| New Reply |
Connecting a pipe line to and existing distribution system |
Share Thread |
| Jun23-11, 05:06 PM | #1 |
|
|
Connecting a pipe line to and existing distribution system
1. The problem statement, all variables and given/known data
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). 2. Relevant 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)) 3. 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 |
| New Reply |
| Tags |
| design, distribution system, energy equation, pipe |
Similar discussions for: Connecting a pipe line to and existing distribution system
|
||||
| Thread | Forum | Replies | ||
| Equation for line connecting 2 planes | Precalculus Mathematics Homework | 4 | ||
| Finding electric field intensity at point outside the line connecting the charges | Engineering, Comp Sci, & Technology Homework | 0 | ||
| pipe line in table. | Math & Science Software | 2 | ||
| Pipe Line Failure | Mechanical Engineering | 11 | ||
| Two cylindrical tanks, connecting pipe and outlet pipe - Differential Equation | Calculus & Beyond Homework | 5 | ||