A brief note on Fluids in Motion and Bernoulli's Principle . Read it before a test. Try each question at the end before you open its answer.
The big ideas Steady flow : the velocity at each point does not change; streamlines never cross. Continuity : A 1 v 1 = A 2 v 2 A_1v_1 = A_2v_2 A 1 v 1 = A 2 v 2 , so narrow means fast. Bernoulli : p + 1 2 ρ v 2 + ρ g h p + \tfrac12\rho v^2 + \rho gh p + 2 1 ρ v 2 + ρ g h is the same along a streamline; faster or higher means lower pressure. Torricelli : a jet from a hole h h h below the surface has v = 2 g h v = \sqrt{2gh} v = 2 g h . Uses : Venturi meters, sprays, wings, swerving balls, roofs in storms. The facts at a glance Situation Formula Flow rate Q = A v Q = Av Q = A v Narrowing pipe v 2 = v 1 A 1 A 2 v_2 = v_1\dfrac{A_1}{A_2} v 2 = v 1 A 2 A 1 Level pipe p 1 − p 2 = 1 2 ρ ( v 2 2 − v 1 2 ) p_1 - p_2 = \tfrac12\rho(v_2^2 - v_1^2) p 1 − p 2 = 2 1 ρ ( v 2 2 − v 1 2 ) Rising pipe, same width p 1 − p 2 = ρ g h p_1 - p_2 = \rho gh p 1 − p 2 = ρ g h Jet from a tank v = 2 g h v = \sqrt{2gh} v = 2 g h Lift on a wing 1 2 ρ ( v top 2 − v bottom 2 ) A \tfrac12\rho(v_{\text{top}}^2 - v_{\text{bottom}}^2)A 2 1 ρ ( v top 2 − v bottom 2 ) A
A worked example A fire hose 6.0 cm 6.0\ \text{cm} 6.0 cm across delivers 12 12 12 litres each second through a nozzle 2.0 cm 2.0\ \text{cm} 2.0 cm across.
In the hose. v 1 = 0.012 m 3 / s π ( 0.030 m ) 2 = 4.24 m/s v_1 = \dfrac{0.012\ \text{m}^3/\text{s}}{\pi(0.030\ \text{m})^2} = 4.24\ \text{m/s} v 1 = π ( 0.030 m ) 2 0.012 m 3 / s = 4.24 m/s . At the nozzle. A third of the diameter, nine times the speed: 38.2 m/s 38.2\ \text{m/s} 38.2 m/s . Pressure needed. 1 2 × 1000 kg/m 3 × ( ( 38.2 m/s ) 2 − ( 4.24 m/s ) 2 ) = 7.2 × 10 5 Pa \tfrac12 \times 1000\ \text{kg/m}^3 \times ((38.2\ \text{m/s})^2 - (4.24\ \text{m/s})^2) = 7.2 \times 10^{5}\ \text{Pa} 2 1 × 1000 kg/m 3 × (( 38.2 m/s ) 2 − ( 4.24 m/s ) 2 ) = 7.2 × 1 0 5 Pa above the air's. How high. Pointed up, the jet could rise v 2 2 g = 74 m \dfrac{v^2}{2g} = 74\ \text{m} 2 g v 2 = 74 m , without air resistance. Practice with new numbers
A program to try The program works out the fire hose of the worked example.
Read it, then run it.
Fit a nozzle 3.0 cm 3.0\ \text{cm} 3.0 cm across. How much lower is the pressure needed?
Mistakes to avoid Diameter ratio for area ratio. Square it. ( v 2 − v 1 ) 2 (v_2 - v_1)^2 ( v 2 − v 1 ) 2 for v 2 2 − v 1 2 v_2^2 - v_1^2 v 2 2 − v 1 2 . Square first, then subtract. Higher pressure where the fluid is fast. It is lower. The density in Torricelli's law. It cancels. The depth of the hole for the height of fall. The speed uses the depth below the surface; the fall uses the height above the ground. Check yourself Water at 1.0 m/s 1.0\ \text{m/s} 1.0 m/s enters a pipe section with a fifth of the area. How fast does it move there? Stuck? Show a hint Show the answerIn a level pipe, water speeds up from 2.0 m/s 2.0\ \text{m/s} 2.0 m/s to 6.0 m/s 6.0\ \text{m/s} 6.0 m/s . By how much does the pressure fall? Stuck? Show a hint Show the answerHow fast does water leave a hole 0.80 m 0.80\ \text{m} 0.80 m below the surface? (Take g = 10 m/s 2 g = 10\ \text{m/s}^2 g = 10 m/s 2 .) Stuck? Show a hint Show the answerWhy do two sheets of paper held a little apart move together when you blow between them? Stuck? Show a hint Show the answerWater rises steadily 5.0 m 5.0\ \text{m} 5.0 m in a pipe of the same width. By how much does its pressure fall? Stuck? Show a hint Show the answer