AP® Physics 1: Algebra-Based review sheet from Aim for Five (aimforfive.com/physics/units/8/8-4)
Unit 8 · Topic 8.4
8.4 Fluids and Conservation Laws
A fluid starts to flow, or speeds up, when the forces on it are unbalanced, usually because the pressure is higher at one end of a pipe than the other. Because an ideal fluid can't be compressed, the volume entering a full pipe each second must leave it, so A₁v₁ = A₂v₂. Bernoulli's equation, P + ρgy + ½ρv² = constant, is energy conservation for a flowing fluid, and it gives Torricelli's theorem, v = √(2gh), for fluid leaving a hole a depth h below the surface of an open tank.
Key terms
- continuity equation
- volume flow rate
- Bernoulli's equation
- Torricelli's theorem
- ideal fluid flow
Why fluids flow
A fluid speeds up when the net force on it is in the direction it's moving. Most often that comes from a pressure difference: water flows out of a hose because the pressure at the tap is higher than the pressure at the open end. Gravity can do it too, as when water pours from a tank or runs downhill. Once an ideal fluid is moving through a level pipe of constant width, it keeps going at the same speed with no pressure difference at all, just as an object with zero net force keeps its velocity (2.4).
Boundary for this course: fluids are ideal (incompressible, no viscosity) and pipes are completely full, unless a problem says otherwise.
Continuity: what goes in must come out
The volume flow rate is the volume passing a point each second: V/t = Av, where A is the pipe's cross-sectional area and v is the fluid's speed. Its unit is m³/s.
In a full pipe, an incompressible fluid can't pile up or thin out. So the flow rate is the same everywhere along the pipe: A₁v₁ = A₂v₂. This is the continuity equation, and it comes from conservation of mass.
Where the pipe narrows, the fluid speeds up. For a round pipe, A = πr², so the speed depends on the diameter squared: half the diameter means a quarter of the area and four times the speed. That's why putting your thumb over a hose makes the water shoot farther.
Bernoulli's equation
Bernoulli's equation applies energy conservation to a flowing ideal fluid: P₁ + ρgy₁ + ½ρv₁² = P₂ + ρgy₂ + ½ρv₂². Each term is energy per cubic meter of fluid (1 J/m³ = 1 Pa). ½ρv² is kinetic energy per volume, ρgy is gravitational potential energy per volume, and P accounts for the work done by the surrounding fluid pushing the parcel along.
The key result: if the height stays the same, faster fluid is at lower pressure. Where a horizontal pipe narrows, the fluid speeds up (continuity) and its pressure drops (Bernoulli). That makes sense with Newton's laws: to speed up, the fluid needs a net push from the higher-pressure region behind it.
Likewise, fluid moving downhill with no change in speed must be gaining pressure, and fluid flowing uphill in a pipe of constant width loses pressure.
Torricelli's theorem
Take a large open tank with a small hole a depth h below the water's surface. Both the surface and the stream from the hole are open to the air, so P is atmospheric at both points and cancels. The surface of a large tank barely moves, so v ≈ 0 there. Bernoulli then reduces to ρgh = ½ρv², so v = √(2gh).
That's the same speed an object gains falling freely through height h. Notice that h is the depth of the hole below the surface, not its height above the ground.
Setting up a flow problem
- Pick two points along the flow and list P, y and v at each.
- Any point open to the air has P = P_atm.
- The surface of a large tank has v ≈ 0.
- If you know one speed and both areas, use continuity first to get the other speed.
- Then use Bernoulli for the unknown pressure, speed or height.
Worked examples
Try each one yourself first, then open the solution.
- Example 1Calculator allowed
Hose and nozzle
Water flows at 1.5 m/s through a hose with an inside diameter of 2.0 cm. It leaves through a nozzle with an inside diameter of 0.50 cm. Find (a) the speed of the water leaving the nozzle, (b) the volume flow rate and (c) how long it takes to fill a 10 L bucket (1 L = 10⁻³ m³).
Show the solutionHide the solution
- Step 1: (a) A₁v₁ = A₂v₂, and A = πd²/4, so v₂ = v₁(d₁/d₂)² = 1.5 × (2.0/0.50)² = 1.5 × 16 = 24 m/s.
- Step 2: (b) Flow rate = A₁v₁ = π(0.010 m)²(1.5 m/s) ≈ 4.7 × 10⁻⁴ m³/s, about 0.47 L/s.
- Step 3: (c) t = volume/flow rate = (0.010 m³)/(4.7 × 10⁻⁴ m³/s) ≈ 21 s.
- Step 4: The nozzle makes the water faster, but the flow rate is the same, so it doesn't fill the bucket any quicker.
Answer: (a) 24 m/s (b) about 4.7 × 10⁻⁴ m³/s (c) about 21 s
- Example 2Calculator allowed
Pressure in a narrowing pipe
Water (ρ = 1000 kg/m³) flows through a horizontal pipe. In the wide section the area is 4.0 × 10⁻³ m², the speed is 2.0 m/s and the pressure is 1.50 × 10⁵ Pa. The pipe narrows to an area of 1.0 × 10⁻³ m². Find the speed and pressure in the narrow section.
Show the solutionHide the solution
- Step 1: Continuity: v₂ = A₁v₁/A₂ = (4.0 × 10⁻³)(2.0)/(1.0 × 10⁻³) = 8.0 m/s.
- Step 2: The pipe is horizontal, so the ρgy terms cancel: P₁ + ½ρv₁² = P₂ + ½ρv₂².
- Step 3: P₂ = P₁ + ½ρ(v₁² − v₂²) = 1.50 × 10⁵ + ½(1000)(4.0 − 64) = 1.50 × 10⁵ − 3.0 × 10⁴ = 1.20 × 10⁵ Pa.
- Step 4: The pressure drops where the water speeds up, as Bernoulli predicts.
Answer: 8.0 m/s and 1.20 × 10⁵ Pa
- Example 3Calculator allowed
A leaking tank (classic trap)
A large open water tank has a small hole in its side. The hole is 1.8 m below the water's surface and 0.80 m above the floor. Use g = 9.8 m/s². Find (a) the speed of the water leaving the hole and (b) how far from the base of the tank the stream hits the floor.
Show the solutionHide the solution
- Step 1: (a) Torricelli: v = √(2gh), with h the depth below the surface, 1.8 m. v = √(2 × 9.8 × 1.8) ≈ 5.9 m/s, horizontally.
- Step 2: The trap is using the hole's height above the floor (0.80 m) here, which gives about 4.0 m/s.
- Step 3: (b) Now the stream is a horizontal projectile (1.5). Fall time from 0.80 m: t = √(2y/g) = √(2 × 0.80/9.8) ≈ 0.404 s.
- Step 4: Horizontal distance: x = vt = (5.94)(0.404) ≈ 2.4 m.
Answer: (a) about 5.9 m/s (b) about 2.4 m from the tank
Common mistakes
- Thinking a narrower pipe makes fluid slower, or that it raises the pressure. In a narrower section the speed goes up and, at the same height, the pressure goes down.
- Using diameter in place of area in A₁v₁ = A₂v₂. Area goes as diameter squared.
- In Torricelli's theorem, measuring h from the ground or from the bottom of the tank. Use the depth of the hole below the fluid's surface.
- Forgetting that points open to the air are both at atmospheric pressure, so those P terms cancel.
On the exam
- Qualitative questions often ask you to rank speed or pressure at several points in a pipe. Use continuity for speed first, then Bernoulli for pressure, and justify each step in words.
- Tank problems often pair Torricelli's theorem with projectile motion: find the exit speed from the depth below the surface, then the landing distance from the hole's height above the ground.
Connected topics
Videos
Check yourself
4 questions on 8.4 Fluids and Conservation Laws. Pick an answer to see if you got it, and why.
Water flows through a completely full pipe whose inner radius narrows from 2.0 cm to 1.0 cm. The water moves at 1.5 m/s in the wide part. How fast does it move in the narrow part?
Water flows at 3.0 m/s through a full pipe with a cross-sectional area of 2.0 × 10⁻⁴ m². What volume of water flows out of the pipe in 10 s?
A smooth stream of water falls from a faucet. The stream gets narrower as it falls. Which best explains this?
A syringe full of water has a barrel with a cross-sectional area of 2.0 cm² and a nozzle opening of 0.020 cm². The plunger is pushed in at a steady 0.010 m/s. How fast does the water leave the nozzle?
0 of 4 answered