8.1 Fluid Dynamics: Continuity and Energy Equations

8.1 Fluid Dynamics: Continuity and Energy Equations

In fluid mechanics, fluid dynamics investigates fluids in motion under the action of unbalanced body and surface forces. In the Philippine Civil Engineering Licensure Examination (CELE), fluid dynamics, closed conduit flow, and open channel hydraulics represent major quantitative components of the Hydraulics and Principles of Geotechnical Engineering (HGE) syllabus. Mastery of these topics requires a firm grasp of the fundamental conservation laws: conservation of mass (Continuity Equation), conservation of energy (First Law of Thermodynamics / Extended Bernoulli Equation), and conservation of linear momentum.

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1. Flow Classifications and Streamline Kinematics

To analyze a moving fluid, engineers classify the velocity vector field $\vec{v}(x, y, z, t)$ according to its variation across time and space:

Streamlines, Pathlines, and Streaklines

• Streamline: An imaginary continuous curve drawn through a flowing fluid such that the velocity vector of fluid particles at every point along the curve is tangent to it at that instant ($dx/u = dy/v = dz/w$). Because velocity is tangent, no fluid can cross a streamline.
• Pathline: The actual trajectory traced by an individual fluid particle over a period of time (Lagrangian description).
• Streakline: The instantaneous locus of all fluid particles that have previously passed through a specific common injection point (e.g., dye injected continuously from a fixed needle).
• Under steady flow conditions, streamlines, pathlines, and streaklines are completely identical.

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2. Conservation of Mass: The Continuity Equation

For a control volume bounded by a control surface, conservation of mass dictates that the net mass flux exiting through the control surface equals the time rate of mass decrease within the control volume.

Volumetric and Mass Flow Rate

For a cross-section of area $A$ with a local velocity profile $v_n$ perpendicular to $dA$: $\text{Volumetric Flow Rate: } Q = \int_A v_n \, dA = A v$ $\text{Mass Flow Rate: } \dot{m} = \int_A \rho v_n \, dA = \rho A v = \rho Q$ where $v = v_{\text{avg}}$ is the area-weighted mean velocity ($\text{m/s}$), $A$ is the flow area ($\text{m}^2$), $\rho$ is mass density ($\text{kg/m}^3$), and $Q$ is discharge ($\text{m}^3/\text{s}$ or $\text{L/s}$, where $1\text{ m}^3/\text{s} = 1,000\text{ L/s}$). The weight flow rate is $W_f = \gamma Q = \rho g Q$ ($\text{N/s}$ or $\text{kN/s}$).

One-Dimensional Continuity for Incompressible Flow

For steady flow of an incompressible fluid ($\rho_1 = \rho_2 = \text{constant}$): $\rho_1 A_1 v_1 = \rho_2 A_2 v_2 \implies A_1 v_1 = A_2 v_2 = Q = \text{constant}$ For circular pipes of internal diameter $D_1$ and $D_2$: $\frac{\pi D_1^2}{4} v_1 = \frac{\pi D_2^2}{4} v_2 \implies v_2 = v_1 \left(\frac{D_1}{D_2}\right)^2$ For a branching pipe junction with $m$ incoming conduits and $n$ outgoing conduits: $\sum_{i=1}^m Q_{\text{in}, i} = \sum_{j=1}^n Q_{\text{out}, j}$

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3. Total Energy Head and Bernoulli's Theorem

Consider an elemental fluid prism moving along a streamline in steady, frictionless (inviscid) flow. Integrating Euler's equation of motion along the streamline yields Bernoulli's Theorem.

Components of Fluid Energy Head

Every unit weight of a flowing fluid possesses three distinct mechanical energy components, each expressed in linear dimensions of length (meters of fluid):

1. Elevation (Potential) Head ($z$): Energy of position above an arbitrary horizontal datum plane (meters, $\text{m}$).
2. Pressure Head ($\frac{P}{\gamma}$): Energy stored in the fluid due to static hydrostatic pressure $P$ relative to specific weight $\gamma = \rho g$ (meters, $\text{m}$).
3. Velocity (Kinetic) Head ($\alpha \frac{v^2}{2g}$): Kinetic energy per unit weight of fluid moving at average velocity $v$ (meters, $\text{m}$).

The Coriolis Kinetic Energy Correction Factor ($\alpha$)

Because the true velocity profile across a conduit is non-uniform (zero at the pipe wall, maximum at centerline), the actual integrated kinetic energy exceeds $\frac{v_{\text{avg}}^2}{2g}$. The kinetic energy correction factor $\alpha$ is defined as: $\alpha = \frac{1}{A v^3} \int_A u^3 \, dA$

• Laminar pipe flow (parabolic profile): $\alpha = 2.0$.
• Turbulent pipe flow (logarithmic profile): $\alpha \approx 1.01\text{ to }1.05$. In civil engineering licensure calculations, $\alpha$ is taken as $1.0$ for turbulent flow unless explicitly instructed otherwise.

Classical Bernoulli Equation (Ideal / Frictionless Flow)

Between any two sections 1 and 2 along a streamline in steady, incompressible, frictionless flow: $z_1 + \frac{P_1}{\gamma} + \frac{v_1^2}{2g} = z_2 + \frac{P_2}{\gamma} + \frac{v_2^2}{2g} = H = \text{constant}$ where $H$ is the total mechanical energy head.

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4. The Extended Energy Equation: Pumps, Turbines, and Head Losses

Real engineering hydraulic systems experience energy inputs from mechanical pumps, energy extractions from turbines, and irreversibilities due to boundary shear friction and turbulence.

$\boxed{z_1 + \frac{P_1}{\gamma} + \alpha_1 \frac{v_1^2}{2g} + H_p - H_t - h_L = z_2 + \frac{P_2}{\gamma} + \alpha_2 \frac{v_2^2}{2g}}$

Mechanical Energy Devices and Power Formulations

1. Pumps ($H_p$): A pump adds mechanical energy head $H_p$ ($\text{m}$) to the fluid, raising its pressure and/or elevation.

   • Water (Fluid) Power Output: The rate of useful energy transferred to the liquid: $P_{\text{water}} = Q \gamma H_p = \rho g Q H_p \quad (\text{Watts or kW})$
   • Brake (Input) Shaft Power: The mechanical power required to drive the pump shaft, considering pump efficiency $\eta_p < 1.0$: $P_{\text{input}} = \frac{Q \gamma H_p}{\eta_p}$ (For $Q$ in $\text{m}^3/\text{s}$, $\gamma$ in $\text{kN/m}^3$, and $H_p$ in $\text{m}$, $P_{\text{input}}$ is in $\text{kW}$. In Imperial units: $\text{HP} = \frac{Q \gamma H_p}{550 \eta}$, with $Q$ in $\text{ft}^3/\text{s}$ and $\gamma = 62.4\text{ lb/ft}^3$).

2. Turbines ($H_t$): A hydraulic turbine extracts energy head $H_t$ ($\text{m}$) from the fluid to generate electrical power.

   • Power Extracted from Fluid: $P_{\text{fluid}} = Q \gamma H_t$.
   • Delivered Output Power: The electrical/mechanical power produced, considering turbine efficiency $\eta_t < 1.0$: $P_{\text{output}} = Q \gamma H_t \eta_t$

3. Total Head Loss ($h_L$): Head lost between sections 1 and 2, consisting of pipe wall friction loss ($h_f$) and minor fitting losses ($h_m$): $h_L = h_f + \sum h_m$

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

The graphical representation of energy along a hydraulic conduit provides vital diagnostic insight into operational pressures and vacuum zones.

Fundamental Geometric Rules for EGL and HGL

1. Vertical Separation: The EGL always lies vertically above the HGL by exactly the velocity head $\frac{v^2}{2g}$. Where the velocity is zero (such as in large storage reservoirs), the EGL and HGL coincide at the free liquid surface.
2. Frictional Slope: In passive conduits (no pumps), the EGL must always slope downward in the direction of flow at a hydraulic slope $S_f = \frac{dh_f}{dL}$.
3. Pump Discontinuity: A pump introduces an instantaneous vertical upward jump in both the EGL and HGL equal to the pump head $H_p$.
4. Turbine Discontinuity: A turbine introduces an instantaneous vertical downward drop in both lines equal to $H_t$.
5. Conduit Contraction: When pipe diameter decreases, velocity $v$ increases, expanding the velocity head $\frac{v^2}{2g}$. Consequently, the HGL drops sharply below the EGL.
6. Conduit Expansion: When pipe diameter increases, velocity $v$ drops. If the reduction in velocity head exceeds the localized expansion head loss, the HGL experiences a localized rise, termed pressure recovery.
7. Sub-Atmospheric (Vacuum) Zones: If the physical pipe profile rises above the HGL, the gauge pressure head $\frac{P}{\gamma} = \text{HGL} - z_{\text{pipe}}$ becomes negative (gauge vacuum).

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6. Siphon Hydraulics and Cavitation Phenomena

A siphon is a closed conduit configured to convey liquid from an elevated reservoir over an intermediate topographic summit to a discharge point at lower elevation without mechanical pumping, relying on atmospheric pressure and gravity.

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The Mechanism of Cavitation

At the summit of a functioning siphon, the physical elevation of the pipe invert ($z_s$) exceeds the Hydraulic Grade Line elevation ($\text{HGL}_s$), producing sub-atmospheric pressure: $\frac{P_{s, \text{gauge}}}{\gamma} = \text{HGL}_s - z_s = z_{\text{res}} - z_s - \frac{v_s^2}{2g} - h_{L, 1\to s} < 0$ If this absolute pressure falls to the saturation vapor pressure of the liquid ($P_v$): $P_{s, \text{abs}} \le P_v \implies \frac{P_{s, \text{gauge}}}{\gamma} \le \frac{P_v - P_{\text{atm}}}{\gamma}$ For water at standard ambient temperature ($20^\circ\text{C}$), $P_v \approx 2.34\text{ kPa abs}$ and $P_{\text{atm}} = 101.325\text{ kPa}$. The maximum permissible negative gauge pressure head before vapor formation is: $\left(\frac{P_{\text{gauge}}}{\gamma}\right)_{\text{limit}} = \frac{2.34 - 101.325}{9.81} = -10.09\text{ m of water}$ When absolute pressure reaches $P_v$, the liquid boils at ambient temperature, generating vapor-filled bubbles (vapor cavities). As these vapor cavities travel into downstream zones of higher pressure, they collapse violently within microseconds, generating localized microjets and shockwave pressures exceeding $500\text{ MPa}$. This causes severe pitting erosion, pipe wall fatigue failure, intense vibration, and immediate de-priming (air locking) of the siphon.

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7. Worked Example: High-Lift Pumping Conduit Analysis

Problem Statement: A municipal water supply pumping station lifts water ($\gamma = 9.81\text{ kN/m}^3$) from a suction well with water surface at $\text{Elev } 18.00\text{ m}$ to a distribution reservoir with water surface at $\text{Elev } 68.00\text{ m}$. The system features:

• Discharge rate: $Q = 0.120\text{ m}^3/\text{s}$ ($120\text{ L/s}$).
• Total pipe length: $L = 550\text{ m}$ of $300\text{ mm}$ internal diameter ductile iron pipe ($D = 0.300\text{ m}$).
• Darcy-Weisbach friction factor: $f = 0.020$.
• Minor losses: square-edged entrance ($K_e = 0.50$), four $90^\circ$ flanged bends ($K_b = 4 \times 0.35 = 1.40$), one swing check valve ($K_v = 2.50$), and submerged pipe exit into the reservoir ($K_{\text{exit}} = 1.00$).
• Combined pump-motor efficiency: $\eta = 76\%$.

Determine (a) the velocity head in the pipe, (b) the total head loss, (c) the total dynamic head developed by the pump ($H_p$), and (d) the electrical input power required in kilowatts.

Step-by-Step Solution:

1. Calculate flow velocity and velocity head: $A = \frac{\pi D^2}{4} = \frac{\pi (0.300\text{ m})^2}{4} = 0.070686\text{ m}^2$ $v = \frac{Q}{A} = \frac{0.120\text{ m}^3/\text{s}}{0.070686\text{ m}^2} = 1.6976\text{ m/s}$ $\frac{v^2}{2g} = \frac{(1.6976\text{ m/s})^2}{2(9.81\text{ m/s}^2)} = \frac{2.8819}{19.62} = 0.1469\text{ m}$

2. Compute major friction head loss ($h_f$): $h_f = f \frac{L}{D} \frac{v^2}{2g} = (0.020) \left(\frac{550\text{ m}}{0.300\text{ m}}\right) (0.1469\text{ m}) = 36.667 \times 0.1469\text{ m} = 5.386\text{ m}$

3. Compute minor head losses ($h_m$): $\sum K = K_e + K_b + K_v + K_{\text{exit}} = 0.50 + 1.40 + 2.50 + 1.00 = 5.40$ $h_m = \sum K \frac{v^2}{2g} = 5.40 \times 0.1469\text{ m} = 0.793\text{ m}$ $\text{Total Head Loss: } h_L = h_f + h_m = 5.386 + 0.793 = 6.179\text{ m}$

4. Apply Energy Equation between suction well surface (1) and upper reservoir surface (2): $z_1 + \frac{P_1}{\gamma} + \frac{v_1^2}{2g} + H_p - h_L = z_2 + \frac{P_2}{\gamma} + \frac{v_2^2}{2g}$ Because both reservoirs are open to atmosphere, $P_1 = P_2 = 0\text{ kPa gauge}$. For large reservoirs, $v_1 \approx 0$ and $v_2 \approx 0$. $18.00 + 0 + 0 + H_p - 6.179 = 68.00 + 0 + 0$ $H_p = (68.00 - 18.00) + 6.179 = 50.00 + 6.179 = 56.18\text{ m}$

5. Compute Water Power and Electrical Input Power: $P_{\text{water}} = Q \gamma H_p = (0.120\text{ m}^3/\text{s})(9.81\text{ kN/m}^3)(56.179\text{ m}) = 66.134\text{ kW}$ $P_{\text{input}} = \frac{P_{\text{water}}}{\eta} = \frac{66.134\text{ kW}}{0.76} = 87.02\text{ kW}$

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8. The Momentum Equation: Forces on Bends, Nozzles, Jets and Vanes

Energy equations give pressures and heads. The impulse-momentum principle gives the forces that moving water exerts on pipes, nozzles and blades. For steady flow through a control volume:

$\sum \vec{F} = \rho Q (\vec{v}_2 - \vec{v}_1)$

The left side includes pressure forces on the inlet and outlet areas, weight if relevant, and the force exerted by the solid boundary on the fluid. The fluid's force on the boundary is equal and opposite.

Free jet striking a fixed flat plate normal to the jet

The jet loses all its velocity in the original direction, so:

$F = \rho Q v = \rho A v^2$

Example. A $50\text{ mm}$ jet moves at $20\text{ m/s}$. $A = 0.0019635\text{ m}^2$ and $Q = 0.03927\text{ m}^3/\text{s}$, so $F = 1000(0.03927)(20) = 785\text{ N}$.

Moving vane

For a single vane moving at speed $u$ in the jet's direction, the relative velocity is $(v - u)$. The mass actually deflected per second is $\rho A (v - u)$. For a flat plate normal to the jet:

$F = \rho A (v - u)^2$

For a series of vanes on a wheel, as in an impulse turbine, all the jet's mass is used, so $F = \rho Q (v - u)(1 - \cos\theta)$ for a deflection angle $\theta$. Power is $F u$, which is maximized near $u = v/2$.

Pipe bend in a horizontal plane

For a bend that turns the flow through angle $\theta$, with pressures $p_1, p_2$ and areas $A_1, A_2$:

$F_x = p_1 A_1 - p_2 A_2 \cos\theta - \rho Q (v_2 \cos\theta - v_1)$ $F_y = - p_2 A_2 \sin\theta - \rho Q v_2 \sin\theta$

Here $F_x$ and $F_y$ are the components the fluid exerts on the bend; the anchor block must resist $R = \sqrt{F_x^2 + F_y^2}$.

Example. A $300\text{ mm}$ horizontal $90^\circ$ bend carries $0.15\text{ m}^3/\text{s}$ at $200\text{ kPa}$, assuming negligible head loss. $A = 0.070686\text{ m}^2$, so $v = 2.122\text{ m/s}$ and $\rho Q v = 1000(0.15)(2.122) = 318\text{ N}$. The pressure force is $pA = 200{,}000(0.070686) = 14{,}137\text{ N}$. Each component is $14{,}137 + 318 = 14{,}455\text{ N}$. The resultant is $\sqrt{2}(14{,}455) = 20{,}443\text{ N} \approx 20.4\text{ kN}$, acting at $45^\circ$ toward the outside of the bend.

9. CELE Exam Traps & Common Computational Errors