Question

Suppose that a tank containing a certain liquid has an outlet near the bottom. Let h(t) denote the height of the liquid's surface above the outlet. Torricelli's principle states that the outflow velocity v at the outlet is equal to the velocity of a particle falling freely (with no drag) from the height h (a) Show that v2gh, where g is the acceleration due to gravity. (b) By equating the rate of outflow to the rate of change of liquid in the tank, show that h(t) satisfies the equation dh dt where A(h) is the area of the cross section of the tank at height h and a is the area of the outlet. The constant α is a "contraction coefficient" that accounts for the observed fact that the cross section of the (smooth) outflow stream is smaller than a. The value of a for water is about 0.6 c) Consider a water tank in the form of a right circular cylinder that is 3m high above the outlet. The radius of the tank is 1m and the radius of the circular outlet is 0.1m If the tank is initially full of water, determine how long it takes to drain the tank down to the level of the outlet

Answer 1

(a) Using Bernoulli's principle between put (1) & (2) p_1 + 1/2 rho v^2_1 + rho gh_1 = p_2 + 1/2 pv^2_2 + rho gh_2 Therefore v_1 = (pi a^2/A) v (using equation of i.e A_1 V_1 = A_2V_2) Therefore A a v_1 alpha 0, v_2 = v, h_1 - h_2 = h p_0 + 1/2 rho times 0^2 + rho gh = p_0 + 1/2 rho v^2 + 0 v = squareroot 2gh (b) Rate of decreases of water equal in tank = -A(h) dh/dt (1) Rate flow of water outside from tank = d/dt [alpha a squareroot 2gh middot t] (2) = alpha a squareroot 2gh [Area of = alpha a] Therefore Rate of decrease of water real in = Rate of flow of outside from tank A(h) dh/dt = -alpha a squareroot 2gh (c) Because A(h) dh/dt = -alpha a squareroot 2gh A(h) = pi(1)^2 = pi, alpha = 0.6 (for) a = pi(0.1)^2 = (0.01)