0
$\begingroup$

Given the following situation: in

Pipe A and B have a known and identical length and diameter, and are in parallel. I'm assuming the flow trough pipes A and B are also identical, as is the roughness of the pipes. Pipe C also has a known length and diameter (which is different from the diameter of pipe A/B).

I'm interested in the total head loss at the end of Pipe C. I am aware of the following Darcy-Weisbach equations that are relevant in this case (parallel and series):

enter image description here

Here the total headloss is the sum of the headloss per pipe, placed in series.

enter image description here

Here the total headloss over set of 2 parallel pipes is equal to the headloss of one of them.

But, how do I combine these two equations to retrieve the total headloss over pipes A, B and C?

$\endgroup$
0
$\begingroup$

The total headloss through pipes A+C is equal to the headloss through pipes B+C. Therefore, you only need to calculate one of the paths, and your first equation gives you the desired total headloss (where 1 and Q in the first term are for pipe A, and 2 and Q in the second term are for pipe C).

Of course, the total headloss is equal to the pressure difference that is driving the flow. For example, if the tank is filled with a liquid to depth H, and if the end of the pipe C is opened to the atmosphere, the pressure driving the flow due to the elevation difference is known. In other words, you usually know the total headloss because you know the pressure difference, and the calculation is to find the flow rate. You can determine that from the first equation because you are assuming that $Q_A=0.5Q_C$.

In addition to the Darcy-Weisbach wall losses, there are also losses due to the inlets into pipes A+B, the tee where A+B combine into C, and the exit loss at the end of C. These may be significant depending on the flow rate and length of pipes.

$\endgroup$
0

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.