Efflux time for draining a conical tank (Fig. 7B.10), a conical tank, with dimensions given in the figure, is initially filled with a liquid. The liquid is allowed to drain out by gravity. Determine the efflux time. In parts (a)-(c) take the liquid in the cone to be the "system." 

(a) First use an unsteady macroscopic mass balance to show that the exit velocity is v₂ = – z²/z²₂ dz/dt 

(b) Write the unsteady-state mechanical energy balance for the system. Discard the viscous loss term and the term containing the time derivative of the kinetic energy, and give reasons for doing so. Show that this leads to v₂ =√2g (z – z₂).

(c) Combine the results of (a) and (b). Solve the resulting differential equation with an appropriate initial condition to get the liquid level z as a function of to From this get the effiux time t_efflux = 1/5(z₀/z₂)² √2z₀/g list all the assumptions that have been made and discuss how serious they are. How could these assumptions be avoided? 

(d) Rework part (b) by choosing plane I to be stationary and slightly below the liquid surface at time t. It is understood that the liquid surface does not go below plane 1 during the differential time interval dt over which the unsteady mechanical energy balance is made. With this choice of plane I the derivative dΦ_tot/dt is zero and there is no work term Win. Furthermore the conditions at plane I are very nearly those at the liquid surface. Then with the pseudo-steady-state approximation that the derivative dKtot/dt is approximately zero and the neglect of the viscous loss term, the mechanical energy balance, with W₁ – W₂ takes the form 0 = ½ (v²₁ – v²₂) + g(h₁ – h₂)

Transcribed Image Text:

-z = Z0 Liquid surface at time t z =Z z= Z2 -z = 0