1
$\begingroup$

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

Single-inflow, multiple-outflow drain

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?

$\endgroup$
5
  • $\begingroup$ Try implementing Bernoulli's equation first and edit the question with any difficulties you have. Bernoulli's equation can account for what you mentioned in the question like pressure drop due to friction factor and head loss due to elevation differences. $\endgroup$ – morristtu Dec 7 '15 at 5:06
  • $\begingroup$ @morristtu, I don't have much background in fluid dynamics, so I will appreciate if you can give me the derivation or at least some pointers on how Bernoulli's equation can be used to determine how much flow in each drain $\endgroup$ – Graviton Dec 7 '15 at 5:31
  • 2
    $\begingroup$ Are there pipes attached to the drains, or do all drain into an empty void? Can you make a sketch? $\endgroup$ – mart Dec 7 '15 at 10:10
  • $\begingroup$ It's basically the analog of resistors in parallel. So you need to know whether the drains are identical, or how they vary (which leads to calculating Bernoulli "stuff" for each drain). $\endgroup$ – Carl Witthoft Dec 7 '15 at 14:26
  • $\begingroup$ @CarlWitthoft, would you like to elaborate more on the answer? $\endgroup$ – Graviton Dec 8 '15 at 2:58
2
$\begingroup$

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).

$\endgroup$
4
  • $\begingroup$ Of course, to solve for f, you need to obtain Q-- I guess you mean to solve for $Pf1$? $\endgroup$ – Graviton Dec 10 '15 at 1:34
  • $\begingroup$ Also, it's quite hard to read without proper notation, especially on the last equation.. would you like to rewrite your equations in mathjax? $\endgroup$ – Graviton Dec 10 '15 at 1:37
  • $\begingroup$ I read your formulation again, in the last parts of the above equations, I know what is $n$, but what is $k$? How is it related to other parameter, or is it just a constant? $\endgroup$ – Graviton Dec 10 '15 at 1:59
  • $\begingroup$ Another issue: we have 5 equations and only 3 unknowns, isn't this an over-determined system, which means it is very possible that all 5 equations might not be satisfied simultaneously? $\endgroup$ – Graviton Dec 10 '15 at 2:01
0
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ I've updated my question. Would you like to show a workout example using the information that I supply? Can this problem be tackled without resorting to computational fluid dynamics simulation? $\endgroup$ – Graviton Dec 9 '15 at 7:27
  • $\begingroup$ Yes you can likely get very close with just the major losses i listed above. Note those are for a fully filled pipe. If by drain pipe you mean a partially filled pipe with an air gap things get a lot more complicated. We may be able to help you more if you are more specific with your application. What are you really trying to do with this? $\endgroup$ – ericnutsch Dec 9 '15 at 16:35
  • $\begingroup$ I am trying to just compute the distribution of the water flow into different outflow. $\endgroup$ – Graviton Dec 14 '15 at 5:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.