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

Density Density 𝜌= 𝑚𝑎𝑠𝑠 𝑣𝑜𝑙𝑢𝑚𝑒 = 𝑚 𝑉 Symbol “rho” ρ Mass kg Volume m3 Density kg/m3 Table 10-1 Example 10-1

Densities of materials 𝑘𝑔 𝑚 3 =1000∙ 𝑔 𝑐𝑚 3 Compare solids, liquids, gasses

Density calculation Show dimensional analysis 𝐼𝑓 𝜌= 𝑚𝑎𝑠𝑠 𝑣𝑜𝑙𝑢𝑚𝑒 𝑡ℎ𝑒𝑛 𝑚𝑎𝑠𝑠=𝜌∙𝑣𝑜𝑙𝑢𝑚𝑒 Show dimensional analysis

Pressure Pressure P= 𝐹𝑜𝑟𝑐𝑒 𝐴𝑟𝑒𝑎 = 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

Pressure variation with Depth Weight of column of water (fluid) 𝑊𝑒𝑖𝑔ℎ𝑡=𝑚𝑔=𝜌𝑉𝑔=𝜌𝐴ℎ𝑔 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒= 𝑤𝑒𝑖𝑔ℎ𝑡 𝑎𝑟𝑒𝑎 =𝜌𝑔ℎ Pressure increases with depth Old sub movies! (Das Boat, Red October, K-19, U-571) Always scene where they approach “crush depth” A h ρghA P = ρgh

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

Example 1– swimming pool (w/o air pressure) Pressure at 2.0 m 𝑃=𝜌𝑔ℎ = (1000 𝑘𝑔 𝑚 3 )(9.8 𝑚 𝑠 2 )(2 𝑚) =19,600 𝑘𝑔 𝑚 𝑠 2 𝑚 2 =19,600 𝑁 𝑚 2 =19,600 𝑃𝑎=19.6 𝑘𝑃𝑎 Total Force on bottom 𝐹=𝑃∙𝐴= 19,600 𝑁 𝑚 2 22 𝑚 ∙8.5 𝑚 =3.655 ∙ 10 6 𝑁 Same pressure on bottom and sides If you add air pressure on top of pool, both go up!

Absolute and gauge pressure Add atmospheric pressure on top of water column 𝐴𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑃𝑟𝑒𝑠𝑠𝑠𝑢𝑟𝑒= 𝑃 𝑎 +𝜌𝑔ℎ 𝐺𝑎𝑢𝑔𝑒 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒=𝜌𝑔ℎ 𝑃 𝑎 =1.013∙ 10 5 𝑃𝑎 =101.3 𝑘𝑃𝑎 Sea level air pressure around 100 kPa Everything has 101.3 kPa acting on it inside and out, we’re usually just interested in differences! Pa h ρghA + PaA P = ρgh + Pa

Not even half air pressure!

Example 2– swimming pool with air pressure Pressure at 2.0 m 𝑃=𝜌𝑔ℎ = 𝑃 𝐴 +(1000 𝑘𝑔 𝑚 3 )(9.8 𝑚 𝑠 2 )(2 𝑚) =101,300 𝑃𝑎+ 19,600 𝑘𝑔 𝑚 𝑠 2 𝑚 2 =120,090 𝑃𝑎 =121 𝑘𝑃𝑎 Total Force on bottom 𝐹=𝑃∙𝐴= 121,090 𝑁 𝑚 2 22 𝑚 ∙8.5 𝑚 =22.6 ∙ 10 6 𝑁

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

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 𝑜𝑢𝑡 𝐴 𝑜𝑢𝑡 Pin Pout

Hydraulic Lift Since Pressures are same 𝐹 𝑖𝑛 𝐴 𝑖𝑛 = 𝐹 𝑜𝑢𝑡 𝐴 𝑜𝑢𝑡 𝐹 𝑖𝑛 𝐴 𝑖𝑛 = 𝐹 𝑜𝑢𝑡 𝐴 𝑜𝑢𝑡 𝐹 𝑜𝑢𝑡 𝐹 𝑖𝑛 = 𝐴 𝑜𝑢𝑡 𝐴 𝑖𝑛

Columns of fluid 𝑃 𝑏𝑜𝑡𝑡𝑜𝑚 = 𝑃 𝑡𝑜𝑝 +𝜌𝑔ℎ 2 cases Top pressure + pressure of column must balance bottom pressure. 𝑃 𝑏𝑜𝑡𝑡𝑜𝑚 = 𝑃 𝑡𝑜𝑝 +𝜌𝑔ℎ 2 cases If top open 𝑃 𝑏𝑜𝑡𝑡𝑜𝑚 = 𝑃 𝑡𝑜𝑝 𝑎𝑛𝑑 ℎ=0 If top closed 𝑃 𝑏𝑜𝑡𝑡𝑜𝑚 =𝜌𝑔ℎ “balance” column of water Finger on the straw trick Ptop Pbottom ρgh

Column of water Calculate maximum height of column of water balanced in a closed straw, when bottom is open at normal air pressure 𝑃 𝑏𝑜𝑡𝑡𝑜𝑚 = 𝑃 𝑡𝑜𝑝 +𝜌𝑔ℎ 1.013∙ 10 5 𝑁 𝑚 2 =(1000 𝑘𝑔 𝑚 3 ) (9.8 𝑚 𝑠 2 ) ℎ ℎ= 1.013∙ 10 5 𝑁 𝑚 2 (1000 𝑘𝑔 𝑚 3 ) (9.8 𝑚 𝑠 2 ) ℎ=10.5 𝑚 Ptop Pbottom ρgh

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 𝑏𝑜𝑡𝑡𝑜𝑚 = 𝑃 𝑡𝑜𝑝 +𝜌𝑔ℎ 1.013∙ 10 5 𝑁 𝑚 2 =(13600 𝑘𝑔 𝑚 3 ) (9.8 𝑚 𝑠 2 ) ℎ ℎ= 1.013∙ 10 5 𝑁 𝑚 2 (13600 𝑘𝑔 𝑚 3 ) (9.8 𝑚 𝑠 2 ) ℎ=0.760 𝑚=760 𝑚𝑚 That’s where 760 mm mercury comes from! Ptop Pbottom ρgh

“Balancing” column of fluid Balancing water with straw 10 m Water Barometer Torricelli Mercury Barometer

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

Example – Problem 16 (cont) Pressure at (a) 𝑃 𝑎 = 𝑃 𝑜 + 𝜌 𝑜𝑖𝑙 𝑔 (0.272 𝑚) Pressure at (b) 𝑃 𝑏 = 𝑃 𝑜 + 𝜌 𝑤𝑎𝑡𝑒𝑟 𝑔 0.272 −0.0941 𝑚 Equating 𝜌 𝑜𝑖𝑙 𝑔 0.272 𝑚 = 𝜌 𝑤𝑎𝑡𝑒𝑟 𝑔 0.272 −0.0941 𝑚 𝜌 𝑜𝑖𝑙 = 0.272 −0.0941 0.272 𝑚 𝜌 𝑤𝑎𝑡𝑒𝑟 𝜌 𝑜𝑖𝑙 = 0.272 −0.0941 0.272 𝑚 1000 𝑘𝑔 𝑚 3 =654 𝑘𝑔 𝑚 3