Question

13) A tank of area Ao is draining in laminar flow through a pipe of diameter D and length L, as shown in Fig. P4.92. Neglecting the exit-jet kinetic energy and assuming the pipe flow is driven by the hydrostatic pressure at its entrance, derive a formula for the tank level h ) if its initial level is ho Area Ao V(t)

Answer 1

Given that Draw the tank and draining system Rightarrow write the equation for pressure drop: Delta p = rho gh Here p is the density of the fluid 'g' is he acceleration due to gravity, and is the height of the fluid in the tank \r\n Rightarrow write equation for volume trick fall rate in each tube. Q_out = pi R^4 Delta R/8 mu L ---(1) Here 'R' = the radius of the tube Delta p is pressure drop mu is viscosity of water L is the length of the tube Rightarrow the volumetric flow is also given by the equation: Q_out = -A dh/dt (- A is the cross sectional area of the tank) Substitute -A dh/dt for Q_out is equation (1) - A dh/dt = pi R^4 Delta p/8 mu L \r\n Rightarrow A_0 for A, 0/2 for R rho gh for Delta p sustain above eq^n - A_0 dh/dt = pi (D/2)^4 (rho gh)/8 mu L dh/h = -pi (D/2)^4 (rho g) dt/8 mu L A_0 Rightarrow Integrate both sides: integral ^h_h_0 dh/h = -integral^t_0 pi (D/2)64 (rho g)/8 mu L A_0 dt [ln h]^h_h_0 = [pi (D/2)^4 (rho g)/8 mu L A_0 times t] [ln h - ln h_0] = pi D^4 rho g/128 mu L A_0 times t ln [h/h_0] = pi D64 rho g/128 mu L A_0 times t \r\n