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Pressure Drops In Pipes: Part 2, Series and Parallel
January 13, 2010Posted by on
This is Part 2 in a series on pressure drops in pipes. Part 1 covered the basic concepts and equations (1-4b). It is recommended to read that before proceeding, or to have it open along side this article for reference.
A single pipe is not usually sufficient to solve a real—life problem. Instead, systems of pipes, often quite complex, are used to achieve all desired results. The math, as a result, becomes more difficult, but where would be the fun if it was easy?
Pipes in series are actually easy. Each length of pipe and each fitting needs to be accounted for, so a sum of all major and minor losses is taken to find the head loss (See Part 1, Eqn. 2), and from that, pressure drop. Parallel pipes, like those depicted in Figure 1, need a bit more consideration.
Fluid flows in along a single pipe before branching off into any number of parallel paths. These paths may be the same, but more often than not, each will be a bit different. Eventually, they rejoin and the fluid exits as one stream again. Finding the overall pressure drop between the entrance and exit isn’t quite as simple as adding everything up.
Consider Figure 1 once more, as the logic will be more easily understood with an example. There is some volumetric flow Q entering the system from the left. It branches, with a flow of Q1 in the upper Path 1 and Q2 in the lower Path 2. The head loss for each particular branch can be found with the unknowns in terms of friction factors (f1, f2) and volumetric flow rates (Q1, Q2), provided the pipe geometry and K factors for fittings are known.
At the point of branching, there is only one pressure. The same applies to the exit, as it would create a contradiction having two different values at the same place. Therefore, the pressure entering each branch and exiting each branch are the same. Similarly, there change in height, Δh, for each will be the same. These provide the key to solving the problem; if the pressure drops and Δh in each branch are equal, then the head loss in each pipe is equal.
In this two branch example, that will mean that h1=h2. Assuming the fluid incompressible and that mass is conserved, the Q entering is equal to Q1+Q2, which is also equal to the Q exiting. Therefore, one can substitute to get rid of Q2 (or Q1). Rearranging for Q1 gives something like Equation 5. For simplicity, due to it’s pseudo-symmetric nature, L1=L2, which cancel. Minor losses were also assumed to cancel.
Now, actually solving for Q1 may be easy or hard depending on the type of flow. If the flow is laminar, Equation 4a can be used for friction factor, where it is an explicit function of Reynolds Number, and therefore volumetric flow. The rest becomes a trivial plug and chug to find Q1, which could then be used to find h1 and pressure drop.
If the flow is turbulent, Equation 4b, the Colebrook-White Equation, must be used. This defines friction factor implicitly, so the solution is not so trivial; an iterative method must be used. The steps are as follows:
- Select a inital guesses for f1 and f2. Literature can help here to find likely values, or at least magnitudes.
- Solve for Q1 using the estimated values for friction factor.
- Find Q2 using Q-Q1
- Calculate Reynolds numbers Re1,Re2 from Q1,Q2
- Calculate new f1 and f2 from the Reynolds calculated. A similar iterative method could be used to solve the implicit Colebrook-White equation, but tools like WolframAlpha can be used.
- Check results. If the magnitude of the difference between the initial guess and calculated friction factor is lower than a threshold required, one can stop here and use these values for the final calculation of head loss and pressure drop with either pipe. Otherwise, use the calculated friction factors as the input for Step 2 and repeat.
As an exercise, try to solve the system shown in Figure 2. It is no longer symmetrical, so lengths and minor losses won’t cancel.