유체역학 환경공학과 20071518 한승희

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유체역학 환경공학과 20071518 한승희. 제 2 장 숙제 및 예제풀이 . 유체내 한점에 작용하는 압력은 모든 방향에 대하여 같다는 것을 증명하라 . 이러한 사실을 입증하기 위하여 정지유체내의 한 점 ( x,y ) 에서 단위 폭을 갖는 쐐기모양의 자유물체를 택하여 이것에 대한 힘의 평형을 생각하여 본다 ( 그림 2.1). 이 경우 전단력은 작용하지 않고 , 다만 수직표면력과 중력만이 작용하므로 . x 와 y 방향의 운동 방정식은 각각 다음과 같이 표시된다 . - PowerPoint PPT Presentation

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20071518

20071518 2 .

.

(x,y) ( 2.1). , . x y .

, . x y .

, px, py, ps , , , ax, ay x, y (x, y) 0

.

() () . yx .

. 2 , 3 1 3 .

. .

0 , .

Newton 2 Euler .

(surface force) (body force) ( 2.2). , y y . (x, y, z) p y

2.2 .. y/2 y . y .

x z ,

2.2 F .

, 0

. , 0 .

(gradient) del . 8.2

p -p f.

. , 0 Newton 2 . a . (f - jr) . .

Pascal . , . p y

. , , .Derive the pressure equation of the compressible fluid for the constant temperature and varying temperature.

.

.

. ln . .

. ()

(-0.00651 K/m). , .

, .

Derive the moments of inertia of simple areas about centroidal axis for the figures such as follows

[ 2.6] (gate) CDE( 2.12) CD (hinge)E P . O.8 . . (a) , (2.5.2) . (b) , (c) P

2.12

, 2.12

x y , y = 4 x = O, y = 6.5 x=3

a, b , a, b 1 x y .

,

(2.5.2)

(b) . (2.5.8)

x , . (2.5.11)

, O.32 ft .(c) , CD

[ 2.7] [ 2.14(a)] . 2,500 . .

. () [ 2.14(b)] , . 0 [ 2.14(c)]. . . 3,000N . 2.14(d) () . y , y , 0

. 2 3 . Newton-Raphson( B.5) ().

y . , y=2.5 y . , y=2.196m .

[ 2.8] (gravity dam) . . 2.15 . 2.5 , . 1ft . , , (foundation pressure) ( )(hydrostatic uplift) . , ()(hydrostatic head) 1/2 , 0 . , . , . () () . , , . 0

. 2.15

. , . ,

, . () . 0 .

. 3 . () 3 .

.

. . 2.18 3 . x . x

x . , x . x x . . x (y ) , x (y ) .()(closed body) , () , () 2.19 , 0 . x B C . B C ,

.

30[ 2.10] . 2m . 8 1 . xz , () y .

yz , 2-(4/3 )(2)m .

[ 2.11] (2.22). . 1m . (a) (b) .

.(CD () A .)

AB

.

(b) ABC CD . BC CD BCD 0.

.

2 9,806 Pa.

.

, () . r () . e .

.

2.25

0 . ,

0 . () . . . ()(center of buoyance) . () , . . ( ) () .

.

, , . 2.26 . , W V

.

()(hydrometer) . 2.27 . a .

(2.72)

, . .

, .

. , (2.7.2) (2.7.3)

. . [ 2.15] 1Mkg 2.32 . 2.0m 0.5m . y-y (rolling) x-x (pitching) .

2.32

[ 2.17] 66, 2 ( 2.35). () . .

. 0 (2.9.2)

p = 0 for y = 0, x = 4.5; so p0 = 2.25. Then, for y = 0, along the bottom . ,

When a free surface occurs in a container that is being rotated, the fluid volume underneath the paraboloid of revolution is the original fluid volume. The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig. 2.37). The rise of liquid from its vertex to the wall of the cylinder is, from Eq. (2.9.6), for a circular cylinder rotating about its axis. Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder, the volume of the liquid above the horizontal plane through the vertex is

. , . . ( 2.37) () (2.9.6)

. 1/2

2.37 ,

. . .