# Velocity of Efflux , Torricelli Theorem

Velocity of Efflux , Torricelli’s Theorem Suppose that a liquid is taken in a container having a large cross-sectional area A1. There is also a small hole of cross-sectional area A2 on the wall of the container.

Suppose that the velocity of efflux of the liquid out of the container is v2.
The velocity of a liquid particle at the surface, at that instant, is v1 . The height of the liquid column at any instant from the position of the hole to the upper end of the tank is h, and the atmospheric pressure is Po

From the principle of continuity, we get

v1A1 = v2A2         ….(1)

Applying Bernoulli’s principle at the free end outside the hole and also at the surface, we get

$\displaystyle P_0 + \frac{1}{2}\rho v_1^2 + \rho g h = P_0 + \frac{1}{2}\rho v_2^2$ …(2)

Therefore, from equations (1) and (2), we get,

$\displaystyle v_2^2 = \frac{2 g h}{[1- (\frac{A_2}{A_1})^2]}$        …(3)

If A2 << A1 then,

$\displaystyle v_2 = \sqrt{2 g h}$         ……(4)

Equations (3) and (4) give the velocity of efflux of a liquid flowing out of a small hole in a tank.

If the surface area of the tank is large compared to the hole

i. e. A1 >> A2 , we can approximate A2/A1 = 0 and the followings conclusions result.

(i) v ∝ √h

(ii) v is independent of the nature of the liquid.

(iii) If the liquid is ejected through the hole horizontally and the hole is at a height h from the upper level of liquid and the total depth of liquid is H, then the time taken by the liquid to reach the ground is ;

$\displaystyle t = \sqrt{\frac{2(H-h)}{g}}$  ……….(5)

Suppose that horizontal distance travelled by the liquid stream is x i.e. the liquid will strikes the ground at a distance x from the base of the container below the hole.

Then, x = v .t

$\displaystyle x = \sqrt{2 g h} . \sqrt{\frac{2(H-h)}{g}}$    …….(6)

This range x will be maximum if $\displaystyle \frac{dx}{dh} = 0$

Now,Squaring eqn(6) ,  x2 = 4h (H – h).

Therefore ,

$\displaystyle 2x \frac{dx}{dh} = 4H – 8h = 0$

Therefore, H = 2h i.e. h = H/2    ……….(7)

Solved Example : Air flows horizontally with a speed v = 108 km/hr. A house has a plane roof of area A = 20 m2. Find the magnitude of aerodynamic lift on the roof.

Solution: Air flows just above the roof and there is no air flow just below the roof within the room.

Therefore v1 = 0 and v2 = v

Applying Bernoulli’s theorem at the points inside and outside the roof, we obtain

$\displaystyle P_1 + \rho g h_1 + \frac{1}{2}\rho v_1^2 = P_2 + \rho g h_2 + \frac{1}{2}\rho v_2^2$

Since h1 = h2 = h, v1 = 0 and v2 = v,

$\displaystyle P_1 = P_2 + \frac{1}{2}\rho v^2$

P1 – P2 = ΔP = (1/2) ρv2

Since the area of the roof is A, the aerodynamic force exerted on it

F = (ΔP) A

F = (1/2) ρAv2

where ρ = density of air = 1.3 kg/m3

A = 20 m2, v = 30 m/sec.

F = {(1/2) × 1.3 × 20 × (30)2} N

F = 1.17 × 104 N