Hidrouliese diameter

Diameter = ${\displaystyle d}$

${\displaystyle A={\frac {\pi }{4}}d^{2}}$

${\displaystyle P=\pi d}$

${\displaystyle D_{H}={\frac {4A}{P}}={\frac {4\times {\frac {\pi }{4}}d^{2}}{\pi d}}=d}$

${\displaystyle D_{ekwivalent}={\sqrt {\frac {4A}{\pi }}}={\sqrt {\frac {4{\frac {\pi }{4}}d^{2}}{\pi }}}=d}$

Lengte van sye = ${\displaystyle L}$

${\displaystyle A=L^{2}}$

${\displaystyle P=4L}$

${\displaystyle D_{H}={\frac {4A}{P}}={\frac {4L^{2}}{4L}}=L}$

${\displaystyle D_{ekwivalent}={\sqrt {\frac {4A}{\pi }}}={\sqrt {\frac {4L^{2}}{\pi }}}=1.128L}$

Lengte = ${\displaystyle L}$ en breedte = ${\displaystyle B}$

${\displaystyle A=LB}$

${\displaystyle P=2(L+B)}$

${\displaystyle D_{H}={\frac {4A}{P}}={\frac {4LB}{2(L+B)}}={\frac {2LB}{L+B}}}$

${\displaystyle D_{ekwivalent}={\sqrt {\frac {4A}{\pi }}}={\sqrt {\frac {4LB}{\pi }}}=1.128{\sqrt {LB}}}$

Ook kanaal.

${\displaystyle A=H\times W}$

${\displaystyle P=2H+W}$

${\displaystyle R_{H}={\frac {A}{P}}={\frac {HW}{2H+W}}}$

Half vol pyp.

Beskou prentjie aan die regterkant vir 'n halfvol pyp:

Area van driehoek:

${\displaystyle r={\frac {D}{2}}}$            ${\displaystyle a=H-r}$

${\displaystyle a^{2}+b^{2}=r^{2}}$ (Volgens Pythagoras)

${\displaystyle \left(H-r\right)^{2}+b^{2}=r^{2}}$

${\displaystyle H^{2}-2Hr+r^{2}+b^{2}=r^{2}}$

${\displaystyle b^{2}=2Hr-H^{2}}$

${\displaystyle b={\sqrt {2Hr-H^{2}}}}$

${\displaystyle Area\ van\ driehoek={\frac {1}{2}}a.2b=ab=\left(H-r\right)\cdot {\sqrt {2Hr-H^{2}}}}$

${\displaystyle Area\ van\ driehoek=\left(H-{\frac {D}{2}}\right)\cdot {\sqrt {HD-H^{2}}}}$

Area van sirkelsegment:

${\displaystyle Area\ van\ sirkelsegment={\frac {\theta }{360}}\cdot {\frac {\pi }{4}}D^{2}}$

${\displaystyle \sin \alpha ={\frac {a}{r}}}$

${\displaystyle \alpha =\arcsin \left({\frac {H-r}{r}}\right)=\arcsin \left({\frac {H-{\frac {D}{2}}}{\frac {D}{2}}}\right)=\arcsin \left({\frac {2H-D}{D}}\right)}$

${\displaystyle \theta +\left(180-2\alpha \right)=360}$            ${\displaystyle \theta =2\alpha +180}$

${\displaystyle \theta =2\alpha +180=2\arcsin \left({\frac {2H-D}{D}}\right)+180}$

${\displaystyle \theta =2\arcsin \left({\frac {2H-D}{D}}\right)+180}$

Totale area:

${\displaystyle Totale\ area\ (A)=\left(H-{\frac {D}{2}}\right)\cdot {\sqrt {HD-H^{2}}}+{\frac {\theta }{360}}\cdot {\frac {\pi }{4}}D^{2}}$

${\displaystyle Benatte\ omtrek\ (P)={\frac {\theta }{360}}\cdot \pi D}$

${\displaystyle \theta =2\arcsin \left({\frac {2H-D}{D}}\right)+180}$

Om ${\displaystyle \theta }$ in Excel te bereken, gebruik die volgende formule:

"=2*DEGREES(ASIN((2*B4-B3)/B3)) + 180"

Hitteruiler vierkantige buiskonfigurasie.

Die hidrouliese diameter word uitgewerk vir hitteruilers aan die mantelkant om die drukval oor die mantelkant en ook die filmkoëffisiënt te bepaal vir die berekening van die warmteoordragskoëffisiënt (U).

Hier volg die berekening vir 'n vierkantige buiskonfigurasie van 'n hitteruilerː

${\displaystyle A=p^{2}-{\frac {4\times 90}{360}}\cdot {\frac {\pi }{4}}d^{2}=p^{2}-{\frac {\pi }{4}}d^{2}}$

${\displaystyle P={\frac {4\times 90}{360}}\pi d=\pi d}$

${\displaystyle d_{e}={\frac {4A}{P}}={\frac {4\left(p^{2}-{\frac {\pi }{4}}d^{2}\right)}{\pi d}}={\frac {4}{\pi d}}\left(p^{2}-{\frac {\pi }{4}}d^{2}\right)}$

${\displaystyle d_{e}={\frac {1.27}{d}}\left(p^{2}-0.785d^{2}\right)}$

Die hidrouliese diameter word uitgewerk vir hitteruilers aan die mantelkant om die drukval oor die mantelkant en ook die filmkoëffisiënt te bepaal vir die berekening van die warmteoordragskoëffisiënt (U).

Hier volg die berekening vir 'n driehoekige buiskonfigurasie van 'n hitteruilerː

Hitteruiler driehoekige buiskonfigurasie.

${\displaystyle p^{2}=h^{2}+\left({\frac {p}{2}}\right)^{2}}$            ${\displaystyle h^{2}=p^{2}-{\frac {p^{2}}{4}}={\frac {3}{4}}p^{2}}$            ${\displaystyle h={\frac {\sqrt {3}}{2}}p}$

${\displaystyle A={\frac {1}{2}}ph-{\frac {3\times 60}{360}}\cdot {\frac {\pi }{4}}d^{2}={\frac {\sqrt {3}}{4}}p^{2}-{\frac {\pi }{8}}d^{2}}$

${\displaystyle P={\frac {3\times 60}{360}}\pi d={\frac {1}{2}}\pi d}$

${\displaystyle d_{e}={\frac {4A}{P}}={\frac {4\left({\frac {\sqrt {3}}{4}}p^{2}-{\frac {\pi }{8}}d^{2}\right)}{{\frac {1}{2}}\pi d}}}$

${\displaystyle d_{e}={\frac {8}{\pi }}{\frac {\sqrt {3}}{4}}{\frac {1}{d}}\left(p^{2}-{\frac {\pi }{8}}{\frac {4}{\sqrt {3}}}d^{2}\right)={\frac {2{\sqrt {3}}}{\pi d}}\left(p^{2}-{\frac {\pi }{2{\sqrt {3}}}}d^{2}\right)}$

${\displaystyle d_{e}={\frac {1.10}{d}}\left(p^{2}-0.907d^{2}\right)}$