Applications of Fluid Dynamics

Chapter 7: Applications of Fluid Dynamics

Chapter 1: Pressure
Chapter 2: Variation of Pressure with Depth
Chapter 3: Pressure Measurement
Chapter 4: Buoyant Forces & Archimedes’ Principle
Chapter 5: Fluid Dynamics
Chapter 6: Bernoulli’s Equation
Chapter 7: Applications of Fluid Dynamics

🚁Overview

This final chapter integrates the principles developed throughout the course into practical, real-world systems involving fluid flow.
Applications include high-speed flow through nozzles, hydrostatic and dynamic pressure loading on large structures like dams, lift generation
on airfoils, and simplified analysis of blood flow in biological vessels. These examples show how fluid dynamics provides critical insight
across engineering and biological domains.

Use continuity and Bernoulli’s principles to model flow through converging/diverging pipes and nozzles.
Compute aerodynamic lift based on pressure differences over a wing surface.
Explain the role of flow velocity and vessel diameter in determining blood pressure and flow rate.
Apply fluid dynamics to analyze forces acting on dam faces and spillways.

🎯Learning Outcomes

By the end of this chapter, students will be able to:

Model flow through nozzles and diffusers using continuity and Bernoulli’s equation.
Understand how velocity and pressure distributions contribute to aerodynamic lift.
Compute lift forces using the pressure differential across airfoil surfaces.
Estimate hydrostatic and dynamic fluid loading on dam walls and gates.
Analyze blood flow in arteries using fluid conservation laws and pressure-flow relationships.
Evaluate energy conversion and pressure drops in real-world pipe systems with elevation changes.

📖Contents

Flow Through Nozzles and Diffusers
Lift on Airfoils and the Bernoulli Principle
Blood Flow in Arteries
Loading on Dams
Hydraulic Lift

7.1 Flow Through Nozzles and Diffusers

When pipe area changes, so does velocity. In a nozzle (converging section) the flow accelerates and static pressure drops; in a diffuser (diverging section) the flow decelerates and pressure recovers. The behavior follows the continuity equation and Bernoulli’s equation for steady, incompressible, inviscid flow.

Continuity: \( Q = A\,v \), so \( A_1 v_1 = A_2 v_2 \) (incompressible flow).
Bernoulli (along a streamline): \( p + \frac{1}{2}\rho v^{2} + \rho g h = \text{constant} \).

Fluid Flow Through a Pipe with Different Diameters demonstrating continuity and Bernoulli effects — Fluid Flow Through a Pipe with Different Diameters

High-speed spray nozzle where viscous losses and mixing invalidate ideal Bernoulli assumptions — Spray Nozzle (viscous losses/mixing → Bernoulli idealization does not apply)

7.2 Lift on Airfoils and the Bernoulli Principle

Faster flow over the cambered upper surface of a wing generally implies lower static pressure than beneath the wing, producing lift. A simple Bernoulli estimate uses the pressure difference between the two sides:

Lift per unit area (pressure difference): \( \Delta p = p_{\text{lower}} – p_{\text{upper}} = \frac{1}{2}\rho\!\left(v_{\text{upper}}^{2} – v_{\text{lower}}^{2}\right) \).

Total lift: \( L \approx \Delta p \, A_{\text{wing}} \). (Real aircraft use circulation/Kutta–Joukowski theory; this is a first‐order estimate.)

Airfoil showing higher speed over the top surface and lower pressure leading to aerodynamic lift — Airfoil Lift from Pressure Difference

7.3 Blood Flow in Arteries

Blood is well approximated as incompressible. In a narrowed artery (stenosis) the reduced area raises velocity and lowers static pressure, consistent with continuity and Bernoulli.

Flow rate: \( Q = A\,v \Rightarrow v_2 = v_1 \frac{A_1}{A_2} \).
Pressure change (horizontal vessel): \( p_2 – p_1 = \frac{1}{2}\rho\!\left(v_1^{2} – v_2^{2}\right) \) (ideal; real blood flow has viscous losses).

Blood flow in artery illustrating increased speed and reduced pressure in a stenosis — Blood Flow in Arteries (stenosis example)

7.4 Loading on Dams

Hydrostatic pressure increases linearly with depth: \( p(h)=\rho g h \). The total horizontal force on a vertical rectangular dam face of width \( b \) and water depth \( h \) is

Total force: \( F=\int_{0}^{h}\rho g y\,b\,dy = \frac{1}{2}\rho g b h^{2} \).

Center of pressure (useful in design): \( y_{\mathrm{cp}}=\frac{\int_0^h y(\rho g y)\,b\,dy}{\int_0^h (\rho g y)\,b\,dy}=\frac{2h}{3} \) below the free surface.

This is for static water. In dynamic conditions (spillways, waves, earthquakes), add kinetic and transient loads; Bernoulli and momentum methods complement structural analysis.

Hydrostatic pressure distribution on a dam wall; higher pressure at greater depth — Pressure and Forces on a Dam Wall

🤔 Puzzle 1: Why does dam-wall thickness grow faster than linearly with depth?

Although \( p=\rho g h \) is linear in \( h \), the bending moment at the base results from the integral of pressure and scales as \( \propto h^{2} \). Designers also include safety factors for waves, uplift, seismic loads, seepage, and material uncertainties, so thickness grows more than linearly.

🤔 Puzzle 2: Why do arch dams curve with the convex side upstream?

Water pressure puts the curved upstream face into compression; concrete is strong in compression and weak in tension. The arch action also redirects load into the abutments, reducing material while maintaining safety.

7.5 Hydraulic Lift

By Pascal’s law, a pressure change applied to a confined incompressible fluid is transmitted undiminished: \( p=\frac{F_1}{A_1}=\frac{F_2}{A_2} \Rightarrow F_2=F_1\frac{A_2}{A_1} \).

Volume continuity sets the motion: if the small piston moves a distance \( s_1 \), the large piston rises \( s_2=s_1\frac{A_1}{A_2} \). Thus the lift provides force amplification at the expense of displacement.

Car hydraulic jack applying Pascal’s law for mechanical advantage — Car Hydraulic Lift Using Pascal’s Law

7.6 Key Formula Recap

Concept	Formula
Continuity (incompressible)	\( Q=A\,v,\quad A_1 v_1 = A_2 v_2 \)
Bernoulli (along a streamline)	\( p + \frac{1}{2}\rho v^{2} + \rho g h = \text{constant} \)
Airfoil lift (Bernoulli estimate)	\( L \approx \left(p_{\text{below}}-p_{\text{above}}\right)A = \frac{1}{2}\rho\!\left(v_{\text{upper}}^{2}-v_{\text{lower}}^{2}\right)A \)
Hydrostatic pressure	\( p(h)=\rho g h \)
Hydrostatic force on dam (width \( b \))	\( F=\frac{1}{2}\rho g b h^{2} \), center of pressure \( y_{\mathrm{cp}}=\frac{2h}{3} \)
Pascal’s law (hydraulic lift)	\( \frac{F_1}{A_1}=\frac{F_2}{A_2} \Rightarrow F_2=F_1\frac{A_2}{A_1} \)

Return to: Prep4Uni Fluid Mechanics 1

📝EXERCISES

20 Questions & Answers (with LaTeX)

What happens to fluid velocity in a converging pipe?
For incompressible flow the continuity equation gives \(Q=Av=\text{const}\). Hence \[ A_1 v_1=A_2 v_2 \;\Rightarrow\; v_2=v_1\,\frac{A_1}{A_2}. \] If \(A_2v_1\). The speed rises because the same volume must pass through a smaller area.
What effect does increasing velocity have on pressure in an ideal fluid?
Along a streamline at the same height \(h\), \[ p+\frac{1}{2}\rho v^{2}=\text{const}. \] Therefore a larger \(v\) implies a smaller static pressure \(p\) (pressure–velocity exchange).
What causes lift on an airplane wing?
A pressure difference between lower and upper surfaces. A simple Bernoulli estimate is \[ \Delta p=p_{\text{lower}}-p_{\text{upper}}\approx \frac{1}{2}\rho\!\left(v_{\text{upper}}^{2}-v_{\text{lower}}^{2}\right), \] so the net lift is \(L\approx \Delta p\,A\). (Full aerodynamics also uses circulation/Kutta–Joukowski.)
What equation is used to estimate lift force?
\[ L \approx \left(p_{\text{below}}-p_{\text{above}}\right)A. \] This models the wing as two pressure panels of area \(A\).
Why is the flow faster at the throat of a Venturi tube?
The throat has minimum area, and with \(Q=Av\) constant, \[ v_{\text{throat}}=v_{\text{inlet}}\frac{A_{\text{inlet}}}{A_{\text{throat}}}. \] Higher speed at the throat corresponds to lower static pressure there.
In which part of a nozzle is the pressure lowest?
At the narrowest section where \(v\) is largest. With height constant, Bernoulli gives lower \(p\) for higher \(v\).
What is dynamic pressure?
Kinetic energy per unit volume: \[ q=\frac{1}{2}\rho v^{2}. \] Pitot tubes measure \(q=p_0-p\) (stagnation minus static) to infer airspeed.
What does Bernoulli’s equation ignore?
Viscosity, turbulence, shaft work (pumps/turbines), heat transfer, and unsteadiness. In real systems add a head-loss term \(h_L\) or pump/turbine work.
How does blood flow faster in narrower arteries?
Assuming incompressible flow, \[ Q=A v=\text{const} \;\Rightarrow\; v\propto\frac{1}{A}. \] When the artery narrows (\(A\) decreases), \(v\) increases to keep \(Q\) the same.
Why do aneurysms form in weakened vessel walls?
Thin-wall hoop stress \(\sigma_\theta\approx p r/t\) increases with radius \(r\). If tissue is weakened (smaller \(t\)), the same \(p\) produces larger stress/strain → further dilation.
How do engineers compute water loading on a dam wall?
Hydrostatic pressure varies with depth \(z\): \(p(z)=\rho g z\). The total force on a vertical rectangular face of height \(h\) and width \(W\) is \[ F=\int_0^{h}\rho g z\,W\,\mathrm{d}z=\frac{1}{2}\rho g h^{2}W, \] acting at \(z=2h/3\) below the surface.
What does the Reynolds number determine?
\[ \mathrm{Re}=\frac{\rho v D}{\mu}. \] It compares inertial to viscous forces; low \(\mathrm{Re}\) → laminar, high \(\mathrm{Re}\) → likely turbulent.
What’s the key assumption when using Bernoulli for blood?
Locally steady, incompressible, inviscid flow along a streamline at roughly constant elevation—reasonable over short, straight segments.
How does flow rate relate to velocity and area?
\[ Q=\frac{\Delta V}{\Delta t}=A v. \] For a given \(Q\), changing \(A\) inversely changes \(v\).
What generates lift on a rotating cylinder (Magnus effect)?
Rotation creates circulation \(\Gamma\) and a pressure asymmetry; lift per unit span is \[ L’=\rho V_\infty \Gamma. \]
Why is dynamic pressure useful in a Pitot tube?
\[ q=p_0-p=\frac{1}{2}\rho v^{2} \quad\Rightarrow\quad v=\sqrt{\frac{2(p_0-p)}{\rho}}. \] It converts a pressure reading into speed.
What causes flow separation behind a bluff body?
An adverse pressure gradient (\(\partial p/\partial x>0\)) decelerates the boundary layer; when wall shear vanishes and reverses, the flow detaches.
How does Bernoulli explain lower pressure beneath a fast stream of water?
At equal elevation, the higher-speed region must have lower static pressure so that \(p+\frac{1}{2}\rho v^{2}\) remains constant.
What happens to pressure near the center of a whirlpool/vortex?
High tangential speed implies a pressure drop: \[ p(r)\approx p_\infty-\frac{1}{2}\rho v_\theta^{2}(r), \] so the core has the lowest \(p\).
What limits Bernoulli’s accuracy in arteries?
Pulsatility (unsteady terms), viscosity (head losses), wall elasticity (compliance), and secondary flows/bends—all violate ideal assumptions.

20 Problems & Solutions (step-by-step)

Pipe narrows from 10 cm to 5 cm; \(v_1=2~\mathrm{m\,s^{-1}}\). Find \(v_2\).
Areas scale with diameter squared: \[ v_2=v_1\left(\frac{D_1}{D_2}\right)^{2} =2\left(\frac{0.10}{0.05}\right)^{2}=8~\mathrm{m\,s^{-1}}. \] Reason: continuity \(Q=Av\).
Wing speeds: \(v_{\text{upper}}=40\), \(v_{\text{lower}}=30~\mathrm{m\,s^{-1}}\); \(\rho=1.2\ \mathrm{kg\,m^{-3}}\). Lift per m\(^2\)?
\[ \Delta p=\frac{1}{2}\rho\!\left(v_{\text{lower}}^{2}-v_{\text{upper}}^{2}\right) =\frac{1}{2}(1.2)(900-1600)=-420~\mathrm{Pa}. \] Magnitude \(=420~\mathrm{N\,m^{-2}}\) upward.
Nozzle: water pressure drops \(300\to240~\mathrm{kPa}\). Change in speed?
\[ \Delta p=60{,}000~\mathrm{Pa},\quad v_2^{2}-v_1^{2}=\frac{2\Delta p}{\rho}=\frac{120{,}000}{1000}=120. \] Thus \(v_2=\sqrt{v_1^{2}+120}\). (A numerical value needs either \(v_1\) or \(v_2\).)
Horizontal artery: velocity doubles; \(\rho=1060\). Pressure change?
\[ \Delta p=p_2-p_1=-\frac{1}{2}\rho\left(v_2^{2}-v_1^{2}\right) =-\frac{1}{2}(1060)(4v_1^{2}-v_1^{2}) =-1590\,v_1^{2}\ \mathrm{Pa}. \]
Dam face \(h=10~\mathrm{m}\), width \(W=20~\mathrm{m}\). Hydrostatic force?
\[ F=\frac{1}{2}\rho g h^{2}W =\frac{1}{2}(1000)(9.81)(100)(20) =9.81\times10^{5}~\mathrm{N}. \]
Venturi \(\Delta p=3000~\mathrm{Pa}\) (water). Speed difference?
\[ v_2^{2}-v_1^{2}=\frac{2\Delta p}{\rho}=6 \;\Rightarrow\; |v_2-v_1|\approx \sqrt{6}=2.45~\mathrm{m\,s^{-1}} \] (if one section is much faster than the other).
Nozzle \(A=0.10~\mathrm{m^{2}}\), \(v=5\). Flow rate?
\[ Q=Av=0.10\times5=0.50~\mathrm{m^{3}\,s^{-1}}. \]
Artery \(v=0.20~\mathrm{m\,s^{-1}}\), diameter \(3~\mathrm{mm}\). Find \(Q\).
\[ A=\pi(0.0015)^{2}=7.07\times10^{-6}\ \mathrm{m^{2}},\quad Q=Av=1.41\times10^{-6}\ \mathrm{m^{3}\,s^{-1}}. \]
Tank orifice \(h=5~\mathrm{m}\) below surface. Exit speed?
\[ v=\sqrt{2gh}=\sqrt{2(9.81)(5)}\approx 9.9~\mathrm{m\,s^{-1}}. \]
Wing \(\Delta p=500~\mathrm{Pa}\), \(A=10~\mathrm{m^{2}}\). Total lift?
\[ L=\Delta p\,A=500\times10=5000~\mathrm{N}. \]
Water speeds up from 1 to 3 m/s. Pressure drop?
\[ \Delta p=\frac{1}{2}\rho (v_2^{2}-v_1^{2}) =\frac{1}{2}(1000)(9-1)=4000~\mathrm{Pa}. \]
Gate \(h=3~\mathrm{m}\), width \(5~\mathrm{m}\), fully submerged. Hydrostatic force?
\[ F=\frac{1}{2}\rho g h^{2}W =\frac{1}{2}(1000)(9.81)(9)(5)=2.207\times10^{5}~\mathrm{N}. \]
Soccer ball: \(\Delta p=100~\mathrm{Pa}\), area \(0.04~\mathrm{m^{2}}\). Lift?
\[ F=\Delta p\,A=100\times0.04=4~\mathrm{N}. \]
Vertical pipe 10 cm → 5 cm, \(Q=0.01~\mathrm{m^{3}\,s^{-1}}\). Find \(v_1,v_2\).
\[ A_1=\pi(0.05)^{2}=7.85\times10^{-3},\quad A_2=\pi(0.025)^{2}=1.96\times10^{-3}\ \mathrm{m^{2}}. \] \[ v_1=\frac{Q}{A_1}=1.27,\quad v_2=\frac{Q}{A_2}=5.10~\mathrm{m\,s^{-1}}. \]
Artery pressure drop \(100~\mathrm{Pa}\). Estimate \(\Delta v\) for \(\rho=1060\).
\[ v_2^{2}-v_1^{2}=\frac{2\Delta p}{\rho}=\frac{200}{1060}=0.188 \quad\Rightarrow\quad \Delta v\approx 0.43~\mathrm{m\,s^{-1}} \] (if initial speed is small).
Pipe \(D=2~\mathrm{cm}\) with \(v=3~\mathrm{m\,s^{-1}}\). Flow rate?
\[ A=\pi(0.01)^{2}=3.14\times10^{-4},\quad Q=Av=9.42\times10^{-4}~\mathrm{m^{3}\,s^{-1}}. \]
How does a Venturi meter yield \(Q\)?
Measure \(\Delta p=p_1-p_2\); from Bernoulli \[ v_2=\sqrt{v_1^{2}+\frac{2\Delta p}{\rho}}, \] and continuity \(A_1 v_1=A_2 v_2\). Solve for \(v_1,v_2\), then \(Q=A_2 v_2\). Include discharge coefficient \(C_d\) in practice.
Nozzle reduces pressure by \(20~\mathrm{kPa}\) (water). Velocity change?
\[ \Delta v=\sqrt{\frac{2\Delta p}{\rho}}=\sqrt{\frac{40{,}000}{1000}} =\sqrt{40}=6.32~\mathrm{m\,s^{-1}}. \]
Why does downward-deflected air imply upward lift?
Momentum balance: the wing imparts a downward momentum flux \(\dot{m}\,\Delta v\) to the air; Newton’s third law gives an equal and opposite (upward) force on the wing.
One reason real flows deviate from Bernoulli?
Viscous dissipation. The extended energy equation adds a head-loss \(h_L\): \[ \frac{p_1}{\rho g}+\frac{v_1^{2}}{2g}+z_1 =\frac{p_2}{\rho g}+\frac{v_2^{2}}{2g}+z_2+h_L\quad(\text{no pumps/turbines}). \] This term accounts for friction and turbulence.