Two big ideas you already know, used on a moving fluid: mass is not created or lost
(the continuity equation) and energy is conserved (Bernoulli's equation).
Six little picture stories
Press Next (or Play) to walk through each story one small step at a time. The numbers come last. After every two stories there is one quick question; if it feels shaky, press Show me another example for one more story. Tap the small round play button next to a caption to hear it read aloud.
1. A thumb on the hose
Read the story as text
You are watering a garden. The water dribbles out of the open hose and lands near your feet.
You press your thumb over the end and leave only a small gap. Now the water shoots across the yard.
Later you notice a rain barrel with a small leak near the bottom. When the barrel is full the jet
squirts far. As the barrel empties, the jet gets weaker and lands closer and closer.
Your thumb did not add any water, and nothing pumps the barrel. So where does the extra speed come from,
and why does the jet slow down as the level drops?
2. What goes in must come out
Quick check
Water flows from a pipe into a section with one third of the area. Its speed in the narrow section is:
Another example: Filling a bucket
3. The leaky rain barrel
4. The river gorge
Quick check
The water in a leaky barrel is 4 times as high above the hole as before. The water leaves the hole:
Another example: Wind in the alley
5. The paper that rises
6. The garden fountain
Quick check
Where a fluid (like air) flows faster, its pressure is:
Another example: The truck and the bike
7. Check yourself
Think of your answer first, then tap to see it.
a) Water flows from a wide pipe into a narrow pipe. In which part does it move faster? Why?
Show answer
In the narrow part. The same volume of water must pass every point each second (A₁v₁ = A₂v₂), so a smaller area needs a bigger speed.
b) Why does the jet from the leaky barrel get weaker as the barrel empties?
Show answer
The water at the top has less height to fall to the hole (the pressure at the hole is lower). Less energy per kilogram turns into motion, so it leaves slower: v = √(2gh).
c) A hose with an opening of 3.0 cm² carries water at 2.0 m/s. A nozzle has an opening of 1.0 cm². How fast does the water leave the nozzle?
Three quick questions. Get all three right on the first try and you can skip ahead to the simulation. Not sure? No problem: just read on.
1) Water flows at 2.0 m/s in a pipe with an area of 6 cm². The pipe narrows to 2 cm². How fast does the water move there?
2) Water leaks from a small hole 0.45 m below the surface of an open tank. About how fast does it come out?
3) Water flows sideways (no change in height) from a wide part of a pipe into a narrow part. The pressure in the narrow part is:
The idea in plain words
1. Continuity: what goes in must come out
Picture watering a garden with a hose. The water flows out gently. Now press your thumb over half the end. The water shoots out much faster and farther. You did not turn the tap up. The same water just had to squeeze through a smaller gap.
An ideal fluid cannot be squeezed (it is incompressible). A pipe is full of it. So every second,
the same volume of water must pass every point of the pipe. If it did not, water would pile up somewhere,
and it has nowhere to go.
The picture shows a pipe that narrows, just like the hose under your thumb. Part 1 is wide, part 2 is narrow. A short arrow in the wide part means slow flow. A long arrow in the narrow part means fast flow.
The volume that passes a point each second is the volume flow rate:
Q = V / t = A v (units: m² × m/s = m³/s)
Here A is the cross-sectional area of the pipe and v is the speed of the fluid. Since Q is the same
everywhere in the pipe:
A₁v₁ = A₂v₂
Narrow part (small A) means fast flow (big v). That is your thumb on the hose.
Trap: using the diameter ratio in A₁v₁ = A₂v₂. Instead: use the AREA ratio. Area goes as diameter squared: half the diameter is ¼ the area, so the speed is 4 times bigger, not 2.
2. A pressure difference is what speeds the fluid up
Water speeds up as it enters the narrow part. Speeding up needs a net force (Newton's second law).
The only thing pushing it forward is the water behind it. So the pressure behind (in the wide part) must be
higher than the pressure ahead (in the narrow part). Fast moving fluid is at lower pressure.
This surprises almost everyone.
Trap: "the narrow part has higher pressure, because the water is squeezed". Instead: in the narrow part the water is faster, so the pressure is LOWER. The higher pressure behind it is what speeds it up.
3. Bernoulli: energy conservation per cubic metre
Follow one small chunk of fluid of volume V. Its kinetic energy per volume is ½ρv². Its gravitational
energy per volume is ρgy. The work done on it by the pressure of the fluid around it, per volume, is the
pressure P. For an ideal fluid nothing is lost to friction, so the total stays the same along the flow:
P₁ + ρgy₁ + ½ρv₁² = P₂ + ρgy₂ + ½ρv₂²
Read it like an energy bar chart. Every term has units of pascals (J/m³). If the fluid speeds up
(½ρv² grows) or climbs (ρgy grows), the pressure must drop to pay for it.
4. Torricelli: the leak in the barrel
Use Bernoulli from the top surface of the water (point 1) to the hole (point 2). Both are open to the air,
so P₁ = P₂ = Patm and they cancel. The top surface of a wide barrel barely moves, so v₁ ≈ 0.
Put the hole at y = 0 and the surface at y = h:
ρgh = ½ρv² so v = √(2gh)
That is the same speed a stone gets by falling a height h. As the barrel empties, h drops, so the jet slows.
Trap: measuring Torricelli's h from the bottom of the tank (or from the hole to the ground). Instead: h is from the hole UP to the water surface. Surface 1.5 m above the floor, hole 0.30 m above it: h = 1.2 m, v = √(2 × 9.8 × 1.2) = 4.9 m/s.
Worked with numbers: the garden hose
Read the steps as text
A hose has an inside diameter of 2.0 cm. Water moves through it at 2.0 m/s. You squeeze the end so the
opening has a diameter of 1.0 cm. The hose is level. Find the exit speed and the pressure in the hose.
Find the area of the hose: A₁ = πr² = π(0.010 m)² = 3.14 × 10⁻⁴ m².
Why: continuity uses area, not diameter. Convert cm to m first: 2.0 cm diameter is a 0.010 m radius.
Find the flow rate: Q = A₁v₁ = (3.14 × 10⁻⁴ m²)(2.0 m/s) = 6.28 × 10⁻⁴ m³/s, which is 0.63 L/s.
Why: this volume per second is the same at every point, including the opening.
Find the opening area: A₂ = π(0.0050 m)² = 7.85 × 10⁻⁵ m².
Why: half the diameter means one quarter of the area, because area goes as diameter squared.
Use continuity: v₂ = Q / A₂ = (6.28 × 10⁻⁴) / (7.85 × 10⁻⁵) = 8.0 m/s.
Why: A₁/A₂ = 4, so the speed is 4 times bigger: 4 × 2.0 = 8.0 m/s. Shortcut and long way agree.
Use Bernoulli with y₁ = y₂ (level hose): P₁ − P₂ = ½ρ(v₂² − v₁²) = ½(1000 kg/m³)(64 − 4.0) m²/s² = 30 000 Pa.
Why: the ρgy terms cancel when the heights are equal. The pressure difference pays for the extra kinetic energy.
The water leaves into the air, so P₂ = Patm = 1.0 × 10⁵ Pa. Then P₁ = 1.0 × 10⁵ + 0.30 × 10⁵ = 1.3 × 10⁵ Pa.
Why: the gauge pressure in the hose is 30 kPa. Check: positive, bigger than the outside air. Makes sense, the water is pushed out.
Bonus: aimed straight up, the water rises h = v²/(2g) = (8.0)² / (2 × 9.8) = 3.3 m.
Why: once out of the hose each bit of water is a projectile; its kinetic energy turns into gravitational energy.
Lab 1: water in a pipe that narrows and rises
Change the two diameters, the speed at point 1, the height of point 2, the fluid density and the pressure
at point 1. Watch the dye drops: they are released at equal times, so where they spread apart the water is
moving faster. The graphs follow the orange drop.
Predict first: with the settings in the lab right now, will the pressure at point 2 be higher,
lower or the same as at point 1? Pick, then play the sim.
Speed vs area for this flow rate: v = Q/A. Halve the area, double the speed.
Dot 1 is point 1, dot 2 is point 2.Pressure along the pipe from Bernoulli. It drops where the pipe narrows (faster flow)
and where it rises (higher up).
Try these: (1) set both diameters equal and raise point 2 by 4 m: the speed stays the same but the
pressure drops by ρgΔy = 1000 × 9.8 × 4 = 39 kPa. (2) Make point 2 wider than point 1: the water slows down
and the pressure goes up. (3) Push the speed and the narrowing to the limit: Bernoulli says the
pressure would go below zero, which cannot happen.
Lab 2: a tank draining through a small hole
Change the tank area, the hole area, the starting depth of water above the hole, and how high the hole is
above the floor. The jet speed is v = √(2gh). The bigger the hole compared with the tank, the faster the tank
empties.
Things to notice: the time to empty depends on the ratio of tank area to hole area. Double the hole area and
the tank empties in half the time. The jet speed falls in a straight line, but the depth h curves.
Examples
Example 1: a pipe narrows basic
Water flows at 3.0 m/s in a pipe with cross-sectional area 0.040 m². The pipe narrows to 0.010 m².
Find the flow rate and the speed in the narrow part.
Show solutionRead the steps as text
Flow rate: Q = Av = (0.040 m²)(3.0 m/s) = 0.12 m³/s. Volume per second through the wide part.
Continuity: v₂ = Q / A₂ = 0.12 / 0.010 = 12 m/s. The area is 4 times smaller, so the speed is 4 times bigger.
Example 2: pressure in a level pipe medium
Water moves at 2.0 m/s through a level pipe where the pressure is 150 kPa. The pipe narrows to half its
area. Find the speed and the pressure in the narrow part.
Show solutionRead the steps as text
Continuity: half the area, so v₂ = 2 × 2.0 = 4.0 m/s. A₁v₁ = A₂v₂ with A₂ = A₁/2.
Bernoulli with equal heights: P₂ = P₁ + ½ρ(v₁² − v₂²) = 150 000 + ½(1000)(4.0 − 16) = 150 000 − 6000 = 144 000 Pa.
The ρgy terms cancel. The fluid gained kinetic energy, so the pressure went down.
Answer: 4.0 m/s and 144 kPa. Check: lower pressure where faster. Good.
Example 3: a pipe that narrows and climbs AP
Water enters a pipe of diameter 4.0 cm at 1.5 m/s and pressure 250 kPa. The pipe rises 5.0 m and
narrows to a diameter of 2.0 cm. Find the speed and pressure at the top.
Show solutionRead the steps as text
Area ratio: A₁/A₂ = (4.0/2.0)² = 4. So v₂ = 4 × 1.5 = 6.0 m/s.
Area goes as diameter squared. Half the diameter is a quarter of the area.
Bernoulli: P₂ = P₁ + ½ρ(v₁² − v₂²) − ρg(y₂ − y₁).
Rearranged so the unknown is alone. y₂ − y₁ = +5.0 m (up is positive).
Numbers: ½(1000)(2.25 − 36) = −16 875 Pa, and (1000)(9.8)(5.0) = 49 000 Pa.
Both terms take pressure away: the water is faster and higher.
P₂ = 250 000 − 16 875 − 49 000 = 184 125 Pa ≈ 1.8 × 10⁵ Pa (184 kPa).
Still positive, so the answer is physical.
Example 4: the leaking rain barrel AP
A rain barrel is open at the top. The water surface is 1.8 m above a small hole. The hole is 0.50 m above
the ground and has an area of 2.0 cm². Find the jet speed, the flow rate and how far from the barrel the
water lands.
Show solutionRead the steps as text
Torricelli: v = √(2gh) = √(2 × 9.8 × 1.8) = √35.3 = 5.9 m/s.
Both the surface and the hole are open to the air, so the pressures cancel. h is measured from the hole to the surface.
Flow rate: Q = av = (2.0 × 10⁻⁴ m²)(5.94 m/s) = 1.2 × 10⁻³ m³/s = 1.2 L/s.
1 cm² = 10⁻⁴ m². 1 m³ = 1000 L.
Fall time from 0.50 m: t = √(2y/g) = √(2 × 0.50 / 9.8) = 0.32 s.
The water leaves horizontally, so it starts with zero vertical speed, like a horizontal projectile.
Distance: x = vt = 5.94 × 0.319 = 1.9 m.
Horizontal speed stays constant during the fall (no air resistance).
Practice (AP style)
1. Water flows through a horizontal pipe. The pipe narrows to one third of its original cross-sectional
area. Compared with the wide part, the speed in the narrow part is
Show answer
C. Continuity: A₁v₁ = A₂v₂. A₂ = A₁/3, so v₂ = 3v₁. (9 times would be the answer if the diameter had been cut to a third.)
2. In the same horizontal pipe, how does the pressure in the narrow part compare with the pressure in the wide part?
Show answer
B. The water speeds up as it enters the narrow part, so there is a net forward force on it. That force comes from
higher pressure behind it. Bernoulli with equal heights: P₂ = P₁ − ½ρ(v₂² − v₁²), which is less than P₁.
3. A pipe's diameter is cut in half. If the flow rate stays the same, the fluid speed becomes
Show answer
C. A = πd²/4, so half the diameter gives one quarter of the area. v = Q/A, so the speed is 4 times as large.
4. Water leaks from a small hole in the side of an open tank. The water level above the hole is made 4 times
as large. The speed of the water leaving the hole becomes
Show answer
B. v = √(2gh). Multiplying h by 4 multiplies v by √4 = 2.
5. A hose delivers water at 0.50 L/s through a nozzle opening of area 1.0 cm². What is the speed of the water leaving the nozzle?
Show answer
B. Convert: 0.50 L/s = 5.0 × 10⁻⁴ m³/s and 1.0 cm² = 1.0 × 10⁻⁴ m². v = Q/A = (5.0 × 10⁻⁴)/(1.0 × 10⁻⁴) = 5.0 m/s.
6. Water flows steadily up a pipe of constant diameter. The top of the pipe is 3.0 m higher than the bottom.
Which statement is correct?
Show answer
C. Same area, so same speed (continuity). Bernoulli with v₁ = v₂: P_bottom − P_top = ρgΔy = (1000)(9.8)(3.0) = 29 400 Pa ≈ 29 kPa.
7. Water flows through a level pipe that narrows in the middle and then widens back to its starting size.
Which describes the pressure from start to end?
Show answer
B. In an ideal fluid nothing is lost. Narrow middle: faster, so lower pressure. At the end the area, speed and height
are the same as at the start, so Bernoulli gives the same pressure. (A real pipe with friction would lose a little.)
8. (Qualitative / quantitative translation) A student says: "Water speeds up in the narrow part of a pipe because
the pressure is higher there, and the higher pressure pushes it faster."
(a) Explain what is wrong with the claim, using forces.
(b) Water in a level pipe moves at 1.0 m/s. The pipe narrows so that the area is 2.5 times smaller.
Calculate the speed in the narrow part and the pressure difference between the two parts.
Show answer
(a) To speed up, a bit of water needs a net force pointing forward. A push from the fluid behind it is bigger than the
push from the fluid in front only if the pressure behind (wide part) is higher. So the narrow, fast part has the lower
pressure. High pressure at the narrow part would push the water backward and slow it down.
(b) v₂ = 2.5 × 1.0 = 2.5 m/s. ΔP = P₁ − P₂ = ½ρ(v₂² − v₁²) = ½(1000)(6.25 − 1.0) = 2625 Pa ≈ 2.6 kPa,
with the wide part at the higher pressure.
9. (Experimental design and analysis) You have a large plastic bottle with a small hole near the bottom, a ruler,
tape, a tray, and water. Design an experiment to test Torricelli's result v = √(2gh) without measuring the speed directly.
Show answer
Plan: Set the bottle on a stand so the hole is a fixed height y above the tray (measure y with the ruler).
Fill the bottle, mark several water levels with tape, and measure h from the hole up to each mark.
As the level passes each mark, measure the horizontal distance R from the hole to where the jet lands.
Repeat each reading several times and average.
Analysis: The jet leaves horizontally at speed v and falls y in time t = √(2y/g), so R = v√(2y/g).
If v = √(2gh), then R² = 2gh × 2y/g = 4yh. So plot R² (vertical) against h (horizontal).
Torricelli is supported if the graph is a straight line through the origin with slope 4y.
For example, with y = 0.30 m the slope should be 1.2.
Sources of error: the hole has real edges (the jet narrows and loses a little speed, so the slope will be a
bit less than 4y); reading the landing point on a moving jet; the level changing while you read.
10. (Mathematical routines) A tank open to the air has water a height H above a small hole of area a.
The tank's cross-sectional area is A, much bigger than a. (a) Derive an expression for the volume flow rate out of the hole.
(b) Derive an expression for how fast the water level drops, in terms of a, A, g and H. (c) Is the level dropping faster or
slower when the tank is nearly empty? Explain.
Show answer
(a) v = √(2gH), so Q = av = a√(2gH).
(b) Continuity between the top surface and the hole: A vtop = a v, so vtop = (a/A)√(2gH).
(c) Slower. vtop grows with √H, so as H gets smaller the level drops more slowly. That is why the
depth vs time graph in Lab 2 curves and flattens out.
Free response (Translation between representations). A level pipe carries water. The wide part has area 4.0 × 10⁻³ m² and the narrow part has area
1.0 × 10⁻³ m². A thin open tube stands up from each part. The water in the tube over the wide part stands 0.30 m higher
than in the tube over the narrow part.
(a) Why is the water higher in the tube over the wide part?
(b) Use the height difference to find the speed in each part and the volume flow rate.
(c) Describe the graph of pressure against position along the pipe, from the wide part, through the narrow part, and into
another wide part of the same size.
Show answer
(a) The height of water in each tube shows the pressure there (P = P0 + ρgh). The wide part is slower, so its pressure is higher.
(c) High and flat in the wide part, drops by 2940 Pa in the narrow part and stays low there, then rises back to the
starting value in the second wide part (ideal fluid, nothing lost).
Point guide (4 points): 1 for "slower means higher pressure"; 1 for ΔP = ρgΔh; 1 for v1, v2 and Q; 1 for the dip-and-recover graph.
Common mistakes
The mistake: "The narrow part has higher pressure, because the water is squeezed."
Why it is wrong: An ideal fluid is not squeezed at all; its density stays the same. The water speeds up in the
narrow part, and that needs a forward net force from higher pressure behind it.
How to spot it: Ask "where is the water faster?" Faster (at the same height) means lower pressure.
The mistake: Using the diameter ratio instead of the area ratio in A₁v₁ = A₂v₂.
Why it is wrong: Area grows with the square of the diameter: A = πd²/4. Half the diameter is a quarter of the area.
How to spot it: If the problem gives a diameter or radius, square the ratio before using it.
The mistake: Mixing units: cm² with m², kPa with Pa, or litres with m³.
Why it is wrong: Bernoulli adds terms. ½ρv² comes out in pascals only if ρ is in kg/m³ and v in m/s.
How to spot it: Convert everything to SI first: 1 cm² = 10⁻⁴ m², 1 L = 10⁻³ m³, 1 kPa = 1000 Pa.
The mistake: In Torricelli's result, measuring h from the floor or from the bottom of the tank.
Why it is wrong: h is the height of the water surface above the hole. Only the drop from surface to hole turns into speed.
How to spot it: Draw a line from the hole straight up to the surface. That is h.
The mistake: Using absolute pressure on one side of Bernoulli and gauge pressure on the other.
Why it is wrong: Gauge pressure is absolute pressure minus 1.0 × 10⁵ Pa. Mixing them adds a false 100 kPa.
How to spot it: Pick one kind and use it for both points. If both points are open to the air, the pressures cancel.
The mistake: Thinking the flow rate drops in the narrow part ("less room, so less water").
Why it is wrong: The pipe is full and the fluid cannot be squeezed, so the volume per second is the same at every point.
Less area is made up for by more speed.
How to spot it: Q = Av is one number for the whole pipe. Only A and v trade off.