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Bernoulli's Applications
Bernoulli's Equation
Bernoulli's equation relates the pressure, speed, and height of an ideal fluid (incompressible, non-viscous, steady flow) at any two points along a streamline.
$$P + \frac{1}{2}\rho v^2 + \rho gh = \text{constant}$$
The total mechanical energy per unit volume is conserved along a streamline.
$P$=Static pressure(Pa)
$\frac{1}{2}\rho v^2$=Dynamic pressure (kinetic energy per unit volume)(Pa)
$\rho gh$=Hydrostatic pressure (potential energy per unit volume)(Pa)
Horizontal pipe ($h_1 = h_2$)
→$P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2$ — height terms cancel
Static fluid ($v = 0$)
→Reduces to $P + \rho gh = \text{constant}$ — the hydrostatic equation
Three Assumptions: The fluid must be (1) incompressible, (2) non-viscous, and (3) in steady (laminar) flow.
Energy Conservation: Derived from the work-energy theorem — net work done by pressure forces equals the change in KE + PE of the fluid element.
Dimensional Check: Each term has dimensions of $[M][L]^{-1}[T]^{-2}$ (pressure), confirming dimensional consistency.
Core Insight: Where the speed is high, the pressure is low — this inverse relationship drives all Bernoulli applications.
Torricelli's Theorem
Torricelli's Theorem states that the speed of efflux from an orifice is equivalent to the velocity gained by an object falling freely from the free surface to the hole's depth.
$$v = \sqrt{2gh}$$
The exit velocity depends solely on the vertical depth of the fluid above the hole.
$v$=Efflux speed(m/s)
$g$=Acceleration due to gravity(m/s²)
$h$=Depth of orifice below surface(m)
Pressurized tank ($P_{top} > P_{atm}$)
→$v = \sqrt{2gh + \frac{2(P_{top} - P_{atm})}{\rho}}$ — gauge pressure adds to velocity
Orifice pointed upward
→The jet rises back to the original water surface level (KE reconverts to PE)
Density Independence: The exit speed is independent of the liquid's density — $\rho$ cancels from both sides of Bernoulli's equation.
Atmospheric Cancellation: Both the top surface and the exit hole are at atmospheric pressure $P_0$, so $P_0$ cancels out.
Free-Fall Equivalence: The efflux speed exactly matches $v = \sqrt{2gh}$ for a ball falling height $h$ — pure PE → KE conversion.
Maximum Range: For a jet from height $y$ above ground in a tank of total height $H$, horizontal range $R = 2\sqrt{y(H - y)}$, maximized when $y = H/2$.
Speed-Pressure Relationship
A direct consequence of Bernoulli's equation for horizontal flow: where the speed of the fluid is high, the static pressure is low, and vice versa.
$$P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2$$
For a horizontal pipe, the height terms cancel, leaving an inverse trade-off between pressure and speed.
$P_1, P_2$=Static pressures at two points(Pa)
$v_1, v_2$=Flow speeds at the same points(m/s)
$v_1 = 0$ (stagnation)
→$P_1 - P_2 = \frac{1}{2}\rho v_2^2$ — all dynamic pressure converts to static
Streamline Crowding: In narrower regions, streamlines bunch together — fluid moves faster, pressure drops.
Danger Near Trains: A person near a fast train feels pulled toward it because the fast-moving air between them and the train has lower pressure than the still air behind.
Two Boats Pulled Together: Parallel boats moving in the same direction create fast-flowing water between them, lowering pressure and pushing them toward each other.
The Venturi Effect & Meter
The Venturi effect is the pressure drop that occurs when fluid flows through a constriction. A Venturi meter uses this effect to measure flow rate by reading the pressure difference via a manometer.
$$P_1 - P_2 = \frac{1}{2}\rho(v_2^2 - v_1^2)$$
The pressure drop between the wide section and the narrow throat equals the gain in dynamic pressure.
$P_1$=Pressure in wide section(Pa)
$P_2$=Pressure in throat(Pa)
$v_1, v_2$=Flow speeds in wide and narrow sections(m/s)
$A_2 \ll A_1$
→$v_1 \approx 0$ and $P_1 - P_2 \approx \frac{1}{2}\rho v_2^2$ — the simplified Venturi relation
Continuity Link: The constriction forces the same volume through a smaller area per second, so $v$ must increase ($A_1 v_1 = A_2 v_2$).
Manometer Reading: The U-tube manometer between wide and narrow sections shows the pressure head difference directly.
Carburetor Application: In a car engine, air rushing through a Venturi constriction creates low pressure that draws petrol vapour into the air stream.
Filter Pump / Aspirator: A fast water jet through a constriction creates low pressure that draws air in from a side tube.
Dynamic Lift & Magnus Effect
Dynamic lift on an aerofoil arises because the wing shape forces air to move faster over the top (lower pressure) than the bottom (higher pressure). The Magnus effect creates lateral force on spinning objects.
$$F_{lift} = \frac{1}{2}\rho(v_{top}^2 - v_{bottom}^2) \times A$$
The net upward force equals the pressure difference between bottom and top surfaces multiplied by the wing area.
$\Delta P$=Pressure difference between bottom and top(Pa)
$A$=Wing (or surface) area(m²)
$v_{top}, v_{bottom}$=Air speeds above and below the wing(m/s)
Camber Design: The wing's curved upper surface forces air to travel a longer path in the same time, increasing $v_{top}$ and lowering pressure above.
Magnus Effect: A spinning cricket ball drags air with it — speed adds on one side, subtracts on the other, creating a lateral pressure difference that produces 'swing'.
Chimney Effect: Wind blowing across a chimney top creates low pressure above, enhancing the upward draft of smoke. Taller chimneys are more effective.
Atomizer / Sprayer: Air blown over the top of a tube dipped in liquid creates low pressure, lifting liquid up and atomizing it (used in perfume bottles and paint sprayers).
A Pitot tube measures fluid velocity by comparing stagnation point pressure (where flow stops) to static pressure in the undisturbed stream.
$$v = \sqrt{\frac{2(P_{stag} - P_{static})}{\rho}}$$
Converts measured dynamic pressure into flow velocity.
$P_{stag}$=Stagnation (total) pressure at the tube opening(Pa)
$P_{static}$=Static pressure from a side tap(Pa)
$\rho$=Fluid density(kg/m³)
Stagnation Point: The point directly facing the flow where fluid velocity drops to zero and all KE converts to pressure energy.
Aircraft Use: Pitot tubes mounted on airplane fuselages measure airspeed by sensing the ram pressure of incoming air.
Dimensional Check: $\sqrt{Pa / (kg/m^3)} = \sqrt{(kg \cdot m^{-1} \cdot s^{-2}) / (kg \cdot m^{-3})} = \sqrt{m^2/s^2} = m/s$ ✓
Blood Pressure & Biological Flow
Blood pressure measurement relies on Bernoulli's principle — when an external cuff pressure equals the systolic pressure, flow resumes through a constricted artery at high speed, producing turbulence detected by a stethoscope.
Systolic vs Diastolic: Systolic (~120 torr) is peak pressure during a heartbeat; diastolic (~75–80 torr) is the minimum between beats. 1 torr = 133.3 N/m².
Sphygmomanometer: The inflatable cuff compresses the artery until flow stops. As pressure is released, the first sounds (Korotkoff sounds) mark systolic pressure; their disappearance marks diastolic.
Continuity in Capillaries: Blood from the aorta (radius ~1 cm, $v \approx 30$ cm/s) flows through millions of capillaries with combined cross-section ~2000 cm², so capillary speed drops to ~0.05 cm/s.
Posture Effect: Blood pressure in the feet is higher than in the head by $\rho g \Delta h$ — lying horizontal minimizes this variation.
Application Strategy
Worked Examples
Problem Statement
Calculate the speed of efflux of kerosene from a tank through a small hole at a depth of 2 m. The tank is pressurized at the top with 2 atm ($2.02 \times 10^5$ Pa). (Density of kerosene = 800 kg/m³, $P_{atm} = 1.01 \times 10^5$ Pa)
Bernoulli's EquationTorricelli's Theorem (Extended)
Strategic Toolkit
$$P_1 + \rho gh_1 + \frac{1}{2}\rho v_1^2 = P_2 + \rho gh_2 + \frac{1}{2}\rho v_2^2$$
Step 1
Setup Bernoulli's Equation
At the surface (1), $v_1 \approx 0$ (large tank). At the orifice (2), set $h_2 = 0$.
$$P_{top} + \rho gh = P_{atm} + \frac{1}{2}\rho v^2$$
Step 2
Solve for Velocity
Rearrange: $P_{exit} = P_{atm}$, $P_{top} = 2.02 \times 10^5$ Pa.
$$v = \sqrt{\frac{2(P_{top} - P_{atm})}{\rho} + 2gh}$$
Step 3
Substitute Values
$\Delta P = 2.02 \times 10^5 - 1.01 \times 10^5 = 1.01 \times 10^5$ Pa, $h = 2$ m.
$$v = \sqrt{\frac{2(1.01 \times 10^5)}{800} + 2(9.8)(2)} = \sqrt{252.5 + 39.2}$$
Step 4
Final Calculation
Calculate the square root.
$$v = \sqrt{291.7} \approx 17.1 \text{ m/s}$$
Problem Statement
Water flows through a horizontal Venturi meter with an inlet diameter of 10 cm and a throat diameter of 5 cm. If the pressure difference between the inlet and throat is 2000 Pa, find the velocity at the inlet.
Continuity EquationBernoulli's Equation
Strategic Toolkit
$$P_1 - P_2 = \frac{1}{2}\rho(v_2^2 - v_1^2)$$
Step 1
Area Ratio from Continuity
From $A_1 v_1 = A_2 v_2$. Since $d_1 = 2d_2$, area scales as $d^2$, so $A_1 = 4A_2$.
$$v_2 = \frac{A_1}{A_2}v_1 = 4v_1$$
Step 2
Apply Bernoulli
Substitute $v_2 = 4v_1$ into the horizontal Bernoulli equation.
$$\Delta P = \frac{1}{2}\rho((4v_1)^2 - v_1^2) = \frac{1}{2}\rho(15v_1^2)$$
Step 3
Solve for Inlet Velocity
Plug in $\Delta P = 2000$ Pa and $\rho = 1000$ kg/m³.
$$2000 = \frac{1}{2}(1000)(15 v_1^2) \implies v_1^2 = \frac{2000}{7500}$$
Step 4
Final Value
Take the square root.
$$v_1 = \sqrt{0.267} \approx 0.52 \text{ m/s}$$
Problem Statement
Water flows through a closed pipe system. At the lower point, the speed is 3.0 m/s and pressure is 80 kPa. At a point 3.0 m higher, the speed is 4.0 m/s. What is the pressure at the upper point?
Bernoulli's Equation with Height
Strategic Toolkit
$$P_1 + \frac{1}{2}\rho v_1^2 + \rho g h_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho g h_2$$
Step 1
Set Reference Level
Let $h_1 = 0$ (lower point) and $h_2 = 3.0$ m (upper point).
$$P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2 + \rho g h_2$$
Step 2
Solve for Upper Pressure
Rearrange to isolate $P_2$.
$$P_2 = P_1 + \frac{1}{2}\rho(v_1^2 - v_2^2) - \rho g h_2$$
Step 3
Substitute Values
$P_1 = 80{,}000$ Pa, $v_1 = 3.0$ m/s, $v_2 = 4.0$ m/s, $h_2 = 3.0$ m.
$$P_2 = 80{,}000 + \frac{1}{2}(1000)(9 - 16) - (1000)(9.8)(3.0)$$
Step 4
Final Calculation
Evaluate each term.
$$P_2 = 80{,}000 - 3{,}500 - 29{,}400 = 47{,}100 \text{ Pa} \approx 47 \text{ kPa}$$
Problem Statement
An airplane wing has air flowing at 450 m/s across the top and 410 m/s below. Find the pressure difference between the surfaces. ($\rho_{air} = 1.29$ kg/m³)
Dynamic LiftBernoulli's Equation
Strategic Toolkit
$$\Delta P = \frac{1}{2}\rho(v_{top}^2 - v_{bottom}^2)$$
Step 1
Identify the Horizontal Form
Both surfaces are at roughly the same height, so height terms cancel.
$$\Delta P = \frac{1}{2}\rho(v_{top}^2 - v_{bottom}^2)$$
Step 2
Calculate Speed Squares
$v_{top}^2 = 450^2 = 202{,}500$, $v_{bottom}^2 = 410^2 = 168{,}100$.
$$v_{top}^2 - v_{bottom}^2 = 202{,}500 - 168{,}100 = 34{,}400 \text{ m}^2/\text{s}^2$$
Step 3
Final Calculation
Multiply by $\frac{1}{2}\rho$.
$$\Delta P = \frac{1}{2}(1.29)(34{,}400) = 22{,}188 \text{ Pa} \approx 22 \text{ kPa}$$
Memorize formulas
Bernoulli's Equation
Energy conservation per unit volume along a streamline.
Formula
$P + \frac{1}{2}\rho v^2 + \rho gh = \text{constant}$
Torricelli's Speed
Efflux speed equals free-fall velocity from depth h.
Formula
$v = \sqrt{2gh}$
Venturi Pressure Drop
Pressure difference in horizontal constriction.
Formula
$P_1 - P_2 = \frac{1}{2}\rho(v_2^2 - v_1^2)$
Dynamic Lift Force
Net upward force from asymmetric airflow over a wing.
Formula
$F_L = \frac{1}{2}\rho(v_{top}^2 - v_{bottom}^2) \times A$
Pitot Tube Velocity
Flow speed from stagnation vs static pressure.
Formula
$v = \sqrt{\frac{2\Delta P}{\rho}}$
Review
1
An airplane wing is designed such that the air speed above the wing is 80 m/s and below the wing is 60 m/s. If the density of air is 1.2 kg/m³ and the wing area is 20 m², what is the net upward lift force?
Ans
33,600 N
Review
Tap to reveal
Find the pressure difference from Bernoulli
The faster flow above the wing lowers pressure relative to the slower flow below it. Use the speed-squared difference, not the simple speed difference.
$$\Delta P = \tfrac{1}{2}\rho(v_{top}^2 - v_{bot}^2) = \tfrac{1}{2}(1.2)(80^2 - 60^2) = 1680\ \text{Pa}$$
Convert pressure difference into force
Lift is the pressure difference multiplied by the wing area.
$$F = \Delta P\,A = 1680 \times 20 = 33{,}600\ \text{N}$$
2
A spinning cricket ball 'swings' in the air. This phenomenon is best explained by:
Ans
The Magnus Effect — spin creates asymmetric air speeds and a lateral pressure difference
Review
Tap to reveal
Identify the pressure imbalance caused by spin
Spin drags air with it, making one side of the ball move faster relative to the air and the opposite side slower. That speed asymmetry creates a pressure difference across the ball.
Faster air on one side means lower pressure there, so the ball is pushed sideways.
3
Water flows through a pipe that goes over a hill. At ground level, the pipe has a cross-sectional area of 0.040 m² and the water speed is 2.0 m/s at a pressure of 300 kPa. At the top of the hill (8.0 m above ground), the pipe narrows to 0.020 m². What is the pressure at the hilltop?
Ans
216 kPa
Review
Tap to reveal
Use continuity to find the hilltop speed
Since the pipe narrows by a factor of 2 in area, the speed doubles to keep flow rate constant.
$$A_1v_1=A_2v_2\Rightarrow v_2=2.0\times\frac{0.040}{0.020}=4.0\ \text{m/s}$$
Apply Bernoulli between the two points
The hilltop pressure is reduced by both the increase in speed and the gain in height.
$$P_2=300{,}000+\tfrac{1}{2}(1000)(2.0^2-4.0^2)-(1000)(9.8)(8.0)=215{,}600\ \text{Pa}$$
Round to the nearest option
The computed pressure is about 216 kPa, which matches the first choice.
$$P_2\approx 216\ \text{kPa}$$
4
Water exits a hole near the bottom of a large open tank and strikes the ground 4.0 m below the hole. If the water lands 3.0 m horizontally from the base of the tank, how far below the water surface is the hole? ($g = 9.8$ m/s²)
Ans
0.57 m
Review
Tap to reveal
Find the time of fall
The stream leaves horizontally, so the vertical motion is free fall from a height of 4.0 m.
$$4.0=\tfrac{1}{2}gt^2\Rightarrow t=\sqrt{\frac{2(4.0)}{9.8}}=0.904\ \text{s}$$
Use horizontal motion to get the exit speed
Horizontal speed stays constant while the water is in flight.
$$v=\frac{x}{t}=\frac{3.0}{0.904}=3.32\ \text{m/s}$$
Convert exit speed to depth below the surface
Torricelli's law gives the relation between exit speed and depth under the surface.
$$h=\frac{v^2}{2g}=\frac{3.32^2}{2(9.8)}\approx 0.57\ \text{m}$$
5
Water flows through a horizontal pipe of internal diameter 1 cm at 1 m/s. If the water must emerge from a nozzle at 21 m/s, what should the nozzle diameter be?
Ans
0.22 cm
Review
Tap to reveal
Use continuity for area and speed
For incompressible flow, the product of area and speed stays constant, so the nozzle area must shrink by a factor of 21.
$$A_1v_1=A_2v_2\Rightarrow A_2=A_1\frac{1}{21}$$
Convert area change into diameter change
Since area scales with the square of diameter, the diameter scales with the square root of the speed ratio.
$$d_2=d_1\sqrt{\frac{v_1}{v_2}}=1\times\sqrt{\frac{1}{21}}\approx 0.22\ \text{cm}$$
6
A Venturi meter is used to measure the flow speed of water in a pipe. The pressure at the wide section (diameter 6 cm) is 250 kPa and at the narrow throat (diameter 3 cm) it is 233.1 kPa. What is the flow speed in the wide section?
Ans
1.5 m/s
Review
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Relate the throat speed to the wide-section speed
Because the diameter halves, the area becomes one quarter as large, so the speed at the throat is four times larger.
$$v_2=v_1\left(\frac{d_1}{d_2}\right)^2=v_1\left(\frac{6}{3}\right)^2=4v_1$$
Write the pressure drop using Bernoulli
The pressure difference comes from the increase in kinetic energy between the wide section and the throat.
$$\Delta P=250{,}000-233{,}100=\tfrac{1}{2}(1000)(16v_1^2-v_1^2)=7500v_1^2$$
Solve for the wide-section speed
Isolate the speed and take the square root.
$$v_1^2=\frac{16{,}900}{7500}=2.253\Rightarrow v_1\approx 1.5\ \text{m/s}$$
7
A fire hose connected to city water mains must shoot water to a vertical height of 15.0 m. What minimum gauge pressure is required in the mains? ($g = 9.8$ m/s²)
Ans
$1.47 \times 10^5$ Pa
Review
Tap to reveal
Use the height requirement directly
At the highest point the water momentarily stops, so the pressure energy must supply the full gravitational potential rise.
$$P_{gauge}=\rho gh=(1000)(9.8)(15)=1.47\times 10^5\ \text{Pa}$$
Match the answer choice
The required minimum gauge pressure is about $1.47\times 10^5$ Pa.
$$P_{gauge}\approx 1.47\times 10^5\ \text{Pa}$$
8
An airplane design requires a lift of 1000 N/m² of wing area. If the speed of flow past the lower wing surface is 160 m/s and air density is 1.29 kg/m³, what speed is required over the upper surface?
Ans
165 m/s
Review
Tap to reveal
Set the pressure difference from the lift requirement
Lift per unit area is the pressure difference between the lower and upper surfaces.
$$\Delta P=1000\ \text{N/m}^2$$
Solve Bernoulli for the upper speed
Rearrange the pressure-speed relation to isolate $v_{top}$ and substitute the lower-surface speed.
$$1000=\tfrac{1}{2}(1.29)(v_{top}^2-160^2)\Rightarrow v_{top}\approx 165\ \text{m/s}$$
9
The radius of the aorta is about 1.0 cm and blood flows through it at about 30 cm/s. If the total cross-sectional area of all capillaries combined is about 2000 cm², what is the average speed of blood in the capillaries?
Ans
0.047 cm/s
Review
Tap to reveal
Find the aorta cross-sectional area
Use the area of a circle from the given radius.
$$A_{aorta}=\pi r^2=\pi(1.0)^2\approx 3.14\ \text{cm}^2$$
Apply continuity to the whole circulatory system
The same flow rate must pass through the capillaries, so the much larger total area makes the speed much smaller.
$$A_{aorta}v_{aorta}=A_{cap}v_{cap}\Rightarrow v_{cap}=\frac{3.14}{2000}\times 30\approx 0.047\ \text{cm/s}$$
10
Water flows downhill through a closed vertical funnel, 40.0 cm long. The flow speed at the top is 12.0 cm/s and at the bottom is 24.0 cm/s. If the pressure at the top is $1.013 \times 10^5$ N/m², what is the pressure at the bottom?
Ans
$1.05 \times 10^5$ N/m²
Review
Tap to reveal
Write Bernoulli for the vertical drop
The bottom point is lower, so it gains pressure from both the height drop and the change in flow speed.
$$P_2=P_1+\rho gh+\tfrac{1}{2}\rho(v_1^2-v_2^2)$$
Substitute the given values
Convert the speeds from cm/s to m/s and use the 0.40 m height difference.
$$P_2=1.013\times 10^5+(1000)(9.8)(0.4)+\tfrac{1}{2}(1000)(0.12^2-0.24^2)\approx 1.05\times 10^5\ \text{N/m}^2$$
Question 1 of 10
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1
An airplane wing is designed such that the air speed above the wing is 80 m/s and below the wing is 60 m/s. If the density of air is 1.2 kg/m³ and the wing area is 20 m², what is the net upward lift force?