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Hydro in the Gutters | 67 comments (67 topical, editorial)
Re: Hydro in the Gutters (3.00 / 0) (#20)
by MattM on Thu Jun 12th, 2008 at 06:26:38 PM MST
(User Info)

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
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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

  1. feet / 2.1593354 seconds = 34.732909 feet / second
  2. 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

  1. feet / 2.47309683 seconds = 40.435133 feet / second
  2. 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
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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
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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
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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
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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 ]


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

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