《流体力学》课程教学资源:(英文版)Fluid statics

Fluid statics Study of fluid at rest i e. not in motion or not flowing On any plane, shear force is zero(no velocity gradient or shear deformation hence only pressure forces exist Sx ss Pasca’sLaw For fluid at rest, pressure at a point is the same in all direction
Fluid Statics • Study of fluid at rest, i.e. not in motion or not flowing • On any plane, shear force is zero (no velocity gradient or shear deformation) hence only pressure forces exist Pascal’s Law • For fluid at rest, pressure at a point is the same in all direction P = Px = Py = Pz

Proving pascal’sLaw P,8 δxδz 6 As element is in equilibrium ∑F,=0 P、δxδz=P6 x SS Sin0 Ss sin e= sz Hence Similarly,forΣF=0
Proving Pascal’s Law As element is in equilibrium : Fy = 0 Py x z = Ps x s Sin s Sin = z Hence : Py = Ps Similarly, for Fz = 0 Pz = Ps

Basic hydrostatic equation for Pressure field Rectangular free body P+↓↑ of top area a W P+ dP Body of fluid in equilibrium ∑F,=0 PA+W=(P+ dP).A .A+pgadh=(P+ dP)a dP/dh=pg=y dP/dy =-pg-Y
Basic Hydrostatic Equation for Pressure Field Body of fluid in equilibrium : Fy = 0 P.A + W = (P + dP) . A P.A + g A dh = (P + dP) A dP/dh = g = dP/dy = - g = - h dh P P + dP W Rectangular free body of top area A y

Incompressible fluid Pressure difference between two points in a body of fluid dP/dh=pg dP=pg dh Integrating from P, to P2 dp=p gdh h If fluid is incompressible and homogeneous p=constant 2-PI=pg(h2-h1=pg(Ah)
Incompressible Fluid Pressure Difference between two points in a body of fluid dP/dh = g dP = g dh Integrating from P1 to P2 : If fluid is incompressible and homogeneous = constant P2 – P1 = g (h2 – h1 ) = g ( h) P1 P2 h1 h2 h = 2 1 2 1 h h P P dP gdh

Incompressible fluid If Pi is on free water surface, i.e. PI=Pat P=P 2-Pi=pg h P P-P atm p g P2.abs =pgh+P atm For gauge pressure measurement, Patm=0 gauge pg h + P ,ga auge Gauge pressure at a point h below free surface =pg h
Incompressible Fluid If P1 is on free water surface, i.e. P1 = Patm : P2 - P1 = g h P2 - Patm = g h P2,abs = g h + Patm For gauge pressure measurement, Patm = 0 P2,gauge = g h P2,abs = P2,gauge + Patm Gauge pressure at a point h below free surface = g h h P2 P1=Patm

Definition of pressure head Pressure=P Datum Pressure head at point=Plpg(m) Piezometric head=p/ g +y (m) Piezometric pressure =P+pgy h In fluid statics, gauge pressure head at depth h below free surface Lpg=pgh/pg-h
Definition of Pressure Head Pressure head at point = P/ g (m) Piezometric head = P/ g + y (m) Piezometric pressure = P + g y In fluid statics, gauge pressure head at depth h below free surface = P/ g = g h / g = h Datum Pressure = P y h

Example I KN 0.2m 0.05m 2.51 dial di lam P1=F/A=1000/(x0.0252)=5093kPa 2=P1-pg(2.5)=48477kPa f=PA =48477x兀0.12=1523kN
Example P1 = F/A = 1000 / ( 0.0252 ) = 509.3 kPa P2 = P1 - g (2.5) = 484.77 kPa F = P2 A2 = 484.77 x 0.12 = 15.23 kN 2.5m 1 kN F ? 0.2 m diam 0.05 m diam P1 P2

Pressure at liquid interface Liquid 1 Liquid 2 p2 Liquid 3 p 1=Pgh 22=pgh+ p2gh2 P3-P1gh1 p2gh2 pgh
Pressure at liquid interface P1 = 1gh1 P2 = 1gh1 + 2gh2 P3 = 1gh1 + 2gh2 + 3gh3 Liquid 1 1 Liquid 2 2 Liquid 3 3 h1 h2 h3 P1 P2 P3

Pressure measurement Absolute Pressure: Measured relative to perfect vacuum Perfect vacuum: 0 absolute pressure Gauge Pressure: Measured relative to local atmospheric pressure If no specified, pressure reading is usually assumed to be gauge Gauge pressure is positive if higher than atmospheric pressure, and negative if lower than atmospheric pressure Negative gauge pressure is also known as suction or Ⅴ acuum pressure
Pressure Measurement • Absolute Pressure : Measured relative to perfect vacuum • Perfect vacuum : 0 absolute pressure • Gauge Pressure : Measured relative to local atmospheric pressure • If no specified, pressure reading is usually assumed to be gauge • Gauge pressure is positive if higher than atmospheric pressure, and negative if lower than atmospheric pressure • Negative gauge pressure is also known as suction or vacuum pressure

Atmospheric Pressure ↓↓↓4↓ Mercury Atmospheric pressure is usually measured using a mercury barometer Patm =phg g ha 101, 300 N/m2 abs 0N/m2 gauge ≈1.013 bar abs ≈760 mm hg abs ≈10.3 m water abs
Atmospheric Pressure Atmospheric pressure is usually measured using a mercury barometer Patm = ρHg g h 101,300 N/m2 abs = 0 N/m2 gauge 1.013 bar abs 760 mm Hg abs 10.3 m water abs
按次数下载不扣除下载券;
注册用户24小时内重复下载只扣除一次;
顺序:VIP每日次数-->可用次数-->下载券;
- 《流体力学》课程教学资源:(英文版)CV2601/G263. Fluid mechanics.ppt
- 郑州大学:《结构力学》课程教学资源(PPT课件)第十九章 结构的极限核载.pps
- 郑州大学:《结构力学》课程教学资源(PPT课件)第十六章 结构的稳定计算.pps
- 郑州大学:《结构力学》课程教学资源(PPT课件)第十五章 结构的动力计算(15.4-15.9).pps
- 郑州大学:《结构力学》课程教学资源(PPT课件)第十七章 结构的动力计算.pps
- 郑州大学:《结构力学》课程教学资源(PPT课件)第十四章 超静定结构总论.pps
- 郑州大学:《结构力学》课程教学资源(PPT课件)第十三章 矩阵位移法.pps
- 郑州大学:《结构力学》课程教学资源(PPT课件)第十二章 渐进法及超静定结构影.pps
- 郑州大学:《结构力学》课程教学资源(PPT课件)第十一章 位移法.pps
- 郑州大学:《结构力学》课程教学资源(PPT课件)第十章 超静定拱.pps
- 郑州大学:《结构力学》课程教学资源(PPT课件)第九章 力法(9.6-9.11).ppt
- 郑州大学:《结构力学》课程教学资源(PPT课件)第九章 力法(9.1-9.5).ppt
- 郑州大学:《结构力学》课程教学资源(PPT课件)第八章 静定位移计算.ppt
- 郑州大学:《结构力学》课程教学资源(PPT课件)第七章 静定结构影响线.ppt
- 郑州大学:《结构力学》课程教学资源(PPT课件)第六章 静定结构总论.ppt
- 郑州大学:《结构力学》课程教学资源(PPT课件)第五章 静定平面桁架.ppt
- 郑州大学:《结构力学》课程教学资源(PPT课件)第四章 三铰拱.ppt
- 郑州大学:《结构力学》课程教学资源(PPT课件)第三章 静定梁与静定刚架.ppt
- 郑州大学:《结构力学》课程教学资源(PPT课件)第二章 几何组成分析.ppt
- 郑州大学:《结构力学》课程教学资源(PPT课件)第一章 绪论(主讲:樊友景).ppt
- 《流体力学》课程教学资源:(英文版)Fluid Flow Concepts.ppt
- 《矩阵位移法》第六章(6-1)矩阵位移法概述.ppt
- 《矩阵位移法》第六章(6-2)矩阵位移法解连续梁.ppt
- 《矩阵位移法》第六章(6-3)矩阵位移法解平面刚架.ppt
- 《矩阵位移法》第六章(6-3)整体分析(后处理法)续.ppt
- 《矩阵位移法》第六章(6-5)非结点荷载处理.ppt
- 《矩阵位移法》第六章(6-4)矩阵位移法解平面桁架.ppt
- 《矩阵位移法》习题讨论.ppt
- 《结构力学》课程教学资源(电子教案)简介.doc
- 《结构力学》课程教学资源(电子教案)第一章 绪论.doc
- 《结构力学》课程教学资源(电子教案)第二章 几何组成分析.doc
- 《结构力学》课程教学资源(电子教案)第三章 静定结构的受力分析.doc
- 《结构力学》课程教学资源(电子教案)第五章 影响线.doc
- 《结构力学》课程教学资源(电子教案)第二章 几何组成分析.doc
- 《结构力学》课程教学资源(电子教案)第六章 结构位移计算与虚功一能量法简述.doc
- 《结构力学》课程教学资源(电子教案)第七章 静定结构总论.doc
- 《结构力学》课程教学资源(电子教案)第四章 静定结构总论.doc
- 《结构力学》课程教学资源(电子教案)第七章 影响线及其应用.doc
- 《结构力学》课程教学资源(电子教案)第九章 静定结构位移计算.doc
- 《结构力学》课程教学资源(电子教案)第十章 力法.doc