Stored Energy In a Watertower

Thanks a million Hero

best regards

Jens

Also the HP will depend on the amount of water being discharged at any given moment, am I right?

Yes, you are!

Also the HP will depend on the amount of water being discharged at any given moment, am I right?

Yes. In fact something that AH did not mention is that the original question might better have asked how many horsepower-hours (or, more commonly, kilowatt hours) are stored. Horsepower is, obviously, a measure of power -- the rate at which work is being done. Horsepower-hour would be a measure of energy -- the capacity for doing work.

So, an average water tower could supply thousands of horsepower for a short time, or a couple horsepower for a very long time.

So, an average water tower could supply thousands of horsepower for a short time, or a couple horsepower for a very long time.

In theory you are correct, since this is a very general statement. However, according to my calculations, for the example given by Jens in his question and with a pressure regulator set to the minimum head when the tank is near empty, a couple horsepower, let's say 2.53 hp, would drain the tank in roughly 16.7 hours at about 100 gpm, not a very long time to me, but perhaps to a housefly it would seem forever! <grin>

BTW, several thousand horsepower, say 2,530 hp, would drain the tank in about one minute, a VERY short time indeed. However, you would need piping and other equipment that would accommodate a flow of 100,000 gpm! As a comparison, the flow rate of Big Spring, near Van Buren, MO, purportedly the largest natural spring in the world, boasts an average flow rate of only about 200,000 gpm (443 ft.³/sec., to be more precise).

There is a heck of a gap between intheory you are correct and In the real world (my post 6).

Good thinking anyhoo.

milo

There is a heck of a gap between intheory you are correct and In the real world (my post 6).

Good thinking anyhoo.

I don't disagree with your "real world" assessment, however, that was not how the question was asked. Perhaps Jens does have a grand plan to subvert the water distribution in his community for his own evil purposes, BUWAAHAAHAAHAA!

Jens himself says, "Don't get me wrong, I'm not trying to steal energy.", but do we believe him?

Obviously, with mechanical and electrical inefficiencies, it takes more energy to pump the water up than you could possibly get back. However, as a large energy storage system, it could make economic sense. Years ago the Union Electric company (now Ameren UE) leveled off the top of a mountain and built a ring reservoir to create the Taum Sauk Power Station, pumping water from a nearby river up at night, during off-peak power usage and reversing the flow to turn the pumps into turbine-generators during the daytime, when energy usage peaked. How was this economically feasible? Energy rates were lower at night during off-peak hours and raised dramatically during peak usage. Essentially, the plant bought energy at a very low price and resold it (less the efficiency factor) at a later time for a much higher price!

Too bad the Dam Engineers didn't prevent a rupture in 2005 that destroyed nearby wildlife, at least one family's home (sending several of the family to the hospital), and a nearby park facility, the Johnson Shut-ins.

"the Dam Engineers!"

milo

Hi STL Engineer.

Here in the UK we made use of water towers to power a lot of machinery from 1880 to 1955. Before the Tower bridge in London was driven by electricity, the bascules were raised by water power from the inbuilt towers, the bascules were raised within 2 minutes! There used to be a huge steam engine that pumped water into the towers, I believe this steam engine still exsists.

In Glasgow in the 1980s they pioneered hydraulics to work Mills, Trams, lifting Bridges and whole factories, I can still remember some of the old water towers when I was 6 or 7 years old.

Nothings new, as the Romans were experts at utilising hydraulics 2,000 years ago!!!

Spencer.

Yes, the awesome power of water flow has great capacity (pardon the pun) for Good or for Evil. For the latter, see my reference to the rupture at the Taum Sauk reservoir in post #11 above.

Time... it's all perspective. Right now I have to get up to go to the bathroom, so 16.7 hours seems like forever, and even one minute seems pretty long.

I don't think "The average is the midpoint of the column if the column is symmetrical in shape"

If you want to calculate the potential energy more accurately you can use the integrated formula:

If the water tower is a cylinder

E=g*p*S*(H1^2-H2^2)/2

Here

p: density of water (1t/m^3)

H1: the height of the top water level.

H2: the height of the bottom water level. (the height of the cylinder h = H1-H2)

S: the base area of the cylinder.

From a real world point of view, The stored energy in the water will be less than the energy used to pump all the water into the tower. The harvesting of this energy will be akin to parastic load, and will subvert the delivery of the water.

milo

A very good read, STL. You should be a teacher -- oh, I just checked and I see you have taught some classes.

I just wanted to add that the flow in a pipe is constant anywhere in the system, but the pressure may vary. It is just like current and voltage in an electric circuit. Jens, you should consider this when designing your system. What you are concerned with is the pressure drop accross the hydraulic motor, not just the pressure head at the base of the tower. Use that in your horsepower calculations. Where will the outlet from the pump discharge to? If it goes right out onto the ground, then STL's response is satisfactory because there is no backpressure. If it goes back into the municipal water supply, then the back pressure will be a big issue. Subtract the outlet pressure head from the inlet pressure head to get the head for STL's formula. Using 100ft is an assumption that there is no backpressure. In either case, you would need to ensure the motor is not contaminating the water supply.

Thanks STL,

I think I can go on my own from here,

And thanks to everyone, who contributed.

All the best from

Jens

Now I just want to design a hydraulic motor to extract the HPs.

But it's not really from a watertower, (I'm not about to steal energy)

but I used it as an example to make it clear to for you guys to see my problem.

However it's a very similar situation.

Now I'll do some thinking and calcs, and come back for some more help.

Again all the best

from Jens

BTW if I can help anyone with electronics (40+ years experience; retired)

I'll be happy to help out.

Horse power is a rate of energy consumption and is 550 ft-lb per second.

If you have water in a tank you have potential energy given in ft-lb. Basically if you know the weight of the water (in lb) and the hight above the point of use (in ft) you can calculate the energy you have available.

The horsepower can only be calculated if you know at which rate you will be using the water (lb per second).

I hope this answers you question.

Thanks Johan

Jens

This may have been mentioned already as I have only skipped through the posts but don't forget about the efficiency of your hydraulic motor, you won't be able to extract 100% of the energy available in the water.

First off, you have the WRONG terminology at hand. You are asking about Potential

Energy not Horsepower. Horsepower involves some type of motion taking place by a

set amount of force over a given amount of time. 1 Horsepower is 550 foot-lbs. per 1

second or 746 Watts = 1 Horsepower. 1 Watt is equal to Force times Distance divided

into Time. In terms of Mechanical Horsepower not Electrical Horsepower, not

Indicated Horsepower for Piston Engines, not Brake Horsepower for both Piston

Engines & Electric Motors. But just straight forward Mechanical Horsepower, 1

Horsepower is equal to a Force in lbs. times Distance in feet divided into Time in

Seconds, then divided into 746. The Metric equivalent to that would be Newtons in

Force times Meters in Distance divided into Time in Seconds, measured as Joules per

second, instead of Horsepower. Whether it be Mechanical Horsepower, or Piston

Engine Horsepower or Jet Engine Thrust Horsepower, you NEED some type of MOTION

in order to calculate Horsepower. Whether it be speed, RPM, forward vehicle speed,

etc. etc. You are wanting to know how much Gravitational Potential Energy exists in

that Water Tower, not Horsepower. Because NO HORSEPOWER exists in that Water

Tower filled up with water. ONLY Potential Energy exists in a Water Tower NOT

Kinetic Energy. Kinetic Energy is the Energy of Motion such as speed or velocity.

Potential Energy is the Energy of Rest, such as inside a Water Tower. I needed to tell

you this, because your question makes absolutely no sense. You are meaning to ask,

'How much Potential Energy exists in a Water Tower, with a Height of 100 feet filled

with 100,000 gallons of water?' In that case, sure, your question can be answered by

the following:

Use the Graviational Potential Energy Formula, which is: Ug = mgh.

But first the mass of the water needs to be calculuated.

1 Gallon of water weighs 8.34 lbs. at 60 Degrees F.

100,000 Gallons of water x 8.34 lbs/gallon = 834,000 lbs.

Ug = Graviational Potential Energy.

Ug = mgh ===> mass x gravitational acceleration constant x height.

mass = 834,000 lbs = [(834,000 / 2.2)] = 379,090.91 Kg.

g = 32.2 feet per second sqaured = 9.8 meters per second squared.

h = 100 feet = [(100 / 3.28)] = 30.49 meters.

Ug = (379,090 Kg.) x (9.8 meters per second squared) x (30.49 meters) =

113,272,850.18 Newton-Meters or Joules of Potential Energy Found in Water

Tower.

For the Standard Conversion For Joules that would be Foot-Lbs.

To get Foot-Lbs. of Gravitational Potential Energy just multiply Joules times

0.737562 ===> 113,272,850.18 x 0.737562 = 83,545,749.92 Ft-Lbs.

The Water Tower Has 83,545,749.92 Foot-Lbs. of Potential Energy Within It.

If there was any motion of the water as in Kinetic Energy as in the water was

coming out of the Water Tower, then Horsepower in Foot-Lbs Per Second or

Watts could be calculated. But as long as there is no motion of the water and it is

static or resting within the Water Tower, there is no Horsepower involved. But

however, you have a great deal of information now about the Potential Energy this

Water Tower has, as in a Water Tower which is 100 feet high loaded with 100,000

gallons of water will have 83,545,749.92 Foot-Lbs of Potential Energy.

Lastly, with the given information, BTU on a Potential Energy basis could also

be used too. Because 1 BTU = 778.26 Foot-Lbs. The BTU Potential is

way to measure the Potential Energy of a Water Tower which is 100 feet

high filled with 100,000 Gallons of water. 83,545,749.92 Ft-Lbs / 778.26 Ft-Lbs =

107,349.41 BTU Potential Energy within that Water Tower.

- Airframe & Powerplant/Propulsion Engineer -

really stupid question but how many kw per hour would you get.

There are no stupid questions.

I'm guessing that you meant kilowatt-hours (kilowatts x hours) rather than kilowatt per hour (kilowatts / hours). I think the latter would be a rarely used concept in this connection: it would be an acceleration, telling you how quickly the kw output changes over time.

If you want to know how long you could power a house, for instance, kilowatt-hours will tell you this. If we assume that the tower is as described above (10 hp for 14,500 seconds) we could first convert the hp to kW: then we'd have about 7.5 kW for 14500 seconds. 14500 seconds is about 4 hours. So if your house draws an average of 7.5 kW you could run your house for 4 hours. If you bought electricity from a utility company a 10 cents per kilowatt-hour, then this amount of energy would cost you about (7.5 x 4 x .10) $3.

More typically, this much energy would be spread out over about two days, unless you heat with electricity. (If you spend about $50/mo on electricity [500 kWh at .10 per kWh] then your average consumption is about 16.7 kWh per day, whereas 7.5kW x 4h is 30 kWh).

Thanks Blink, for your answer.

Jens

Figuring out how much energy potential is straight forward. Energy potential in watts can be calculated by knowing the head (height of the water) and how much water can be delivered in GPM. There is all sorts of limiting criteria but a good rule of thumb would be HeadXFlow/10=watts. However, the only real reason to pump water up to a tower is to pressurize that water for later use, like what comes out of your faucet. Remember the 1st law of thermodynamics, energy can not be created or destroyed. So if you pass all that water through a generator you are robbing it of its energy (pressure). You are left with a dead pool of water that would not have enough pressure to pass through your faucets. Also, you will never reap more energy than it took to pump up in the first place, you will use more energy than you produced. Even if you used solar, wind, or some other free source to pump the water you are still left with a dead pool of water that cannot be easily used.