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:

$\dot{m} = \iint_A \rho (\mathbf{V} \cdot \mathbf{n}) dA = \text{constant}$

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

$\frac{P_1}{\gamma} + \frac{V_1^2}{2g} + z_1 = \frac{P_2}{\gamma} + \frac{V_2^2}{2g} + z_2 = H$

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:

$P_0 = P + \frac{1}{2} \rho V^2$

• 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: $V = \sqrt{\frac{2(P_0 - P)}{\rho}} = \sqrt{\frac{2 \Delta P}{\rho}}$

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

$\frac{P_1}{\gamma} + \alpha_1 \frac{V_1^2}{2g} + z_1 + h_p = \frac{P_2}{\gamma} + \alpha_2 \frac{V_2^2}{2g} + z_2 + h_t + h_L$

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}).

$\frac{P_1}{\gamma} + \frac{V_1^2}{2g} + z_1 + h_p = \frac{P_2}{\gamma} + \frac{V_2^2}{2g} + z_2 + h_L$ $0 + 0 + 15.0 + h_p = \frac{150.0 \text{ kPa}}{9.81 \text{ kN/m}^3} + 0 + 55.0 + 12.4$ $15.0 + h_p = 15.290 + 55.0 + 12.4$ $15.0 + h_p = 82.690 \implies h_p = 82.690 - 15.0 = 67.69 \text{ m}$

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}).