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Columns of fluid Density Pressure Pressure variation with depth
Pascalβs principle Absolute pressure vs. gauge pressure Columns of fluid
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Density Density π= πππ π π£πππ’ππ = π π Symbol βrhoβ Ο Mass kg Volume m3
Density kg/m3 Table 10-1 Example 10-1
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Densities of materials
ππ π 3 =1000β π ππ 3 Compare solids, liquids, gasses
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Density calculation Show dimensional analysis
πΌπ π= πππ π π£πππ’ππ π‘βππ πππ π =πβπ£πππ’ππ Show dimensional analysis
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Pressure Pressure P= πΉππππ π΄πππ = πΉ π΄ 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
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Pressure variation with Depth
Weight of column of water (fluid) ππππβπ‘=ππ=πππ=ππ΄βπ ππππ π π’ππ= π€πππβπ‘ ππππ =ππβ Pressure increases with depth Old sub movies! (Das Boat, Red October, K-19, U-571) Always scene where they approach βcrush depthβ A h ΟghA P = Οgh
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Properties of pressure
Pressure pushes equally all directions. Perpendicular to surface restraining it. Equal at common depth
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Example 1β swimming pool (w/o air pressure)
Pressure at 2.0 m π=ππβ = (1000 ππ π 3 )(9.8 π π 2 )(2 π) =19,600 ππ π π 2 π 2 =19,600 π π 2 =19,600 ππ=19.6 πππ Total Force on bottom πΉ=πβπ΄= 19,600 π π π β8.5 π =3.655 β π Same pressure on bottom and sides If you add air pressure on top of pool, both go up!
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Absolute and gauge pressure
Add atmospheric pressure on top of water column π΄ππ πππ’π‘π ππππ π π π’ππ= π π +ππβ πΊππ’ππ ππππ π π’ππ=ππβ π π =1.013β ππ =101.3 πππ Sea level air pressure around 100 kPa Everything has kPa acting on it inside and out, weβre usually just interested in differences! Pa h ΟghA + PaA P = Οgh + Pa
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Not even half air pressure!
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Example 2β swimming pool with air pressure
Pressure at 2.0 m π=ππβ = π π΄ +(1000 ππ π 3 )(9.8 π π 2 )(2 π) =101,300 ππ+ 19,600 ππ π π 2 π 2 =120,090 ππ =121 πππ Total Force on bottom πΉ=πβπ΄= 121,090 π π π β8.5 π =22.6 β π
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Various units of pressure
We always use Pascals (Pa or kPa)
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Pascalβs 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) πΉ ππ π΄ ππ πΉ ππ’π‘ π΄ ππ’π‘ Pin Pout
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Hydraulic Lift Since Pressures are same πΉ ππ π΄ ππ = πΉ ππ’π‘ π΄ ππ’π‘
πΉ ππ π΄ ππ = πΉ ππ’π‘ π΄ ππ’π‘ πΉ ππ’π‘ πΉ ππ = π΄ ππ’π‘ π΄ ππ
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Columns of fluid π πππ‘π‘ππ = π π‘ππ +ππβ 2 cases
Top pressure + pressure of column must balance bottom pressure. π πππ‘π‘ππ = π π‘ππ +ππβ 2 cases If top open π πππ‘π‘ππ = π π‘ππ πππ β=0 If top closed π πππ‘π‘ππ =ππβ βbalanceβ column of water Finger on the straw trick Ptop Pbottom Οgh
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Column of water Calculate maximum height of column of water balanced in a closed straw, when bottom is open at normal air pressure π πππ‘π‘ππ = π π‘ππ +ππβ 1.013β π π 2 =(1000 ππ π 3 ) (9.8 π π 2 ) β β= 1.013β π π 2 (1000 ππ π 3 ) (9.8 π π 2 ) β=10.5 π Ptop Pbottom Οgh
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Thatβs 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 π πππ‘π‘ππ = π π‘ππ +ππβ 1.013β π π 2 =( ππ π 3 ) (9.8 π π 2 ) β β= 1.013β π π 2 ( ππ π 3 ) (9.8 π π 2 ) β=0.760 π=760 ππ Thatβs where 760 mm mercury comes from! Ptop Pbottom Οgh
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βBalancingβ column of fluid
Balancing water with straw m Water Barometer Torricelli Mercury Barometer
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Example β Problem 16 Equate pressure along line a---b
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Example β Problem 16 (cont)
Pressure at (a) π π = π π + π πππ π (0.272 π) Pressure at (b) π π = π π + π π€ππ‘ππ π β π Equating π πππ π π = π π€ππ‘ππ π β π π πππ = β π π π€ππ‘ππ π πππ = β π ππ π 3 =654 ππ π 3
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