Chapter 38: Bernoulli's Principle

38.1 Why Fluid Speed and Pressure Are Connected

Consider a horizontal pipe that narrows. Chapter 37 showed that by continuity (A₁v₁ = A₂v₂), the fluid must speed up in the narrow section. What causes this acceleration? The fluid element entering the narrow section must be pushed from behind and not fully supported from in front: the pressure behind it (upstream) must be greater than the pressure ahead of it (downstream, in the narrow section). The pressure drops wherever the fluid accelerates. This is Newton’s second law for a fluid: a fluid parcel accelerates only if there is a net pressure force acting on it. Acceleration requires a pressure gradient in the direction of acceleration. When a fluid speeds up, pressure decreases in the direction of flow. When it slows down, pressure increases in the direction of flow. Bernoulli’s equation makes this precise by tracking the energy of a fluid element as it moves along a streamline. The three forms of energy — pressure energy (stored in the compressed fluid), kinetic energy (due to motion), and gravitational potential energy (due to elevation) — trade off with each other, with their sum remaining constant Physics: Foundations of Physical Science Chapter 38 — Bernoulli's Principle along a streamline in ideal flow.

38.2 Review of Continuity in Fluid Flow

Before deriving Bernoulli’s equation, recall the continuity result from Chapter 37: for

steady, incompressible flow, the volume flow rate Q = Av is constant along any

unbranched streamtube.

A₁ v₁ = A₂ v₂

Continuity tells us how speed changes with geometry. Bernoulli’s equation will tell us how pressure changes with speed and elevation. The two equations together allow complete analysis of ideal fluid flow: given the geometry and flow conditions at one point, both speed and pressure at any other point along the streamline can be determined.

38.3 Energy in Moving Fluids

Consider a small parcel of fluid with volume ΔV, density ρ, and mass Δm = ρΔV,

moving along a streamline. The fluid parcel possesses three forms of energy:

Kinetic energy: KE = ½(Δm)v² = ½ρΔV v². A fluid parcel moving at speed v has kinetic

energy proportional to the square of its speed.

Gravitational potential energy: PE = (Δm)gh = ρΔV g h. A fluid parcel at elevation h has

gravitational potential energy proportional to h. Pressure energy (flow work): The pressure in a fluid represents the ability of the fluid to do work by pushing against other fluid parcels. The work done by the fluid at pressure P to push the parcel through area A a distance Δx is P × A × Δx = PΔV. This is often called flow work or pressure energy per unit volume.

Dividing all three energy terms by the volume ΔV gives the energy density (energy per

unit volume) associated with each: Physics: Foundations of Physical Science Chapter 38 — Bernoulli's Principle Energy form Per parcel Per unit volume (energy density) Kinetic energy

½ρΔV v²

½ρv² (Pa)

Gravitational PE

ρΔV g h

ρgh (Pa)

Pressure (flow work)

PΔV

P (Pa)

Table 38.1

Note that all three energy densities have units of Pa = N/m² = J/m³ — pressure units

and energy-per-volume units are the same. This is not a coincidence: pressure is mechanically equivalent to energy per unit volume. Bernoulli’s equation asserts that the sum of these three energy densities is constant along a streamline.

38.4 Pressure Energy, Kinetic Energy, and Gravitational Potential Energy

The derivation of Bernoulli’s equation follows from applying the work-energy theorem to a fluid parcel moving along a streamline. Imagine tracking a small parcel from position 1 (wide, lower section) to position 2 (narrow, higher section). The work done by pressure forces on the parcel as it moves equals the change in its kinetic plus gravitational potential energy. At position 1, the pressure P₁ pushes the parcel forward; at position 2, the pressure P₂ pushes it backward (opposing further entry into the narrow section). Working through the algebra of force times displacement and using the incompressibility condition (so

that the volume ΔV of the parcel is conserved), one obtains:

P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂

38.5 Bernoulli's Principle

Bernoulli's Principle: In steady, incompressible, frictionless (ideal) flow, the sum of static pressure, dynamic pressure, and hydrostatic pressure is constant along a streamline. Where flow speed is high, static pressure is low; where flow speed is low, static pressure is high. This is the qualitative statement. The quantitative form is Bernoulli's equation (Section Physics: Foundations of Physical Science Chapter 38 — Bernoulli's Principle 38.6). The principle is most strikingly illustrated in horizontal flow (h constant), where it reduces to: higher speed →lower pressure; lower speed →higher pressure. This is the form most often quoted and most often misapplied. The key word is ‘along a streamline.’ Bernoulli's equation is a statement about the energy of a fluid parcel as it travels its path. It does not compare pressures at different elevations in a static fluid (that is the hydrostatic equation, Chapter 35). It does not compare pressures on different streamlines unless the flow is irrotational (a more advanced concept). Applied within its scope, it is exact for ideal flow.

38.6 Bernoulli's Equation

P + ½ρv² + ρgh = constant

(along a streamline) Applied between two points 1 and 2 on the same streamline:

P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂

Step 1: Choose streamline:

Select two points 1 and 2 on the same streamline. Identify what is known and unknown at each.

Step 2: Apply continuity:

If areas differ, use A₁v₁ = A₂v₂ to relate speeds. This gives

one equation relating v₁ and v₂.

Step 3: Apply Bernoulli's:

Write P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂. Substitute

known values and solve.

Step 4: Check units:

All terms must be in Pa. Verify: ½ρv² →(kg/m³)(m²/s²) =

kg/(m s²) = Pa ✓. ρgh →(kg/m³)(m/s²)(m) = Pa ✓.

Water flows in a horizontal pipe

(h₁ = h₂). Wide section: diameter 8 cm, P₁ = 200 kPa, v₁ = 1.0 m/s. Narrow section:

diameter 4 cm. Find P₂. Continuity: v₂ = v₁(A₁/A₂) = 1.0 × (8/4)² = 4.0 m/s. Bernoulli:

P₂ = P₁ + ½ρ(v₁²

v₂²) = 200 000 + ½(1000)(1²

16) = 200 000

7500 = 192.5 kPa.

Pressure drops by 7.5 kPa as flow speeds up. Physics: Foundations of Physical Science Chapter 38 — Bernoulli's Principle

38.7 Fast Flow and Low Pressure

The most quoted form of Bernoulli’s principle — fast flow has low pressure — is the

horizontal (constant h) special case. Setting h₁ = h₂ and rearranging:

P₁ −P₂ = ½ρ(v₂² −v₁²)

This explains the operation of many devices. A Venturi meter (Section 38.10) constricts a pipe to speed up flow and measures the resulting pressure drop to infer flow rate. An atomiser nozzle blows air over the top of a liquid-filled tube; the fast air creates low pressure at the tube opening, and atmospheric pressure in the reservoir pushes liquid up and out. Carburettors work the same way — air speeds through a venturi throat, pressure drops, and fuel is drawn in from a bowl maintained at atmospheric pressure.

The pressure ½ρv² is called the dynamic pressure or velocity pressure. It represents the

kinetic energy per unit volume of the flowing fluid and is the pressure associated with the directed motion of the fluid. It is always additive to static pressure when the fluid is brought to rest (stagnated), which is the basis of Pitot tube speed measurement.

38.8 Flow Through Narrow Openings: Torricelli's Theorem

Consider a large open tank with a small opening (orifice) in its side at a depth h below the surface. What is the speed of water flowing out? Apply Bernoulli’s equation from the surface (point 1) to the orifice (point 2). At the surface, the pressure is P₀ (atmospheric) and the speed is approximately zero (large

tank). At the orifice, the pressure is also P₀ (open to atmosphere). Setting h₂ = 0 (orifice

as datum) and h₁ = h:

P₀ + 0 + ρgh = P₀ + ½ρv² + 0

v = √(2gh)

surface. v = √(2 × 9.81 × 3.0) = √58.86 = 7.67 m/s. If the hole has area 0.5 cm² = 5 × Physics: Foundations of Physical Science Chapter 38 — Bernoulli's Principle 10 ⁵ m², the flow rate is Q = Av = 5 × 10 ⁵ × 7.67 = 3.84 × 10 ⁴ m³/s = 0.384 L/s.

Evangelista Torricelli (1608–1647), a student of Galileo, discovered this result and also invented the mercury barometer. He correctly identified that atmospheric pressure supports a column of mercury about 76 cm tall, and by analogy recognised that air has weight and exerts pressure. Torricelli ’s theorem was one of the first quantitative results in fluid dynamics and predates Newton’s mechanics by several decades.

38.9 Lift and Airfoils

An airfoil (the cross-sectional shape of a wing) is designed so that air flows faster over the upper curved surface than over the lower flatter surface. By Bernoulli’s principle, faster flow over the top creates lower pressure on the upper surface than on the lower surface. The pressure difference produces a net upward force — lift. The lift force per unit span on an airfoil is given (in ideal flow) by the Kutta-Joukowski theorem, which involves the flow speed and the circulation (a measure of the airfoil’s tendency to induce rotation in the flow). For a first approximation: if the pressure

above the wing is P_upper and below is P_lower, and the wing area is A_wing:

F_lift = (P_lower −P_upper) × A_wing

A common misconception is that lift arises purely because the upper path is longer and air must “meet up” with air taking the lower path (the “equal transit time” fallacy). This is incorrect: air over the upper surface does travel faster, but not because of equal transit times. The correct explanation involves the angle of attack (the angle between the wing chord and the oncoming airflow), the wing’s camber (curvature), and the Kutta condition (that the flow must leave smoothly from the trailing edge). Bernoulli’s equation correctly relates the speed increase to the pressure decrease; what requires more careful analysis is why the flow speeds up. Lift also depends on angle of attack: tilting the wing leading edge upward increases lift (up to the stall angle, beyond which the flow separates from the upper surface and lift collapses dramatically). This is why aircraft rotate (pitch up) on takeoff — increasing angle of attack to generate sufficient lift at the available speed. Physics: Foundations of Physical Science Chapter 38 — Bernoulli's Principle

Aerodynamic lift is real and quantitatively described by Bernoulli’s principle (for the pressure-speed relationship) combined with circulation theory (for the explanation of why the speed difference arises). The "equal transit time" explanation commonly taught at elementary level is incorrect and should be discarded. Air over the upper surface arrives at the trailing edge before air over the lower surface, not at the same time.

38.10 Venturi Effect

The Venturi effect is the pressure drop that occurs when a fluid is forced through a constricted section of pipe. It is named after Giovanni Battista Venturi (1746–1822), who demonstrated the effect experimentally. The Venturi meter exploits this pressure drop to measure flow rate. In a Venturi meter, a fluid flows through a converging section (reducing area), a throat (minimum area), and a diverging section (expanding area). By continuity, flow speed is highest at the throat. By Bernoulli’s equation (horizontal flow), pressure is lowest at the throat. The pressure difference between the inlet and the throat is measured (typically with a U-tube manometer or a differential pressure transducer). This pressure difference can be directly related to the flow rate:

ΔP = P₁ −P₂ = ½ρ(v₂² −v₁²)

Q = A₁ A₂ √(2ΔP / (ρ(A₁² −A₂²)))

A horizontal water pipe (inlet diameter 100 mm, throat diameter 50 mm) has a pressure difference of 40 kPa between inlet and throat. Find the flow rate. A₁ = π(0.05)² = 7.854 × 10 ³ m²; A₂ = π(0.025)² = 1.963 × 10 ³ m². v₂ =

v₁(A₁/A₂) = 4v₁. Bernoulli: 40 000 = ½(1000)(16v₁²

v₁²) = 7500 v₁². v₁ = √(40

000/7500) = 2.31 m/s. Q = A₁v₁ = 7.854 × 10 ³ × 2.31 = 0.0181 m³/s = 18.1 L/s. Venturi meters are used in municipal water supply monitoring, oil pipeline flow measurement, medical oxygen delivery systems, and aircraft fuel systems. Their advantage over orifice plates is lower permanent pressure loss (the diverging section recovers most of the pressure drop); their disadvantage is larger size and higher cost. Physics: Foundations of Physical Science Chapter 38 — Bernoulli's Principle

38.11 Applications in Medicine, Engineering, and Weather

Bernoulli’s principle appears across many fields wherever fluids flow through varying geometries. Medicine: The Bernoulli effect in blood flow is clinically important. Atherosclerotic plaques narrow arterial cross-sections, increasing blood flow speed through the constriction (by continuity) and decreasing pressure (by Bernoulli). The reduced pressure can cause the artery walls to be drawn inward, potentially worsening the narrowing — a positive feedback effect called the Venturi effect in haemodynamics.

Echocardiography uses the modified Bernoulli equation (ΔP = 4v², where v is the peak

flow speed in m/s and ΔP is in mmHg) to non-invasively estimate pressure gradients

across narrowed heart valves. Engineering: Carburettors mix fuel and air using a Venturi throat. Aircraft Pitot-static systems measure airspeed using the stagnation pressure at a forward-facing orifice

minus the static pressure at a side-facing orifice: ΔP = ½ρv², giving v = √(2ΔP/ρ).

Injectors in diesel engines use high-speed fuel jets to entrain (suck in) air. Ejector pumps in submarines and industrial equipment use fast jets to draw in and move surrounding fluid with no moving parts. Weather: Pressure variations in the atmosphere are driven partly by Bernoulli effects. Wind blowing over a hill speeds up over the crest (reduced cross-sectional area of the air column above terrain), creating a pressure drop. In extreme cases (mountain passes, building corners), localised high-speed flow and low pressure can strip roofs, overturn vehicles, or accelerate erosion. The Venturi effect between closely spaced buildings creates dangerous wind corridors in urban environments — a key consideration in architectural fluid dynamics.

38.12 Limits of Bernoulli's Principle

Bernoulli's equation is derived under four assumptions: steady flow, incompressible flow, inviscid (frictionless) flow, and application along a single streamline. Each assumption fails in certain situations: Unsteady flow: When flow conditions change with time (pumps starting, valves closing, pulsatile blood flow), the steady-flow Bernoulli equation is inapplicable. The unsteady Bernoulli equation includes an additional term involving the time derivative of velocity, which accounts for the energy stored in accelerating the fluid mass. Compressibility: At high speeds (Mach > 0.3 for gases), density changes significantly with pressure. Bernoulli's equation for compressible flow has a more complex pressure- Physics: Foundations of Physical Science Chapter 38 — Bernoulli's Principle velocity relationship, and at supersonic speeds, shock waves create sudden pressure jumps that the equation cannot describe at all. Viscosity and friction: Real fluids lose energy to viscous friction, especially near walls (boundary layers) and in turbulent eddies. This energy is not recovered — it is converted to heat. A more general energy equation includes a loss term (head loss) that accounts for friction. In a long pipe, most of the pressure drop is due to friction, not to Bernoulli effects. Ignoring friction in a long pipe and applying Bernoulli gives dramatically incorrect results. Different streamlines: Bernoulli's equation relates conditions on the same streamline. In rotational flows (where fluid elements spin as they move), conditions on different streamlines cannot be compared using Bernoulli's equation without additional assumptions. In irrotational flows (common around streamlined bodies at high Reynolds number), the Bernoulli constant is the same on all streamlines, and inter- streamline comparison is valid.

Do NOT apply Bernoulli's equation: (1) across a pump, fan, or turbine (which add or remove energy from the flow), (2) when significant friction losses occur (long pipes, rough surfaces, sharp bends), (3) at supersonic conditions, or (4) to compare points on different streamlines in rotational flow. In all these cases, Bernoulli gives the wrong answer.

38.13 Common Mistakes in Bernoulli Problems

• Applying Bernoulli across a pump or turbine. Bernoulli conserves energy along a streamline; pumps and turbines add or remove energy. If the flow path includes

a pump, use the extended Bernoulli equation: P₁ + ½ρv₁² + ρgh₁ + W_pump =

P₂ + ½ρv₂² + ρgh₂ + W_losses.

• Using gauge pressure in Bernoulli's equation when absolute pressure is needed. Bernoulli's equation is consistent whether you use absolute or gauge pressure, as long as you are consistent: the atmospheric P₀ appears on both sides and cancels. The mistake is mixing absolute pressure at one point with gauge pressure at another. Use one system throughout. •

Forgetting to include the elevation term ρgh when height changes are

significant. For water at height changes of more than ∼0.5 m, the gravitational term is comparable to typical dynamic pressure changes and cannot be neglected. • Not applying continuity first. Bernoulli’s equation has two unknowns (P and v) at the second point if the speed is not known. Continuity gives the speed. Physics: Foundations of Physical Science Chapter 38 — Bernoulli's Principle Forgetting to use continuity before Bernoulli leads to under-determined problems. •

Using the wrong fluid density. Bernoulli's ρ is the flowing fluid density, not the

density of a manometer fluid used to measure the pressure difference. If a

mercury manometer measures ΔP, use ρ_Hg g h to find ΔP, then use this ΔP in

Bernoulli with ρ_water (or whichever fluid is flowing).

• Confusing total pressure with static pressure. Static pressure P is the pressure measured by a gauge moving with the fluid (or a side-facing orifice). Total

(stagnation) pressure is P_total = P + ½ρv², measured at a point where the flow

is brought to rest. Bernoulli’s equation says total pressure is constant along a streamline (in horizontal flow with no elevation change). A Pitot tube measures stagnation pressure; a static port measures static pressure; their difference is the dynamic pressure.

Bernoulli's equation is a statement about energy conservation in ideal flow. It is not a magical rule that 'fast flow always causes low pressure everywhere.' The pressure reduction requires the flow to be along the same streamline, with no energy additions, subtractions, or losses. Always check whether the ideal flow conditions are met before applying Bernoulli. Physics: Foundations of Physical Science Chapter 38 — Bernoulli's Principle

Chapter Summary

Bernoulli's equation is the energy conservation equation for ideal (steady, incompressible, inviscid) flow. Along any streamline, the sum of static pressure, dynamic pressure, and hydrostatic pressure is constant.

P + ½ρv² + ρgh = constant

• Static pressure P decreases where flow speed increases (horizontal flow); this is the quantitative form of "fast flow, low pressure." •

Dynamic pressure = ½ρv². Total (stagnation) pressure = P + ½ρv² = constant

along a streamline in horizontal ideal flow. •

Torricelli's theorem: exit speed through orifice at depth h is v = √(2gh).

• Venturi effect: pressure drop in constriction can be measured to infer flow rate Q. • Lift on airfoils arises from faster flow (lower pressure) on the upper surface than the lower surface. • Bernoulli's equation fails: across pumps/turbines, with significant friction, at high Mach numbers, or when comparing different streamlines in rotational flow. •

Always combine Bernoulli with continuity (A₁v₁ = A₂v₂) to solve for both speed

and pressure at unknown points. Physics: Foundations of Physical Science Chapter 38 — Bernoulli's Principle

Key Terms

Bernoulli's equation

P + ½ρv² + ρgh = constant; the energy conservation equation

for steady, incompressible, inviscid flow along a streamline.

Bernoulli's principle

In ideal flow, regions of high speed have lower static pressure;

regions of low speed have higher static pressure (for flow at constant elevation).

static pressure (P)

The pressure in a fluid measured by a gauge moving with the fluid

or by a side-facing port; represents the pressure energy per unit volume.

dynamic pressure

½ρv²; the kinetic energy per unit volume of a moving fluid; the

pressure recovered when the fluid is brought to rest.

total pressure (stagnation pressure)

P + ½ρv²; the pressure at a point where the

fluid is brought to rest isentropically; constant along a streamline in ideal horizontal flow.

Torricelli's theorem

The exit speed of a fluid through an orifice at depth h below the

free surface is v = √(2gh); a special case of Bernoulli’s equation.

Venturi effect

The pressure drop accompanying flow speed increase in a constricted

section; used in Venturi meters to measure flow rate.

Pitot tube

A device that measures flow speed by comparing stagnation pressure

(forward orifice) and static pressure (side orifice)

v = √(2ΔP/ρ).

lift

The aerodynamic force perpendicular to the oncoming flow direction on an airfoil;

arises from the pressure difference between faster-flow upper surface and slower- flow lower surface.

angle of attack

The angle between the chord line of an airfoil and the direction of

oncoming airflow; increasing angle of attack increases lift until the stall angle.

stall

The condition in which the angle of attack exceeds the critical value, causing flow

separation from the upper wing surface and sudden collapse of lift.

head loss

Energy dissipated per unit weight of fluid by viscous friction in a real flow

system; the correction added to Bernoulli’s equation for real pipes.

Physics

Foundations of Physical Science

Chapter 38 — Bernoulli's Principle

Concept Review Questions

Answer in complete sentences. No calculations required unless stated.

1. State Bernoulli's principle in your own words. What three forms of energy per

unit volume does it balance?

2. Explain why, in a horizontal pipe that narrows, the pressure drops in the narrow

section. Begin with Newton's second law, not Bernoulli's equation.

3. Derive Torricelli's theorem from Bernoulli's equation, showing all steps. What

approximation is made about the surface speed?

4. A Pitot tube on an aircraft measures stagnation pressure P_stag and static

pressure P_static. Write the formula for airspeed v in terms of these pressures

and air density ρ.

5. Explain why the "equal transit time" explanation of airfoil lift is incorrect. What is

the correct physical explanation?

6. List four conditions under which Bernoulli's equation should NOT be applied. For

each, state what goes wrong physically.

7. A doctor uses the simplified Bernoulli equation ΔP = 4v² (with ΔP in mmHg and

v in m/s) to estimate pressure drop across a heart valve. What assumptions underlie this formula?

8. Water flows from a large reservoir through a pipe that descends 10 m and then

rises 5 m. At the lowest point, the pipe narrows. Using Bernoulli’s equation, qualitatively describe how pressure changes at each stage of this flow path. Physics: Foundations of Physical Science Chapter 38 — Bernoulli's Principle Bernoulli Concept Questions

9. Two identical tubes are immersed in water side by side. Tube A has air blown

across its top at high speed; Tube B is open to still air. In which tube does water rise higher? Explain.

10. An aircraft wing generates more lift at lower altitude than at higher altitude for

the same speed. Explain in terms of air density and Bernoulli’s equation.

11. A constriction in a horizontal pipe causes water to speed up. The pressure drops.

If the constriction is followed by a gradual expansion (as in a Venturi meter), what happens to pressure after the constriction? Does it return to its original value in ideal flow? In real flow?

12. A 1 m² plate is held perpendicular to an airstream moving at 20 m/s. On the

upstream face, air is slowed to nearly zero (stagnation). Estimate the force on the

plate using dynamic pressure. (ρ_air = 1.2 kg/m³)

Bernoulli Equation Problems

13. Water flows horizontally from a large tank through a pipe of diameter 6.0 cm at

P₁ = 180 kPa and v₁ = 1.5 m/s. The pipe narrows to diameter 3.0 cm. Find (a) v₂

by continuity, (b) P₂ by Bernoulli, and (c) the pressure difference P₁ P₂.

14. A pipe descends 4.0 m (h₁ = 4.0 m above h₂ = 0). The diameter is constant at

5.0 cm. Flow rate Q = 2.0 × 10 ³ m³/s. P₁ = 120 kPa. Find v₁ = v₂ (constant diameter) and P₂.

15. A horizontal Venturi meter (inlet diameter 80 mm, throat diameter 40 mm) is

connected to a mercury U-tube manometer. The manometer reads a height

difference of 150 mm. (ρ_Hg = 13 600 kg/m³, ρ_water = 1000 kg/m³). Find (a) ΔP,

(b) inlet speed, (c) flow rate.

16. A small hole (area 0.80 cm²) is drilled in the side of a large water tank 2.5 m

below the surface. (a) Find the exit speed using Torricelli’s theorem. (b) Find the

flow rate through the hole. (c) If the tank has a cross-sectional area of 1.0 m²,

how fast is the surface dropping? Is the Torricelli approximation (surface speed 0) justified?

17. An aircraft (mass 15 000 kg) has wings of total area 55 m². Air density at cruising

altitude is 0.90 kg/m³. For level flight, lift must equal weight. (a) Find the

required lift force. (b) Find the required pressure difference between lower and Physics: Foundations of Physical Science Chapter 38 — Bernoulli's Principle upper wing surfaces. (c) If the speed below the wing is 220 m/s, find the speed above the wing using Bernoulli.

Venturi and Lift Analysis

18. A Venturi meter in an oil pipeline (ρ_oil = 850 kg/m³) has inlet diameter 200 mm

and throat diameter 100 mm. A differential pressure gauge reads 25 kPa. Find the volume flow rate and the mass flow rate.

19. A household tap (faucet) has a 20 mm diameter opening. When the valve is 80%

open, the flow rate is 8 L/min at a supply pressure of 300 kPa gauge. (a) Find the exit speed of water from the tap. (b) Using Bernoulli, estimate the pressure just inside the tap opening (treating the tap as a horizontal Venturi). (c) Comment on whether ideal flow is a good model here.

20. An airfoil in a wind tunnel has upper surface speed 80 m/s and lower surface

speed 60 m/s. Air density = 1.2 kg/m³. (a) Find the pressure difference between

lower and upper surfaces. (b) If the wing chord (width) is 0.5 m and span (length) is 3.0 m, find the lift force. (c) Express lift as a weight equivalent (kg) that the wing can support.

21. A small Pitot tube on a boat measures stagnation pressure P_stag = 112 500 Pa

and static pressure P_static = 101 325 Pa (atmospheric). Find the boat’s speed.

(ρ_water = 1000 kg/m³; note: the Pitot tube is submerged).

Physics: Foundations of Physical Science Chapter 38 — Bernoulli's Principle

Applied Fluid Dynamics Problems

22. A hospital oxygen delivery system uses a Venturi injector: pure oxygen flows

through a narrow throat (diameter 3.0 mm, pressure 200 kPa absolute, speed 40

m/s) and entrains room air through a side port at atmospheric pressure. ρ_O2

ρ_air = 1.3 kg/m³ for this estimate. Find (a) the throat dynamic pressure, (b) the

minimum pressure in the throat, and (c) whether the throat pressure is below atmospheric (enabling air entrainment).

23. A water main runs horizontally at ground level (P = 400 kPa, diameter 150 mm,

speed 1.2 m/s). It rises 8.0 m vertically and continues horizontally at the same diameter. Find the pressure at the top of the rise. Is the pressure sufficient to reach a tap on the second floor?

24. A large dam has a penstock (supply pipe) of diameter 1.5 m and height 80 m.

Water exits at the turbine inlet (diameter 1.5 m, height 0). (a) Using Torricelli (zero speed at reservoir surface, speed at penstock exit, equal pressures), find the ideal exit speed and flow rate. (b) If the turbine extracts 85% of the water’s

kinetic energy, find the power output in MW. (ρ = 1000 kg/m³)

25. A house roof (area 120 m², flat) is exposed to a wind of 30 m/s. Under the roof,

air is still. (ρ_air = 1.2 kg/m³). (a) Find the dynamic pressure of the wind. (b) Find

the pressure difference between below and above the roof. (c) Find the net upward force on the roof. (d) Express this as the equivalent weight in kg-force. Why must roofs be firmly attached to walls? Physics: Foundations of Physical Science Chapter 38 — Bernoulli's Principle

Chapter 38 Checkpoint

Before completing Part VII, confirm you can do each of the following. I can… See Section Explain the connection between fluid speed and pressure using Newton’s second law 38.1 State the three energy densities in Bernoulli’s equation and their units 38.3–38.4 Write Bernoulli's equation and identify each term 38.6 Apply the two-point Bernoulli equation combined with continuity 38.6 Derive and apply Torricelli's theorem for flow from an orifice 38.8 Explain aerodynamic lift in terms of pressure difference (not equal transit times) 38.9 Describe the Venturi effect and solve Venturi meter problems 38.10 Apply Bernoulli to medical, engineering, and weather contexts 38.11 List conditions under which Bernoulli's equation does not apply 38.12 Identify and avoid common errors in Bernoulli problems 38.13