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

where

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?