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Hello. I am trying to model a thermosiphon flat-plate solar thermal collector with a horizontal storage tank at the top. I have selected Type 45a for this purpose. When it comes to the friction head loss in the pipes, in the mathematical references (page 4-511), Eq. 4.9.2-5 indicates that the friction factor is: f = 64/ Re for Re < 2000 and f = 0.032 for Re > 2000. Now, ten (10) lines below, on the same page, it is indicated:
f = f[1 + .038/(L/Re/d)^964]
1/ This is confusing for two reasons. First, f appears twice in the equation, meaning that [1 + .038/(L/Re/d)^964] equals one, which is probably not correct. In addition, is this friction factor f different from the one of the Eq. 4.9.2-5 given above 64/Re and 0.032?
2/ Finally, if I want to use another correlation, for instance f = (0.79*ln(Re- 1.64)^(-2), as recently did Kalogirou et al. (2017), what would be the easiest way to proceed?
Thank you.
I believe that equation 4.9.2-5 applies to the collector inlet and outlet pipes whereas the other equation applies to the pipes and risers that connect the inlet to the outlet. In other words the second equation accounts for additional friction beyond that accounted for by 4.9.2-5.
In order to change the algorithm that the component uses you would need to modify the Type45 source code and either recompile the TRNDll.dll. Issues of changing source code and recompiling are treated in the documentation set in 07-ProgrammersGuide.pdf
kind regards,
David
Ken Topic starter
28/11/2023 1:47 am
@davidbradley Thank you for your feddback. I will try to modify the code source. I know it is sometimes very challenging.
But definitely, there is a problem with the equation f = f[1 + .038/(L/Re/d)^964] where f appears on both side of the equation. Thanks
A_Weiss 30/11/2023 4:35 am
@ken I think David is suggesting that, in the Fortran code, the value f is calculated first as either 64/Re or 0.032, then it is further modified by the second equation. In other words, f = (64/Re)[1 + .038/(L/Re/d)^964] when Re is less than 2000, and f = (0.032)[1 + .038/(L/Re/d)^964] when Re is greater than 2000. It's no different than incrementing an integer i from 1 to 2 by writing i=1, then i=i+1 in successive lines of code.
Ken Topic starter
06/12/2023 6:12 pm
@A_Weiss. Thank you for your feedback. Actually, when Re is less than 2000, f = Re/64 for friction factor calculation for laminar flow in a circular pipe. Equation f = (64/Re)[1 + .038/(L/Re/d)^964], to my opinion, could not be correct, at least for Re less than 2000. For sure, there is a problem with the equations in the section. I do not mean it to be solved here but it could be worth looking closely at it for correction later. Thanks
