Question

A large tank is filled with three liquids of densities ρ, 2ρ and 3ρ up to a height of ^h ⁄ 3 each. What will be the expression for the instantaneous velocity of discharge through a small opening at the base of the tank? (assume that the diameter of the opening is negligible compared to the height of the liquid column)

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My question is from Hydrostatic Force on Plane Area topic in division Hydrostatic Forces on Surfaces of Fluid Mechanics

Answer 1

The instantaneous velocity of discharge through a small opening at the base of the tank can be found using Torricelli's law, which is derived from Bernoulli's principle. The velocity is determined by the potential energy of the liquid column above the opening.

Given:

• The tank is filled with three layers of liquids.
• The densities of the liquids are ρ\rhoρ, 2ρ2\rho2ρ, and 3ρ3\rho3ρ, and each liquid has a height of h3\frac{h}{3}3h​.
• The total height of the liquid column is hhh.
• The diameter of the opening is negligible.

Steps:

1. Total Pressure at the Base: The total pressure at the base of the tank is the sum of the pressures exerted by each liquid layer.

   The pressure exerted by a liquid column is given by:

   P=ρghP = \rho g hP=ρgh

   where:

   • ρ\rhoρ is the density of the liquid,
   • ggg is the acceleration due to gravity,
   • hhh is the height of the liquid column.

   For the three layers:

   • The pressure due to the first liquid layer (density ρ\rhoρ, height h/3h/3h/3) is: P1=ρg(h3)P_1 = \rho g \left(\frac{h}{3}\right)P1​=ρg(3h​)
   • The pressure due to the second liquid layer (density 2ρ2\rho2ρ, height h/3h/3h/3) is: P2=2ρg(h3)P_2 = 2\rho g \left(\frac{h}{3}\right)P2​=2ρg(3h​)
   • The pressure due to the third liquid layer (density 3ρ3\rho3ρ, height h/3h/3h/3) is: P3=3ρg(h3)P_3 = 3\rho g \left(\frac{h}{3}\right)P3​=3ρg(3h​)

   Total pressure at the base is the sum of these pressures:

   Ptotal=P1+P2+P3=ρg(h3)+2ρg(h3)+3ρg(h3)P_{\text{total}} = P_1 + P_2 + P_3 = \rho g \left(\frac{h}{3}\right) + 2\rho g \left(\frac{h}{3}\right) + 3\rho g \left(\frac{h}{3}\right)Ptotal​=P1​+P2​+P3​=ρg(3h​)+2ρg(3h​)+3ρg(3h​)

   Simplifying:

   Ptotal=ρg(h3)(1+2+3)=ρg(h3)×6=2ρghP_{\text{total}} = \rho g \left(\frac{h}{3}\right) (1 + 2 + 3) = \rho g \left(\frac{h}{3}\right) \times 6 = 2 \rho g hPtotal​=ρg(3h​)(1+2+3)=ρg(3h​)×6=2ρgh

2. Instantaneous Velocity of Discharge: According to Torricelli's law, the instantaneous velocity of discharge vvv from the opening at the base is given by:

   v=2Ptotalρv = \sqrt{\frac{2 P_{\text{total}}}{\rho}}v=ρ2Ptotal​​​

   Substituting the value of PtotalP_{\text{total}}Ptotal​:

   v=2×2ρghρ=4ghv = \sqrt{\frac{2 \times 2 \rho g h}{\rho}} = \sqrt{4 g h}v=ρ2×2ρgh​​=4gh​

   Therefore, the expression for the instantaneous velocity of discharge is:

   v=2ghv = 2 \sqrt{g h}v=2gh​

This is the velocity of discharge through the small opening at the base of the tank.