Connecting a pipe line to and existing distribution system

[MikEngineer]

Homework Statement

I am connecting a 4" diameter PVC pipe to an existing distribution system. there will be a single check valve (K=2), eight 90 elbows (K=.9) and ten gate valves (K=0.2). I am trying to find the flow rate at the end of a 1500' run and what pressure I would see if I were tie this into a house. Max Pressure in system in 120 psi. z1 = 0 and z2 = 45 feet
y = 62.4 lbf/ft^3 specific gravity; u = 1.41 x 10^-5 ft^2/s (kinematic viscosity).

Homework Equations

Energy Equation: gamma = y and alpha = a;
P1/y + a1(V1^2/2g) + z1 + hp = P2/y + a2(V2^2/2g) + z2 + ht + hL

hL= sum(f(L/D)(V^2/2g)+sum(K(V^2/2g))

The Attempt at a Solution

P1/y + a1(V1^2/2g) + z1 + hp = P2/y + a2(V2^2/2g) + z2 + ht + hL

Assume a1=a2; turbulent flow
v1 = v2; incompressible flow, same diameter pipe and point 2 is taken just before water exits open valve at the end of 1500'
hp = ht = 0; no pump or turbine. Therefore
p2 = 0; open to atm.

P1/y + z1 = z2 + hL => hL = P1/y + z1 - z2 = (120 psi * 144 in^2/ft^2)/(62.4 lb/ft^3) + 0 - 45 feet = 232 feet = hL

hL= sum(f(L/D)(V^2/2g)+sum(K(V^2/2g)) = (V^2/2g)(f*L/D + sum[K])

solve for V in terms of f => V = (2g*hL)/(f*L/D + sum[K]) > [14940.8 / (4500f + 11.2)]^(1/2) = 4.05

now we need to iterate using the Moody Diagram or the Swaame-Jain equation.

let f = 0.2 => V = 4.05 => Re# = (4.05*(4/12))/1.41x10^-5 = 1.35x10^6 = 9.5 x 10^5

Iteration eventually yields f = 0.0142 and V = 14.1 ft/s

So my question is this, am I correct in my assumption that V1 = V2 = 14.1 ft/s when the end of the 1500 extension is left opened to the atmosphere?

I can get my flow rate by Q = AV = Pi/4 (4/12)^2 * 14.1 = 1.23 ft^3/s => 9.2 gps => 552.2 gpm.

So, assuming I am correct up this point, How do I now figure out what pressure could be delivered to two houses, if the end of the pipe is closed with a valve and two TEE's are placed one at 1200 feet and the other at 1280 feet.
l TEE to one house
It would look something like this -----------------l-------l-----GV closed
l TEE to house 2

Check out my spreadsheet attached

 

[Valkarie]

for calculations. The pressure at the end of the pipe is calculated by:P2 = P1 + (V1^2 - V2^2)/(2g) where P1 is the pressure at the inlet, V1 is the velocity at the inlet, V2 is the velocity at the outlet, and g is the acceleration due to gravity. The pressure at the two TEE's is then calculated by P1 = P2 + (V2^2 - V3^2)/(2g) where V3 is the velocity at the TEEs. The flow rate can then be calculated from Q = AV, where A is the cross sectional area of the pipe, and V is the velocity at the TEEs.