- #1

sowinski

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## Homework Statement

"Consider the tank and water supply system as shown in the figure below. The diameter

of the supply pipe, D1 = 20 mm, and the average velocity leaving the supply pipe is

V1 = 0.595 m/s. A shut-off valve is located at z = 0.1 m, and the exit pipe, D2 = 10 mm.

The tank diameter is Dt = 0.3 m. The density of the water is uniform at 998 kg/m3."

"At the instant the water level reaches H0 = 1 m the shut-off valve is opened. The

instantaneous average velocity of the outflow depends on the water depth above z

and can be expressed by, V2=0.85*sqrt(g*[H(t)-z]) where g is the acceleration due to

gravity. Determine whether the tank continues to fill or begins to empty immediately

after the valve is opened." (Found that it will drain when valve opens)

Determine the steady-state value of the water depth.

(found to be 0.9m)

Determine the time required to achieve steady state after the valve is opened.

(This is where I'm getting stuck)

## Homework Equations

definition of volumetric flow

V_dot=Area*Velocity

continuity for incompressible flow:

V_dot_in-V_dot_out=dV/dt

relationship between tank volume and water level

V=pi*(d/2)^2*H

## The Attempt at a Solution

V_dot_in-V_dot_out=dV/dt

V_dot_in=pi*(d1/2)^2*v1=1.869e-4 m^3/s

V_dot_out=A2*v2=7.854e-5*v2

substituting into continuity equation...

(1.869e-4)-(7.854e-5)(0.85)sqrt(9.81(H(t)-0.1))=dV/dt

now using the relationship between V and H...

(1.869e-4)-(7.854e-5)(0.85)sqrt{9.81[(4*V/0.09*pi)-0.1]}=dV/dt

now simplifying

(1.869e-4)-(6.676e-5)sqrt(139V-0.98)=dV/dt

now I believe I need to move the dt to the other side and move V terms to the RHS

with dV then integrate and solve for t... but I'm not sure how to split everything up

Sorry for the long post, and thanks in advance!