Suppose that a liquid is taken in a container having a large cross-sectional area A_{1}. There is also a small hole of cross-sectional area A_{2} on the wall of the container.

Suppose that the velocity of efflux of the liquid out of the container is v_{2}.

The velocity of a liquid particle at the surface, at that instant, is v_{1} . 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 P_{o}

From the principle of continuity, we get

v_{1}A_{1} = v_{2}A_{2} ….(i)

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 $ …(ii)

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

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

If A_{2} << A_{1} then,

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

Equations (iii) and (iv) 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. A_{1} >> A_{2} , we can approximate A_{2}/A_{1} = 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}} $ ……….(v)

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}} $ …….(vi)

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

Now,Squaring eqn(vi) , x^{2} = 4h (H – h).

Therefore ,

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

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

Solved Example : Air flows horizontally with a speed v = 108 km/hr. A house has a plane roof of area A = 20 m^{2}. 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 v_{1} = 0 and v_{2} = 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 h_{1} = h_{2} = h, v_{1} = 0 and v_{2} = v,

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

P_{1} – P_{2} = ΔP = (1/2) ρv^{2}

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

F = (ΔP) A

F = (1/2) ρAv^{2}

where ρ = density of air = 1.3 kg/m^{3}

A = 20 m^{2}, v = 30 m/sec.

F = {(1/2) × 1.3 × 20 × (30)^{2}} N

F = 1.17 × 10^{4} N