Question

Draining of cylindrical tank. You have a cylindrical tank full of water with a diameter =Dtank. The height (htank) is changing with time. You are draining the tank through a hole in the bottom. The hole has a diameter Dhole. The velocity of the water leaving the tank depends on the height of the water and can be given as: v2 = 2 g htank. When the hole is first opened, the height of the water is ho. Draw and label the diagram and write out the initial conditions. as a function of time. Write out the material balance on the system. Write an equation for the height Assume that the diameter of the tank is 10 times the diameter of the hole. How many seconds does it take for the height change from 1 m to 0.5 m?

Answer 1

Rough fig ater in tank velocity ef Let the to , hzho Lef the any wefer in tnk h X Hme t-t hight of t-t, hih V2 - dx dt Bal ance Equation -- PA2 V Let Dia of hole: Di-Dhole aMass PA, Vi Dia. of tank: D- Dtank A: Area of hole A2 Area of tank D2 Mass Given DI Balan ce E tonk From e ii) fn e i put the value -dx ct d t tt ntet it! dx at x: WĻ ho t-o function of hight f time - then ho

t ht ho 20D at ht 20て tこ 13.225 sec)

So, Equation of hight as a function of time is given as-

(h_t)^0.5=(h₀)^0.5 - (D_hole/D_tank)²×(t/2)×(2×g)^0.5

Where, 'g' is the acceleration dur to gravity which is taken as 9.81 m/s²