传递过程原理习题讲解

传递过程原理习题动量传递

例1-1

以两圆盘中心为 Z轴，在中间建立柱坐标 系.稳态层流， / 0, 轴对称无旋运动 u 0; 流体只沿径向流动， u z 0; 流体不可压缩，

const; 对连续性方程及运动方 程化简。

柱坐标系下： 1 1 ( ru r ) ( u ) ( u z ) 0 r r r z (ru r ) 0 r u r u r (r , z ), ru r 只与r , z有关. (ru r ) 又 0, r ru r 仅为z的函数，与r无关。

ur ur u ur u 2 ur ur uz r r r z 2 2 u 1 p 1 1 u 2 ur r Xr (ru r ) 2 2 2 2 r r r r r z r ur 1 p 2u r ur 2 r r z u u u u ur u u 1 1 p ur uz X r r r z r 2 1 2 u 2u 1 u (ru ) 2 2 2 2 r z r r r r

p 0

Z方向

u z u z u u z u z ur uz r r z 1 u z 1 2u z 2u z 1 p Xz r 2 2 2 z z r r r r 1 p g 0 z运动方程化简得到:

ur 1 p 2u r ur 2 r r z p 0 p g z

u r 2 ur ( ) 3 r r r r r 2u r 1 d 2 2 z r dz2 2 p d 2 3 r r r dz2

1 p 2u r 0 r z 2 p 0 p 0 z p仅与r有关，与 , z无关dp d 2 dr r dz2 r r1 , p p1 ; r r2 , p p2 , 积分上式 r2 d 2 p p1 p2 ln r dz2 1 r2 d 2 p ( ln ) 2 0 r1 dz

r2 d 2 r2 d 2 ( ru r ) p ( ln ) ( ln ) 2 r1 dz r1 dz2 u r d z 0, 0, 0; z b, ru r 0 z dz p ru r z 2 C1 z C2 r2 2 ln r C1 0, C2 p ( 2 ln r2 ) r1 b2

b 2 p z 2 ur 1 ( ) r2 b ( 2r ln ) r1

b p z 2 ur 1 ( ) r2 b (2r ln ) r12

b 2 p Q ur dA b r2 2r1 ln r1b

z 2 1 ( ) 2 r1dz b

4 b 3 p r2 3 ln r1

例1-2

SOLUTION: Cylindrical coordinates are appropriate for this problem, and the equations of change are given as follows

Equation of continuity

1 1 ( ru r ) ( u ) ( u z ) 0 r r r z u 0

r-componet ur ur u ur u 2 ur

ur uz r r r z 2 2 1 1 p 1 ur 2 u ur Xr (ru r ) 2 2 2 2 r r z r r r r

θ-component u u u u u r u u 1 1 p ur uz X r r r z r 2 2 1 1 u 2 u u (ru ) 2 2 2 2 r z r r r r

1 (ru ) 0 r r r

z-componentu u z u z ur r r 1 1 p Xz z r 1 p g 0 z u z u z uz z u z 1 2u z 2u z r 2 2 r r r z 2

At steady state we postulate that ur, and uz are both zero and that uθdepends only on r. We also postulate that p depends on z because of the gravitational force and on r because of the centrifugal force but not on θ. These postulates give 0 = 0 for the equation of continuity, and the equation of motion gives:r com ponent2 u 1 p r r 1 0 ( (ru )) r r r 1 p g 0 z

com ponentz com ponent

The θ-component of the equation of motion can be integrated to giveC1 C2 u r 2 r

in which C1 and C2 are constants of integration. Because uθcannot be infinite at r = 0, the constant C2 must be zero. At r = R the velocity u is RΩ. Hence C, = 2 Ω and

u r