MATLAB: How is the equivalent length contributing to the pressure losses in a pipe in Simscape Fluids

aggregateequivalentfluidsgasheadlengthlocallossespipepressureresistancesSimscape

There is something in the following page from the documentation that I would like to know more about:
https://www.mathworks.com/help/releases/R2018b/physmod/simscape/ref/pipeg.html?s_tid=mwa_osa_a#bvd_i7y-1
Normally, the equation for the pressure loss in a pipe (I will use the turbulent case as an example) should look like:
Delta_p = f * ( (m_dot^2) / (2 * rho * D * S^2) ) * Length
(I am putting the Length in the end for convenience)
In the case of Simscape Fluids, according to the formula in the doc:
Length = (L + L_eqv) / 2
So my question is, does L_eqv stand only for the equivalent length of resistances, or the actual length plus the equivalent length?
If it only stands for the length of resistances, it seems that the length used for the calculations is being reduced by roughly a factor of two (with respect to what I have seen in the literature) by doing the mean of both, thus I would be seeing less pressure losses than he expected.
Is this correct?

Best Answer

L_eqv is defined to be the aggregate equivalent length of local resistances. It stands for the combined length of all local resistances present in the pipe (these including fittings, bends, armatures, pipe inlets and outlets).
The effect of the local resistances is to increase the effective length of the pipe segment. This length is added to the geometrical pipe length only for friction calculations. The gas volume depends only on the pipe geometrical length, defined by the pipe length parameter.
Therefore, the total length would be:
total length = pipe length + aggregate equivalent length = L + L_eqv
The reason that the total length seems to be instead half of it (apparently divided by the factor of 2 that we were observing) is that the pressure difference equation between inlet and outlet ports is also divided into two parts, since the pipe block has volume of fluid:
  1. Pressure difference between port A and the internal node I: p_A - p_I = (L+L_eqv)/2 * ...
  2. Pressure difference between port B and the internal node I: p_B - p_I = (L+L_eqv)/2 * ......
This is used for the calculations in the node "I", so it goes from A to I first, and then from I to B, let us say. The length is split into two in a sense.
So the total pressure difference across the ports will include the pipe total length, i.e., L+L_eqv, therefore not "losing" that half pipe that seemed to happen from the factor of 2 issue.
Related Question