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Columns of fluid Density Pressure Pressure variation with depth

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1 Columns of fluid Density Pressure Pressure variation with depth
Pascal’s principle Absolute pressure vs. gauge pressure Columns of fluid

2 Density Density 𝜌= π‘šπ‘Žπ‘ π‘  π‘£π‘œπ‘™π‘’π‘šπ‘’ = π‘š 𝑉 Symbol β€œrho” ρ Mass kg Volume m3
Density kg/m3 Table 10-1 Example 10-1

3 Densities of materials
π‘˜π‘” π‘š 3 =1000βˆ™ 𝑔 π‘π‘š 3 Compare solids, liquids, gasses

4 Density calculation Show dimensional analysis
𝐼𝑓 𝜌= π‘šπ‘Žπ‘ π‘  π‘£π‘œπ‘™π‘’π‘šπ‘’ π‘‘β„Žπ‘’π‘› π‘šπ‘Žπ‘ π‘ =πœŒβˆ™π‘£π‘œπ‘™π‘’π‘šπ‘’ Show dimensional analysis

5 Pressure Pressure P= πΉπ‘œπ‘Ÿπ‘π‘’ π΄π‘Ÿπ‘’π‘Ž = 𝐹 𝐴 Symbol P (capital) Force Newtons
Area m2 Pressure N/m2 = Pascals = Pa (small often use kPa) Example 10-2 Same units as stress - Problem 9-39

6 Pressure variation with Depth
Weight of column of water (fluid) π‘Šπ‘’π‘–π‘”β„Žπ‘‘=π‘šπ‘”=πœŒπ‘‰π‘”=πœŒπ΄β„Žπ‘” π‘ƒπ‘Ÿπ‘’π‘ π‘ π‘’π‘Ÿπ‘’= π‘€π‘’π‘–π‘”β„Žπ‘‘ π‘Žπ‘Ÿπ‘’π‘Ž =πœŒπ‘”β„Ž Pressure increases with depth Old sub movies! (Das Boat, Red October, K-19, U-571) Always scene where they approach β€œcrush depth” A h ρghA P = ρgh

7 Properties of pressure
Pressure pushes equally all directions. Perpendicular to surface restraining it. Equal at common depth

8 Example 1– swimming pool (w/o air pressure)
Pressure at 2.0 m 𝑃=πœŒπ‘”β„Ž = (1000 π‘˜π‘” π‘š 3 )(9.8 π‘š 𝑠 2 )(2 π‘š) =19,600 π‘˜π‘” π‘š 𝑠 2 π‘š 2 =19,600 𝑁 π‘š 2 =19,600 π‘ƒπ‘Ž=19.6 π‘˜π‘ƒπ‘Ž Total Force on bottom 𝐹=π‘ƒβˆ™π΄= 19,600 𝑁 π‘š π‘š βˆ™8.5 π‘š =3.655 βˆ™ 𝑁 Same pressure on bottom and sides If you add air pressure on top of pool, both go up!

9 Absolute and gauge pressure
Add atmospheric pressure on top of water column π΄π‘π‘ π‘œπ‘™π‘’π‘‘π‘’ π‘ƒπ‘Ÿπ‘’π‘ π‘ π‘ π‘’π‘Ÿπ‘’= 𝑃 π‘Ž +πœŒπ‘”β„Ž πΊπ‘Žπ‘’π‘”π‘’ π‘ƒπ‘Ÿπ‘’π‘ π‘ π‘’π‘Ÿπ‘’=πœŒπ‘”β„Ž 𝑃 π‘Ž =1.013βˆ™ π‘ƒπ‘Ž =101.3 π‘˜π‘ƒπ‘Ž Sea level air pressure around 100 kPa Everything has kPa acting on it inside and out, we’re usually just interested in differences! Pa h ρghA + PaA P = ρgh + Pa

10 Not even half air pressure!

11 Example 2– swimming pool with air pressure
Pressure at 2.0 m 𝑃=πœŒπ‘”β„Ž = 𝑃 𝐴 +(1000 π‘˜π‘” π‘š 3 )(9.8 π‘š 𝑠 2 )(2 π‘š) =101,300 π‘ƒπ‘Ž+ 19,600 π‘˜π‘” π‘š 𝑠 2 π‘š 2 =120,090 π‘ƒπ‘Ž =121 π‘˜π‘ƒπ‘Ž Total Force on bottom 𝐹=π‘ƒβˆ™π΄= 121,090 𝑁 π‘š π‘š βˆ™8.5 π‘š =22.6 βˆ™ 𝑁

12 Various units of pressure
We always use Pascals (Pa or kPa)

13 Pascal’s Principle External pressure distributed uniformly at uniform depth. Pressure is same at common level. Thus Fin/Ain = Fout/Aout (if A small F small). Hydraulic lift. (make up example) 𝐹 𝑖𝑛 𝐴 𝑖𝑛 𝐹 π‘œπ‘’π‘‘ 𝐴 π‘œπ‘’π‘‘ Pin Pout

14 Hydraulic Lift Since Pressures are same 𝐹 𝑖𝑛 𝐴 𝑖𝑛 = 𝐹 π‘œπ‘’π‘‘ 𝐴 π‘œπ‘’π‘‘
𝐹 𝑖𝑛 𝐴 𝑖𝑛 = 𝐹 π‘œπ‘’π‘‘ 𝐴 π‘œπ‘’π‘‘ 𝐹 π‘œπ‘’π‘‘ 𝐹 𝑖𝑛 = 𝐴 π‘œπ‘’π‘‘ 𝐴 𝑖𝑛

15 Columns of fluid 𝑃 π‘π‘œπ‘‘π‘‘π‘œπ‘š = 𝑃 π‘‘π‘œπ‘ +πœŒπ‘”β„Ž 2 cases
Top pressure + pressure of column must balance bottom pressure. 𝑃 π‘π‘œπ‘‘π‘‘π‘œπ‘š = 𝑃 π‘‘π‘œπ‘ +πœŒπ‘”β„Ž 2 cases If top open 𝑃 π‘π‘œπ‘‘π‘‘π‘œπ‘š = 𝑃 π‘‘π‘œπ‘ π‘Žπ‘›π‘‘ β„Ž=0 If top closed 𝑃 π‘π‘œπ‘‘π‘‘π‘œπ‘š =πœŒπ‘”β„Ž β€œbalance” column of water Finger on the straw trick Ptop Pbottom ρgh

16 Column of water Calculate maximum height of column of water balanced in a closed straw, when bottom is open at normal air pressure 𝑃 π‘π‘œπ‘‘π‘‘π‘œπ‘š = 𝑃 π‘‘π‘œπ‘ +πœŒπ‘”β„Ž 1.013βˆ™ 𝑁 π‘š 2 =(1000 π‘˜π‘” π‘š 3 ) (9.8 π‘š 𝑠 2 ) β„Ž β„Ž= 1.013βˆ™ 𝑁 π‘š 2 (1000 π‘˜π‘” π‘š 3 ) (9.8 π‘š 𝑠 2 ) β„Ž=10.5 π‘š Ptop Pbottom ρgh

17 That’s where 760 mm mercury comes from!
Column of mercury Calculate maximum height of column of mercury (denser) balanced in a closed tube, when bottom is open at normal air pressure 𝑃 π‘π‘œπ‘‘π‘‘π‘œπ‘š = 𝑃 π‘‘π‘œπ‘ +πœŒπ‘”β„Ž 1.013βˆ™ 𝑁 π‘š 2 =( π‘˜π‘” π‘š 3 ) (9.8 π‘š 𝑠 2 ) β„Ž β„Ž= 1.013βˆ™ 𝑁 π‘š 2 ( π‘˜π‘” π‘š 3 ) (9.8 π‘š 𝑠 2 ) β„Ž=0.760 π‘š=760 π‘šπ‘š That’s where 760 mm mercury comes from! Ptop Pbottom ρgh

18 β€œBalancing” column of fluid
Balancing water with straw m Water Barometer Torricelli Mercury Barometer

19 Example – Problem 16 Equate pressure along line a---b

20 Example – Problem 16 (cont)
Pressure at (a) 𝑃 π‘Ž = 𝑃 π‘œ + 𝜌 π‘œπ‘–π‘™ 𝑔 (0.272 π‘š) Pressure at (b) 𝑃 𝑏 = 𝑃 π‘œ + 𝜌 π‘€π‘Žπ‘‘π‘’π‘Ÿ 𝑔 βˆ’ π‘š Equating 𝜌 π‘œπ‘–π‘™ 𝑔 π‘š = 𝜌 π‘€π‘Žπ‘‘π‘’π‘Ÿ 𝑔 βˆ’ π‘š 𝜌 π‘œπ‘–π‘™ = βˆ’ π‘š 𝜌 π‘€π‘Žπ‘‘π‘’π‘Ÿ 𝜌 π‘œπ‘–π‘™ = βˆ’ π‘š π‘˜π‘” π‘š 3 =654 π‘˜π‘” π‘š 3


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