To determine the instantaneous velocity of discharge through a small opening at the base of the tank, when the tank is filled with three different liquids of densities ρ1\rho_1ρ1, ρ2\rho_2ρ2, and ρ3\rho_3ρ3 up to heights h1h_1h1, h2h_2h2, and h3h_3h3, we can use Torricelli’s Law and Bernoulli’s principle.
Step 1: Understanding the problem
The liquid column consists of three layers, and the total pressure at the opening at the base of the tank is the sum of the pressures exerted by each liquid layer. The pressure at the base of the liquid column is the cumulative effect of the pressures from all the individual liquid columns. The pressure at the base due to a liquid column is given by the equation:
P=ρghP = \rho g hP=ρgh
where:
- PPP is the pressure at the base due to a liquid column,
- ρ\rhoρ is the density of the liquid,
- ggg is the acceleration due to gravity,
- hhh is the height of the liquid column.
Step 2: Calculate the total pressure at the base
For three layers of liquid, the total pressure at the base is the sum of the pressures due to each layer:
- Pressure from liquid 1 (with density ρ1\rho_1ρ1 and height h1h_1h1):
P1=ρ1gh1P_1 = \rho_1 g h_1P1=ρ1gh1
- Pressure from liquid 2 (with density ρ2\rho_2ρ2 and height h2h_2h2):
P2=ρ2gh2P_2 = \rho_2 g h_2P2=ρ2gh2
- Pressure from liquid 3 (with density ρ3\rho_3ρ3 and height h3h_3h3):
P3=ρ3gh3P_3 = \rho_3 g h_3P3=ρ3gh3
Thus, the total pressure at the base PtotalP_{\text{total}}Ptotal is:
Ptotal=P1+P2+P3=ρ1gh1+ρ2gh2+ρ3gh3P_{\text{total}} = P_1 + P_2 + P_3 = \rho_1 g h_1 + \rho_2 g h_2 + \rho_3 g h_3Ptotal=P1+P2+P3=ρ1gh1+ρ2gh2+ρ3gh3
Step 3: Use Torricelli’s Law to find the velocity
The instantaneous velocity of discharge through the opening at the base can be determined using Torricelli’s law, which is derived from Bernoulli’s principle:
v=2Ptotalρv = \sqrt{\frac{2 P_{\text{total}}}{\rho}}v=ρ2Ptotal
In this case, the velocity of discharge is:
v=2(ρ1gh1+ρ2gh2+ρ3gh3)ρeffv = \sqrt{\frac{2 (\rho_1 g h_1 + \rho_2 g h_2 + \rho_3 g h_3)}{\rho_{\text{eff}}}}v=ρeff2(ρ1gh1+ρ2gh2+ρ3gh3)
where ρeff\rho_{\text{eff}}ρeff is the density of the liquid. If the tank is considered to be filled with a uniform liquid of density ρ\rhoρ, the velocity will be based on the combined effects of the different layers, and the combined density will influence the effective velocity.
For a single liquid of uniform density (approximating the system in the absence of complex density variations), you could simplify the equation, but in general, you would use the specific liquid densities involved, as we did above.
Final Expression:
The final expression for the instantaneous velocity of discharge vvv is:
v=2g(ρ1h1+ρ2h2+ρ3h3)ρeffv = \sqrt{\frac{2 g (\rho_1 h_1 + \rho_2 h_2 + \rho_3 h_3)}{\rho_{\text{eff}}}}v=ρeff2g(ρ1h1+ρ2h2+ρ3h3)
This equation considers the total pressure exerted by the three liquid layers, and provides the instantaneous discharge velocity based on the height and density of each liquid layer.