유체역학 3판(한티미디어)chapter02

소개글

유체역학 3판(한티미디어)chapter02에 대한 보고서 자료입니다.

목차

없음

본문내용

2.1 S D D
D D
F ma p z p s
y z
y = y : y - sina = r ay
2
S D D
D D D D
F ma p y p s
y z
a g
y z
z z z z = : - cosa = r + r
2 2
Since Dscosa = Dy and Ds sin a = Dz, we have
p p
y
a y y - = r
D
2
and p p ( )
z
a g z z - = r +
D
2
Let Dy ® 0 and Dz ®0:
p p
p p
y
z
- =
- =
üýþ
0
0
\\ p = p = p y z .
2.2 p = gh. a) 9810 ´ 10 = 98 100 Pa or 98.1 kPa
b) (0.8 ´ 9810) ´ 10 = 78 480 Pa or 78.5 kPa
c) (13.6 ´ 9810) ´ 10 = 1 334 000 Pa or 1334 kPa
d) (1.59 ´ 9810) ´ 10 = 155 980 Pa or 156.0 kPa
e) (0.68 ´ 9810) ´ 10 = 66 710 Pa or 66.7 kPa
2.3 h = p/g. a) h = 250 000/9810 = 25.5 m
b) h = 250 000/(0.8 ´ 9810) = 31.9 m
c) h = 250 000/(13.6 ´ 9810) = 1.874 m
d) h = 250 000/(1.59 ´ 9810) = 16.0 m
e) h = 250 000/(0.68 ´ 9810) = 37.5 m
2.4 (C) p = g Hg h = (13.6´ 9810)´ (28.5´ 0.0254) = 96600 Pa
2.5 S =
20 144
62.4 20
p
g h
´
=
´
= 2.31. r = 1.94 ´ 2.31 = 4.48 slug/ft3.
2.6 a) p = gh = 0.76 ´ (13.6 ´ 9810) = 9810 h. \\h = 10.34 m.
b) (13.6 ´ 9810) ´ 0.75 = 9810 h. \\h = 10.2 m.
c) (13.6 ´ 9810) ´ 0.01 = 9810 h. \\h = 0.136 m or 13.6 cm.
2.7 a) p = g1h1 + g2h2 = 9810 ´ 0.2 + (13.6 ´ 9810) ´ 0.02 = 4630 Pa or 4.63 kPa.
b) 9810 ´ 0.052 + 15 630 ´ 0.026 = 916 Pa or 0.916 kPa.
c) 9016 ´ 3 + 9810 ´ 2 + (13.6 ´ 9810) ´ 0.1 = 60 010 Pa or 60.0 kPa.

키워드

유체,   역학,   유체역학,   유체역학 3판(한티미디어)chapter02