北京航空航天大学：《工程流体力学》课程PPT教学课件（双语版）Chapter 2 Pressure Distribution in a fluid

Chapter 2 Pressure distribution in a fluid (Fluid Statics Basic) White Chapter 2 潘錦珈:第一章

1 White: Chapter 2 潘锦珊： 第一章 Chapter 2 Pressure Distribution in a fluid (Fluid Statics Basic)

Pressure △F Definition: p=lim 0△A Unit Pa=N/m2(sl) kg/cm" atm mmHg mmHo PSI ( Pound per Squire Inch) Properties of Pressure u Vertical to the surface and point into it O At any point, pressure is independent of orientation

2 A F p A   =  → lim 0 Definition: Unit: 2 Pa = N m (SI) 2 kg cm atm PSI mmHg mmH2 O (Pound per Squire Inch) ❑ Vertical to the surface and point into it. ❑ At any point, pressure is independent of orientation. Properties of Pressure Pressure

Verification: At any point in a static fluid, pressure is independent of orientation (up) Forces on left and up surface G=Pg(-bAxAz) P=p,b△y △ P=p,b sin e △ Px=pn bAy=p,bsin 0 sin e p2=pn+pg△ When)0→P1=p2=pn=p

3 At any point in a static fluid, pressure is independent of orientation. Verification:     = +  = p p g z p p z n x n  2 1 When z ⎯→0 px = pz = pn = p pn y x z x pz x p  ) 2 1 G = g( bxz (up)   sin sin y p b y p b x n   = sin  y P p b n n  = P p b y x = x  Forces on left and up surface: x : 

Fluid mechanics Fluid Dynamics Fluid statics Fluid in motion Fluid at rest Aerodynamics

4 Fluid Mechanics Fluid at rest Aerodynamics Fluid Statics Fluid Dynamics Fluid in motion

§21 Fluid@rest u Pressure is the only surface force O Pressure distribution relates to body force only Applications 口Dams(水坝) 口 Buoyancy related instrument(利用浮力的装置) 口 Fluid power system(液压驱动系统 口 Connected vessel(连通器)

5 ❑ Pressure is the only surface force. ❑ Pressure distribution relates to body force only. ❑ Dams（水坝） ❑ Buoyancy related instrument（利用浮力的装置） ❑ Fluid power system（液压驱动系统） ❑ Connected vessel （连通器） ❑ …… Applications: § 2.1 Fluid @ rest

82.2 Equilibrium of a fluid element Consider a cube in a static fluid Pressure at the center is p, Body forces are R=Xi+Yj+Zk ap dx Pleft=P Ox ap dx Pleft=p+ OX left d dy dx

6 Consider a cube in a static fluid Pressure at the center is p; Body forces are R Xi Yj Zk     = + + dx dy dz right p left p p        = +   = − 2 2 dx x p p p dx x p p p left left § 2.2 Equilibrium of a Fluid Element

Pressure Op dx d dx Body force: )Zdxdyd pkdxdydz dx

7 dx dy dz Xdxdydz ZdxdydzYdxdydz dx dy dz 2 dx xp p  + 2 dx xp p  − p Pressure: Body force:

In x direction plhh-、parh+phe=0 Ox 2 个 Force on Force on Body force in left surface right surface direction Op- pX Euler Equilibrium Equations ax (Euler 1775) P qp=p(kax+Ya小y+z) z

8 0 2 2  + =         − +        − dydz Xdxdydz dx x p dydz p dx x p p  In x direction: Force on left surface Force on right surface Body force in x direction X x p =             =   =   Z z p Y y p   Euler Equilibrium Equations (Euler 1775) dp = (Xdx +Ydy + Zdz)

pi+ Opk=p(Xi +Yj+Zk) ax Our O p=0 Pressure increase in the direction of body force Equipressure surface(等压面) Surfaces in fluid with same pressure, vertical to body force everywhere, in gravity field it is a horizontal plane

9          =   =   =   Z z p Y y p X x p    Pressure increase in the direction of body force. Surfaces in fluid with same pressure, vertical to body force everywhere, in gravity field it is a horizontal plane. Equipressure surface（等压面） p R   =  k (Xi Yj Zk ) z p j y p i x p       = + +   +   +   

8 2 3 Pressure Distribution under Gravity Basic rule P d=p(女x++Z) dp=pzdz =-pgdz(z=-g) x p=-pg=+C General solution 2 Boundary 2=20,P=P0 C=p0+p2=0 condition p=p0+g(=0-2) p=po+pgh= po+rh

10 dp = (Xdx +Ydy + Zdz) x z z h z0 p0 dp = Zdz = −gdz (Z = −g) 1 Basic rule: p = −gz +C General solution 2 Boundary condition: 0 0 z = z , p = p 0 0 C = p + gz ( ) 0 0 p = p + g z − z p = p + gh = p +h 0 0 § 2.3 Pressure Distribution under Gravity