Jeepers

In article <241120041442412259%RhmpL33@nospam.net>,
Rich Hampel <RhmpL33@nospam.net> wrote:

> If the fill hose touches the bottom of the barrel that is being
> filled..... then each system will require the same exact pressure to
> fill: The pressure required is the weight of the column of water
> (static head); but, since the hose that goes over the top will partly
> siphon (the distance from the top of the barrel to the bottomof the
> barrel) the pressure requirements will be the same. The only
> difference is that for 'starting' the hose that goes over the top, will
> require more pressure to lift the water to the top of the bend, but
> once the water entirely fills the hose and the sipon effect starts,
> then either system will require the same pressure. whew!

I was clear, no hose over the top. 3 bungs on side of barrel, two at
top, one at bottom. One top one is vent the other is fill.

Why isn't the additional water in the barrel not included as the static
head on the lower fill? There is tremendously more volume in the barrel
at, say half way.

--
Member AAAAAAAA
American Association Against Acronym Abuse And Also Ambiguity.

----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==----
http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups
----= East and West-Coast Server Farms - Total Privacy via Encryption =----

Jeepers

In article <241120041442412259%RhmpL33@nospam.net>,
Rich Hampel <RhmpL33@nospam.net> wrote:

> If the fill hose touches the bottom of the barrel that is being
> filled..... then each system will require the same exact pressure to
> fill: The pressure required is the weight of the column of water
> (static head); but, since the hose that goes over the top will partly
> siphon (the distance from the top of the barrel to the bottomof the
> barrel) the pressure requirements will be the same. The only
> difference is that for 'starting' the hose that goes over the top, will
> require more pressure to lift the water to the top of the bend, but
> once the water entirely fills the hose and the sipon effect starts,
> then either system will require the same pressure. whew!

I was clear, no hose over the top. 3 bungs on side of barrel, two at
top, one at bottom. One top one is vent the other is fill.

Why isn't the additional water in the barrel not included as the static
head on the lower fill? There is tremendously more volume in the barrel
at, say half way.

--
Member AAAAAAAA
American Association Against Acronym Abuse And Also Ambiguity.

----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==----
http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups
----= East and West-Coast Server Farms - Total Privacy via Encryption =----

L.W.(=?iso-8859-1?Q?=DFill?=) Hughes III

The common denominator is one square inch. One pound per square
inch pressure equals 2.31 feet of water. The energy required to create a
pressure of 43 psi is the same as the energy required to lift water 100
feet.
God Bless America, ßill O|||||||O
mailto:-------------------- http://www.----------.com/

Jeepers wrote:
>
> O.K. my lame-*** attempt at physics argument:
>
> Isn't part of the column the barrel? The barrel has a bigger column in
> it's part, than the hose. So the weight of the water column is GREATLY
> larger than the one in the hose going to the top. Less water - less
> weight, right?
>
> --
> Member AAAAAAAA
> American Association Against Acronym Abuse And Also Ambiguity.
>
> ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==----
> http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups
> ----= East and West-Coast Server Farms - Total Privacy via Encryption =----

L.W.(=?iso-8859-1?Q?=DFill?=) Hughes III

The common denominator is one square inch. One pound per square
inch pressure equals 2.31 feet of water. The energy required to create a
pressure of 43 psi is the same as the energy required to lift water 100
feet.
God Bless America, ßill O|||||||O
mailto:-------------------- http://www.----------.com/

Jeepers wrote:
>
> O.K. my lame-*** attempt at physics argument:
>
> Isn't part of the column the barrel? The barrel has a bigger column in
> it's part, than the hose. So the weight of the water column is GREATLY
> larger than the one in the hose going to the top. Less water - less
> weight, right?
>
> --
> Member AAAAAAAA
> American Association Against Acronym Abuse And Also Ambiguity.
>
> ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==----
> http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups
> ----= East and West-Coast Server Farms - Total Privacy via Encryption =----

L.W.(=?iso-8859-1?Q?=DFill?=) Hughes III

The common denominator is one square inch. One pound per square
inch pressure equals 2.31 feet of water. The energy required to create a
pressure of 43 psi is the same as the energy required to lift water 100
feet.
God Bless America, ßill O|||||||O
mailto:-------------------- http://www.----------.com/

Jeepers wrote:
>
> O.K. my lame-*** attempt at physics argument:
>
> Isn't part of the column the barrel? The barrel has a bigger column in
> it's part, than the hose. So the weight of the water column is GREATLY
> larger than the one in the hose going to the top. Less water - less
> weight, right?
>
> --
> Member AAAAAAAA
> American Association Against Acronym Abuse And Also Ambiguity.
>
> ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==----
> http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups
> ----= East and West-Coast Server Farms - Total Privacy via Encryption =----

Tom Quackenbush

Jeepers wrote:
<snip>

>O.K. my lame-*** attempt at physics argument:
>
>Isn't part of the column the barrel? The barrel has a bigger column in
>it's part, than the hose. So the weight of the water column is GREATLY
>larger than the one in the hose going to the top. Less water - less
>weight, right?

Nope. When we speak of a column in this context, the column has the
same area as the hose through which you're pumping fluid. If you
wanted to know the pressure in pounds per square inch, you'd consider
an imaginary coulumn with a cross-sectional area of 1 square inch.

The volume of fluid outside the column doesn't matter.

Consider this:

If you submerge yourself one foot under water in your swimming pool,
you're subject to the same pressure as if you were submerged one foot
under water in Lake Superior, even though the volume of Lake Superior
is a gazillion times larger than your swimming pool (gazillion being a
highly technical term used for these sorts of comparisons).

R,
Tom Q..



Remove bogusinfo to reply

Tom Quackenbush

Jeepers wrote:
<snip>

>O.K. my lame-*** attempt at physics argument:
>
>Isn't part of the column the barrel? The barrel has a bigger column in
>it's part, than the hose. So the weight of the water column is GREATLY
>larger than the one in the hose going to the top. Less water - less
>weight, right?

Nope. When we speak of a column in this context, the column has the
same area as the hose through which you're pumping fluid. If you
wanted to know the pressure in pounds per square inch, you'd consider
an imaginary coulumn with a cross-sectional area of 1 square inch.

The volume of fluid outside the column doesn't matter.

Consider this:

If you submerge yourself one foot under water in your swimming pool,
you're subject to the same pressure as if you were submerged one foot
under water in Lake Superior, even though the volume of Lake Superior
is a gazillion times larger than your swimming pool (gazillion being a
highly technical term used for these sorts of comparisons).

R,
Tom Q..



Remove bogusinfo to reply

Tom Quackenbush

Jeepers wrote:
<snip>

>O.K. my lame-*** attempt at physics argument:
>
>Isn't part of the column the barrel? The barrel has a bigger column in
>it's part, than the hose. So the weight of the water column is GREATLY
>larger than the one in the hose going to the top. Less water - less
>weight, right?

Nope. When we speak of a column in this context, the column has the
same area as the hose through which you're pumping fluid. If you
wanted to know the pressure in pounds per square inch, you'd consider
an imaginary coulumn with a cross-sectional area of 1 square inch.

The volume of fluid outside the column doesn't matter.

Consider this:

If you submerge yourself one foot under water in your swimming pool,
you're subject to the same pressure as if you were submerged one foot
under water in Lake Superior, even though the volume of Lake Superior
is a gazillion times larger than your swimming pool (gazillion being a
highly technical term used for these sorts of comparisons).

R,
Tom Q..



Remove bogusinfo to reply

Bartolomeo Cristofori

The physics is simple.

If the hose is of the proper diameter for the flow rate generated by
the pump, then the pressure at the pump will be about equal to the
vertical distance (the height) between the pump and the point where
the water becomes free-flowing, times a density value of about .45
PSI/ft (which is derived from the weight of a cubic foot of water
divided by 144 to convert from square feet to square inches). If you
are pumping the water to the top of the barrel and letting it fall,
then that would be the height difference between the pump and the top
of the barrel. If you are pumping the water into the bottom bung,
then you would use the height difference between the pump and the top
of the water in the barrel.

Therefore, the pump would encounter slightly less pressure if the tank
were filled through the bottom bung, at least until it became nearly
full. The difference is small enough that it would make only a small
change in the amount of time it would take to fill the tank.

The fact that there may be several hundred pounds of water in the
barrel is immaterial, since pressure is pounds per square inch. You
could calculate the pressure at the bottom of the barrel by dividing
the weight of the water in the barrel by the area of the barrel head
in square inches (assuming straight sides, for a wooden barrel you
would have to use an average diameter to get an exact figure).

Bartolomeo

Bartolomeo Cristofori

The physics is simple.

If the hose is of the proper diameter for the flow rate generated by
the pump, then the pressure at the pump will be about equal to the
vertical distance (the height) between the pump and the point where
the water becomes free-flowing, times a density value of about .45
PSI/ft (which is derived from the weight of a cubic foot of water
divided by 144 to convert from square feet to square inches). If you
are pumping the water to the top of the barrel and letting it fall,
then that would be the height difference between the pump and the top
of the barrel. If you are pumping the water into the bottom bung,
then you would use the height difference between the pump and the top
of the water in the barrel.

Therefore, the pump would encounter slightly less pressure if the tank
were filled through the bottom bung, at least until it became nearly
full. The difference is small enough that it would make only a small
change in the amount of time it would take to fill the tank.

The fact that there may be several hundred pounds of water in the
barrel is immaterial, since pressure is pounds per square inch. You
could calculate the pressure at the bottom of the barrel by dividing
the weight of the water in the barrel by the area of the barrel head
in square inches (assuming straight sides, for a wooden barrel you
would have to use an average diameter to get an exact figure).

Bartolomeo