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| Author |
Water generator and Solar power questions
|
|
|
| Can anyone suggest any good sources for water powered
electric generators like this one:
http://tinyurl.com/akthl
and/or other groups to ask in? How about solar generators?
How practical is it to use either or both?
Note: I'm aware of Google, and Dogpile, etc, but would still
appreciate any specific sites and groups people are willing
to recommend.
| |
|
|
| The Watcher 2005-10-20, 4:21 pm |
| On Thu, 20 Oct 2005 12:37:24 -0400, dh@. wrote:
>Can anyone suggest any good sources for water powered
>electric generators like this one:
>
>http://tinyurl.com/akthl
>
>and/or other groups to ask in? How about solar generators?
>How practical is it to use either or both?
Depends on your site. There are devices you can use to check your location to
measure how appropriate it is for either type.
>
>Note: I'm aware of Google, and Dogpile, etc, but would still
>appreciate any specific sites and groups people are willing
>to recommend.
| |
| Eric Sears 2005-10-20, 5:21 pm |
| On Thu, 20 Oct 2005 12:37:24 -0400, dh@. wrote:
>Can anyone suggest any good sources for water powered
>electric generators like this one:
>
>http://tinyurl.com/akthl
As far as I know this unit is not especially efficient - just
convenient in some situations.
"good sources"? - there are many water powered electric generators
(also called hydropower units) - but availability might depend on what
country you are in .
>
>and/or other groups to ask in? How about solar generators?
>How practical is it to use either or both?
It really depends what you mean by "practical".
If you can manage your house/home with 2.4 Kwh of electrical energy a
day - well yes! - I suppose its practical (I run a holiday home on
about that amount with lpg cooking and wood heating.)
You might need about six time the solar generation capacity in most
situations, to equal the output of hydro - and probably a bigger
battery as well - if it is stand-alone.
>
>Note: I'm aware of Google, and Dogpile, etc, but would still
>appreciate any specific sites and groups people are willing
>to recommend.
Steve Spence' site is very useful, also homepower.com
Eric Sears
| |
| Derek Broughton 2005-10-21, 10:21 am |
| Eric Sears wrote:
> On Thu, 20 Oct 2005 12:37:24 -0400, dh@. wrote:
>
....[color=darkred]
[color=darkred]
> It really depends what you mean by "practical".
> If you can manage your house/home with 2.4 Kwh of electrical energy a
Luxury! We live well on 1.5kWh / day.
> day - well yes! - I suppose its practical (I run a holiday home on
> about that amount with lpg cooking and wood heating.)
Propane cooking, hot water & fridge and oil/solar heat. An electric fridge
would probably get us close to that 2.4kWh/day (we already have a small
electric freezer).
>
> You might need about six time the solar generation capacity in most
> situations, to equal the output of hydro - and probably a bigger
> battery as well - if it is stand-alone.
Typical solar gain in the US/southern Canada is in the realm of 4 hours/day
or less, so at least six times the capacity.
That particular generator seems pretty expensive for so little power - my
Air-X (marine) can put out more power than that for the same sort of money.
I've never really looked into hydro generators because my stream is
seasonal, but I always expected them to be cheaper per watt.
--
derek
| |
|
| On Fri, 21 Oct 2005 09:30:09 -0300, Derek Broughton <news@pointerstop.ca> wrote:
>Eric Sears wrote:
>
>...
>
>
>Luxury! We live well on 1.5kWh / day.
>
>
>Propane cooking, hot water & fridge and oil/solar heat. An electric fridge
>would probably get us close to that 2.4kWh/day (we already have a small
>electric freezer).
>
>Typical solar gain in the US/southern Canada is in the realm of 4 hours/day
>or less, so at least six times the capacity.
>
>That particular generator seems pretty expensive for so little power - my
>Air-X (marine) can put out more power than that for the same sort of money.
>I've never really looked into hydro generators because my stream is
>seasonal, but I always expected them to be cheaper per watt.
I don't even understand how it all works. I have 3 115 amp hour trolling
batteries I use with an inverter to run my TV etc when I take the boat out.
But I don't understand how to calculate how long they would run a 1.5KW
space heater, or how many more I'd need to make it practical if three isn't
enough, or how to calculate what it would take to run it for how long, or
how to monitor the batteries so I know not to overcharge them, etc.
I really need to learn those basics before I'll be able to think about it in
any detail. The idea is to save money, but if I do it wrong it will probably
cost more. My original idea was to get water to turn an alternator, since
alternators are already set up to charge 12V batteries...and supposedly
set up to not overcharge them...
| |
| Mike Swift 2005-10-23, 3:21 pm |
| In article <29Q5f.30213$Ge5.14561@fe10.lga>,
Steve Spence <sspence@green-trust.org> wrote:
> dh@. wrote:
>
> http://www.realgoods.com/renew/shop...1201/ts/1017104
>
> if you have apprpriate conditions, hydro is most practical. wind is
> next, followed by solar.
The system you show is used in a flowing stream. For hydropower the
first thing you need to do is find out what you have available in the
way of water. Until you have the velocity (in meters per second) or
head (fall in meters) and flow (in L per second) you can't find out how
much energy is present. When you can come back with these numbers some
of us can help you in what type of system you need.
| |
| nicksanspam@ece.villanova.edu 2005-10-23, 3:21 pm |
| Mike Swift <tomswift@cruzio.com> wrote:
>...For hydropower the first thing you need to do is find out what you have
>available in the way of water. Until you have the velocity (in meters per
>second) or head (fall in meters) and flow (in L per second) you can't find
>out how much energy is present.
Head and flow are almost enough to know power.
Nick
| |
| Dan Bloomquist 2005-10-23, 5:21 pm |
|
Mike Swift wrote:
> Steve Spence <sspence@green-trust.org> wrote:
>
> The system you show is used in a flowing stream. For hydropower the
> first thing you need to do is find out what you have available in the
> way of water. Until you have the velocity (in meters per second) or
> head (fall in meters) and flow (in L per second) you can't find out how
> much energy is present. When you can come back with these numbers some
> of us can help you in what type of system you need.
Another thing is that the above turbine is extremely wasteful. It may be
suitable if you have a lot of hydro power you can do without. The first
thing is to know the head from a barrier to the generator. Low head,
high volume and you would use a turbine. High head and low volume and
you are probably better off with a pelton wheel.
In the coastal foothills of CA where realgoods is based, most resources
are better met with a pelton wheel from what I've seen.
Best, Dan.
| |
| Steve Spence 2005-10-23, 7:21 pm |
| Dan Bloomquist wrote:
>
>
> Mike Swift wrote:
>
>
> Another thing is that the above turbine is extremely wasteful. It may be
> suitable if you have a lot of hydro power you can do without. The first
> thing is to know the head from a barrier to the generator. Low head,
> high volume and you would use a turbine. High head and low volume and
> you are probably better off with a pelton wheel.
>
> In the coastal foothills of CA where realgoods is based, most resources
> are better met with a pelton wheel from what I've seen.
>
> Best, Dan.
>
It may be wasteful, but if you have lots of flow without much head, it's
a simple, appropriate solution.
--
Steve Spence
Dir., Green Trust, http://www.green-trust.org
Contributing Editor, http://www.off-grid.net
http://www.rebelwolf.com/essn.html
| |
| dances_with_barkadas@yahoo.com 2005-10-24, 3:21 am |
| nicksanspam@ece.villanova.edu wrote:
> Head and flow are almost enough to know power.
respectfully request you strike the word "almost"
| |
| nicksanspam@ece.villanova.edu 2005-10-24, 5:21 am |
| <dances_with_barkadas@yahoo.com> wrote:
>
>respectfully request you strike the word "almost"
Velocity matters too, no? A 10 L/s sheet of water that meanders over rocks for
a kilometer at 0.1 m/s while falling 1 meter and losing power by friction has
less available power than a 10 L/s water column that drops 1 m from a pipe.
Nick
| |
| Tim May 2005-10-24, 5:21 am |
| In article <dji1pr$mj@acadia.ece.villanova.edu>,
<nicksanspam@ece.villanova.edu> wrote:
> <dances_with_barkadas@yahoo.com> wrote:
>
>
> Velocity matters too, no? A 10 L/s sheet of water that meanders over rocks for
> a kilometer at 0.1 m/s while falling 1 meter and losing power by friction has
> less available power than a 10 L/s water column that drops 1 m from a pipe.
Technically, this is "speed," not velocity. (Basic physics.)
But head and flow are by far the most important things, especially
within a defined region. For example, within someone's property.
(Unless someone is wealthy enough to own an entire stream course, the
relevant details are nearly all captured in head and flow.)
There are a bunch of equivalent ways of viewing energy...power times
time, force times distance, etc., but, basically, it comes down to
"mgh," the mass of the falling water (flow times time) times g times h
(the height, or head).
(Energy is not power, of course. Energy is power times time, the
integral of power. So the "mgh" translates to power in the obvious
ways.)
What is lost in frictional heating of the rocks, blah blah blah, is a
small but easily calculated loss.
Relating to what someone said earlier about stream power being the
first choice, this is obviously not so. Only a small number of people
live near appropriate streams, with appropriate head and flow, and
which they have legal access to. Wind and solar are less encumbered by
"water rights." Solar is the most predictable, for most people. Wind is
mostly OK for pumping water, where it is most often used.
--Tim May
| |
| nicksanspam@ece.villanova.edu 2005-10-24, 5:21 am |
| Tim May <timcmay@removethis.got.net> wrote:
[color=darkred]
>What is lost in frictional heating of the rocks, blah blah blah, is a
>small but easily calculated loss.
I disagree, and invite you to calculate the difference.
Nick
| |
| Tim May 2005-10-24, 5:21 am |
| In article <dji46s$o4@acadia.ece.villanova.edu>,
<nicksanspam@ece.villanova.edu> wrote:
> Tim May <timcmay@removethis.got.net> wrote:
>
>
>
> I disagree, and invite you to calculate the difference.
>
>
I have, several times. Most recently, for my hill's water system.
Without a lot more details about the rocks you talk about, no basis for
a detailed calculation.
Fact is, the OP was correct in saying that "head" and "flow" encompass
most of the critical issues.
And as I said, this is even more obviously so over the necessarily
small run of a private piece of land. Gibbering about the energy losses
from the top of the Himalayas to the outlet of the Ganges is not
relevant to the situation of a stream with a dam and a wheel or turbine
in a length that one would find in a reasonablel set-up.
You're a buffoon.
--Tim May
| |
| nicksanspam@ece.villanova.edu 2005-10-24, 8:21 am |
| Tim May <timcmay@removethis.got.net> wrote:
>
>Without a lot more details about the rocks you talk about, no basis for
>a detailed calculation.
0.1 m/s.
>You're a buffoon.
You mispelled "I don't know how" :-)
Nick
| |
| Dan Bloomquist 2005-10-24, 1:21 pm |
|
Steve Spence wrote:
> Dan Bloomquist wrote:
>
> It may be wasteful, but if you have lots of flow without much head, it's
> a simple, appropriate solution.
Geee Steve, what part of, 'It may be suitable if you have a lot of hydro
power you can do without', didn't you understand?
Best, Dan.
| |
|
| dh:
I have 3 115 amp hour trolling
batteries I use with an inverter to run my TV etc when I take the boat
out.
But I don't understand how to calculate how long they would run a 1.5KW
========================================================
Amp hrs is only used with batteries.... so if you mult by 12 you get
watt-hrs 3x115=4140.. thats 4KWhrs. So they wouldnt last long putting
running 1.5KW.. you dont want to discharge the batts too deep.
| |
| nicksanspam@ece.villanova.edu 2005-10-24, 3:21 pm |
| Tim May <timcmay@removethis.got.net> wrote:
>
>0.1 m/s.
P = 1/2(10)0.1^2 = 0.05 kW for the sheet and 10x9.8x1 = 98 kW for the pipe.
Would you call this a small difference? :-)
Nick
| |
| Steve Spence 2005-10-24, 4:21 pm |
| BobG wrote:
> dh:
> I have 3 115 amp hour trolling
> batteries I use with an inverter to run my TV etc when I take the boat
> out.
> But I don't understand how to calculate how long they would run a 1.5KW
>
> ========================================================
> Amp hrs is only used with batteries.... so if you mult by 12 you get
> watt-hrs 3x115=4140.. thats 4KWhrs. So they wouldnt last long putting
> running 1.5KW.. you dont want to discharge the batts too deep.
>
a typical deep cycle battery wants to be discharged at 50% or less depth
of discharge. So only pull out 2 kWh before recharging.
--
Steve Spence
Dir., Green Trust, http://www.green-trust.org
Contributing Editor, http://www.off-grid.net
http://www.rebelwolf.com/essn.html
| |
| Derek Broughton 2005-10-24, 4:21 pm |
| nicksanspam@ece.villanova.edu wrote:
> Tim May <timcmay@removethis.got.net> wrote:
>
>
> P = 1/2(10)0.1^2 = 0.05 kW for the sheet and 10x9.8x1 = 98 kW for the
> pipe.
>
> Would you call this a small difference? :-)
That doesn't look reasonable to me. What's the 1/2?? looks like (1/2)mv^2
- energy, not power.
Then you're using m*a*d (I think) for the pipe. You couldn't just state
your formulas and give two related equations, could you?
--
derek
| |
| nicksanspam@ece.villanova.edu 2005-10-24, 5:21 pm |
| Derek Broughton <news@pointerstop.ca> wrote:
>
>That doesn't look reasonable to me. What's the 1/2?? looks like (1/2)mv^2
>- energy, not power.
>
>Then you're using m*a*d (I think) for the pipe. You couldn't just state
>your formulas and give two related equations, could you?
You would think regular alt.ENERGY.homePOWER readers would know more about
energy and power...
OK. Spoon feeding: 1 kilojoule = 1 kg-m^2/s^2; 1 joule/second = 1 watt and
kinetic energy E = 1/2mv^2 and m = 10 kg, so energy = 0.05 kJ every second,
ie 50 watts in the sheet. The potential energy in the falling water is mgh
= 10kgx9.8m/s^2x1m = 98 kgm^2/s^2, ie 98 kJ, and this is converted to kinetic
energy when 10 kg falls every second, so we have 98 kW of power.
From another point of view, v = sqrt(2ad) as an apple falls to the ground,
so the final falling water velocity sqrt(2x9.8x1) = 4.43 m/s, so it has
1/2(10)4.43^2 = 98 kJ of kinetic energy or 98 kW of power. It's no accident
that 50W(4.43/0.1)^2 = 98 kW, given the velocity ratios.
Nick
| |
| dances_with_barkadas@yahoo.com 2005-10-25, 7:21 am |
| > Velocity matters too, no?
velocity of water falling down under gravitational pull is
invariant, depends only upon the duration of it's free fall.
> A 10 L/s sheet of water that meanders over rocks for
> a kilometer at 0.1 m/s while falling 1 meter and losing power by friction
that's not the definition of "head".
Tell me the flow, I can tell you the number of pounds of water that
move in freefall per minute.
Tell me the head, I can tell you the number of feet through which
gravity did work and became converted potential to kinetic energy.
so now I can calculate foot-pounds per minute.
The dimension is power. For example, 33,000 foot-pounds/minute is one
horsepower.
Please explain to us, which part of that you don't understand?
| |
| nicksanspam@ece.villanova.edu 2005-10-25, 8:21 am |
| <dances_with_barkadas@yahoo.com> wrote:
> velocity of water falling down under gravitational pull is
>invariant, depends only upon the duration of it's free fall.
If it falls freely. IIRC, the OP didn't say that.
>
> that's not the definition of "head".
Agreed. A definition might begin "head is..."
>Tell me the flow, I can tell you the number of pounds of water that
>move in freefall per minute.
Sounds like you know some basic physics.
>Please explain to us, which part of that you don't understand?
Have you stopped beating your wife? :-)
Nick
PS: Can you find the huge flaw in my last posting?
| |
| Derek Broughton 2005-10-25, 10:21 am |
| nicksanspam@ece.villanova.edu wrote:
> Derek Broughton <news@pointerstop.ca> wrote:
>
>
> You would think regular alt.ENERGY.homePOWER readers would know more about
> energy and power...
First, you were posting to misc.survivalism, too, and I have no idea what
they'd know :-)
I am a regular on a.e.homepower, and I _do_ know the difference, but if you
insist on throwing out numbers specifically intended to baffle with
bullshit, I'll call you on it (my original post tried to be polite).
> OK. Spoon feeding: 1 kilojoule = 1 kg-m^2/s^2; 1 joule/second = 1 watt and
> kinetic energy E = 1/2mv^2 and m = 10 kg, so energy = 0.05 kJ every
> second, ie 50 watts in the sheet.
Was that so hard to do? Now he knows where you came up with those numbers.
> The potential energy in the falling
> water is mgh = 10kgx9.8m/s^2x1m = 98 kgm^2/s^2, ie 98 kJ, and this is
> converted to kinetic energy when 10 kg falls every second, so we have 98
> kW of power.
You still failed to actually explain _that_ formula. E = force (F) x
distance; F = mass x acceleration.
>
> From another point of view, v = sqrt(2ad) as an apple falls to the ground,
> so the final falling water velocity sqrt(2x9.8x1) = 4.43 m/s, so it has
> 1/2(10)4.43^2 = 98 kJ of kinetic energy or 98 kW of power. It's no
> accident that 50W(4.43/0.1)^2 = 98 kW, given the velocity ratios.
Thank you
--
derek
| |
| nicksanspam@ece.villanova.edu 2005-10-25, 11:21 am |
| Derek Broughton <news@pointerstop.ca> wrote:
>I am a regular on a.e.homepower, and I _do_ know the difference, but if you
>insist on throwing out numbers specifically intended to baffle with bullshit,
That was not my intent, and anyone with a smattering of this 300-year-old
physics would know where these numbers come from :-)
>I'll call you on it (my original post tried to be polite).
Get a life.
Nick
| |
| Derek Broughton 2005-10-25, 1:21 pm |
| nicksanspam@ece.villanova.edu wrote:
> Derek Broughton <news@pointerstop.ca> wrote:
>
>
> That was not my intent, and anyone with a smattering of this 300-year-old
> physics would know where these numbers come from :-)
Nick, the majority of posters on a.e.homepower do _not_ have much physics -
and a smattering _isn't_ enough. It's fair to assume that the posters on
sci.energy _do_, so I removed them from the crossposts. Is it really too
much to ask you to try to explain how you arrive at a solution? btw, a
smiley is completely inappropriate at that point.
>
> Get a life.
Ooh! A hit! Get a comeback...
--
derek
| |
| Eric Gisin 2005-10-25, 4:21 pm |
| Looks like a toy. Would not survive spring melt around here.
A more reliable system for mountain areas uses a pipe and turbine.
A flow of 10L/s with 10m of head gives you 9.8*10*10 = 980W.
If the land slope is 1:10, you need 100m of 2" pipe.
<dh@.> wrote in message news:rrhfl11e98oaq8b40jdrrbdvflea7dufjf@4ax.com...
> Can anyone suggest any good sources for water powered
> electric generators like this one:
>
> http://tinyurl.com/akthl
>
> and/or other groups to ask in? How about solar generators?
> How practical is it to use either or both?
>
| |
| nicksanspam@ece.villanova.edu 2005-10-25, 5:21 pm |
| Derek Broughton <news@pointerstop.ca> wrote:
BTW, that's basic high-school physics.
[color=darkred]
>Nick, the majority of posters on a.e.homepower do _not_ have much physics...
It's time to learn. Otherwise, we are an ignorant cargo cult.
BTW, have you found my huge error yet?
Nick
It's a snap to save energy in this country. As soon as more people
become involved in the basic math of heat transfer and get a gut-level,
as well as intellectual, grasp on how a house works, solution after
solution will appear.
Tom Smith, 1980
| |
| daestrom 2005-10-25, 6:21 pm |
|
<nicksanspam@ece.villanova.edu> wrote in message
news:djjejj$1kp@acadia.ece.villanova.edu...
> Derek Broughton <news@pointerstop.ca> wrote:
>
>
> You would think regular alt.ENERGY.homePOWER readers would know more about
> energy and power...
>
> OK. Spoon feeding: 1 kilojoule = 1 kg-m^2/s^2;
Oops. 1 Joule = 1 kg-m^2/s^2. 1kiloJoule is of course 1000 Joules, or 1000
kg-m^2/s^2
http://en.wikipedia.org/wiki/Joule
Looks like your 'spoon' was mistaken for a 'shovel'.
> ... 1 joule/second = 1 watt and
> kinetic energy E = 1/2mv^2 and m = 10 kg, so energy = 0.05 kJ every
> second,
With V of 0.1 m/s, that would be E = (1/2)* 10 kg * (0.1 m/s)^2 = 0.05 J in
every 10 L of water. Since that flows by every second, then yes, the POWER
of that flow is 0.05 Joule/sec or 0.05 watts. Not 50 watts.
> ie 50 watts in the sheet. The potential energy in the falling water is mgh
> = 10kgx9.8m/s^2x1m = 98 kgm^2/s^2, ie 98 kJ, and this is converted to
> kinetic
> energy when 10 kg falls every second, so we have 98 kW of power.
Again, wrong units. 10kg*9.8 m/s^2 * 1m = 98 kg-m^2/s^2 = 98 Joules, not 98
kJ. Thus, 98 W of power.
Ah if only we could get 98 kW out of such a small stream....
daestrom
| |
| daestrom 2005-10-25, 6:21 pm |
|
<nicksanspam@ece.villanova.edu> wrote in message
news:djls27$2jl@acadia.ece.villanova.edu...
> Derek Broughton <news@pointerstop.ca> wrote:
>
>
> BTW, that's basic high-school physics.
>
>
> It's time to learn. Otherwise, we are an ignorant cargo cult.
>
Speaking of 'learning' Nick....
> BTW, have you found my huge error yet?
>
Yes.
daestrom
| |
| nicksanspam@ece.villanova.edu 2005-10-25, 9:21 pm |
| daestrom <daestrom@NO_SPAM_HEREtwcny.rr.com> wrote:
>
>Oops. 1 Joule = 1 kg-m^2/s^2...
Right. I just picked up a book (Kreider and Rabl's 1994 Heating and Cooling
of Buildings) and read "1 kJ = 1 kg x m^2/s^2" on page 17 and didn't think
about that. Page 17 of the 1994 2nd edition repeats the mistake...
Nick
| |
| Mike Swift 2005-10-26, 3:21 am |
| In article <djgj8g$bi@acadia.ece.villanova.edu>,
nicksanspam@ece.villanova.edu wrote:
> Mike Swift <tomswift@cruzio.com> wrote:
>
>
> Head and flow are almost enough to know power.
>
> Nick
True Nick, however he was talking about flow in a stream. In this
environment head is not possible to measure, however the velocity or
speed is measurable. This can be converted to effective head.
Wake generators (name from the boating industry) are not very efficient,
but can be effective is applications where little civl works are
available and stream velocity is high.
Mike
| |
| John Phillips 2005-11-22, 4:21 pm |
| On Sun, 23 Oct 2005 19:36:57 GMT, Dan Bloomquist
<public21@lakeweb.com> wrote:
>
>Another thing is that the above turbine is extremely wasteful. It may be
>suitable if you have a lot of hydro power you can do without. The first
>thing is to know the head from a barrier to the generator. Low head,
>high volume and you would use a turbine. High head and low volume and
>you are probably better off with a pelton wheel.
>
>In the coastal foothills of CA where realgoods is based, most resources
>are better met with a pelton wheel from what I've seen.
>
Dan,
I have been in the hydro industry for more than 35 years. My friends
in California still call Pelton wheels "turbines". My other
professional friends also call Francis and Kaplan wheels "turbines".
Where have we gone wrong since we do not conform with your definition?
In point of fact, while the boundaries are very blurred, most low head
conventional hydro turbines employ the Kaplan (variable blade) or
propeller design (fixed blade), most moderate head turbines (above 100
feet) a Francis design, and high head (above 1,000 feet) a Pelton
design. Peltons are not reversible so pumped turbines are always
Francis at all but low heads. Ontario Hydro at Niagara Falls uses a
variable vane pumped turbine similar to a Kaplan but called a "Deriaz"
after the original manufacturer. Its maximum head is about 75 feet but
it is much more efficient than its counterparts across the River at
Lewiston that are of the Francis design.
Regards,
John Phillips
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