Flowing in pipe orce analySIS (1)the pressure on two end faces (p, -p2)zr (2 )the friction force on cylinder face T2Trl So, from),=0, obtain (p1-p2)m2-z2ml=0 Simplify it and quote Newton's internal friction law [=-u d r the result is P1=P2 (5-3) b 2ul 2 ul This method can choose the above balance cylinder to found the balance equation only on the circumstance of steady unilateral flow axial symmetry, equal diameter uniform flow etc 21
21 (1) the pressure on two end faces (2)the friction force on cylinder face ( ) 2 1 2 p − p r 2rl So，from = 0 ，obtain Fy ( ) 2 0 2 p1 − p2 r − rl = Simplify it and quote Newton’s internal friction law dr d y = − This method can choose the above balance cylinder to found the balance equation only on the circumstance of steady ,unilateral flow , axial symmetry, equal diameter uniform flow etc . r l p r l p p dr duy 2 2 1 2 = − − = − （5—3） force analysis： the result is
着空 受力分析: 、两端面上的压力(p1P2)厘m2 、圆柱面上的摩擦力z2元Z 于是,由∑F=0,可得 P1-P2)m2-2m=0 化简并引用牛顿内摩擦定律τ=-4,,可得 P1=P2 (5-3) b 2ul 2 ul 该方法只有在定常、单向流动、轴对称、等径均匀流等情况 下才能取出上述平衡圆柱体,建立该平衡方程。 22
22 一、两端面上的压力 ； 二、圆柱面上的摩擦力 。 ( ) 2 1 2 p − p r 2rl 于是，由 Fy = 0 ，可得 ( ) 2 0 2 p1 − p2 r − rl = 化简并引用牛顿内摩擦定律 ，可得 dr d y = − 该方法只有在定常、单向流动、轴对称、等径均匀流等情况 下才能取出上述平衡圆柱体，建立该平衡方程。 r l p r l p p dr duy 2 2 1 2 = − − = − （5—3） 受力分析：
管空流动 2. Velocity analysis and flux Integral the formula (5-3), then b)<<4 r+c 4ul the boundary condition of round pipe as r=R, Dy=0,So C R R2一r (5-4) 4 The above formula shows that the velocity on cross section of pipe u has 2th power spinning paraboloidal relation with the radius r, as shown in Figure 5-4 Figure 5-4 velocity and shearing force distribution of laminar flow in round pipe 23
23 r C l p y + = − 2 4 Integral the formula（5—3），then the boundary condition of round pipe : as , r = R y = 0 ，so R l p C 4 = ( ) 2 2 4 R r l p y − = （5—4） The above formula shows that the velocity on cross section of pipe has 2th power spinning paraboloidal relation with the radius , as shown in Figure 5—4. y r d 0 max y r dr R Figure 5—4 velocity and shearing force distribution of laminar flow in round pipe 2. Velocity analysis and flux
中 速度分析与流量 对(53)式积分,则D=-4n2+C 圆管边界条件r=R时,U=0,于是C=,R,所以 . R-r (54) 4ul 上式说明过流断面上的速度D与半径r成二次旋转抛物 面关系,如图54所示。 max 图54圆管层流的速度分布与切应力分布4
24 r C l p y + = − 2 4 对（5—3）式积分，则 圆管边界条件 r = R 时， y = 0 ，于是 R ，所以 l p C 4 = ( ) 2 2 4 R r l p y − = （5—4） 上式说明过流断面上的速度 与半径 成二次旋转抛物 面关系，如图5—4所示。 y r d 0 max y r dr R 图 5—4 圆管层流的速度分布与切应力分布 二、速度分析与流量
Flowing in pipe Choose a micro round area whose width is r on the position where the radius is dr, then the flux is 0=Ju,d4=AP(R-Pprmdr-Td d--Tapd' 8128l (5-5) This formula is called Hagen- Poisuille law shows When the flow is laminar flow the flux in the pipe is in direct ratio with fourth power of the radius of pipe or diameter 25
25 Choose a micro round area whose width is on the position where the radius is ，then the flux is r dr ( ) l pd l pd R r rdr l p Q dA A R y 8 128 2 4 4 4 2 0 = − = = = （5—5） This formula is called Hagen- Poisuille law . shows： When the flow is laminar flow the flux in the pipe is in direct ratio with fourth power of the radius of pipe or diameter