The Circular Pipe component models a circular pipe with losses. The pressure drop is computed with the Darcy equation, with the friction factor determined using the Haaland approximation for turbulent flow. When the Reynolds number is greater than the maximum value for laminar flow but less than the minimum value for turbulent flow, the friction factor is determined using linear interpolation.

$\mathrm{\Re }\=\frac{q\mathrm{D}}{a\mathrm{\nu }}{\mathrm{D}}_h\=4\frac{A}{U}A\=\frac{\mathrm{\pi }{\mathrm{D}}^2}{4}$

$f_L\=\frac{64}{\mathrm{\Re }}{f}_T\=f_{\mathrm{Colebrook}}\left({\mathrm{\Re }}_T\,\frac{\mathrm{\epsilon }}{{\mathrm{D}}_h}\right)$

$\mathrm{mode}\&equals;\{\begin{array}{cc}{\mathrm{pos}}_{\mathrm{turbulent}}& {\mathrm{\Re }}_T<\mathrm{\Re }\\ {\mathrm{neg}}_{\mathrm{turbulent}}& {\mathrm{\Re }}_T<-\mathrm{\Re }\\ {\mathrm{pos}}_{\mathrm{mixed}}& {\mathrm{\Re }}_L<\mathrm{\Re }\\ {\mathrm{neg}}_{\mathrm{mixed}}& {\mathrm{\Re }}_L<-\mathrm{\Re }\\ \mathrm{laminar}& \mathrm{otherwise}\end{array}$

$p\&equals;p_A-p_B\&equals;\frac{1}{2}L\mathrm{\rho }\frac{{\mathrm{\nu }}^2}{{\mathrm{D}}_h^3}\mathrm{\Re }\{\begin{array}{cc}{f}_{\mathrm{Colebrook}}\left(\left|\mathrm{\Re }\right|\&comma;\frac{\mathrm{\epsilon }}{{\mathrm{D}}_h}\right)\left|\mathrm{\Re }\right|& \mathrm{mode}\&equals;{\mathrm{pos}}_{\mathrm{turbulent}}\vee \mathrm{mode}\&equals;{\mathrm{neg}}_{\mathrm{turbulent}}\\ \left(f_L\&plus;\frac{f_T-f_L}{{\mathrm{\Re }}_T-{\mathrm{\Re }}_L}\left(\left|\mathrm{\Re }\right|-{\mathrm{\Re }}_L\right)\right)\left|\mathrm{\Re }\right|& \mathrm{mode}\&equals;{\mathrm{pos}}_{\mathrm{mixed}}\vee \mathrm{mode}\&equals;{\mathrm{neg}}_{\mathrm{mixed}}\\ 64& \mathrm{otherwise}\end{array}$

$q\&equals;q_A\&equals;-q_B\&equals;\mathrm{\Re }A\frac{\mathrm{\nu }}{{\mathrm{D}}_h}$

$f_{\mathrm{Colebrook}}\&equals;\left(\mathrm{\Re }\&comma;{\mathrm{\epsilon }}_{\mathrm{D}}\right)\to {\left(1.8{\log }_{10}{\left(\frac{6.9}{\mathrm{\Re }}\&plus;\left(\frac{{\mathrm{\epsilon }}_{\mathrm{D}}}{3.7}\right)\right)}^{1.11}\right)}^{\mathrm{-2}}$