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Bernoulli Equation Reservoir Flow Example

Medium DifficultyFE Fluid Mechanics

Calculate exit velocity and flow rate from a reservoir using Bernoulli's equation.

[Diagram: Reservoir with pipe and nozzle - to be added]

Concept

Bernoulli's equation is a statement of energy conservation for flowing fluids. It relates pressure, velocity, and elevation at different points along a streamline: z_{1} + \frac{V _{1}^{2}}{2 g} + \frac{P _{1}}{ρ g} = z_{2} + \frac{V _{2}^{2}}{2 g} + \frac{P _{2}}{ρ g} z = elevation above datum (m) V^{2} / (2 g) = velocity head (m) P / (ρ g) = pressure head (m) Torricelli's theorem: For flow from a reservoir to atmosphere, exit velocity depends only on elevation difference: V = 2 g h

Water flows from a large reservoir through a pipe and exits to the atmosphere through a nozzle. The water surface in the reservoir is h = 15 m above the nozzle exit. The pipe diameter is D_{1} = 100 mm and the nozzle exit diameter is D_{2} = 50 mm. Find: Exit velocity V_{2} (m/s) Volumetric flow rate Q (m³/s) Neglect friction losses.

h = 15 m D_{1} = 100 mm = 0.1 m D_{2} = 50 mm = 0.05 m Atmospheric pressure at reservoir surface and nozzle exit Negligible friction losses g = 9.81 m/s^{2} ρ = 1000 kg/m^{3}

Between reservoir surface (point 1) and nozzle exit (point 2): z_{1} + \frac{V _{1}^{2}}{2 g} + \frac{P _{1}}{ρ g} = z_{2} + \frac{V _{2}^{2}}{2 g} + \frac{P _{2}}{ρ g}

Apply the following simplifications: P_{1} = P_{2} = P_{a t m} (both exposed to atmosphere, cancel out) V_{1} \approx 0 (large reservoir, negligible velocity) z_{1} - z_{2} = h = 15 m The equation simplifies to: h = \frac{V _{2}^{2}}{2 g}

Rearrange to solve for V_{2} (Torricelli's theorem): V_{2} = 2 g h V_{2} = 2 \times 9.81 \times 15 V_{2} = 294.3 V_{2} = 17.15 m/s

Flow rate equals velocity times cross-sectional area at the exit: Q = V_{2} \times A_{2} A_{2} = \frac{π D _{2}^{2}}{4} A_{2} = \frac{π ( 0.05 ) ^{2}}{4} = 0.00196 m^{2} Q = 17.15 \times 0.00196 Q = 0.0336 m^{3} / s = 33.6 L/s

Check that our assumption V_{1} \approx 0 was reasonable using continuity: A_{1} V_{1} = A_{2} V_{2} V_{1} = (\frac{A _{2}}{A _{1}}) V_{2} = (\frac{0.05}{0.1})^{2} \times 17.15 V_{1} = 0.25 \times 17.15 = 4.29 m/s Since V_{1} = 4.29 m/s is much smaller than V_{2} = 17.15 m/s, neglecting V_{1} introduces only about 6% error, which validates our simplification for this problem.

V_{2} = 17.15 m/s Q = 0.0336 m^{3} / s = 33.6 L/s Note: Exit velocity depends only on elevation difference when friction is neglected (Torricelli's theorem).

Bernoulli's Equation: z_{1} + \frac{V _{1}^{2}}{2 g} + \frac{P _{1}}{ρ g} = z_{2} + \frac{V _{2}^{2}}{2 g} + \frac{P _{2}}{ρ g} Torricelli's Theorem (special case): V = 2 g h where h = elevation difference from reservoir surface to exit