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Hydro in the Gutters
By wdyasq, Section Hydro Posted on Sun Jun 8th, 2008 at 03:54:13 PM MST
Check out the math

Occasionally, one gets what they perceive as a brilliant idea. An idea that comes up regularly is 'Can I put hydro-turbines in my downspouts and make power'. I answered one of these on the IRC channel recently and thought I should post the math here. In this manor, my math, it it is wrong, will get corrected and Wooferhound can stick this thread in the unseen FAQ section.
We will start win a 1000SF roof and a 12 inch/1 foot rain in an hour. I realize this is absurd but, putting turbines in gutters is in the same league so just bear with this a bit. We will also use a 10' drop or 'head' for our figures.
On to the numbers:
We have 1000SF area getting 1 foot of water so we have 1000 cubic feet of water at 62.4 pounds a cubic foot or 62,400 pounds of water. This will fall 10 feet so we have 624,000 foot pounds of energy in one hour time. As power is in 'foot pounds a minute', we divide this by 60 and get 10,400 foot pounds of energy a minute.
We take the 10,400 foot pounds of energy and divide by 33,000 foot pounds of energy in a horsepower and get .31HP. At this time we have no inefficiencies or loses due to friction or anything else. I think if we got 20% efficiency out of our turbine for such a small one we would be doing very well. We end up producing .062HP for the hour.
This we need to convert to our International Watt or Kilowatt so we can converse with the world. 1HP = 746 Watts and .062 X 746 = 46.25W. Since we got this over an hour this is .046kWh for every foot of rain.
Just to finish the thinking, I'll take an example of a 3500SF roof here in North Central Texas where we get about 30" (2-1/2') of rain per year. In a year, if one had a place to store the 63,000 gallons of water, the electrical production would be ~.4kW. That is $0.07 worth of electricity/year at $0.16 kWh rate (one of the highest in the nation).
Ron
Yeah, woofer should stuff this in the often unread FAQ / WIKI.

Hydro in the Gutters | 67 comments (67 topical, 0 editorial)

Re: Hyro in the Gutters (3.00 / 0) (#1)
by spinningmagnets (velmis1450bc(at)aol(dot)com) on Sun Jun 8th, 2008 at 10:49:27 AM MST
(User Info)

Can this be grid-tied?

"All I asked for was a few sharks with freakin lasers mounted to their heads, is that really too much to ask for, people?" -Dr Evil (Austin Powers)

Re: Hyro in the Gutters (3.00 / 0) (#2)
by TomW on Sun Jun 8th, 2008 at 11:36:41 AM MST
(User Info)

Golly, Ron.

No fair injecting reality into a really good idea.

Tom.

"Education consists mainly of what we have unlearned."--Mark Twain

Re: Hydro in the Gutters (3.00 / 0) (#3)
by PeterAVT on Sun Jun 8th, 2008 at 03:53:39 PM MST
(User Info)

I gotta admit, the math looks good. Even up here in the Northeast, it isn't really feasible. But I would save the basic math ideas for a real FAQ.
AKA "inode_buddha" Power to the people!

Re: Hydro in the Gutters (3.00 / 0) (#4)
by DamonHD (d@hd.org) on Sun Jun 8th, 2008 at 05:23:06 PM MST
(User Info) http://www.earth.org.uk/

Now you got me thinking... that's 1Wh for every mm of rain on my roof (~50m^2)!

Rgds

Damon

Re: Hydro in the Gutters (3.00 / 0) (#5)
by elvin1949 (elvin1949@yahoo.com) on Mon Jun 9th, 2008 at 01:12:58 AM MST
(User Info)

Ron
About 1 time every 50 years i get slam dunked by a Hurricane here.
So you can multiply that by 50 on those rare occasions. HEHEHE
later
Elvin

Re: Hydro in the Gutters (3.00 / 0) (#6)
by tecker on Mon Jun 9th, 2008 at 05:50:01 AM MST
(User Info)

If you get enough rain to warrent the expense you can collect water in a tank and work out the details using a down spoute will back up water in the gutter and overflow to the soffit .

think water instead (3.00 / 0) (#7)
by Norm (peppysue@suite224.net) on Mon Jun 9th, 2008 at 06:19:47 AM MST
(User Info)

If only the ones that thought up this brilliant
idea would look at their water bill?
Roughly speaking 100 gallons of water cost me
a $1 (our new sewerage is included....ever gallon
of their water you use goes down their drains)
We use 200 gallons a day.
Now if I put a (free)55 gallon plastic barrel
at the 4 corners of my 36' x 24' house, they would
be filled in a couple of hours during just an average rainy day....
So just flushing the toilet using 50 gallons
of water a day and it rains just enough so I use
this amount 10 days out of a month....
I could be saving $5 a month...double that
by learning to save on the water.
( :>) Norm

Re: think water instead (3.00 / 0) (#9)
by wdyasq on Mon Jun 9th, 2008 at 09:16:52 AM MST
(User Info)

Sorry Norm,

I haven't figured out how to make people think.....

Ron
Adventure is just bad planning." -- Roald Amundsen
[ Parent ]

Re: Hydro in the Gutters (3.00 / 0) (#8)
by wooferhound (tim((NoSpamAt))wooferhound.com) on Mon Jun 9th, 2008 at 06:27:49 AM MST
(User Info) http://wooferhound.com

so payoff of the $600 system would be in 8571 years
or
it would never pay to even keep batteries updated

this has been discussed briefly several times, along with ideas like,
I want to pump water uphill using my dumpload power . . .
W o o f -={(

Re: Hydro in the Gutters (3.00 / 0) (#10)
by Norm (peppysue@suite224.net) on Mon Jun 9th, 2008 at 03:16:11 PM MST
(User Info)

That's okay Ron...no one is perfect...
Ah well....???
( :>) Norm

Re: Hydro in the Gutters (3.00 / 0) (#11)
by rossw on Mon Jun 9th, 2008 at 03:45:43 PM MST
(User Info) http://house.albury.net.au/

Ahh, Ron, you missed a key ingredient. Head.

If you dug a deeep hole (say, 90' deep) you would increase (substantially) the pressure of your water flow and increase your yield. With a mere 90' deep hole, you now have 100' head - or 10 times your output power (neglecting losses)

If you happen to live really high up say on a mountain......

(grin)

Re: Hydro in the Gutters (3.00 / 0) (#16)
by MattM on Wed Jun 11th, 2008 at 07:13:52 PM MST
(User Info)

Actually, if you calculate the gravity and velocity of the water in 100' its a little more than 3 times the power of a 10' drop.
----------------------------- Go Huskers!
[ Parent ]

Re: Hydro in the Gutters (3.00 / 0) (#17)
by joestue on Wed Jun 11th, 2008 at 07:27:32 PM MST
(User Info)

I hope you were being sarcastic. ROFLMAOAAPMP!
what is the definition of one joule? hs physics. one newton meter.

I suppose since the water is traveling three times faster it's going to give the waterwheel three times the momentum too..

[ Parent ]

Re: Hydro in the Gutters (3.00 / 0) (#20)
by MattM on Thu Jun 12th, 2008 at 06:26:38 PM MST
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No, I was calculating power - a product of time - whereas you are calculating work potential.

Maybe we had different physics classes.
----------------------------- Go Huskers!
[ Parent ]

Re: Hydro in the Gutters (3.00 / 0) (#21)
by joestue on Thu Jun 12th, 2008 at 06:43:20 PM MST
(User Info)

it would still be 10 times the power

[ Parent ]

Re: Hydro in the Gutters (3.00 / 0) (#23)
by MattM on Fri Jun 13th, 2008 at 08:35:42 PM MST
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Using math we can figure the time necessary for the water to fall 75 feet by using the equation for the distance that the object falls in a given time. This equation is d = gt²/2

where
    d is the distance the object falls
    g is the acceleration due to gravity
    t is the time considered
    t² is t times t or t-squared
    gt²/2 is g times t-squared divided by 2

so 75 feet = (32.17 feet/seconds²)t²/2

    (32.17 feet/seconds²)t² = 150 feet
    t² = 150 feet / 32.17 feet/seconds²
    t² = 4.6627293 seconds²
    t = 2.1593354 seconds

feet / 2.1593354 seconds = 34.732909 feet / second
333 pounds / second * 34.732909 feet / second = 602 foot-pounds per second, which is also 1.09 horsepower.

Correct my math where it went wrong.
----------------------------- Go Huskers!
[ Parent ]

Re: Hydro in the Gutters (3.00 / 0) (#24)
by MattM on Fri Jun 13th, 2008 at 08:40:42 PM MST
(User Info)

Ooopps, you were saying 10 times for 100 feet and I calculated 75 foot. Here's 100 foot.

100 feet = (32.17 feet/seconds²)t²/2

    (32.17 feet/seconds²)t² = 200 feet
    t² = 200 feet / 32.17 feet/seconds²
    t² = 6.116208 seconds²
    t = 2.47309683 seconds

feet / 2.47309683 seconds = 40.435133 feet / second
333 pounds / second * 40.435133 feet / second = 700 foot-pounds per second, which is also 1.27 horsepower.

----------------------------- Go Huskers!
[ Parent ]

Re: Hydro in the Gutters (3.00 / 0) (#25)
by joestue on Fri Jun 13th, 2008 at 09:34:09 PM MST
(User Info)

man, i really don't know what you are trying to figure out.

time to fall 100 feet will be square root of ten times as long as the time to fall 10 feet.

energy is 1/2 Kg (m/s)^2

gravitational potential energy is mgh.

[bastardization] gravitational potential power would be (kilograms/second)x(9.8m/s^2)x(height in meters)

[ Parent ]

Re: Hydro in the Gutters (3.00 / 0) (#26)
by MattM on Sat Jun 14th, 2008 at 07:50:07 AM MST
(User Info)

joestue said, "it would still be 10 times the power"

I have shown that the power for 100 foot drop is not 10 times more than for a 10 foot drop. Are you arguing about the math or the whole concept in general?
----------------------------- Go Huskers!
[ Parent ]

Re: Hydro in the Gutters (3.00 / 0) (#28)
by DamonHD (d@hd.org) on Sat Jun 14th, 2008 at 10:19:56 AM MST
(User Info) http://www.earth.org.uk/

I think it's pretty certain that 10 times the height in a constant gravitational field has 10 times the potential energy.

Rgds

Damon

[ Parent ]

Re: Hydro in the Gutters (3.00 / 0) (#29)
by joestue on Sat Jun 14th, 2008 at 12:22:28 PM MST
(User Info)

The flow rate didn't change - a logical assumption, water flow = available source.
Further more the time required to fall doesn't matter.

If i drop a 10 Kg ball one meter, just before it touches the ground, it's moving 4.427 meters per second and it has 98 joules of kinetic energy.

If i drop a 10 Kg ball ten meters, just before it touches the ground, it's moving 14 meters per second and it has 980 joules of kinetic energy.

If i drop [one] 10 Kg ball per second, then i have an average energy dissipated at the bottom of 98 watts, and 980 watts respectively.

The definition of work is Fd force x distance.

ask yourself where did the energy go,
it is obvious it takes 10 times the energy to lift that water 10X height.

[ Parent ]

Re: Hydro in the Gutters (3.00 / 0) (#30)
by MattM on Sat Jun 14th, 2008 at 04:01:29 PM MST
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The energy did not go anywhere, joe, you're intermixing "energy" with "power".

Horsepower is a unit of power, not energy. Pound-force per foot is the relative unit of energy.
----------------------------- Go Huskers!
[ Parent ]

Re: Hydro in the Gutters (3.00 / 0) (#31)
by joestue on Sat Jun 14th, 2008 at 06:12:09 PM MST
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Please try to piss me off, it's not going to work.

I am not mixing up energy and power

energy is mgh
power is mgh/seconds

one thousand kilograms of water falling 100 meters is 980,000 joules.

I don't care how long it takes to fall.

you can figure the math out.

100 feet = (32.17 feet/seconds²)t²/2

(32.17 feet/seconds²)t² = 200 feet
t² = 200 feet / 32.17 feet/seconds²
t² = 6.116208 seconds²
t = 2.47309683 seconds

100. feet / 2.47309683 seconds = 40.435133 feet / second [average speed!]
17. 333 pounds / second * 40.435133 feet / second = 700 foot-pounds per second, which is also 1.27 horsepower.

If the energy didn't go anywhere, then it should work backwards.

700 foot pounds/second will lift 7 pounds of water 100 feet per second.

you started with 17.33 pounds per second.

KISS 17.33 pound/second x 100 feet. = 1733 foot pounds per second or 3.15 horse power.

[ Parent ]

Re: Hydro in the Gutters (3.00 / 0) (#32)
by MattM on Sun Jun 15th, 2008 at 12:36:03 AM MST
(User Info)

But when you fall its not at just one static velocity. Your height displacement affects the overall velocity, which then makes the time nonlinear relative to the displacement. If the velocity is constant your formula works. Once you add in the acceleration due to gravity things are not so cut and dry.

Piss you off? Please. If pointing out the facts piss you off then so be it.
----------------------------- Go Huskers!
[ Parent ]

Re: Hydro in the Gutters (3.00 / 0) (#33)
by rossw on Sun Jun 15th, 2008 at 01:58:23 AM MST
(User Info) http://house.albury.net.au/

I think "someone" has a fundamental oopsee in thinking.

We're not dropping a given mass of water 100 feet.

We've got a continuous column of water, 100 feet high.

The PRESSURE at the bottom of a 100' column of water, times the flow, not the kinetic energy of a unit of water dropped 100' and being allowed to accelerate.

Yes, I started out tongue-in-cheek, just like the OP.

[ Parent ]

Re: Hydro in the Gutters (3.00 / 0) (#36)
by joestue on Sun Jun 15th, 2008 at 02:11:53 AM MST
(User Info)

I was using kinetic energy as a way of explaining it, but you are absolutely right

[ Parent ]

Re: Hydro in the Gutters (3.00 / 0) (#47)
by MattM on Sun Jun 15th, 2008 at 06:28:03 PM MST
(User Info)

That would make sense if your home could support 62400 pounds of water on a rooftop. Otherwise you better be draining off at the same rate the water is collecting. If you are draining at the same rate you are collecting then your water is in freefall. Once your restrict the flow until its backed up all the way down that 100' drop then you now lose energy from the pressure of going through a pipe. Now you just added in atmospheric pressure into the equation. This is all going to get a lot more complicated then a simple 100' drop.
----------------------------- Go Huskers!
[ Parent ]

Re: Hydro in the Gutters (3.00 / 0) (#35)
by joestue on Sun Jun 15th, 2008 at 02:08:08 AM MST
(User Info)

Velocity does not matter.

Furthermore, you calculated average velocity, a worthless number, the RMS velocity would be a better number.

force x distance/time = power.
mass x gravitational acceleration x distance[height] / time = power

state your math

[ Parent ]

Re: Hydro in the Gutters (3.00 / 0) (#43)
by thirteen on Sun Jun 15th, 2008 at 08:37:52 AM MST
(User Info)

where does the pipe restriction work into this varying water supply problem?

[ Parent ]

Re: Hydro in the Gutters (3.00 / 0) (#12)
by vawtman (vawtman(at)charter(dot)net) on Mon Jun 9th, 2008 at 04:12:17 PM MST
(User Info)

Hi Ron,
These homes can't use the scheme anymore.
http://www.620wtmj.com/news/local/19662404.html

Re: Hydro in the Gutters (3.00 / 0) (#13)
by richsuka on Mon Jun 9th, 2008 at 06:52:45 PM MST
(User Info)

hydro from city water supply 70-100psi every time u turn on the water.
Needs to be setup before it come into home.
At lest you get head pressure (head ache)all the time.
File this one tooo.

Rich
If it's free it cost too much!

Re: Hydro in the Gutters (3.00 / 0) (#14)
by MattM on Mon Jun 9th, 2008 at 09:00:12 PM MST
(User Info)

I think your figure for total foot pounds is too generalized. Gravity accelerates at a negative 32 odd feet a second square. So its not really 10 foot pounds for a single pound dropping 10 feet, more like 25 foot pounds if my finger counting is right and so that should be more like 3/4 hp.

If your house is on the top of a 65' cliff and you put an extension on your 10' downspout down the cliff (for a 75' total) then you only triple the kinetic energy, right?
----------------------------- Go Huskers!

Re: Hydro in the Gutters (3.00 / 0) (#15)
by wdyasq on Tue Jun 10th, 2008 at 07:40:20 AM MST
(User Info)

Feel free to recalculate and post your findings.

Ron
Adventure is just bad planning." -- Roald Amundsen
[ Parent ]

Re: Hydro in the Gutters (3.00 / 0) (#22)
by MattM on Fri Jun 13th, 2008 at 08:25:49 PM MST
(User Info)

My math was probably off because I was only figuring velocity.

Here's a new calculation.

foot-pound per second = 0.001818 horsepower
cubic feet of water is equal to 62,400 pounds
,400 pounds over one hour is 17.333 pounds per second
333 gallons per second times 0.001818 horsepower is 0.031512 horsepower

Now this is for one foot displacement per pound without the effects of gravity. Let's factor in gravity... Using math we can figure the time necessary for the water to fall 10 feet by using the equation for the distance that the object falls in a given time. This equation is d = gt²/2

where
    d is the distance the object falls
    g is the acceleration due to gravity
    t is the time considered
    t² is t times t or t-squared
    gt²/2 is g times t-squared divided by 2

so 10 feet = (32.17 feet/seconds²)t²/2

    (32.17 feet/seconds²)t² = 20 feet
    t² = 20 feet / 32.17 feet/seconds²
    t² = .6216972 seconds²
    t = .78847795 seconds

Now we need to figure the average velocity for the object in order to figure out the force it will have at the 10 feet mark. We know the distance and the time travelled, therefore we can calculate the distance over the time. At the .78847795 seconds mark the water is travelling at 12.68266 feet/second.

17.333 pounds per second times 12.68266 feet per second is equal to approximately 220 foot-pounds per second, which is also .4 horsepower.

Is this a little better or did I fudge something here?
----------------------------- Go Huskers!
[ Parent ]

Re: Hydro in the Gutters (3.00 / 0) (#38)
by joestue on Sun Jun 15th, 2008 at 02:27:11 AM MST
(User Info)

"Now we need to figure the average velocity for the object in order to figure out the force it will have at the 10 feet mark."

lets not invent new math here.

[ Parent ]

Re: Hydro in the Gutters (3.00 / 0) (#52)
by MattM on Mon Jun 16th, 2008 at 07:53:42 PM MST
(User Info)

You're right, I was calculating the velocity delta not the average velocity. Once again you point my grammar out to be wrong but totally ignore the reality of the proof. The grammar was wrong but the result was still correct.
----------------------------- Go Huskers!
[ Parent ]

Re: Hydro in the Gutters (3.00 / 0) (#53)
by joestue on Mon Jun 16th, 2008 at 11:12:59 PM MST
(User Info)

omg, no, the result is not correct, average velocity is of no interest to anyone here.and i did not correct your grammer, it is perfectly understandable at the moment.

furthermore, if the energy and power are as you suggest, then we can get more out by cascading n systems. this is obviously not true. do you know what this means?

_________________________

MATTM said "17.333 pounds per second times 12.68266 feet per second is equal to approximately 220 foot-pounds per second, which is also .4 horsepower."

I see what you are doing, and i'm trying to explain it to you without re-proving newton's laws, you evidently don't understand them.

calculate the kinetic energy: no, i'm not going to derive it for you, but i did in physics class...

33 pounds/second x (12.7 ft/sec)^2 x .5 = you figure this out and tell me what you get in imperial units, then compare to mks...
87 kg/s * 3.87m/s^2 x .5 = 58.99 watts.
foot pounds/s = 298 watts.

answer this question: how much power does it take to lift 17.33 pounds of water 10 feet per second?

DEFFINITION: work is done then a force acts over a distance. one horse power is lifting 550 pounds one foot off the ground in one second
lifting 33000 pounds one foot off the ground in one minute
lifting 55 pounds of water 10 feet off the ground in one second,
do you see where this is going?
pushing a car down the road at 15 miles per hour. ~approx~
pushing a train at a few inches per second...~approx~

pounds is a unit of force, i ignored the kinetic energy still in the water, after it got lifted off the ground..
it's a squared function of speed , so we can ignore it for relatively slow moving objects.

OVERUNITY [10 feet, 220 out, 173.33 in], [40.3% at 100, 700 out, 1733 in], what do you get for 10,000 feet?

using you math,
ttf 10,000 feet is 24.99 seconds[i agree, this is elementary]
average speed is 400 feet/second [i agree, this is also elementary]
400 feet/second x 17.33 pounds/second is 6932 foot pound/sec or 12 horsepower
speed at bottom [800ft/sec conveniently always twice the average speed for acc.= constant and V_i = zero] x 17.33 pounds/sec is 24 horsepower.

power needed to lift 17.33 pounds of water per second to 10,000 feet: 173,300 foot pounds/sec or 315 horsepower!

your math results in a 3.8 % out using average speed, and 7.6 % out using peak speed

in fact, if we cascade one thousand of your ten foot drops, and drop 17.33 pounds of water out of an air plane... we get 2.2 million foot pounds out of the system, and we put 173 thousand in.

whereas if we dropped it all the way at the bottom we'd get 13.86 thousand foot pounds by your 'math'

__________________________

17. 333 gallons per second times 0.001818 horsepower is 0.031512 horsepower

you invented more math!

assuming you meant 17.33 pounds/sec

17.33 pounds/sec is a flow rate, a unit of force exerted by the earth, on the water, not mass.
in theory, it would take NO energy to move 17.33 pounds of water per second.
in practice, it takes almost no energy. (joules or less, we don't have to move it again, after it's moving_ Newtons first law__)

multiply this by height to get work.

divide by time to get power, but since we already did that (pounds/sec), we DO NOT do it again.

[ Parent ]

Re: Hydro in the Gutters (3.00 / 0) (#54)
by MattM on Tue Jun 17th, 2008 at 05:58:45 AM MST
(User Info)

Once again you brought up the "average velocity" comment which I already said was not correct. The water settles on the roof in the form of rain. According to the op you have 62,400 pounds of water over the course of an hour, which is 17.33333333333333 pounds/second. If your home can support 62,400 pounds then great, store it all at once. Otherwise you have to drain it as it comes, which is what a gutter system does. So you are starting with a velocity of zero and end up with a change after 10 feet, 75 feet, 100 feet, etc.

Now you're perverting the whole argument when you know that air resistance and terminal velocity were not part of the original equation for a reason; I kept it simple. Perhaps now you want to get more granular to make a point that originally was only to show the amount of energy in the venture is still trivial even if you lived next to a 75' cliff to generate power from the rainwater.

But whoa is me because somehow I proclaimed over-unity.....
----------------------------- Go Huskers!
[ Parent ]

Re: Hydro in the Gutters (3.00 / 0) (#55)
by joestue on Tue Jun 17th, 2008 at 02:27:05 PM MST
(User Info)

You did achieve over unity!!! you got 220 foot pounds out when it only took 173 to lift the water back up the 10 foot drop!

not only that, you obviously didn't read this:

furthermore, if the energy and power are as you suggest, then we can get more out by cascading n systems. this is obviously not true. do you know what this means?

or this:
DEFFINITION: work is done then a force acts over a distance. one horse power is lifting 550 pounds one foot off the ground in one second
lifting 33000 pounds one foot off the ground in one minute
lifting 55 pounds of water 10 feet off the ground in one second,
do you see where this is going?
pushing a car down the road at 15 miles per hour. ~approx~
pushing a train at a few inches per second...~approx~

pounds is a unit of force!!!!

You were the one who added terminal velocity, this is what you wrote:

"Now we need to figure the average velocity for the object in order to figure out the force it will have at the 10 feet mark. We know the distance and the time travelled, therefore we can calculate the distance over the time. At the .78847795 seconds mark the water is travelling at 12.68266 feet/second."

"17.333 pounds per second times 12.68266 feet per second is equal to approximately 220 foot-pounds per second, which is also .4 horsepower."

and i proved it is incorrect:

lifting 17.33 pounds of water 10 feet per second required only 173 foot pounds per second

[ Parent ]

Re: Hydro in the Gutters (3.00 / 0) (#56)
by TomW on Tue Jun 17th, 2008 at 03:44:45 PM MST
(User Info)

Geeze, guys. Give it up.

Nobody is gaining anything by this constant math exercise. Anyone with half a brain gets that there is very little power from your roof. Period. End Quote.

Why argue it beyond that?

Take it to a physics forum if you must decide who is cracked and who is merely arrogant about their math skills. The feedback I am getting off board is that its just a dead end waste of board space.

Then, again, I have no interest in wannabe armchair arguing from the academics. We are here for builders, you know.

Just my opinion.

Tom

"Education consists mainly of what we have unlearned."--Mark Twain
[ Parent ]

Re: Hydro in the Gutters (3.00 / 0) (#57)
by joestue on Tue Jun 17th, 2008 at 04:16:51 PM MST
(User Info)

I was about to email my high school physics teacher with the statement:

"(force/time) times velocity equals energy"

and Then I though the better of it.

peace Matt.., please post this on www.physicsforums.com

[ Parent ]

Re: Hydro in the Gutters (3.00 / 0) (#58)
by MattM on Tue Jun 17th, 2008 at 10:54:20 PM MST
(User Info)

Not a problem, Tom. I think that Joe was just stuck on the 10 foot drop being over a 1 second time frame and doesn't realize gravity pulls it down in a fraction of a second, making the power required for lifting and the power generated by dropping different although the actual energy in both exercises is the same. The differences between energy (W or F) and power (W/t or F/t) are ideas people confuse.
----------------------------- Go Huskers!
[ Parent ]

Re: Hydro in the Gutters (3.00 / 0) (#59)
by joestue on Wed Jun 18th, 2008 at 03:47:21 PM MST
(User Info)

everything i said was correct for a constant flow of water. post your math on a physics forum and you will be forced to explain things in concepts, not specific examples.

"making the power required for lifting and the power generated by dropping different although the actual energy in both exercises is the same"

constant flow of water = constant power out MattM

further more we have a simple concept known as conservation of energy....

[ Parent ]

Re: Hydro in the Gutters (3.00 / 0) (#60)
by MattM on Wed Jun 18th, 2008 at 04:54:32 PM MST
(User Info)

Unfortunately, joe, it still would not convince you even when the people in the physics forum agreed with me. You are not understanding that the power to lift 17.333 pounds of water 10 feet in 1 second is NOT the same amount of power you generate in dropping 17.333 pounds of water 10 feet in .78847795 seconds. (Thus you do not understand where I came from saying 173.33 pounds/second is less than 220 pounds/second.) Power is simply an amount of energy exerted over a time frame and if that time frame is less than 1 second then the relative units for power will grow accordingly (i.e. 220 pounds/second in that 10 foot, .78847795 seconds drop) although the amount of energy (173.333 foot-pounds) does not change.

If you simply cannot grasp these points then arguing with you is pointless.
----------------------------- Go Huskers!
[ Parent ]

Re: Hydro in the Gutters (3.00 / 0) (#61)
by TomW on Wed Jun 18th, 2008 at 05:11:37 PM MST
(User Info)

What part of "give it up" don't you get?

Tom

"Education consists mainly of what we have unlearned."--Mark Twain
[ Parent ]

Re: Hydro in the Gutters (3.00 / 0) (#63)
by joestue on Wed Jun 18th, 2008 at 05:32:32 PM MST
(User Info)

please delete the thread

[ Parent ]

Re: Hydro in the Gutters (3.00 / 0) (#62)
by joestue on Wed Jun 18th, 2008 at 05:25:44 PM MST
(User Info)

I completely understand that Matt!

but that wasn't your original claim.

when you originally replied to Rossw's post that ten times the head would give 10 times the energy and i said power as well. you said more like 3 times the power

Both he and I assumed we aren't quantifying packets of water free falling, and capturing bursts of energy at the bottom, but rather a steady stream of water measured in pounds per second. also a common error, water is measured in mass/second.

the same flow of water at 10 times the pressure, head etc gives ten times the energy, and therefore power, because time is relative to the rate of falling water.

the irony here is my statement "(force/time) times velocity equals energy" was correct, and a typo, i intended to quote your statement:
"Z pounds per second times Y feet per second is equal to approximately W foot-pounds per second:

producing the following: FV/s^2 = power

a simple concept, but you got the physics all wrong

FV=power
1/2mV^2 = energy
and 1/2mV^2/s = power.
mgh= gravitational potential energy and divide by time to get power.

don't even try arguing with those concepts, derived from newton's three laws.

if you have 17.33 pounds of water falling at 12.68266 feet per second. and you attain 220 foot pounds/sec , the obviously the flow of water isn't continuous, and this is a point you failed to explain in the beginning.

[ Parent ]

Re: Hydro in the Gutters (3.00 / 0) (#64)
by MattM on Thu Jun 19th, 2008 at 05:31:37 AM MST
(User Info)

You want to argue mass flow rate with a sheet metal guy? Seriously. You aren't even filling in the right factors.
----------------------------- Go Huskers!
[ Parent ]

Re: Hydro in the Gutters (3.00 / 0) (#65)
by DamonHD (d@hd.org) on Thu Jun 19th, 2008 at 06:29:18 AM MST
(User Info) http://www.earth.org.uk/

Hi Guys,

Absolutely none of my business not being a mod (thank goodness for us all!) but could you both drop it?

I have a fairly firm view of which one of you is right but that's by-the-by.

The "miss-miss-miss-he-said-it-wrong" tone is quite tiresome.

The point is it's pointless anyway, so why burn all this mental energy and goodwill in willy-waving for the rest of us? You may be confusing your audience with PeopleWhoGiveAMoneysCuss(TM). Please.

Rgds

Damon

[ Parent ]

Re: Hydro in the Gutters (3.00 / 0) (#66)
by TomW on Thu Jun 19th, 2008 at 08:32:43 AM MST
(User Info)

Damon;

Agreed.

Next one to post reopening this useless sub thread arguing this will get some time in the read only club.

Seriously.

I tried to give them the chance to drop it like good forum citizens. It is no longer a suggestion. It will cease as of now.

Tom

"Education consists mainly of what we have unlearned."--Mark Twain
[ Parent ]

Re: Hydro in the Gutters (3.00 / 0) (#18)
by pyrocasto (pyrocasto at hotmail dot com) on Wed Jun 11th, 2008 at 10:32:32 PM MST
(User Info)

Well if any of you would use your brain you'd realize why not use some of the power you make, to pump the water back up, and generate more. You wouldnt even need a constant rain. Just one to get you started, recycle the water, and then you could power some of your house with the extra!

I think you could do the same with solar. Just use some power to power a lightbulb in front of the panels. You could make power at night!

I think I might start prototyping this sometime this month.

Re: Hydro in the Gutters (3.00 / 0) (#19)
by richsuka on Wed Jun 11th, 2008 at 10:50:22 PM MST
(User Info)

Using the rain water can save you money ,not in making energy but used as a supply a solar water heater. Or it can be diverted back to your water supply line to the house, with the simple filter. or a new water line just for washing cloth or bath water, lawn and garden watering cost me 25.00 monthly. If you are still thinking on using it as a power source, watch TV because it would take the rain fall that has cause flooding in the north over the last week or so to give you any measure or energy vs dollars spent. Save you money on this one.

rich
If it's free it cost too much!
[ Parent ]

Re: Hydro in the Gutters (3.00 / 0) (#27)
by thirteen on Sat Jun 14th, 2008 at 09:53:41 AM MST
(User Info)

You mentioned holding water, a couple of large buried tanks might help you. Maybe you could collect all of the rain water into large tanks and have a longer running flow of water using some floats inside the tank. When full it opens valves and thus letting the water out for generating purposes. Also depends on the slope of the land you have to work with. If you could keep all of the water maybe you could make a pond and raise fish. Just a idea

Re: Hydro in the Gutters (3.00 / 0) (#34)
by BruceDownunder (brucedownunder@hibis.wbs.net.au) on Sun Jun 15th, 2008 at 02:04:48 AM MST
(User Info)

How many times have I told you before,,,,

If you don't stop that ,,you'll go blind

Bruce

Re: Hydro in the Gutters (3.00 / 0) (#37)
by ruddycrazy on Sun Jun 15th, 2008 at 02:20:43 AM MST
(User Info)

Ron You've done it again mate, raise a simple discussion then all ya get da gutter talk... Well done mate this thread might even beat the Global Climate Change thread.

Re: Hydro in the Gutters (3.00 / 0) (#39)
by jacquesm (j@ww.com- I run a whitelist, add 'stjoes' to msg) on Sun Jun 15th, 2008 at 04:25:04 AM MST
(User Info) http://www.greenbits.com/

some grade school kid actually won a prize for this 'original idea' and implementation about 5 years ago.

It was just about enough to light a small bicycle lamp during a downpour.
www.greenbits.com

Re: Hydro in the Gutters (3.00 / 0) (#40)
by wdyasq on Sun Jun 15th, 2008 at 07:48:50 AM MST
(User Info)

The point is/was it is absurd to do this, even at 100 foot drop, it is still a stupid idea. No amount of R squaring pie and cubing the inverse will change that fact.

The minuscule amount of energy in even a torrential rain over a small area is the reason I started this mess. Even though we perceive our energy bills high, purchased energy is really a substitute for labor. I would not care to pay a person for the work a band saw will do in just a few minutes.

This cheap supply of 'help' has lead to a lot of waste. It has also allowed the mechanized farmer to feed multiple times what the farmer could do only a century ago. It allows the metropolitan lifestyle, suburbia, factory farming, the internet and cheap personal transportation.

I like the lifestyle it affords. I think we are all in for a great change in the ways we purchase and treat energy. I believe the religion of global warming is just a way for politicians to gain more control.

Before we endorse any policies, or install gutter turbines, we should take a good look at the project.

Ron
Adventure is just bad planning." -- Roald Amundsen

Re: Hydro in the Gutters (3.00 / 0) (#41)
by DamonHD (d@hd.org) on Sun Jun 15th, 2008 at 07:54:41 AM MST
(User Info) http://www.earth.org.uk/

I predict that fewer combines will get airborne when energy gets more expensive. B^>

Rgds

Damon

[ Parent ]

Re: Hydro in the Gutters (3.00 / 0) (#42)
by wdyasq on Sun Jun 15th, 2008 at 08:36:46 AM MST
(User Info)

That is a 'plug-in Hybrid' Combine Damon......

geeze ... don't they teach anything in schools anymore ....

Ron
Adventure is just bad planning." -- Roald Amundsen
[ Parent ]

Re: Hydro in the Gutters (3.00 / 0) (#44)
by DamonHD (d@hd.org) on Sun Jun 15th, 2008 at 09:58:10 AM MST
(User Info) http://www.earth.org.uk/

Don't know: I'll ask my daughter when she goes! B^> B^>

Rgds

Damon

[ Parent ]

Re: Hydro in the Gutters (3.00 / 0) (#45)
by DrDave on Sun Jun 15th, 2008 at 03:24:30 PM MST
(User Info)

What if instead of catching the rain, you catch just one lightning bolt?
Put that in a battery. How many watts is that?

Re: Hydro in the Gutters (3.00 / 0) (#46)
by ghurd on Sun Jun 15th, 2008 at 05:39:33 PM MST
(User Info)

My calculator says exactly 10x12.667^Boo-Koo.
G-

[ Parent ]

Re: Hydro in the Gutters (3.00 / 0) (#48)
by joestue on Sun Jun 15th, 2008 at 08:00:34 PM MST
(User Info)

10,000 amps x 100Mv = About one TeraWatt.

[ Parent ]

Re: Hydro in the Gutters (3.00 / 0) (#49)
by TomW on Mon Jun 16th, 2008 at 10:37:38 AM MST
(User Info)

I say direct the bolt to a small pressure vessel with water in it super heat the water flash to steam drive a weight upwards on a ratchet system

Actually think this is doable but at what cost I don't know.

Lets see, flash 1 cup [8 ounces] water into steam creates X cubic inches / feet of steam?

Tom

"Education consists mainly of what we have unlearned."--Mark Twain
[ Parent ]

Re: Hydro in the Gutters (3.00 / 0) (#50)
by joestue on Mon Jun 16th, 2008 at 05:26:31 PM MST
(User Info)

TomW, the details are numerous.

Electrically, it is possible to store the energy, basically make a gigantic tapped tank circuit, a lightning spike will charge it, then over the next few minutes rectify it slowly.
The energy available is proportional to the height of the antenna. so when we say 10 kiloamps, at 100 mega volts, that's for the 5 mile long strike.
To be practical, this needs to to be in the middle of a well insulated sky scraper.

There is a thread on sciencemadness about using a glass tube full of carbon powder in order to make a sh*t ton of diamond powder, but no one is sure if that will even work, buckyballs, nanotubes and graphite more likely...

only one small issue: containment... a glass carbon thermal bomb...

Same issue applies to the water, and the added effect of water breaking down at 500v/mm into a relatively low resistance short circuit, not something we want.

basically it will take just as much effort to extract any energy out of the water as it would be to store it electrically, also, supposing you are able to match the water at 2000K/10,000Psi for a normal lighting strike, what happens when a lightning bolt makes this pressure/temperature 10 times higher than normal?
electrically the voltage would be 3.2 times higher...
if your response to this is to design it for 800K at 1,000 psi... well, not much usable energy there...

It is an absolute waste to even think about recovering energy from lighting bolts, because the materials invested could be used to make thousands of times more energy somewhere else.

I grew up in a lighting devoid area, but i have always been interested in testing lighting's effect on things, and certainly if you could find a lighting attractor, you have at least one buyer.

[ Parent ]

Re: Hydro in the Gutters (3.00 / 0) (#51)
by wdyasq on Mon Jun 16th, 2008 at 07:16:31 PM MST
(User Info)

I can't believe a moderator just steered this thread further off topic!

If you folks want to start your own thread on absurd ideas, the posting button is on the right near the top of the page..... THIS absurd idea is my post...

Ron
Adventure is just bad planning." -- Roald Amundsen
[ Parent ]

Re: Hydro in the Gutters (3.00 / 0) (#67)
by interestingstuff on Mon Jul 14th, 2008 at 11:53:21 AM MST
(User Info)

Okay, so I get it.. the whole point of this math exercise is to point out how unlikely it is for turbines in the gutters to produce viable electricity production..

that being said... there must be some situation at which it does become viable.. so then what is that situation?

Since there's a lot of debate on the theoretical numbers here, and no one formula.. it becomes harder to figure out which numbers to plug in where.

So, lets take a real world example...

9,000 sq. ft. roof - average yearly rainfall about 42" - six 3" or 4" interior drainpipes, a solid enough building (poured concrete) strong enough to hold a significant amount of cached water on an upper floor. A drop in the drainpipes of at least 85'. If one wanted to, one could even hook up each of the drainpipes so that two go into one, three go into one... or even all six into one so that the volume of water, etc. is greater..

Would this be more viable if you had more square footage of roof? 15,000? 20,000? 100,000? At what point does it become feasible? Or is there a need for better, or cheaper technology?

Hydro in the Gutters | 67 comments (67 topical, 0 editorial)

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