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Re: Diesel from new/used wast oil...? Eeyore 07-03-2008
Posted by daestrom on July 5, 2008, 11:34 am
Please log in for more thread options
 Piccolo Pete wrote:
>> Piccolo Pete wrote:
>>>>
>>>>> (crap... here comes today's storm...)
>>>>
>>>> Yep, it ran me inside. It was like a cloud busted open.
>>>>
>>>> Vaughn
>>>
>>> I've often thought about getting a bit of hydro power from these
>>> storms in Florida. Unfortunately, this place doesn't have any
>>> gutters to channel the flow.
>>>
>>
>> 2 inches of rain (a lot for even Florida in one storm) over 1600
>> ft^3 of a single story house (say, 10 ft) would be about 166 000
>> ft-lb of energy or about 213 BTU.
>>
>> A bit of fun, but not really enough energy to bother with. More
>> valuable as freshwater I would think.
>>
>> daestrom
>
> The house I had in Lakeland was a two story. It produced about 55
> gallons in 5 minutes. I'm not a big hydro expert, so could you do
> the math on that?
> I agree that most of that would just be fun stuff - which is why I
> don't mess with it. Still, that is a lot of energy that is just
> going down the drain.
>
> Oh, and I wouldn't trust any rainwater in Florida as fresh water
> unless I was dying of thirst.
>

To find the *theoretical* energy, convert the gallons to pounds
55 gal * 8.33 = 458 lb of water

Then figure out how far you could let the water 'fall'. Two story, let's
say 20 ft for 'round numbers'. Multiply the weight times the height to get
ft-lb of work

458lb * 20ft = 9163 ft-lb of work

There are 778 ft-lb in a BTU, or 2655200 ft-lb in a kWhr. So we're not
getting a lot here with only 9163. About 12 BTU.

daestrom




Posted by Piccolo Pete on July 5, 2008, 4:50 pm
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> Piccolo Pete wrote:
>>> Piccolo Pete wrote:
>>>>>
>>>>>> (crap... here comes today's storm...)
>>>>>
>>>>> Yep, it ran me inside. It was like a cloud busted open.
>>>>>
>>>>> Vaughn
>>>>
>>>> I've often thought about getting a bit of hydro power from these
>>>> storms in Florida. Unfortunately, this place doesn't have any
>>>> gutters to channel the flow.
>>>>
>>>
>>> 2 inches of rain (a lot for even Florida in one storm) over 1600
>>> ft^3 of a single story house (say, 10 ft) would be about 166 000
>>> ft-lb of energy or about 213 BTU.
>>>
>>> A bit of fun, but not really enough energy to bother with. More
>>> valuable as freshwater I would think.
>>>
>>> daestrom
>>
>> The house I had in Lakeland was a two story. It produced about 55
>> gallons in 5 minutes. I'm not a big hydro expert, so could you do
>> the math on that?
>> I agree that most of that would just be fun stuff - which is why I
>> don't mess with it. Still, that is a lot of energy that is just
>> going down the drain.
>>
>> Oh, and I wouldn't trust any rainwater in Florida as fresh water
>> unless I was dying of thirst.
>>
>
> To find the *theoretical* energy, convert the gallons to pounds
> 55 gal * 8.33 = 458 lb of water
>
> Then figure out how far you could let the water 'fall'. Two story, let's
> say 20 ft for 'round numbers'. Multiply the weight times the height to
> get ft-lb of work
>
> 458lb * 20ft = 9163 ft-lb of work
>
> There are 778 ft-lb in a BTU, or 2655200 ft-lb in a kWhr. So we're not
> getting a lot here with only 9163. About 12 BTU.
>
> daestrom

Thanks. I'm just not use to doing this kind of power stuff yet. Seems like
something that could easily crush me from 20 feet up should have more power
than that.

By the way, a typical storm in central Florida can last as long as an hour
with this kind of flow. I only had one 55 gallon drum to play with. So,
while in Lakeland, I was probably getting about 144 or more BTUs. Wouldn't
this equate to about 42 watt hours?

Sorry if I'm coming off as a complete idiot with many of my posts. I'm just
trying to learn.

Bart



Posted by daestrom on July 7, 2008, 7:30 pm
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Piccolo Pete wrote:
<snip>
>> To find the *theoretical* energy, convert the gallons to pounds
>> 55 gal * 8.33 = 458 lb of water
>>
>> Then figure out how far you could let the water 'fall'. Two story,
>> let's say 20 ft for 'round numbers'. Multiply the weight times the
>> height to get ft-lb of work
>>
>> 458lb * 20ft = 9163 ft-lb of work
>>
>> There are 778 ft-lb in a BTU, or 2655200 ft-lb in a kWhr. So we're
>> not getting a lot here with only 9163. About 12 BTU.
>>
>> daestrom
>
> Thanks. I'm just not use to doing this kind of power stuff yet. Seems
> like something that could easily crush me from 20 feet up
> should have more power than that.
>
> By the way, a typical storm in central Florida can last as long as an
> hour with this kind of flow. I only had one 55 gallon drum to play
> with. So, while in Lakeland, I was probably getting about 144 or
> more BTUs. Wouldn't this equate to about 42 watt hours?
>
> Sorry if I'm coming off as a complete idiot with many of my posts. I'm
> just trying to learn.
>

Remember a few 'conversions' and become really good at 'unit analysis' and
learn how to multiple a lot of things by '1'.

1054.8 watt-seconds
---------------------- = 1
1 BTU

1 Minute
----------- = 1
60 seconds


1 hour
------------ = 1
60 minutes


So you take 144 BTU and multiply that by '1'. We all know that multiplying
anything times '1' doesn't change the value of the 'thing' we multiplied by,
right? But the trick is to use the right form of '1'. In this case, I'd
use the form that related BTU to watt-seconds, and then the form that
relates seconds to minutes and minutes to hours

144 BTU * (1) * (1) * (1) = ??watt-hours
144 BTU * (1054.8 watt-seconds / 1 BTU ) * ( 1 minute / 60 seconds) * ( 1
hour / 60 minutes) = 42.192 watt-hours

Notice that if you write all those 'fractions' of units with a horizontal
line instead of the slashes I used here, you can 'cancel' all the ones in
the numerators with ones in the denominator except for 'watt' and 'hour'
(there is a 'BTU' in a numerator and a 'BTU' in a denominator, same with
'minute' and 'second'). When you get done 'canceling' all the units
through, the units on the left should match up with those on the right. If
you don't get the same 'units', you forgot something or messed up somewhere.

daestrom

> Bart


Posted by Eeyore on July 5, 2008, 5:20 pm
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daestrom wrote:

> To find the *theoretical* energy, convert the gallons to pounds
> 55 gal * 8.33 = 458 lb of water
>
> Then figure out how far you could let the water 'fall'. Two story, let's
> say 20 ft for 'round numbers'. Multiply the weight times the height to get
> ft-lb of work
>
> 458lb * 20ft = 9163 ft-lb of work
>
> There are 778 ft-lb in a BTU, or 2655200 ft-lb in a kWhr. So we're not
> getting a lot here with only 9163. About 12 BTU.

How you manage to do these sums in mixed American and metric units never fails
to amaze me.

Graham <amazed>


Posted by Piccolo Pete on July 5, 2008, 6:06 pm
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>
>
> daestrom wrote:
>
>> To find the *theoretical* energy, convert the gallons to pounds
>> 55 gal * 8.33 = 458 lb of water
>>
>> Then figure out how far you could let the water 'fall'. Two story, let's
>> say 20 ft for 'round numbers'. Multiply the weight times the height to
>> get
>> ft-lb of work
>>
>> 458lb * 20ft = 9163 ft-lb of work
>>
>> There are 778 ft-lb in a BTU, or 2655200 ft-lb in a kWhr. So we're not
>> getting a lot here with only 9163. About 12 BTU.
>
> How you manage to do these sums in mixed American and metric units never
> fails
> to amaze me.
>
> Graham <amazed>
>

Something like:

http://www.onlineconversion.com/

Possibly? Or maybe "he, she, it" is simply not human and has to convert all
this stuff from Zeta Reticuli math anyway.

;-)

Bart



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