Bernoulli's equation Knowledge-Based Study

Study notes

1. Bernoulli's equation is a mathematical statement of the work-energy theorem applied to fluid flow. For the equation to hold strictly, the fluid must satisfy the conditions of an ideal fluid along a streamline:

   • Incompressible: The fluid density $\rho$ remains constant throughout the flow field.
   • Non-viscous (Inviscid): Internal fluid friction is negligible, meaning no mechanical energy is dissipated into thermal energy.
   • Steady Flow: The velocity, pressure, and density at any fixed spatial coordinate do not change with time.
   • Irrotational Flow: Individual fluid elements possess no net angular velocity about their own centers of mass.

   Under these conditions, mechanical energy is conserved along any continuous streamline.

   Consider an ideal fluid moving between two points along a streamline, labeled 1 and 2. The total mechanical energy per unit volume remains constant:

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

   Each term in this equation represents an energy density (energy per unit volume, with SI units of $\text{J/m}^3$ or $\text{N/m}^2 = \text{Pa}$):

   • Static Pressure ($P$): The thermodynamic pressure exerted by the fluid on its surroundings.
   • Dynamic Pressure ($\frac{1}{2}\rho v^2$): The kinetic energy per unit volume associated with the bulk fluid motion.
   • Hydrostatic Potential Term ($\rho g h$): The gravitational potential energy per unit volume relative to an arbitrary reference height.

   When combined with the continuity equation ($A_1 v_1 = A_2 v_2$), Bernoulli's equation explains fundamental fluid phenomena:

   1. The Venturi Effect (Horizontal Flow)

   If the flow occurs along a horizontal pipe ($h_1 = h_2$), the gravitational terms cancel out:

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

   When fluid enters a constricted section ($A_2 < A_1$), mass conservation requires its speed to increase ($v_2 > v_1$). Consequently, the static pressure must decrease ($P_2 < P_1$). A region of higher fluid speed corresponds to a region of lower static pressure.

   2. Torricelli's Law

   Consider a large open reservoir filled to a depth $h$ with a small orifice at the bottom. Since both the top surface and the exiting jet are open to atmospheric pressure ($P_1 = P_2 = P_{\text{atm}}$) and the surface drops at a negligible speed ($v_1 \approx 0$), Bernoulli's equation reduces to:

   $v = \sqrt{2gh}$

   This indicates that the efflux speed from the orifice equals the speed an object would attain in free fall from height $h$.

   Problem: Water with density $\rho = 1000\text{ kg/m}^3$ flows steadily through a horizontal pipe. At section 1, the cross-sectional area is $A_1 = 4.0 \times 10^{-2}\text{ m}^2$, the flow speed is $v_1 = 2.0\text{ m/s}$, and the static pressure is $P_1 = 1.5 \times 10^5\text{ Pa}$. At section 2, the pipe constricts to $A_2 = 1.0 \times 10^{-2}\text{ m}^2$. Determine the static pressure $P_2$ at section 2.

   Step 1: Determine the velocity at section 2 using the continuity equation.

   $A_1 v_1 = A_2 v_2 \implies v_2 = \frac{A_1}{A_2} v_1 = \left(\frac{4.0 \times 10^{-2}}{1.0 \times 10^{-2}}\right)(2.0\text{ m/s}) = 8.0\text{ m/s}$

   Step 2: Apply Bernoulli's equation for horizontal flow ($h_1 = h_2$).

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

   Rearranging for $P_2$:

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

   $P_2 = 1.5 \times 10^5 + \frac{1}{2}(1000)(2.0^2 - 8.0^2)$

   $P_2 = 1.5 \times 10^5 + 500(4 - 64) = 1.5 \times 10^5 - 3.0 \times 10^4 = 1.2 \times 10^5\text{ Pa}$

   Conclusion: As expected from the Venturi effect, the fluid speed increases fourfold and the static pressure drops from $150\text{ kPa}$ to $120\text{ kPa}$.