Applications of Bernoulli's equation Knowledge-Based Study

Study notes

1. Conservation of Mechanical Energy in Fluids

   Bernoulli's equation is a statement of the work-energy theorem applied to steady, non-viscous, incompressible fluid flow along a streamline. For an ideal fluid of uniform density $\rho$, the total mechanical energy remains constant between any two points along a streamline:

   $P_1 + \frac{1}{2}\rho v_1^2 + \rho g h_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho g h_2$

   The three terms represent distinct energy contributions per unit volume:

   • Static Pressure ($P$): The thermodynamic pressure of the fluid.
   • Dynamic Pressure ($\frac{1}{2}\rho v^2$): The kinetic energy per unit volume of the moving fluid.
   • Hydrostatic Pressure ($\rho g h$): The potential energy per unit volume associated with elevation $h$.

   When viscous forces (internal friction) are negligible and the flow is irrotational, this sum remains invariant throughout the entire fluid body.

   Flow Through a Constriction

   When an incompressible fluid moves horizontally through a pipe with varying cross-sectional area, elevation remains constant ($h_1 = h_2$). Under this condition, Bernoulli's equation simplifies to:

   $P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2$

   According to the continuity equation, $A_1 v_1 = A_2 v_2$, a fluid must accelerate when entering a constriction (where $A_2 < A_1$, so $v_2 > v_1$). Because kinetic energy density increases, the static pressure must drop ($P_2 < P_1$). This phenomenon is known as the Venturi effect.

   A Venturi meter exploits this pressure difference to determine volume flow rate $Q$:

   $P_1 - P_2 = \frac{1}{2}\rho (v_2^2 - v_1^2) = \frac{1}{2}\rho v_1^2 \left[\left(\frac{A_1}{A_2}\right)^2 - 1\right]$

   By measuring the pressure differential $P_1 - P_2$ using a manometer, the flow velocity and discharge rate can be computed directly without inserting moving mechanical parts into the stream.

   Discharge from an Orifice

   Torricelli's law describes the speed at which liquid escapes from a small aperture in an open tank. Consider a large open reservoir of liquid filled to a depth $h$ above a small drain orifice.

   Applying Bernoulli's equation between the top surface (point 1) and the exit orifice (point 2):

   • Both the upper liquid surface and the exit jet are exposed to the atmosphere, so $P_1 = P_2 = P_{\text{atm}}$.
   • Because the surface area of the reservoir is vastly larger than the orifice area ($A_1 \gg A_2$), the continuity equation dictates that the surface drops negligibly slow: $v_1 \approx 0$.
   • Setting the datum at the orifice level ($h_2 = 0$), the surface elevation is $h_1 = h$.

   Substituting these boundary conditions yields:

   $P_{\text{atm}} + \rho g h = P_{\text{atm}} + \frac{1}{2}\rho v_2^2$

   $v_2 = \sqrt{2gh}$

   The efflux speed $v_2$ is identical to the speed acquired by an object in free fall from height $h$. If the tank is pressurized from above with a gauge pressure $P_g$, the efflux velocity becomes $v_2 = \sqrt{2gh + \frac{2P_g}{\rho}}$.

   Measuring Flow and Airspeed

   A Pitot-static tube measures fluid velocity by determining both static pressure and stagnation pressure.

   • Static Port: Positioned flush with the tube surface parallel to the flow, sensing only the ambient static pressure $P_s$.
   • Stagnation Port: Positioned facing directly into the oncoming stream. At the tip, fluid particles are brought completely to rest ($v = 0$) isentropically.

   Applying Bernoulli's equation along the central streamline between the undisturbed flow and the stagnation point gives:

   $P_{\text{stagnation}} = P_s + \frac{1}{2}\rho v^2$

   The differential pressure between the stagnation and static chambers is directly proportional to the dynamic pressure:

   $\Delta P = P_{\text{stagnation}} - P_s = \frac{1}{2}\rho v^2$

   Solving for the flow speed relative to the instrument:

   $v = \sqrt{\frac{2(P_{\text{stagnation}} - P_s)}{\rho}}$

   Aircraft use heated Pitot-static tubes to provide real-time indicated airspeed measurements across varying altitudes.

   Aspirators and Atomizers

   The local drop in static pressure caused by high flow speed can be harnessed to draw another fluid into the main stream—a process called fluid entrainment.

   • Perfume Atomizers and Spray Bottles: Squeezing an air bulb forces air rapidly over the top of a vertical draw tube submerged in liquid. The reduced pressure at the nozzle tip drops below atmospheric pressure, causing the liquid to rise and be dispersed into fine droplets by the passing air stream.
   • Aspirators (Filter Pumps): High-speed water running through a constriction creates a low-pressure region that draws air out of an attached vessel, generating a partial vacuum.
   • Chimneys: Wind blowing across the top of an open chimney reduces the external pressure relative to the quiescent air inside the building, enhancing the upward draft of exhaust gases.

   Note on Aerodynamic Lift: While pressure differences across curved streamlines contribute significantly to lift on an airfoil, real lift involves downward momentum deflection and fluid circulation, rather than the common misconception of equal transit times across upper and lower surfaces.