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Home > Archive > Electrical Engineering > June 2006 > An electical question
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An electical question
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| Student 2006-06-06, 9:21 pm |
| I'm a HS sophomore, and I am hoping to become an engineer of some sort
someday. I'm working on a homework problem which I can't seem to solve.
I'll post the question and how I went about trying to solve it. Please
let me know if I've done anything wrong.
"Assume a wind turbine with a hub 50 m above ground, a rotor diameter
of 25 m and a wind-conversion efficientcy of 25%. The turbine will
operate in an area w/ an average windpower density of 600 watts per
square meter at 50m altitude. How many kwh can the turbine generate per
year?"
I started out by using
pi x r^2
pi x (12.5)^2=490.8738521 square meters
then I multiplied by .25 (the efficiency) and multiplied by 600.
That gives me 73631.07782 watts, which equals 73.632 Kilwatts
I've been told that 73.632 is incorrect... why?
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| Salmon Egg 2006-06-06, 10:21 pm |
| On 6/6/06 5:12 PM, in article
1149639178.052829.308330@j55g2000cwa.googlegroups.com, "Student"
<studentzr@hotmail.com> wrote:
> I'm a HS sophomore, and I am hoping to become an engineer of some sort
> someday. I'm working on a homework problem which I can't seem to solve.
> I'll post the question and how I went about trying to solve it. Please
> let me know if I've done anything wrong.
>
>
> "Assume a wind turbine with a hub 50 m above ground, a rotor diameter
> of 25 m and a wind-conversion efficientcy of 25%. The turbine will
> operate in an area w/ an average windpower density of 600 watts per
> square meter at 50m altitude. How many kwh can the turbine generate per
> year?"
>
> I started out by using
> pi x r^2
>
> pi x (12.5)^2=490.8738521 square meters
>
> then I multiplied by .25 (the efficiency) and multiplied by 600.
>
> That gives me 73631.07782 watts, which equals 73.632 Kilwatts
>
> I've been told that 73.632 is incorrect... why?
>
First of all, do not take all those significant figures seriously.
It looks like your arithmetic is OK. Nevertheless, you did not answer the
question. It asked for the energy obtainable during a whole year
Of course the problem is greatly oversimplified. I know of no place that
will give uniform wind.
Bill
-- Ferme le Bush
| |
| chuck 2006-06-06, 10:21 pm |
| Salmon Egg wrote:
> On 6/6/06 5:12 PM, in article
> 1149639178.052829.308330@j55g2000cwa.googlegroups.com, "Student"
> <studentzr@hotmail.com> wrote:
>
>
> First of all, do not take all those significant figures seriously.
>
> It looks like your arithmetic is OK. Nevertheless, you did not answer the
> question. It asked for the energy obtainable during a whole year
>
> Of course the problem is greatly oversimplified. I know of no place that
> will give uniform wind.
>
> Bill
> -- Ferme le Bush
>
>
The wind is not given as uniform. 600 watts per
square meter is given as the "average" windpower
density.
Chuck
| |
| Bill Kaszeta / Photovoltaic Resources 2006-06-06, 11:21 pm |
| You need to understand the difference between power and energy.
600 watts per square meter is a power
Energy is the product of power times time.
Look up the defination of kWh, then you should be able to calculate it from
your power calculations. Do not expect someone to do your homework.
On 6 Jun 2006 17:12:58 -0700, "Student" <studentzr@hotmail.com> wrote:
>I'm a HS sophomore, and I am hoping to become an engineer of some sort
>someday. I'm working on a homework problem which I can't seem to solve.
>I'll post the question and how I went about trying to solve it. Please
>let me know if I've done anything wrong.
>
>
>"Assume a wind turbine with a hub 50 m above ground, a rotor diameter
>of 25 m and a wind-conversion efficientcy of 25%. The turbine will
>operate in an area w/ an average windpower density of 600 watts per
>square meter at 50m altitude. How many kwh can the turbine generate per
>year?"
>
>I started out by using
>pi x r^2
>
>pi x (12.5)^2=490.8738521 square meters
>
>then I multiplied by .25 (the efficiency) and multiplied by 600.
>
>That gives me 73631.07782 watts, which equals 73.632 Kilwatts
>
>I've been told that 73.632 is incorrect... why?
>
Bill Kaszeta
Photovoltaic Resources Int'l
Tempe Arizona USA
bill@pvri-removethis.biz
| |
| Student 2006-06-06, 11:21 pm |
|
Salmon Egg wrote:
> On 6/6/06 5:12 PM, in article
> 1149639178.052829.308330@j55g2000cwa.googlegroups.com, "Student"
> <studentzr@hotmail.com> wrote:
>
>
> First of all, do not take all those significant figures seriously.
>
> It looks like your arithmetic is OK. Nevertheless, you did not answer the
> question. It asked for the energy obtainable during a whole year
>
> Of course the problem is greatly oversimplified. I know of no place that
> will give uniform wind.
>
> Bill
> -- Ferme le Bush
Oh, it looks like you are correct. I answered in kilowatts, not KWH.
How do i interconvert the two?
| |
| Student 2006-06-06, 11:21 pm |
| So I need to multiply the number of watts generated by the number of
hours in a year?
so the final answer should be 24 x 365 x 73.632 = 645016.32 KWh
Is this right?
btw, I'm not looking to have someone else complete my homework. I just
don't have anyone else to learn from, and it seems like the people in
this group are quite smart.
Bill Kaszeta / Photovoltaic Resources wrote:
> You need to understand the difference between power and energy.
>
> 600 watts per square meter is a power
>
> Energy is the product of power times time.
>
> Look up the defination of kWh, then you should be able to calculate it from
> your power calculations. Do not expect someone to do your homework.
>
>
> On 6 Jun 2006 17:12:58 -0700, "Student" <studentzr@hotmail.com> wrote:
>
>
> Bill Kaszeta
> Photovoltaic Resources Int'l
> Tempe Arizona USA
> bill@pvri-removethis.biz
| |
|
| Student wrote:
> I'm a HS sophomore, and I am hoping to become an engineer of some sort
> someday. I'm working on a homework problem which I can't seem to solve.
> I'll post the question and how I went about trying to solve it. Please
> let me know if I've done anything wrong.
>
>
> "Assume a wind turbine with a hub 50 m above ground, a rotor diameter
> of 25 m and a wind-conversion efficientcy of 25%. The turbine will
> operate in an area w/ an average windpower density of 600 watts per
> square meter at 50m altitude. How many kwh can the turbine generate per
> year?"
>
> I started out by using
> pi x r^2
>
> pi x (12.5)^2=490.8738521 square meters
>
> then I multiplied by .25 (the efficiency) and multiplied by 600.
>
> That gives me 73631.07782 watts, which equals 73.632 Kilwatts
>
> I've been told that 73.632 is incorrect... why?
>
No where in your calculation do you take time
into account. The question asks about kwh - which
needs time to be factored into the calculation.
(h = hour)
Ed
| |
| thrugoodmarshall@hotmail.com 2006-06-07, 1:21 pm |
| Student wrote:
> I'm a HS sophomore, and I am hoping to become an engineer of some sort
> someday. I'm working on a homework problem which I can't seem to solve.
> I'll post the question and how I went about trying to solve it. Please
> let me know if I've done anything wrong.
>
>
> "Assume a wind turbine with a hub 50 m above ground, a rotor diameter
> of 25 m and a wind-conversion efficientcy of 25%. The turbine will
> operate in an area w/ an average windpower density of 600 watts per
> square meter at 50m altitude. How many kwh can the turbine generate per
> year?"
>
> I started out by using
> pi x r^2
>
> pi x (12.5)^2=490.8738521 square meters
>
> then I multiplied by .25 (the efficiency) and multiplied by 600.
>
> That gives me 73631.07782 watts, which equals 73.632 Kilwatts
>
> I've been told that 73.632 is incorrect... why?
Holy cats!
You read the instructions!
http://www.catb.org/~esr/faqs/smart-questions.html
Good job.
The rest of you, take a lesson from this student.
| |
| Paul Hovnanian P.E. 2006-06-08, 12:21 am |
| Student wrote:
>
> So I need to multiply the number of watts generated by the number of
> hours in a year?
>
> so the final answer should be 24 x 365 x 73.632 = 645016.32 KWh
>
> Is this right?
Good.
One hint: carry the units through the equation with you. That way,
problems will jump out if the units don't match up. Like this:
PI * (12.5 m)^2 * 0.25 * (600 W / m^2) * (24 h/day) * (365 day/year) * (
1/1000 k ) = 645016* kWh/year
Cancel out the units:
(m)^2 * ( W / m^2 ) * (h/day) * (day/year) * ( k )= kWh / year
If they don't come out the way they're supposed to, you forgot
something.
> btw, I'm not looking to have someone else complete my homework. I just
> don't have anyone else to learn from, and it seems like the people in
> this group are quite smart.
Yes. Strange, isn't it? ;-)
[color=darkred]
> Bill Kaszeta / Photovoltaic Resources wrote:
--
Paul Hovnanian mailto:Paul@Hovnanian.com
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Why are so many towns named after water towers?
| |
| Dean Hoffman 2006-06-08, 1:21 am |
| In article <1149645311.233338.52650@i39g2000cwa.googlegroups.com>,
"Student" <studentzr@hotmail.com> wrote:
> btw, I'm not looking to have someone else complete my homework. I just
> don't have anyone else to learn from, and it seems like the people in
> this group are quite smart.
Did you notice the "PE" behind some of their names? It stands
for "pretty expensive".
Dean
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| |
| Paul Hovnanian P.E. 2006-06-08, 4:21 pm |
| Dean Hoffman wrote:
>
> In article <1149645311.233338.52650@i39g2000cwa.googlegroups.com>,
> "Student" <studentzr@hotmail.com> wrote:
>
>
> Did you notice the "PE" behind some of their names? It stands
> for "pretty expensive".
Pizza Eater
--
Paul Hovnanian mailto:Paul@Hovnanian.com
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The world is coming to an end ... SAVE YOUR BUFFERS!!!
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