How much pressure is lost when a fire hose is at the top of a long 100 meter ladder? If it comes out of the hose at 63 bar (at street level) how much pressure is lost by raising it 100 meters?

I'm confused about a few things: in a sealed container the pressure is the same everywhere - up down left right - in a hose (when the tap is ON but the nozzle is OFF (water (pressure) builds up in the hose but no water appears) when I twist the nozzle ON, water bursts out generally at a constant pressure. It doesn't matter where I am in the garden it's consistent - but what happens if I raise the hose - does the pressure drop? Or is the pressure maintained because it is a (nearly) sealed container?


I found this Fire apparatus - Wikipedia, the free encyclopedia (http://en.wikipedia.org/wiki/Fire_apparatus)
Telescopic aerial platforms can reach heights of over 328 feet (over 100 meters). However, most of them are designed to reach the height of approx. 100 feet (33 meters)


Firehose - Wikipedia, the free encyclopedia (http://en.wikipedia.org/wiki/Firehose)
The usual working pressure of a firehose can vary between 8 bar and 20 bar ((0.8 to 2.0 MPa), while its bursting pressure can be up to 63 bar (6.3 MPa). (This level of pressure emitted by the hose can actually break in a weaker brick wall.)

Thank you
John

Hi and thanks for the question

I believe you need to consider hydrostatic pressure (http://en.wikipedia.org/wiki/Hydrostatic_pressure)

If you assume there is 63bar at street level, you can use this equation to work out the loss in pressure as you raise it up by 100 meters.

Hope this helps! :)

Pascal's law
From Wikipedia, the free encyclopedia

In the physical sciences, Pascal's law or Pascal's principle states that the fluid pressure at all points in a connected body of an incompressible fluid at rest, which are at the same absolute height, are the same, even if additional pressure is applied on the fluid at some place.

The hose with the nozzle shut is fluid at rest, and at all points the pressure is the same (Pascals law). What I have learned that I didn't understand before is: "which are at the same absolute height".

This "height" of 100 meters is a water column which must take into account hydrostatic pressure - hence we will have a drop in pressure. The fire hose must have a lot more pressure to squirt the water out the other end of the hose which is 100 meters up, than it would to project it sideways at street level.

That's what I didn't understand and I've muddled my way to find: pressure is equal at all points "up down left right" (Pascal's law) but does NOT include "height" because height introduces another factor called Hydrostatic pressure.

Am I in the ball park:-)

Regards
John

You are understanding it well, which is absolutely fantastic to hear.

Now if you want a numerical estimate, you can go one step further and use the hydrostatic formula from the article:

P = \rho g h

here you have:
\rho = 1000kg.m^{-3} for water (at 4 degrees C, but it doesn't change very rapidly above this)
g = 9.8 m.s^{-1}
h = 100 m

putting this in we find:

P = 1000*9.8*100 = 980000 Pa = 0.98 MPa

This will be your decrease in pressure from ground level (h=0m) to h=100m

Your pressure at ground level is 63 bar.
1 bar = 10^5 Pa = 0.1 MPa
so
63 bar = 6,300,000 Pa = 6.3 MPa.

So your pressure at the top of the hose is
6300000 Pa - 980000 Pa
or
6.3 MPa - 0.98 MPa = 5.32 MPa

This is equal to 53.2 bar.

I hope you follow this and you find it helpful. Please feel free to ask if there is anything you haven't understood.
There are few other factors in real life that would affect this, so this should be close to what you would experience in real life :)

wow

After reading about a million times I started to see the pattern of formula but I'm not used to looking at that language, so I usually avoid it. Numerical.

I'm very impressed with your answer - it is definitely little g. It's nearly midnight here in Sydney so it's late but I'm urged on.

63 bar at the bottom
53.2 bar at the top 100 meters
approx. 10 bar loss

hope you don't mind I'd like to talk psi

1 bar = 14.5 psi
10 bar = 145 psi
10 bar is approx. 100 meters
145 psi = 100 meters
145 psi loss
therefore to start with on the ground we need more, if we are going to lose 145 psi, and we still want it to squirt and break down light brick walls - then the pressure to start with would be what, 5 times... 5 x 145 psi = 725 psi :: 10 times = 1450 psi

so
63 bar at the bottom (14.5 x 63) = 913 psi at the bottom
913 -
53 bar 100 meters up (14.5 x 53) = 768.5 psi at the top
913 - 768 = 145 psi
hmm
it's still 145 psi loss
that's always encouraging - I'm simply making the same mistake over :)
so does that mean, to raise water 100 meters = 145 psi = 10 bar = 100 meters


wow thanks
John

It will always be the same loss, the loss is due to the gravitational pull on the body of water (which is always the same). You can see this by looking at the formula

\rho is constant for water
g is constant
h is constant as long as you are always talking about 100 meters
so P, the drop in pressure, is always constant, independent of your pressure at the bottom of the pipe.

I hope that the equation is satisfactorily correct for real life situations.

You're not making any mistake, it is a 145 psi loss.

You want to take the psi you want at the top, add 145 psi to it and that's the psi yuou need at the bottom.

Wow again

Or in another way
It takes 145 psi to raise water 100 meters

Yes that will probably be the best way of looking at it for you. :)

Hope I have been helpful (you can always click the rate this answer button, if I have been helpful)

can't be that easy...

If

1 litre of water = 1 Kg
and it takes 145 psi to raise water 100 meters
then does it take 145 psi to raise 1 Kg of water 100 meters

You can't think of it like that really, you have to think of it like this:

If you have a length of hose, say 100m.

you hang it vertically, plug the bottom and put water in the top until it it full.

when you release the plug, the water will come out at 145psi (at the instant you release the plug, of course when water escapes, the height of water is less and so the pressure is less). This is how the equation is designed to be used.

But I don't see why you can't do the same thing for your situation of pushing water upwards.

But the way you're thinking about it is wrong, yes.

for your situation of pushing water upwards

What do you mean "for your situation of pushing water upwards":-)

Well, you're wanting to push some water upwards, right? :p

Good
Yep
Talk later
Off to slepp
Thanks heaps
John