Section 1.8: Advanced Problems (Fluids)
These problems are designed to stretch your understanding and integrate multiple principles of fluid mechanics. Expect multi-step reasoning and careful application of formulas.
Problem 1
A large tank is filled with water to a depth of 4.0 m. A hole of radius 0.01 m is opened at a depth of 1.0 m from the bottom. Find the speed and volume flow rate of water exiting the hole.
Height of water above hole: \( h = 3.0 \,\text{m} \).
Torricelli’s law: \( v = \sqrt{2gh} = \sqrt{2 \cdot 9.8 \cdot 3.0} \approx 7.67 \,\text{m/s} \).
Area of hole: \( A = \pi (0.01)^2 = 3.14 \times 10^{-4} \,\text{m}^2 \).
Flow rate: \( Q = Av = 3.14 \times 10^{-4} \cdot 7.67 \approx 2.41 \times 10^{-3} \,\text{m}^3/s \).
Problem 2
An airplane wing of area \( 20 \,\text{m}^2 \) experiences air moving over the top surface at 250 m/s and below at 230 m/s. Assume air density is \( 1.2 \,\text{kg/m}^3 \). Estimate the lift force.
Bernoulli: \( \Delta P = \tfrac{1}{2}\rho (v_\text{top}^2 - v_\text{bottom}^2) \).
\( = 0.5 \cdot 1.2 \cdot (250^2 - 230^2) = 0.6 \cdot (62500 - 52900) = 0.6 \cdot 9600 = 5760 \,\text{Pa} \).
Lift = \( \Delta P \cdot A = 5760 \cdot 20 \approx 1.15 \times 10^5 \,\text{N} \).
Problem 3
A cube of edge 0.40 m floats in water with half its volume submerged. Then oil of density \( 800 \,\text{kg/m}^3 \) is poured on top, fully covering the cube. How much of the cube’s height is submerged now?
Weight = \( \rho_w g V/2 = 1000 \cdot 9.8 \cdot (0.40^3)/2 \).
Let submerged fraction = \( f \). Buoyant force = \( (\rho_w h_w + \rho_o h_o) g A \).
Careful balancing gives \( f \approx 0.625 \). Thus 62.5% of cube’s height is submerged.
Problem 4
Water flows in a horizontal pipe with pressure 2.0 × 105 Pa at speed 2.0 m/s. The pipe narrows and rises 3.0 m higher, where the pressure is 1.5 × 105 Pa. Find the speed at the higher point. (\( \rho = 1000 \,\text{kg/m}^3 \))
Bernoulli: \( P_1 + \tfrac{1}{2}\rho v_1^2 + \rho g y_1 = P_2 + \tfrac{1}{2}\rho v_2^2 + \rho g y_2 \).
Substitute: \( 2.0\times10^5 + 0.5(1000)(2^2) + 0 = 1.5\times10^5 + 0.5(1000)v_2^2 + 1000\cdot9.8\cdot3 \).
\( 200000 + 2000 = 150000 + 500v_2^2 + 29400 \).
\( 202000 - 179400 = 500v_2^2 \Rightarrow 22600 = 500v_2^2 \).
\( v_2 \approx 6.73 \,\text{m/s} \).