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Topic: Flow from a raised tank  (Read 3847 times)

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Offline NPSH_is_fun

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Flow from a raised tank
« on: November 21, 2007, 01:14:39 PM »
Hey I am new to this board, and have a relatively simple problem that is throwing me for a loop for some reason.  I have a 4 ft high open head tank raised up with its bottom outlet sitting 120 ft above the ground.  The outlet is 2".  Water is added to the head tank via a pump on the ground whenever there is a demand through the 2" outlet, to keep the water level in the head tank relatively constant at 4 ft.  If I use the Bernoulli equation to calculate the velocity of the water coming out of the outlet, assuming its going to atmospheric pressure, then it simplifies to Toricelli's equation, v2 = sq. rt. (2gH) with H= 4 ft of water in the tank.  This gives a velocity of 16 ft/sec and corresponding volumetric flow rate of 40 gpm.

But what if I attach the outlet to a 2" pipe, and drop it straight down to the ground 120 ft?  I calculate the head loss due to friction of 120 ft. of pipe to be 7.9 ft of H20 using a Hazen-Williams calculator I found online.  I use the Bernoulli equation as follows:

(P1-P2)/ rho + (v1^2-v2^2)/2*gc + (z1-z2)*g/gc + F = 0
P1= 4 ft
P2 = 0 ft (atmosphere)
rho = 62.3 lb/sq. ft.
v1 = 16 ft/sec
v2 = ?
z1 = 120 ft
z2 = 0 ft
g/gc =1 lbf/lbm
gc= 32.174 lbm*ft*lbf^-1*s^-2

I come up with v2= 92.1 ft/sec.  That doesn't seem right at all.  What am I doing wrong here?

Offline technologist

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Re: Flow from a raised tank
« Reply #1 on: December 30, 2007, 03:35:45 AM »
Why R u doubting on it? If your numbers & calculation is correct its fine.
Bcoz now U have Head of 120 ft available as a driving force SO dont worry

In fact, in actual plant operations, it is used as a good measure  for emptying out a tank at certain height; by adding hoses to drop it to bottom gorund so that higher head provide more velocity.

Its OK

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