Calculating Water Pressure: Hazen-Willams Formula for 24' Tank & 650' Distance

[Steve E]

I have a water tank that is 24' tall. Water flows from the bottom of the tank thru a 8" plastic pipe for a distance of 650' down a hill for a elevation drop of 70'. What would be the expected water pressure at the bottom of the pipe given a flow of 20 gallons per minute? I've researched several books and they say to use the Hazen-Willams formula. Can somebody help me work thru this? Thanks!

 

[redargon]

Hazen williams will give you the velocity in the pipe due to pipe friction and head loss. You can rearrange it to give a pressure drop per length of pipe (so psi/ft for example)

More accurately you can also use the darcy weisbach equation but hazen williams seems to be the right match for the question you're asking.

Here is some relatively good info on Hazen williams http://en.wikipedia.org/wiki/Hazen-Williams_equation and here http://www.engineeringtoolbox.com/hazen-williams-water-d_797.html

Remember that your answer for pressure loss or head loss will be per foot of pipe length, so you'll have to multiply your answer by you total pipe length.

 

[redargon]

If you show me a couple of attempts or steps, I'll help you through the rest

 

[Steve E]

Thanks redargon,

To figure the pressure at the bottom of the tank: Pressure = 62.4lb/cf x 24ft = 1497.6 lb/sf
To convert to square inches: 1497.6 lb/sf / 144 = 10.4 psi

Then to figure the pressure at the end of the 8" plastic water line for 20 gpm:

(4.52 x 20 (raised to 1.85 power)) divided by (130 raised to the 1.85 power x 8 raised to the 4.87 power)

That equals 0.000005665.

Is that how much pressure is lost per foot of pipe under that level of flow just based on the pipe fricton itself?

How do I account for the lower elevation at the end of the line and how much pressure that adds.

Thanks again for your help redargon.

 

[minger]

You can use this pressure loss and Bernoulli's equation.

[tex] \left( p + \frac{\rho v^2}{2} + \rho gz \right)_1 = \left( p + \frac{\rho v^2}{2} + \rho g z+ h_l \right)_2 [/tex]
Where [tex]h_l[/tex] is your pressure loss you just calculated. You should have everything here except outlet pressure [tex]p_2[/tex]