12.3 Continuity Equation, Bernoulli Equation, and Energy Equation

12.3 Continuity Equation, Bernoulli Equation, and Energy Equation

Core FE Exam Principle: Fluid dynamics governs fluids in motion using three fundamental conservation laws: Conservation of Mass (Continuity Equation), Conservation of Linear Momentum, and Conservation of Energy (Bernoulli and General Energy Equations).

Conservation of Mass and the Continuity Equation

For a fixed control volume under steady-state conditions, the rate of mass entering the system equals the rate of mass leaving:

One-Dimensional Flow Formulations

• Compressible Steady Flow:

  $$\dot{m} = \rho_1 A_1 V_1 = \rho_2 A_2 V_2 \quad [\text{kg/s} \text{ or } \text{lbm/s}]$$

• Incompressible Steady Flow ($\rho_1 = \rho_2 = \rho$):

  $$Q = A_1 V_1 = A_2 V_2 \quad [\text{m}^3/\text{s} \text{ or } \text{ft}^3/\text{s} \text{ or } \text{gpm}]$$

  where:

  • (Q) = Volumetric flow rate
  • (A) = Cross-sectional conduit area ($\frac{\pi D^2}{4}$ for circular pipes)
  • (V) = Average flow velocity across the section

Key Relation: Velocity varies inversely with the square of conduit diameter ($V_2 = V_1 \cdot (D_1 / D_2)^2$). Halving the pipe diameter increases average velocity by a factor of 4.

The Bernoulli Equation

The Bernoulli equation represents Euler's equation integrated along a streamline for ideal fluid flow.

Four Governing Assumptions

1. Steady Flow: Flow parameters at any point do not change with time ($\frac{\partial}{\partial t} = 0$).
2. Incompressible Flow: Fluid density remains constant ($\rho = \text{const}$).
3. Frictionless / Inviscid Flow: Viscous shear stresses are negligible ($\mu = 0$).
4. Flow Along a Streamline: Applied between two points along the same flow path.

Head Form of the Bernoulli Equation

Each term possesses dimensions of length (meters or feet) representing specific energy per unit weight:

• Pressure Head ($\frac{P}{\gamma}$): Height of fluid column required to produce static pressure (P).
• Velocity Head ($\frac{V^2}{2g}$): Vertical distance fluid would fall under gravity to attain velocity (V).
• Elevation Head ((z)): Potential energy head above an arbitrary horizontal reference datum.
• Total Hydraulic Head ((H)): Constant sum of all three energy heads along an ideal streamline.

Stagnation Pressure and Velocity Measurement

When fluid is brought to rest ($V_0 = 0$) isentropically at a stagnation point:

• Static Pressure ((P)): Pressure exerted by fluid on a surface moving with the flow.
• Dynamic Pressure ($\frac{1}{2} \rho V^2$): Pressure rise caused by bringing fluid kinetic energy to rest.
• Stagnation (Total) Pressure ($P_0$): Pressure at stagnation point.

Pitot-Static Tube Equation:

Extended General Energy Equation

Real engineering fluids experience head losses due to wall friction, fittings, and valves, while turbomachinery adds or extracts energy.

Extended Energy Equation

where:

• $h_p$ = Net useful head added by a pump ($\text{m}$ or $\text{ft}$)
• $h_t$ = Net head extracted by a turbine ($\text{m}$ or $\text{ft}$)
• $h_L$ = Total head loss between sections 1 and 2 ($h_L = h_f + h_m$)
• $\alpha$ = Kinetic energy correction factor ($\alpha = 1.0$ for fully turbulent flow; $\alpha = 2.0$ for laminar flow)

Power Relations

• Hydraulic Power Delivered by Pump ($W_{\text{hyd}}$):

  $$W_{\text{hyd}} = \gamma Q h_p = \rho g Q h_p \quad [\text{Watts or } \text{ft}\cdot\text{lbf/s}]$$

  Note: $1 \text{ HP} = 550 \text{ ft}\cdot\text{lbf/s} = 745.7 \text{ W}$.

• Brake Horsepower / Input Shaft Power ($P_{\text{shaft}}$):

  $$P_{\text{shaft}} = \frac{W_{\text{hyd}}}{\eta_{\text{pump}}} = \frac{\gamma Q h_p}{\eta_{\text{pump}}}$$

Hydraulic Grade Line (HGL) and Energy Grade Line (EGL)

Visualizing hydraulic head profiles along a pipeline highlights system pressure variations, energy drops, and potential flow hazards.

• Energy Grade Line (EGL): Plots total available head along the pipeline flow path: $$\text{EGL} = \frac{P}{\gamma} + \frac{V^2}{2g} + z$$
• Hydraulic Grade Line (HGL): Plots piezometric head along the pipeline flow path: $$\text{HGL} = \frac{P}{\gamma} + z$$
• Vertical Clearance: The vertical distance between EGL and HGL at any point equals the local velocity head ($\frac{V^2}{2g}$).

 Energy 
 Head [m]  ^      [EGL] = P/gamma + V^2/2g + z
          |      *---------------------------\\  (Slopes downward due to h_L)
          |      | <--- V^2/2g --->          
          |      *----------------------------\\ [HGL] = P/gamma + z
          |      |                             
          |      | <--- P/gamma --->            
          |     ---                              
          +---------------------------------------> Pipeline Distance [x]

Key EGL/HGL Behavior Rules

1. For an open reservoir surface, both EGL and HGL coincide at the liquid free surface.
2. A pump causes an immediate vertical step-up in both EGL and HGL equal to pump head $h_p$.
3. A turbine causes an immediate vertical step-down equal to turbine head $h_t$.
4. Frictional head losses cause EGL and HGL to slope continuously downward in the direction of flow.
5. If the HGL drops below the physical pipe centerline, local static pressure $P_{gage}$ becomes negative (sub-atmospheric vacuum).
6. Cavitation Danger: If absolute static pressure falls to the fluid's vapor pressure ($P_{abs} \le P_v$), the liquid vaporizes into gas bubbles, causing severe erosion, pitting, and noise when the bubbles collapse downstream.

Comprehensive Worked Engineering Example

Problem Statement

Water ($\rho = 1000 \text{ kg/m}^3$, $\gamma = 9.81 \text{ kN/m}^3$, vapor pressure $P_v = 2.34 \text{ kPa abs}$) is pumped from a lower open storage reservoir (Surface Elevation $z_1 = 15.0 \text{ m}$) to an elevated pressurized tank (Surface Elevation $z_2 = 55.0 \text{ m}$) at a volumetric flow rate $Q = 0.080 \text{ m}^3/\text{s}$.

The discharge tank carries a compressed air headspace gage pressure $P_2 = 150.0 \text{ kPa gage}$. The suction line is a $200 \text{ mm}$ diameter pipe and the discharge line is a $150 \text{ mm}$ diameter pipe. The total combined friction and minor head losses across the entire piping system are calculated to be $h_L = 12.4 \text{ m}$. Local atmospheric pressure is $P_{atm} = 101.3 \text{ kPa}$.

Calculate:

1. The average fluid flow velocity in the discharge pipe $V_2$.
2. The net pump head $h_p$ required to maintain the flow rate.
3. The total electrical power consumed by the motor if pump efficiency is $\eta_{\text{pump}} = 80\%$ and motor efficiency is $\eta_{\text{motor}} = 90\%$.

Step-by-Step Solution

Step 1: Compute Flow Velocities

• Discharge pipe diameter $D_2 = 0.150 \text{ m}$, area $A_2 = \frac{\pi (0.150)^2}{4} = 0.017671 \text{ m}^2$: $$V_2 = \frac{Q}{A_2} = \frac{0.080 \text{ m}^3/\text{s}}{0.017671 \text{ m}^2} = 4.527 \text{ m/s}$$
• Discharge velocity head: $$\frac{V_2^2}{2g} = \frac{(4.527)^2}{2 \times 9.81} = \frac{20.495}{19.62} = 1.045 \text{ m}$$

Step 2: Apply Extended Energy Equation Between Reservoir 1 and Tank 2

• Datum at sea level ((z = 0)). Point 1 is at lower open surface; Point 2 is at elevated surface inside pressurized tank.
• Point 1 conditions: $P_1 = 0 \text{ (gage)}$, $V_1 \approx 0$ (large surface), $z_1 = 15.0 \text{ m}$.
• Point 2 conditions: $P_2 = 150.0 \text{ kPa gage}$, $V_2 \approx 0$ (large surface in tank), $z_2 = 55.0 \text{ m}$.

Step 3: Compute Hydraulic Power, Shaft Power, and Total Electrical Input Power

• Hydraulic power delivered to water: $$W_{\text{hyd}} = \gamma Q h_p = 9810 \text{ N/m}^3 \times 0.080 \text{ m}^3/\text{s} \times 67.69 \text{ m} = 53,123 \text{ W} = 53.12 \text{ kW}$$
• Shaft power required from motor: $$P_{\text{shaft}} = \frac{W_{\text{hyd}}}{\eta_{\text{pump}}} = \frac{53.123 \text{ kW}}{0.80} = 66.40 \text{ kW}$$
• Total electrical power consumed by motor: $$P_{\text{elec}} = \frac{P_{\text{shaft}}}{\eta_{\text{motor}}} = \frac{66.404 \text{ kW}}{0.90} = 73.78 \text{ kW}$$

Final Answer: Discharge velocity $V_2 = 4.53 \text{ m/s}$, required pump head $h_p = 67.7 \text{ m}$, and total electrical input power $P_{\text{elec}} = 73.8 \text{ kW}$.