Pipe Design - One Inlet, 3 Outlets

[TBen24]

Any tips for solving this would be greatly appreciated. Thank you all for your time and help

The schematic of the problem is in the attached PDF

PROBLEM INFORMATION
I’m trying to determine the equations to solve for the conditions at 1, 2, and 3 when they are at a given length in their respective pipe. I need to solve for the pressures and velocities at states 0 – 3. Also, the gaps between the pipe branches are considerably larger than the drawing shows. In other words, the drawing is not to scale. I don’t need numbers specifically, just help on setting up the system of equations.

KNOWN
Gravity, density, roughness, kinetic energy correction factor, size of duct (all areas, lengths, and diameters)

SOLUTION ATTEMPT
Notes: Q is volumetric flow rate, epsilon is roughness, mu is viscosity, alpha is kinetic energy correction factor
Assumptions: Steady flow, inviscid flow, incompressible flow, and irrotational flow.

Converse mass
Q0 = V0*A0, Q1 = V1*A1, Q2 = V2*A2, Q3 = V3*A3
Q0 = Q1+Q2+Q3

Friction Factors
Re0 = (rho*V0*D0)/mu, Re1 = (rho*V1*D1)/mu, Re2 = (rho*V2*D2)/mu, Re3 = (rho*V3*D3)/mu
1/sqrt(f0) = -2.0*log((epsilon/D0)/3.7+2.51/(Re0*sqrt(f0)))
1/sqrt(f1) = -2.0*log((epsilon/D1)/3.7+2.51/(Re1*sqrt(f1)))
1/sqrt(f2) = -2.0*log((epsilon/D2)/3.7+2.51/(Re2*sqrt(f2)))
1/sqrt(f3) = -2.0*log((epsilon/D3)/3.7+2.51/(Re3*sqrt(f3)))

Head Losses
HL1 = f0*(L01/D0)*(V0^2/(2*g))+(f1*(L1/D1)+K1)*(V1^2/(2*g))
HL2 = f0*(L02/D0)*(V0^2/(2*g))+(f2*(L2/D2)+K2)*(V2^2/(2*g))
HL3 = f0*(L03/D0)*(V0^2/(2*g))+(f3*(L3/D3)+K3)*(V3^2/(2*g))
K1 is caused by branched tee flow in tee 1
K2 is caused by straight flow through tee 1 and branched flow through tee 2
K3 is caused by straight flow through tees 1 and 2 and branched flow in tee 3

Analysis
Based on the current setup, there will most likely be more unknowns than equations. In addition, pressure is not solved for at any of the states. I still need to relate the states.

Concerns and Questions
1) Is the head loss done correctly?
2) Can Bernoulli be applied between the various states to conserve energy? Specifically can it be applied between states 0 --> 1, 0 --> 2, and 0 --> 3 using the balances shown below despite the flow branching in several pipes?
P0/(rho*g)+alpha*(V0^2/(2*g))+L1 = P1/(rho*g)+alpha*(V1^2/(2*g))+HL1
P0/(rho*g)+alpha*(V0^2/(2*g))+L2 = P2/(rho*g)+alpha*(V2^2/(2*g))+HL2
P0/(rho*g)+alpha*(V0^2/(2*g))+L3 = P3/(rho*g)+alpha*(V3^2/(2*g))+HL3
3) Are the head losses equal because they are parallel? I personally do not think that they are because there are large gaps separating the branches?
4) Should I solve for pressure loss using head loss if the energy balance is not valid?

Thank you all again

 

[richpaulcol]

As far as I can tell everything you have there is fine. The one thing you are missing are the boundary conditions. As you have 4 boundaries (ends of pipes) to calculate a unique solution to your problem you will need 4 boundary conditions (flows or heads).

Rich

 

[firavia]

One of the best ways to calculate complicated pipe networks is to use Newton-Raphson method >> http://www.firavia.com/Newton.html
this will help you hopefully .