How to calculate how much flow there is from a single source to multiple drains

Question

Assuming that I have a water source ($Q$) that can flow into 3 drains ( $Q_1$, $Q_2$, $Q_3$):

Assume that we know that

1. The drain geometry is predetermined—we know the $x,y,z$ coordinates of the start and end node of the river/drain and we know the cross sectional information and length of the river/drain
2. The velocity is approximated by manning formula, $$V=\frac{k}{n}R_h^\frac{2}{3} S^\frac{1}{2}$$
3. The flow is incompressible flow, so $Q=VA$

where

1. $R_h$ is the hydraulic radius of the river/drain
2. $S$ is the slope
3. $k$ conversion between the SI and English units
4. $n$ the manning coefficient
5. $A$ is the cross section area of the river/drain

For one, I know that the water flow volume must be conserved

$Q=Q_1+Q_2+Q_3+...$

But I don't know what are the other factors that could help us determine $Q_i$.

Edit: After some research, I think that I need use Bernoulli equation to include headloss in this calculation, but I have no idea how to do it, how to proceed?

Answer 1

This is essentially explaining how to solve parallel flow problems using the fundamentals on fluid dynamics - let me know if you need additional information.

Let's assign some variables. Let's let:

• Q1, Q2, Q3, and Q be the flow rates in each of the pipes (Q is the overall flow). Note Q = V * A, where V1, V2, and V3 are the fluid velocities. A1, A2, A3 being the channel area used.
• Z1, Z2, Z3 and Z are the heights above sea level of each of the outlets (Z is the cross point).
• Let's assume you are working with manning formula - in this case, convert the Zs into different slopes - S1, S2, and S3. If your drains slope changes along the way, you can model it using an average slope, or you can model using several pieces of pipe in parallel. The average slope is easier, and generally accurate.
• While X, Y, and Z are nice - let's use L1, L2, and L3 for the lengths of each pipe.
• Finally, we need the energy in each of these points. Note that power is used because for a steady state assumption, the energy would be infinite. So, Power = Flow Rate * Pressure (Check out the units!). Here, we use P1, P2, P3 and P for the powers at each ending - P Being the power at the cross point. For the power consumed by friction along the length, let's use Pf1, Pf2, and Pf3.
• Rho is the density of the fluid. D is hydraulic diameter (D1, D2, D3 for consistency).

Now the total power at the cross point must be the total power - expended via friction along the pipes or delivered at the outlet. That's just conservation of energy. For a bunch of reasons though, power is not in units of Watts like it should be for fluid dynamics. It's in units of length - the height of a reservoir of energy. It's called head. I'm explaining this in terms of power, and it's valid - but it's not historically accurate.

The power expended by the pipes:

Pf1 = rho*Q1*f*(L1/D1)*(Q1/A1)^2/2

f is of course the Darcy Friction Factor. Of course, to solve for f, you need to obtain Q. But Q is unknown! So, we guess at an f, check the problem via iteration, solve for Q, then check that the guess for f was valid. If we're close, move on - otherwise guess at a new f and keep going.

With this friction formula in mind, then the power consumed by each of the channels is:

P1 = (Z-Z1) * rho * Q1 + Q1^3*rho/(2*g*A1) + Pf1

With this in mind, at last we have the equations needed to solve this problem:

Q1 + Q2 + Q3 = Q P1 + P2 + P3 = P Q1 = A1*(k/n)*D1^(2/3)S1^(1/2) Q2 = A2(k/n)*D2^(2/3)S2^(1/2) Q3 = A3(k/n)*D3^(2/3)*S3^(1/2)

With this, we have five equations and 3 unknowns - the various Q1, Q2, and Q3. The simplest solution is to iterate through and find 3 Qs to solve all 5 equations, disregarding the most empirical assumptions first (in this case, I'd disregard the f's - followed by the Manning equations. Energy and Mass conservation is top priority).

Answer 2

Volumetric flowrate is conserved; giving the equation:

Q.soruce = Q.drain1 + Q.drain2 + Q.drain3

Now to determine how much flow goes where you have to determine the flow resistance of each. To reduce complexity (especially the first time around) you should hold the upstream pressure constant (this is actually the pressure in the center of your fork. For example:

P = 700 kPa Gauge (above atmospheric)

There are many factors that affect flow resistance. The primary factors are pipe diameter and length. Obviously the most simple case is all three resistances being equal and all 3 pressure drops being equal, so consequently all flows would also be equal.

To solve something more complicated we first need to know the pressures at all 3 outlets. If it is to atmosphere the gauge pressure is 0kPa.

Once pressure drops are established we need to calculate resistance either theoretically for simple geometry or empirically for complex geometry. There is also the option of Computational Fluid Dynamics for theoretically solving complex geometry. Some good places to start are major losses and nozzle pressure drop. If you are dealing with low pressure or big changes in height you will also have to account for gravity.

Once you have all the outlet resistances and all the outlet pressure drops, you can solve for each outlet flow; then sum them to get the total flow. You can then remove one variable and solve for it under different conditions or divide the flows over the total flow to get ratios.