Question

EXERCISE NR. 2 A constant density fluid flows into a perfect cone tank as pictured below. Showing all steps, derive the dynamic ordinary differential equation (ODE) de- scribing liquid height in the tank. Express your result in the form da. Exit flow, F(t), should be assumed to be proportional to the square root of the hydrostatic head, and the geometry of the tank indicates that = 0 Page 1 of 3 Dr. Jawhar Ghommam 22-04-2018 Fo(t) - h(t) ► F(t)

Answer 1

@ solution: L filt) Given that Change in volume perunit time will be the difference of input & exit flow Hence dult Hence duct? – F, (+) – f (+) Here Fo (t) input flowe f (t) = Enit flow but Filt) & Jolt) = F, (t) = k/h(t) (given) Hince dv (+) Elt) - k hlt) d we know the volume of cone = 4TR #

alow, volume of water at any time It) ir VC). " (ow))"m[] in the above equation both re) 4 h (t) are Manžiables related as >> (lt) h48) XR Λίσω (1). (heo)”. so v(e) becomes V(t) = _ " (Mt 17x " x hee vet) * (heeyyen differentiate on both sides dre - 4 1 2 [h ces de dule) = H (hCAJPR" dhe at " HV de putting in eruation

why we foll). A vot © ((49ces) fole? - KC647 Y> V 12 to (t)- k Th (t)]"2 te) H x [5 Ct)]" to (t) HN - KH a ah(t)