4.2 Fluid Kinematics, Dynamics & Bernoulli Equation

4.2 Fluid Kinematics, Dynamics & Bernoulli Equation

Fluid kinematics describes fluid motion without considering the forces causing that motion, whereas fluid dynamics analyzes the relationship between forces, fluid mass, and resulting acceleration. The Bernoulli equation is one of the most powerful and widely applied principles in fluid mechanics, forming the foundation for flow measurement, piping system design, and turbomachinery analysis.

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1. Eulerian vs. Lagrangian Flow Descriptions

• Lagrangian Approach: Tracks individual fluid particles as they move through space and time. Particle position is defined as $\vec{r}_p(t) = (x(t), y(t), z(t))$, and velocity is $\vec{V}_p = d\vec{r}_p/dt$. This approach is standard in solid mechanics but cumbersome for fluid continua.
• Eulerian Approach: Focuses on fixed spatial points in the flow field. Velocity, pressure, and density are defined as field functions of position and time: $\vec{V} = \vec{V}(x,y,z,t) = u\hat{i} + v\hat{j} + w\hat{k}$. This is the standard framework in fluid engineering.

Flow Lines:

• Streamline: A continuous curve drawn through the flow field such that the velocity vector of every fluid particle on the curve is everywhere tangent to the curve at a given instant: $\frac{dx}{u} = \frac{dy}{v} = \frac{dz}{w}$
• Pathline: The actual trajectory path traversed by a specific fluid particle over a time interval.
• Streakline: The locus of all fluid particles that have passed sequentially through a single fixed spatial point.
• Note: In steady flow, streamlines, pathlines, and streaklines are completely identical.

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2. Conservation of Mass & The Continuity Equation

The fundamental law of conservation of mass states that mass can neither be created nor destroyed. For a control volume, the net rate of mass outflow equals the rate of mass decrease within the control volume.

General 3D Differential Form:

$\frac{\partial \rho}{\partial t} + \nabla \cdot (\rho \vec{V}) = 0 \implies \frac{\partial \rho}{\partial t} + \frac{\partial (\rho u)}{\partial x} + \frac{\partial (\rho v)}{\partial y} + \frac{\partial (\rho w)}{\partial z} = 0$

For steady flow ($\partial \rho / \partial t = 0$) and incompressible fluid (constant density $\rho$): $\nabla \cdot \vec{V} = \frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} + \frac{\partial w}{\partial z} = 0$

1D Integral Continuity Equation:

For flow through a stream tube or pipe bounded by sections 1 and 2:

• Mass Flow Rate ($\dot{m}$): $\dot{m} = \rho_1 A_1 V_1 = \rho_2 A_2 V_2 = \text{constant} \quad [\text{kg/s}]$
• Volumetric Flow Rate ($Q$): For incompressible fluids ($\rho_1 = \rho_2$): $Q = A_1 V_1 = A_2 V_2 = \text{constant} \quad [\text{m}^3/\text{s} \text{ or } \text{ft}^3/\text{s}]$ where $V_1, V_2$ represent average cross-sectional velocities.

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3. Euler's Equation of Motion & Bernoulli's Equation

Applying Newton's second law ($\sum dF_s = dm \cdot a_s$) to a frictionless fluid element moving along a streamline coordinate $s$: $-\frac{\partial P}{\partial s} ds \cdot dA - \rho g dA ds \frac{dz}{ds} = \rho (dA ds) \left(V \frac{\partial V}{\partial s}\right)$

Dividing by $\rho dA ds$ yields Euler's Equation of Motion along a streamline: $\frac{dP}{\rho} + V dV + g dz = 0$

Integrating Euler's equation for an incompressible fluid (constant density $\rho$ and specific weight $\gamma = \rho g$): $\int \frac{dP}{\rho} + \int V dV + \int g dz = \text{constant}$ $\frac{P}{\rho} + \frac{V^2}{2} + gz = C$

Dividing by gravitational acceleration $g$ yields the standard Bernoulli Equation in head units (meters or feet of fluid column): $\frac{P_1}{\gamma} + \frac{V_1^2}{2g} + z_1 = \frac{P_2}{\gamma} + \frac{V_2^2}{2g} + z_2 = H = \text{constant}$

Physical Interpretation of Energy Heads:

• Pressure Head ($\frac{P}{\gamma}$): Height of fluid column required to produce static pressure $P$.
• Velocity Head ($\frac{V^2}{2g}$): Height to which fluid would rise if kinetic energy were converted entirely into potential energy.
• Elevation Head ($z$): Potential energy per unit weight of fluid above an arbitrary datum plane.
• Total Dynamic Head ($H$): Sum of pressure, velocity, and elevation heads.

Essential Assumptions & Limitations of Bernoulli's Equation:

1. Inviscid Flow: Viscous friction forces are neglected ($h_L = 0$).
2. Steady Flow: Flow parameters at any point do not vary with time ($\partial / \partial t = 0$).
3. Incompressible Flow: Density remains constant along the streamline.
4. Along a Streamline: Applies strictly along the same streamline (unless flow is irrotational).
5. No Shaft Work or Heat Transfer: No energy added by pumps or extracted by turbines between points 1 and 2.

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

Graphical representations of energy distribution along a piping flow path provide vital insight into hydraulic performance:

• Energy Grade Line (EGL): Plot of total mechanical head available at each point along the pipeline: $EGL = \frac{P}{\gamma} + \frac{V^2}{2g} + z$
• Hydraulic Grade Line (HGL): Plot of piezometric head (static pressure head plus elevation head): $HGL = \frac{P}{\gamma} + z$
• Key Relationship: The EGL is always located vertically above the HGL by a distance exactly equal to the velocity head $\frac{V^2}{2g}$: $EGL - HGL = \frac{V^2}{2g}$

Extended Energy Equation for Real Fluid Systems:

Accounting for mechanical energy inputs (pumps $h_p$), outputs (turbines $h_t$), and friction/minor losses ($h_L$): $\frac{P_1}{\gamma} + \frac{V_1^2}{2g} + z_1 + h_p = \frac{P_2}{\gamma} + \frac{V_2^2}{2g} + z_2 + h_t + h_L$

• Water Horsepower ($WHP$): Useful hydraulic power delivered to the fluid: $WHP = \gamma Q h_p = \rho g Q h_p \quad [\text{Watts or kW}]$ $\text{In English Units: } WHP = \frac{\gamma Q h_p}{550} = \frac{(62.4 \text{ lbf/ft}^3)(Q \text{ ft}^3/\text{s})(h_p \text{ ft})}{550} \quad [\text{hp}]$
• Brake Horsepower ($BHP$): Input shaft power required by the pump driven with efficiency $\eta_p$: $BHP = \frac{WHP}{\eta_p}$

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5. Flow Measurement Devices

1. Venturi Meter

A converging-diverging pipe flow meter that measures volumetric discharge by creating a pressure drop across a throat restriction.

• Applying Bernoulli between inlet (1) and throat (2): $Q_{ideal} = A_2 V_2 = A_2 \sqrt{\frac{2g (HGL_1 - HGL_2)}{1 - (A_2/A_1)^2}} = A_2 \sqrt{\frac{2g \Delta h}{1 - (d_2/d_1)^4}}$
• Actual Flow Rate ($Q_{actual}$): $Q_{actual} = C_d A_2 \sqrt{\frac{2g \Delta h}{1 - (d_2/d_1)^4}}$ where $C_d$ is discharge coefficient ($0.95 - 0.98$ for smooth Venturis), and $\Delta h = \frac{P_1 - P_2}{\gamma} + (z_1 - z_2)$.

2. Orifice Plate

A thin plate with a sharp-edged concentric circular hole inserted into a pipe. It causes flow separation and forms a narrowed stream jet called the vena contracta. Due to turbulence losses, $C_d \approx 0.60 - 0.65$.

3. Pitot-Static Tube

Measures local flow velocity by comparing total stagnation pressure ($P_{stag}$) with local static pressure ($P_{static}$): $P_{stag} = P_{static} + \frac{1}{2} \rho V^2 \implies V = \sqrt{\frac{2(P_{stag} - P_{static})}{\rho}} = \sqrt{2g h_v}$

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6. Worked Numerical Examples

Example 1: Pipe Contraction & Pressure Drop

Water ($\rho = 1000\text{ kg/m}^3$, $\gamma = 9.81\text{ kN/m}^3$) flows through a horizontal pipe contraction from diameter $D_1 = 300\text{ mm}$ to $D_2 = 150\text{ mm}$ at a volumetric flow rate of $Q = 0.10\text{ m}^3/\text{s}$. If static gage pressure at Section 1 is $P_1 = 250\text{ kPa}$, determine:

1. Flow velocities $V_1$ and $V_2$.
2. Static gage pressure $P_2$ at Section 2 assuming ideal frictionless flow.

Solution Step-by-Step:

• Step 1: Calculate areas and flow velocities: $A_1 = \frac{\pi (0.30)^2}{4} = 0.070686\text{ m}^2$ $A_2 = \frac{\pi (0.15)^2}{4} = 0.017671\text{ m}^2$ $V_1 = \frac{Q}{A_1} = \frac{0.10}{0.070686} = 1.4147\text{ m/s}$ $V_2 = \frac{Q}{A_2} = \frac{0.10}{0.017671} = 5.6588\text{ m/s}$

• Step 2: Apply horizontal Bernoulli equation ($z_1 = z_2$): $\frac{P_1}{\gamma} + \frac{V_1^2}{2g} = \frac{P_2}{\gamma} + \frac{V_2^2}{2g}$ $\frac{250}{9.81} + \frac{(1.4147)^2}{2(9.81)} = \frac{P_2}{9.81} + \frac{(5.6588)^2}{2(9.81)}$ $25.4842 + 0.1020 = \frac{P_2}{9.81} + 1.6321$ $25.5862 = \frac{P_2}{9.81} + 1.6321 \implies \frac{P_2}{9.81} = 23.9541\text{ m}$ $P_2 = 23.9541 \times 9.81 = 234.99\text{ kPa}$

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Example 2: Horizontal Venturi Meter Flow Rate

A horizontal Venturi meter with inlet diameter $D_1 = 200\text{ mm}$ and throat diameter $D_2 = 100\text{ mm}$ is used to measure oil flow ($SG = 0.88$, $\gamma_o = 8.633\text{ kN/m}^3$). A differential mercury manometer connected between inlet and throat shows a deflection of $h_m = 200\text{ mm}$ of mercury ($SG_m = 13.6$). If discharge coefficient is $C_d = 0.97$, calculate the actual volumetric flow rate $Q_{actual}$.

Solution Step-by-Step:

• Step 1: Calculate differential head $\Delta h$ in terms of oil column: $\Delta h = h_m \left(\frac{SG_m}{SG_o} - 1\right) = 0.20\text{ m} \left(\frac{13.6}{0.88} - 1\right) = 0.20 \times (15.4545 - 1) = 2.8909\text{ m of oil}$

• Step 2: Calculate cross-sectional throat area and area ratio: $A_2 = \frac{\pi (0.10)^2}{4} = 0.007854\text{ m}^2$ $\frac{d_2}{d_1} = \frac{100}{200} = 0.50 \implies \left(\frac{d_2}{d_1}\right)^4 = (0.50)^4 = 0.0625$

• Step 3: Calculate actual discharge $Q_{actual}$: $Q_{actual} = C_d A_2 \sqrt{\frac{2g \Delta h}{1 - (d_2/d_1)^4}}$ $Q_{actual} = (0.97)(0.007854) \sqrt{\frac{2(9.81)(2.8909)}{1 - 0.0625}}$ $Q_{actual} = 0.007618 \times \sqrt{\frac{56.719}{0.9375}} = 0.007618 \times \sqrt{60.500} = 0.007618 \times 7.7782 = 0.05925\text{ m}^3/\text{s}$ $Q_{actual} = 59.25\text{ liters/sec}$

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Example 3: Pump Power & Total Dynamic Head (TDH)

A pump draws water ($\gamma = 9.81\text{ kN/m}^3$) from a lower reservoir at elevation $z_1 = 15.0\text{ m}$ and delivers it to an elevated tank at $z_2 = 45.0\text{ m}$ at a flow rate of $Q = 0.050\text{ m}^3/\text{s}$. Total friction head loss in suction and discharge piping is $h_L = 4.80\text{ m}$. Water surface in both tanks is open to atmosphere. If pump efficiency is $\eta_p = 78%$, compute:

1. Total Dynamic Head ($TDH = h_p$) supplied by the pump.
2. Water Horsepower ($WHP$).
3. Brake Power ($BHP$) required to drive the pump.

Solution Step-by-Step:

• Step 1: Apply Extended Energy Equation from surface 1 to surface 2: $\frac{P_1}{\gamma} + \frac{V_1^2}{2g} + z_1 + h_p = \frac{P_2}{\gamma} + \frac{V_2^2}{2g} + z_2 + h_L$ Since $P_1 = P_2 = P_{atm} = 0\text{ gage}$, and free surface velocities $V_1 \approx V_2 \approx 0$: $z_1 + h_p = z_2 + h_L \implies h_p = (z_2 - z_1) + h_L$ $h_p = (45.0 - 15.0) + 4.80 = 30.0 + 4.80 = 34.80\text{ m}$

• Step 2: Calculate Water Power ($WHP$): $WHP = \gamma Q h_p = (9.81\text{ kN/m}^3)(0.050\text{ m}^3/\text{s})(34.80\text{ m}) = 17.069\text{ kW}$

• Step 3: Calculate Brake Power ($BHP$): $BHP = \frac{WHP}{\eta_p} = \frac{17.069\text{ kW}}{0.78} = 21.883\text{ kW}$ In mechanical horsepower ($1\text{ hp} = 0.7457\text{ kW}$): $BHP = \frac{21.883}{0.7457} = 29.35\text{ hp}$