Water Distribution

What is the diameter of the pipe?

You will also need the static height difference between the 2 points.

Example

Tank 12m ,

fall to end point x m,

diameter and lengths of pipe with an indication of any takeoff.

minimum flow required at points.

You would also be limited to the volume available at the tank.

(The demand on the tank cannot exceed the supply.)

If the 2 points are at the same level you may have to increase the pipe size to an uneconomical size and a booster may be the better option.

You are going to need provide some more information. What are the rough alignments and estimate of controls, e.g valves, for the conveyance system (to account for any minor losses due to valves, elbows, transitions etc.)? Also as previously mentioned, What is the elevation difference between the bottom of the tank to the distribution point? What is the peak hour flow rate (pressure losses relate to the flow rate)? What type and size of conveyance (s) would you propose to use (different materials can have substantially different friction losses)? How much pressure do you want to have at the delivery connection to the community? Finally, What is the depth and volume of the tank, and the relative elevation of the bottom of the tank? You can estimate the frictional losses due to pipe using Hazen-Williams, or Darcy-Weisbach (Hazen-williams is easier to start with and get an estimate of pipe size needed, it gives a reasonable estimate for turbulent flow at 60 degrees, and provides pressure loss in feet, so be aware).

The pressure is due to the water column and will not change unless the elevation changes at the point at which the water is being drawn. If you take a long hose 3km long and ran it out over flat ground the pressure will be the same as at he tank.

If the hose is going down hill the pressure will be higher if up it will be lower.

Pressure will be loss if the hose is of too small of a diameter. The volume of water you need to supply will determine the size. This is at a time of max peak usage. Which usually from supper time to bed time. May change some what depending on life style of the area.

Only the pressure in a static column of water will be related to the difference in elevation. However, the pressure in a water system while water is moving will always be subjected to frictional losses as the water traverses the system. There will always be some pressure losses due to friction no matter the conveyance system size. However, the trick is to reduce the frictional losses to something that allows acceptable pressure ranges at the point of delivery. Static pressure with a full tank will be the highest end of the range. Also note, if the tank is full and you begin pulling a large volume of water from the tank (relative to the tank volume) the available pressure head will fall until the tank is dry, or the demand decreases to a rate below the filling rate of the tank. For a 20 ft high tank set at a bottom elevation of 1 foot above the delivery point this could mean a substantial change in head, unlike a 5 ft high tank set 16 ft above the delivery point.

I think you are forgetting the common possibility that the tank is considerably higher than the outlet. In this case you will need a series of pressure reducing valves or tanks along the line to the town so the pressure is not exorbitant. Don't forget the pipes will be there a long time and there will be frequent and significant shock loadings as taps are turned on and off in town.

As far as low pressure goes I'd use a hydraulic ram. They can be noisy, they go Bang Bang all day,but if you can utilise the overflow water or water overflows anyway they will keep the water flowing nicely. The Ram is essentially two differently sized pistons in two cylinders connected directly by a rod. The smaller one is a simple pump which pumps your water, the larger one is the 'motor' as the water fills its champer it pushes the rod pumping the water in the smaller one. When the large pistons bottoms out a valve opens dumping the water out of the large cylinder and the water pressure in the little cylinder pushes the pistons up till the valve shuts and the inlet valve opens to the large cylinder. The cycle then repeats all day. you will have to find a use for the spill water but otherwise these pumps are great for low flow, low pressure systems such as long distance water distribution where you only need to overcome pipe friction. They are energy free if you can utilise the spill water

In a water reticulation system like this there MUST be provision for different flow conditions including zero flow and sometimes water hammer. Flow and pressure restrictors are seldom required but boosters and secondary tanks may be required.

The real life situation is also that users don't behave according to a text book.

You need to give following details also,

1) Tank Cpacity.

2) Size of discharge pipe.

3) Dischrage leve ie. from the bottom of tank.

For example if you have a tank height of 1m^3 i.e 1000 ltr of capacity. You can dischrge water through gravity incase the discharge point is below the bottom of suction.Approx pressure at bottom would be 1Kg/Cm2 for 1 meter height. no need to install booster pump in between.

Chandra Shekhar Pant

There will not be a pressure loss but flow will be reduced by drag in the pipe This might seem like a pressure loss but it is not. Close the end and the pressure will be restored - bigger the pipe the less the loss. A pump would definitely help if you cannot use a bigger bore pipe. If you reduce the bore just at the delivery end you will find a better pressure. You could also put a bit of air pressure on the tank - not more than 5psi!!! A normal oil drum will stand this.

Pressure loss is going to be related to pipe size, length and volume. If you go to www.elkhartbrass.com they have a chart showing pressure drop per 100 ft of hose. There is a formula for determining feet of head to psi, but I do not know it. Your pressure without a pump will be very low, and therefor volume will be low. If volume is going to be an issue, a larger pipe is more practical than trying to use a pump. I hope this helps.

Do the right thing and hire a Civil Engineer.

In case of water flowing by gravity from an elevated storage tank through a piping system (without a pump) into air, it is easy to derive the following equation to calculate the velocity inside the pipe: V = [2 g Z / f (L/d) + k + 1]^0.5

If there is a pump, the equation shall be: V = [2 g (Z+Hp) / f (L/d) + k + 1]^0.5

where

L = Length of pipe (ft)

d = Inside diameter (ft)

V = Velocity (ft /sec)

g = Gravitational constant (ft/sec²)

f = DARCY-WEISBACH friction loss coefficient

f = function (Re & e/d)

H_f = Head loss due to friction (ft)

Z = Elevational Head (ft)

Hp = Pump head (ft)

K = Resistance coefficients for pipe entrance

e = Absolute roughness (ft)

e/d = Relative roughness (dimensionless)

By knowing the pipe diameter d and total length of pipe L (including the equivalent lengths of fittings and valves), you can assume the water velocity to calculate the Reynolds No. Re and calculate the friction loss f, and substitute into equation until the assumed velocity equal to the calculated velocity or what we called solution has converged. The calculated velocity is preferred to equal or close to the recommended velocity located from famous hydraulic or fluid mechanics handbooks.

Important note. You can increase the pipe diameter to minimize the friction loss.

Wouldn't be wise to consider a secondary buffer tank at the receiving end for the smaller community?

Normally the output pressure for the system is limited to 2 to 4 bar.In your case you need to check the friction loss for the entire pipe routes using the design books or manufacturers catalogues.Diffinitely the tank outlet cannot produce the required flow with available 12 mtr head. so you have to have the booster pump selected so that the friction losses are overcame.

As others have said (guests #6 and #13): use a secondary smaller header tank at the receiving end. Make sure that the valve which allows this tank to fill closes in a very slow and controlled way to avoid "water hammer": think about the energy in a 3 Km cylinder of water travelling at almost any speed.

I have never known so many make a simple thing so complex. Read a basic Physics book,

The Pressure depends solely upon the head of liquid.

The flow depends upon the diameter of the pipe and the viscosity of the liquid.

Obviously your book has some shortcomings. Based your statement, the material composing the inside wall of the pipe would not be a factor, nor would the pressure differential across the pipe (when flowing). However, since all the licensed Civil Engineers already know that the flow rate is a function of the pressure differential across the conveyance length and a friction from the pipe material (as well as pipe factors). Since the Energy and Hydraulic Grade lines decrease in pipes with flowing water, this implies that energy is being translated from the flowing water into heat through energy losses. This energy corresponds to the total energy available from static head, pressure head, and velocity head. In order to maintain one of these forms of energy, the others must be adjusted to compensate for energy losses due to the conveyance. These energies can be translated, as long as the total energy minus losses due to flow remains the same. In otherwords, flowing water in a pipe would have pressure, but at an orifice this is translated in to elevation and velocity. A municipal water system does not drop to zero pressure because one sprinkler head (small orifice) is in operation. It would be bad if it did, because pressure tanks would never refill when one sprinkler head was in operation.

Common error - confusion of flow and pressure - same problem with electricuty

OHMS law spells it out amps X volts = resistance & pressure (atm) x flow (cu ft/min)= volume

No, you cannot do any of this. We don't have the diameter among many other things as enumerated in the other posts. In addition to Bernoulli's equation you will need the equation of continuity which states that Q=VA and assumes laminar flow in the pipe.

Now, if the diameter is known or assumed, then the area is calculated. Please notice that the velocity head is a function of (D/2) SQUARED. Losses add up quickly due to this NON-LINEAR function. What size pipe did you say it is? Oh, I forgot, we don't know.

Typically, it is this variable that the engineer is solving for.

Yes, the poster who mentioned fire flow was correct. The main is worthless if it cannot put out the fire. You are going to need sufficient pressure at the nozzle end. Consider a booster pump at the other side.

The Fire Flow rates vary by community, but 3000 gpm for 2 hours is not unusual for residential development in California (commercial development can be much higher). The minimum pressure during fire flow on peak demand day is at least 20 psi everywhere that utilizes AWWA standards.

You are correct in the fact that the pipe size is necessary. I assumed that was obvious in order to calculate flow velocity from a flow rate. Mostly I was just addressing the idea that if you know elevations at two ends of a pipe water pressure doesn't change, even during flow, and is somehow similar to Ohms law when applied to a very simple single resistor circuit. It is true that Civil Engineers, myself included, usually, when designing new facilities, are utilizing flow, materials, and pressure requirements to determine applicable sizes of pipe to utilize. However, sometimes I also evaluate flow capacities for required pressures and existing facilities, or Pressure losses for projected flows and existing facilities (usually because standards change over time and older facilities may need to be updated).

A booster pump may be necessary if flow or pressure requirements can not be met, A small storage tank could be useful to shorten the distance over which pressurized flow must occur,and a regulating tank (pressure tank), may be useful to equalize pressure of flow variations (spikes) in the system.

You can utilize high volume/pressure booster pumps after a storage tank, or if there is sufficient elevation head or use a smaller pump to increase head at lower flows, you can fill an elevated tank and use local gravity feed to supply peak demands during the day. You size the tank to supply flow over a period of 8 hours to 24 hours for a peak day, and usually you can accomodate peak hour demands during the day. This mean you can fill the tank slower and more consistently over the day, and draw from it as needed. However, tanks and storage in general tends to require more land than pumping stations. There a numerous solutions to such a generalized problem, more detail about the specific situation would be needed to evaluate the problem, and usually cost, maintenance/management capabilities, and construction capabilities are going to be determining factors in the selection of the final solution.

Small storage tank - I would consider the 12m tank as the reservoir or source water supply, the "small storage tank" at the other side to be "distribution storage", and the mystery pipe to be a transmission main. Downstream of the distribution reservoir would be located the distribution pipelines, smaller in diameter, closer together, and shorter in length. Off of these would be the hydrants, and service lines to the individual houses or industries.

The booster pump would be an alternative to the distribution storage reservoir.

The transmission line would be of relatively large diameter to keep the HGL or TEGL high while the "network" at the other side could be smaller as the distances and flows (and therefore velocities) are less. A six to 8 inch line will probably be minimum in order to carry the fire flow.

No hydrants would come off the transmission line directly.

But again, the OP has long since gone leaving behind few clues. We seem to be left contemplating a hypothetical water supply problem.

Someone mentioned prv's which we cannot dismiss outright. For all we know, these may be necessary we have yet to learn whether the 12m tank is high on the hill (and the village down in the valley).

This problem cannot be solved because we do not know the piezometric pressure in the tank. We need the elevation above datum to the tank base. Add this to the tank height and you have it. Assume the absolute pressure in the tank is 0 (tank water level is exposed to the atmosphere (i.e. tank is vented).

Now we look at the other point. Again, no elevation is provided. Assume this point is at the bottom of the hill. The pipe diameter, also unknown, determines the pipe velocity. The velocity squared divided by 2g gives the velocity head per foot of length. The smaller the pipe the greater the velocity head loss. This has to be multiplied by the pipe length to get total head loss from the fluid motion.

If at the end of this pipe, there is a nozzle, i.e. the fluid is exposed again to the atmosphere, the total energy remaining may be calculated by the Bernoulli equation.

So, the water may make it from the tank to the other community by gravity, but this will depend on:

pipe diameter

pipe roughness

pipe velocity

elevation above datum tank 1

elevation of nozzle point 2

distance between tank and nozzle

all bends, valves, pipe entrances, exits

As others have mentioned, the fluid motion creates friction and energy is lost in the form of heat. This means that the energy grade line may have a steep negative slope as we move from point 1 to point 2 depending on pipe diamter. This may not be a problem if the distance is small, but as the hose gets longer and longer, more and more energy is lost. By the time we get to the nozzle (end of pipe), there may be no pressure or little pressure, it really depends.

Goggle Bernoulli's equation, Darcy Weisbach, etc. or hire a civil engineer, or look at a topomap.

Along with minimum pressure at peak hour flow, you should check for minimum pressure during fire flow requirements. For very small comunities fire flows during peak day can be larger than peak hour demands.

Suggest everyone goes back and reads the original question - this is not a major hydraulics study! It is only water - Viscosity is a minor considerstion - the pressure at the delivery end will depend solely on the head. Question implies that this is static so any loss will depend upon that lost in the pipe and flow will dictate. if, for example, a 3" pipe is used and a 1/2" tap put on the end then there should be adequate pressure. probably get away with a 2" pipe. Trick is to reduce the outlet - and use the venturi effect.

Suggest you go back and read the posts in this thread.

We still do not know the nozzle height above datum at point 2, or the tank height above datum at point 1. Trust me, these must be known.

Maybe we are to assume that the pipe is slope is level?

i GIVE UP!!!!

Me too! Do your own homework!

Did you mean this original Post?

" want to distribute water from a 12m high water tank from one community to another at a distance of 3Km.

I fear there might be a problem of pressure loss.

Do i have to install a booster pump, and what could likely be the output pressure at the far end? Thanks."

This is a basic hydraulics problem, yes, but it still requires information about elevations, conveyance size, fittings, and conveyance materials at a minimum to evqluate the pressure loss. No where does this indicate that the water is static, though the 12 m high tank could be assumed to have the water surface pressure at atmospheric (unless it is a pressurized tank). To address the question of needing a booster pump, there are other alternatives that may be evaluated, and cost would determine the most appropriate choice.

RCE

I did not mean to imply that you did not understand the question. My frustration was with the OP. He is long gone. As is quite common on this forum, the original question seems like it was "homework" and by now he has passed something in and is now busy with the next assignment.

I think you and I both shared the hope that more information would make it into the discussion, but in the meantime, we began expanding the original question into something likely to be found "in the real world".

Then, others began to slam in reminding us that we are bound to respond only to the OP's abstract "riddle". This becomes the classic Catch 22. You are not allowed to call the problem a "riddle" (we all did) while as an alternative suggesting solutions to "non-riddle" likely real-world problems.

So the bottom line is, there never was a real problem brought to the forum. Case closed without a finding.