Then take this weight and divide by the area covered - 5.45 lbs per foot/12.56 sq. in. and you get 0.433 lbs/sq in or psi/ft. So as has been said, the head over the bottom fill pipe will always be 0.433 psi/ft no matter how big the fill pipe is or how big the tank is.

ff26 there is one other issue I would consider. You mention that the new pump will flow 500 gpm. At what net pressure? A centrifugal pump balances itself out between the suction and discharge conditions and will produce a flow and pressure along its performance curve balanced between the suction and discharge conditions. If you want to try to max out the volume of any centrifugal pump, you make the discharge conditions look like the pump is discharging to atmosphere (no "back pressure"). This will essentially take a centrifual pump out on its curve where it is essentially flowing the max it can flow and developing 0 pressure in doing it. In most applications, there really isn't much sense in running a pump at this flow because it is producing 0 pressure (the whole reason you install a pump - convert the energy of electricilty of diesel fuel into pressure ("lift" if your suction is below the pump centerline)). So you really need to understand your pump and its pump curve. This is a case where "bigger" isn't always better and trying to completely eliminate any "back pressure" in the discharge of the pump isn't always best because you may cause a situation where the pump is trying to flow a maximum amount of gpm and can't generate any net pressure at that flow. Which means it can "lift" any water in a suction condition or pump against any discharge head. Typically, a little back pressure on a centrfiugal pump is a good thing. The other issue with running the pump so far out on its pump curve (high flow/low pressure) is that this is usually way off from the pumps best efficiency point. At these high flows/low pressures, a vey large % of the energy delivered to the pump goes into vibration, pump shaft deflection, etc. and not into generating water pressure. Typically the BEP is around 80% or so of shut off head (no flow pressure). So if you have pump that has a max pressure of 100 psi, it will be the most efficient (and cause the least damage to itself) when the flow/pressure combination produces about 80 psi of pressure. As you can see, this is way off from the pump maxing out it's flow with 0 pressure.

Take a look at the whole suction/pump/discharge conditions and not just on minimizing discharge head. Just some thoughts.

Devoe's math is off a bit, but he has the correct idea. If you fill from the bottom, you're average flow will be greater than if you fill from the top.

While theoretically true in the absolute, for all practical purposes it doesn't matter.

Do you flow more water through a nozzle if you lay on your belly, or if you stand up with the hose line? If you're on your belly, the water has three feet less gravity to fight and has about 1psi more pressure.

If you want to try to max out the volume of any centrifugal pump, you make the discharge conditions look like the pump is discharging to atmosphere (no "back pressure"). This will essentially take a centrifual pump out on its curve where it is essentially flowing the max it can flow and developing 0 pressure in doing it.

I guess I just don't fundementally understand what you're trying to say here, at least related to fire service pumps.

A properly designed, built, and operated pump will always build pressure. The spinning impeller will always add energy. The manufacturer, in designing the pump should have encased the impeller properly so that added energy is translated to added pressure being discharged. If the manufacturer has a four inch discharge, the pump should put out it's rated capacity through that discharge into the air -- a small amount of back pressure will only neglibly affect it.

If you were designing/building your own pump this is a concern -- for cilfd's reasons a small impeller in a big housing won't be too effective, since the energy the impeller adds will "churn" instead of going out the discharge.

Matt

Duh, 26 I take it your tank has a relief valve or overflow at the top of the tank?

otherwise the air pressure in the tank comes into play.

The math isn't bad, just that some dope forgot to drop the "si". The 12.56 square inches is the cross sectional area for a 4 inch pipe. That is pi X r X r. Where pi is 3.14 and r is the radius. So for a 4 inch pipe, you will need to overcome 5.45 pounds of pressure for each 1 foot in rise.

However, this is not just a basic physics problem. This is a problem in fluid dynamics, as well as system engineering. One also has to consider the venturi effects. Remember, water is an incompressible fluid. To maximize the flow rate you must minimize the resistive forces. These forces come in the forms of resistance from the pipes, hydralic turbulence, and the force of the column of water pushing back aganst the pump. Consider this. I have water flowing in a 2 inch pipe. I expand it to a 4 inch pipe. The flow rate must remain constant. Therefor, if I have 100 psi in the 2 inch pipe, I will have less than 100 psi in the 4 inch pipe.

Now consider the turbulence in the pipe. This created by making sharp bends and changing pipe sizes. The smoother and straighter the pipe, the lower the loss from turbulence. If you can, consider a funnel type transition in pipe sizes. This will reduce the turbulence.

As dalmation90 states
A 4" pipe has a surface area of pi(radius squared) or 12.56 square inches.

A 1' long section of 4" pipe has 12" x 12.56 square inches or 150.72 cubic inches.

There's 231 cubic inches in a gallon (remember, a gallon is measure of volume). So our 1 foot section of 4" pipe has 150.72/231 = 0.65 gallons of water.

Water weighs 8.34 pounds per gallon, so 0.65 gallon x 8.34 pounds per gallon = 5.42 pounds. Where have we seen that number earlier? Within a reasonable rounding error, that's 5.45 pounds!


So for each 1 foot of rise we have 5.45 pounds.

To understand the physics, draw a force diagram of the tank. Assume a square tank for simplicity. You will see that the bottom of the tank (assuming it is level) sees the same force across the entire surface.

The force that the pump must overcome is dependent on the depth of the water.

If the objective is to fill the tank as fast as possible then one fills from the bottom. This would even hold true if the water source was above the tank.

Damn, now your saying I can't have my pi and eat it too.:rolleyes:

Dalmation,
For the size tanks (really height of tanks) we are talking about on apparatus, I agree with you - for practical purposes the bottom and top fill will be essentially the same. If we were talking about a tank 50 ft. tall, a bottom vs. top fill would make a difference. Similarly, 3" vs. 4" for 500 gpm over what are going to be relatively short piping lengths - very little difference. You might not be able to even pick up with guages to due guage accuracy.

With regard to pump performance, the link below is off the CET portable pump web site. It is a performance curve for a centrifugal portable pump. As you can, the pump curve runs from 700 gpm @ 0 psi to 0 gpm @ 125 psi (churn). There are an infinite number of flow and pressure combinations from this pump. It will produce 500 gpm @ 50 psi, no more, no less. You can flow 600 gpm, but at a lower pressure, or you can develop more pressure than 50 psi, but you will not get 500 gpm out of it. A pump rating really doesn't have all that much to do with the actual pump curve. NFPA stds define how a pump will be rated (ie 1000 gpm @ 150 psi for a 1000 gpm pump) but that is just one point on a pump curve and it is somewhat arbitrary. The CET pump could be rated as: 500 gpm @ 50 psi, 350 gpm @ 75 psi, 150 gpm @ 100 psi. From just a pump performane standpoint, this pump could be rated at any of these flow/pressure combos.

It is not true that at spinning impellor in a centrifugal pump will always produce pressure. The impellor may be producing some pressure (potential energy), but the losses within the overall pump will negate that pressure increase. Impellors are not rated, pumps are. Look at the CET curve. At 700 gpm, this pump will produce 0 pressure. This is what will happen if you take a centrifual pump and open up enough discharges. The guage on the discharge side of the pump will eventually read the same as the suction side and you will be producing 0 net pressure. The inefficiencies associated with the pump (trying to shove 700 gpm through this pump) will cause vibratiom, shaft deflection, etc. - this is where your energy is going, not into water pressure. Plus, your friction loss through the pump casing will be increasing by almost the square of the flow. So in the end, the net effect is that this CET pump trying to pump 700 gpm will produce no net pressure on the water coming into and leaving this pump. All of the energy input by the engine will either be converted to kinetic energy associated with flow of water (tied up in gpm)or will be lost to mechanical vibration, shaft deflection, friction loss with the pump casing, heat, etc. The 700 gpm leaving the CET pump has essentially no potential energy ("pressure") as we normal consider it.

http://www.fire-pump.com/firepump_vi...mp=PFP-25hp-HV

"If the manufacturer has a four inch discharge, the pump should put out it's rated capacity through that discharge into the air -- a small amount of back pressure will only neglibly affect it."
Not really, a centrifugal pump will only produce its rated capacity when the upstream and downstream conditions are balanced by the pump producing its rated flow and pressure. Look at the CET pump: what's its rating? 500 @ 50, 350 @ 75, or 150 @ 100? I guess if CET were to "rate" this pump at 700 gpm @ 0 pressure then I guess you would be right. It would produce 700 gpm @ 0 discharging to atmosphere. I don't know if CET actually makes the pumps themselves, but whoever does designs the pump to produce a pump curve that pass through either rating points or operating point,etc. The same pump could cross many different ratings or operating points - still same exact pump. You say I want a pump that produces 150 gpm @100 psi. they will send you this pump; ask for a pump that produces 500 gpm @ 50 psi - they will send you the exact same pump unless you request some other operating characteristic (chrun pressure no greater than x, or +/- 10% of the pumps Best Efficiecny Point BEP, etc.)

An example of a pump balancing suction and discharge conditions. You want this CET pump to pump 500 gpm @ 50 psi net on it curve. You are supplying 500 gpm @ 50 psi at the pump suction. That means that you have the potential to have 500 gpm @ 100 psi at the pump discharge. But the only way for you to end up with 500 gpm @100 psi at the pump discharge (or 500 gpm @ 50 psi net) is that you have to have a disharge configuration - pipe, hose, nozzles, whatever, that will disharge exaclty 500 gpm. And that means that you have to balance friction loss and nozzle pressure and nozzle size so that you just flow 500 gpm. And on this CET pump, if you try to flow more than 500 gpm, your net pressure will drop until you get to 700 gpm at which time you will producing 0 net pressure and in fact, if you try to flow more than 700 gpm through this pump, the pump will essentially just become a source of net pressure loss as you try to jam more than 700 gpm through this pump.

The simplest example is just straight pipe off the discharge. If you have 330 ft. of 3" pipe coming off the CET pump discharge with no nozzle, just the 3" open pipe, the CET pump will produce 500 gpm @ 100 psi and you will lose 100 psi in friction loss flowing 500 gpm through 330 ft. of pipe (no elevation change). At the end of that 330 ft. of pipe, you will have 500 gpm coming out and if I put a guage just before the end of the pipe, it will read essentially 0 static pressure within the pipe.

Then as I decrease the length of the discharge pipe from 330 ft., I will start flowing more water, when I get to 90 ft. of 3" pipe, I will be flowing 700 gpm. At 700 gpm, the CET pump produces 0 net pressure, so the pump discharge will be reading 50 psi just from my suction water supply (actually it would be less because a typcial water supply with a flow/pressure of 500 gpm @ 50 psi, will produce 700 gpm at a lower pressure - but lets just use 50 psi suction to make it simple). And 700 gpm through 90 ft. of 3" pipe will use up 50 psi of friction loss. The system will try to use up all of its energy discharging as much volume as possible. So with this water supply to to this CET pump, 90' of 3" pipe is essentially discharing to atmosphere - certainly discharing directly off the pump disharge with no pipe or hose is discharging to atmoshere and no net pressure will be developed by the pump. Basically, at 700 gpm, all of the energy input into the pump is either lost to mechanical vibration, heat generation, etc. or converted to kinetic energy which is included in your flow(gpm) and there is no potential energy left over (what we think of as "pressure").

You take this CET pump, supply 700 gpm @ 50 psi to it, stick 90' of 3" pipe on the discharge and you will be flowing 700 gpm and you will read 50 psi on the suction side or the pump and 50 psi on discharge side for a net pressure of 0 psi and this will fall on the very end of the CET pump curve. The pump has to operate on its curve and if it is flowing 700 gpm, you will see a net pressure across the pump of 0 psi.

I am going with #3) Something you have not considered.
Run the discharge line to the top of your tank from the pump for the tank fill. Run that 3” in the bottom of the tank to the intake on the pump. Do a little more plumbing and now you can pump off the load of water into your engine. Use the “cam - lock” type fittings when you connect to the pump and you can now quickly remove the pump an use it to draft from a water source.

Everything sounds good so far, and it actually almost makes sense.

For those playing at home, this is how the old system looked.

The old pump would move 195gpm at 100psi, it was plumbed with 3inch pipes, camlocks on the pump connections and hard suction connections (so that the pump could be removed at set up at a static water source)

sorry, this is how it looks

Dalmation,
Your comment below hit me this morning:
"If the manufacturer has a four inch discharge, the pump should put out it's rated capacity through that discharge into the air -- a small amount of back pressure will only neglibly affect it."

That CET pump I referred to earlier has 2 - 2-1/2" oulets and a 4" inlet. This pump produced 500 gpm @ 50 psi. If you take a look at you smooth tip selection chart, a tip size in the 1.5" to 1.75" @ at nozzle pressure of 50 psi will flow 500 gpm. There is absolutely no way that two 2-1/2" outlets discharging to atmosphere will flow 500 gpm @ 50 psi for this pump. You open both of those discharges wide open and you will take this pump right out to 0 psi net on its curve. Pretty much every centrifugal pump will operate this way. 3 and 4" suction/discharge connections are common for pumps in the 500 gpm range.

This is a big misconception in the fire service with regard to pumps. Pumps add energy, they don't create flow. Flow is determined by the suction and discharge system. If the suction and discharge config for this CET pump will flow 700 gpm or greater, this pump will add no pressure. If the suction/discharge config will flow 500 gpm, this pump will add 50 psi. If you try to flow let's say 1000 gpm throught this pump, then you have made a very bad pump selection choice.

You select a pump based on the flow you desire from the discharge and is capable of being supplied by the suction config. You then choose a pump that adds the desired amount of pressure to the water stream at this desired flow. The suction/discharge drives the flow, not the pump rating, curve, etc.

All a pump manufacturer will ensure is that if you flow in this example 500 gpm through this pump, that the pump will add 50 psi to water pressure and it will provide you very large suction and discharge connections. If you put this pump into a system that will flow 1000 gpm, the pump manufacturer ensures nothing, you have improperly sized and selected the appropriate pump.

Quote:

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Originally posted by cdevoe
So for each 1 foot of rise we have 5.45 pounds.

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Basic pump operator's knowlege, roughly 1psi for every 2 feet of elevation. By your math, a human would be ok in a small pool but be crushed diving to the same depth in the ocean.

You're right that there is 5.45 lbs of water in the pipe, but you have to finish your math...
5.45lbs of water /12.56sqin of bottom = .4339psi

Dal's right, it's .434lbs per foot of elevation no matter if its a standpipe in a high rise or the bottom of the Atlantic Ocean.

WOW....I didn't think it ever got warm enough up there to keep water in a tank....doesn't it freeze like...364 days out of the year??:D :rolleyes: