ページ3:

Ex.
Driver
(Engine)
712
Priven
AD
[had]
角速度
W₁
W₂
wr
(Tire)
⇒ need T = ?
Z= M
Wi> Wo
JA = 2πurdr
du
力距
dy-h
=
②
JF = ZJA
W₁-Wz
dr
=
③LT = df.r.
=
④
=
(u W₁-We z
h
Sat Safir s (u W.-W= 2π) j³ dr
=
Tα W₁ - Wz
h
•
Cu
Wi-Wz
h
.2π)
27
(*)*
4
4
= M. Tv. d (W₁-W=)
32 h

ページ4:

毛細管上升
Capillary rise ₤0±4
Jow
(-10)
C:上升高度
:
Jaw tension force / unit length
(air, water)
r:
管子半徑
Σ Jaw cost = W
FP
Jow (aπr) cas &= (TLP) zu eg
Zc=
2 Jances $0.02 T
Boiling $175
egr
GSD + 1 egr
external absolute pressure = vapor pressure
Cavitation 穴蝕
因為V太快, VPL
液體pe飽和蒸氣壓
Cavitation Phenomenon
What is Cavitation and its Causes?
→產生氣泡,造成機器損壞
Surrounding liquid
Increased static pressure
Cavitation bubble imploding close
to a fixed surface generating a jet (4)
of the surrounding liquid.

ページ5:

Fluid Statics 流體靜力學
P.
FP
P. pressure stress (normal)
A
7 = Fs
·Fp pressure force
Fr
A
2 shear stress.
(")
Fs-
A
Ps shear force
=
· Prelative Pgage = Pabsolute - Putm
基本方程式推倒只
P-PLAR
P-p(JR-37)
dx
p+p (op d²)
dz
6設為0
P+P(學業)
12 P+p (JR Ja
p-p (of 12) p-p (+1 d²)
dz
P
× A
1ΣF=0 (P- (2) Jysk - (p+ (+ 1/2)) dy dz=0
- (+47) dx dy dz=0
JP
=>
-=0
+15=0 £140
=

ページ6:

-evg =
1+ΣF = 0 p - (of d² ) dx dy-(P + (1±²))
- (+12) dx dydz - eg dx dy dz.o
-(+)-eg=0
If = -eg
=
P2
P₁
=
72
dpegdz
-
P₂- P₁ = eg (32-3.) =-egh
=
P₁ - P₂ egh
結論:同一水體中,同一水平面,壓力一樣
Fr. SJF,
Fp = SdFp
SP. JA
合力作用點
Jp
=
SAYP JA
Fp
D

ページ7:

Ex.
水面
✓
M
b
▼
P=0
L
==
h = Lsno
=
P₂ = eg Hsin
Fp Sd Fp Sp. dA
=
H
- Segh) (bal)
H
Seg Lsin @bdL
Fp= ?
Lp?
W = ?
=
1961 sino
S. eg L'sin O b JL
57 - Fp x Lp SdFpx L
-
3
egbH Sino Lp = eg b H³ sino
X
Lp = =² H
3

ページ8:

為了讓門不要轉
+27=0
Fx Up - Wx
eg b H³ sin o
3
Hase
2
=
W
W = zeg bhi tano
3
0
Hoso
=
W

ページ9:

合力
3-69 Determine the resultant force acting on the 0.7-m-high
and 0.7-m-wide triangular gate shown in Fig. P3-69 and its
line of action.
作用線
0.9 m
Water
hc
FIGURE P3-69
0.3 m
0.7 m
L= 0x0.3m
10.3
0.7 m
Using Using
Peg Lsing
@ da = BidL.
• JFp= PJA
=
=
(YL sin ẞ) (BIL)
= r L x 0.9 ( 1-0.3) dL.
=
·Fp = ?
Lp = ?
1-0.3
=0.3-0.3=0
dl
· L-0.3
-11-03-07
·B-L-0.3
Sing=0.9
Fp SdFp (r10.9) SL1, L203 LIL
+J T = Fp · Lp
辈
1.658 x 4p
L=0.3
1.658 (KN)
#
= (ring) fl\13 (²=-0.3 L) LIL
=
Lp
=
1.33
0.8 (m).

ページ10:

Ex. Direct intergration 直接積分法用在曲面
1-O!
水面
h₂-R
L
dé
·A
W = ?
F p Z E D F F A TZ Y ?
R=1mL=5m
0
5
=兀
* r = 9.81 KN
P = egh = rhr (R· Ruso)
egh-rh
h = R - Rus
A=L(R↓o)
1+Σ F₁ = 0, dFP+ + W = 0.
JFp =Px dA cso
= [r (R-Rus)] [L (Rdo)]
=
(VRL - VR as OL do
rŔi (|- Coso) Cso do
Coso

ページ11:

- W = SFp₁ = √ √ RL (wo-we
=
0
TV
9.81 × 1.5 S caso -50 do
x
0
49.05x (-3.36)
=
-164.62 (kN)
W = 164.62 kN
Fp sind - √ & ~ [r (R-Ruso)] [L (Rdo)] sinodo
=
0
tan O
Fp sin
Fp CSO
命力
Sing
O tan o

ページ12:

Free body diagram 水體受力圖
D
Perak)
0
P=r (R)
FX 互相抵消
Fz P= (³R)
P= rh P₁ = r(R); P₂ = ricr
1.
F = AP, F₁ = (RL) R.
YR'L
=
2
2
24.5(kN)
(kN)
F2 = (2RL) ri²R = 4rR²L = 4×9.81×5-196.2
9.81×5
=
=
2
F₂ = FP - WFB
WFB = [(+R) (R) L - TUR³ L] = r = (4-7) L = Y = 31.58
W= Fps = 196.2-31:58
= 196.2-31.58 = 164.62 (KN).
·FP, = F₁ = 24.5 (kN)
(KN)
FP
24.5
tang=
164.62 = 8.4°

ページ13:

慣性坐標
L.
Fpx
P=0
I inertial
ax
¨p + (JP) dx
±2Fx =0. Fpx Fix = max
m
(dk) Jx dj dz = f (dx dy Jz) ax
y
ap
dx.
=
-lax
↑ + Σ F₂ =0, Fpz - Fiz - W = maz
- (*) dxdydz - eg (dx dydz) = ( (dxdy dz) az
JR = 1 (9+α²)
JZ
dz
ax
=
dx
g+az
非慣性坐標
把(Fpx-Fax).(Fp2-F13)分别看成一個力就好

ページ14:

Ex. 3
Az
a
A
Ax = a cosα
ax
oh = ?
PA=?
ax
slope = | d3 | = | (gar-122) |
g+az
L
acosa
oh=
2
9+asina
Az = a sin α
-
a cosa
-
oh
9+asina
L
PA = 0 + (2/2) (-07) = e (g+az) oz
= 0 + ( 38 ) (-*x)
(袋)(-x)
e ax ox

ページ15:

Ex. 4
水面線方程式?
300'
Ac
M
=
V = wr
r
Prdod rdz
F
dr
rdo
dz
rdp P+ (JP) dr
dr rdg
F₁ = ma c
=
Idz.
w² re r d o d r d z
ai
dz
Fp = pro dz - p + (JP) dy do da
Jr
rdOdr dz
Fp + F₁ = 0
Fc
-
ΣFr=0
,
因為軸對稱0
=0
(173) (1909)
,
m
- ³r + w² er = 0

ページ16:

P₁ = Pr Yi + Pz Z t + Pp Or
t
=
dp JP dr + JP dz
+
JP 20
0
=0水面是等雁,沒邺
=>
= ( w² (r) dr + (- (g) dz = 0 * & x x x p
》
werdr= eg dz
dz
13 war wy
eg
Sidz = y²
2
w²
rdr
2.2. № 1
。
=
g
2 = Zo +
W.
2
=
g
是g + az,但這題沒az
(↑)

ページ17:

Lau of Buoyancy 阿基米德浮力定律
D
上半部
1
自由體
下半部
4
FB = W total - (W+W)
FB = r Vb
水比重x水裡體後
自由體
Fp-rh A

ページ18:

Fluid kinematics 流體運動學
Lagrangian viewpoint 跟縱同一個粒子
流場特性B:
X (t) y (t) Z (t).
√ (t). α H). P. (...
Eulerian Viewpoint 埋伏看通過的粒子
·B-B (x, y, z, t)
觀察點
位置
e.g.
連著飛機,
日
(x,y,z)
不是跟著流體
Eulerian

ページ19:

(substantial or material derivative ***)
DB Total derivative (全導數)
D.t
B
dt
JB
JB + v dB + wa
JB
=
+4
ot
xx
dx = u di local derivative
dy=v.dt
V
ay
Convective derivative
*14Q
B=Buda de B = B (x, y, z, t) Eulerian:
dz = w
Lagrangian
Steady flow 恆定流
JB
ot
= 0
31 x (+), V(t). P (t). (It)
都不隨時間改變
流線、徑線、煙線會重疊

ページ20:

dt
=
Ex. 1. - (u, v, w) » * a (ax, aj, az) = ?
DV
bt
=
Jū
Jū
u
+ V
dt
dt
+ W
ot
ax
Du
Ju
Ju
Ju
=
=
+ u
+ V
Ju
bt
dt
dt
+ W
ot
=
Dv
ay D+
d.v.
Ju
dv
Jv
+
+ V
dt
dt
dt
+ W
ot
Dw
d
dw
az
=
+
u
+ V
& w
bt
dt
dt
dt
+ W
ot
Ex. 2
DT
bt
=
x = 4000 km
T₁ = -0.5°C / day
Tx = -1 °C/100 km
u = 200
Jt
200 km/day
JT
+ u
= - 0.5 + 200
溫度變化率
DI
?
Dt
(-1)
= -2.5 °C / day
100
W
+(x-22) 2
Ex. 3 v = (3x-y) + (by +37) 3 +
①跟隨流場
② 固定 (2,1,1)
i
→>>>
③相對V14,2,3)
Lagrangian
Eulerian
=-
→ relative (3xy, 2y+32, x-22)
Eulerian
沒跟隨流場

ページ21:

Streamline 流線
某瞬間速度(切線)方向:拍不出來
a.不含相交
b.很多流線→流管 streamtube
流線方程式
dx
dy
紫::赀→積分
=
Pathline 徑線
連續曝光,得到「單一」粒子
9.一段時間才拍得出圓圈
b. 7A 3D +7 7
Streakline 煙線
瞬間曝光,得到『所有」粒子
a. 看得出擴散
&

ページ22:

2D flow field U-3y²
xy
dt: dx dy
①
u
dy
V=-2X
dx
dy
3y²
-2X
-2x dx = √3y² dy
=
3
①求流線方程式
② (5,2),(4,-1)是否在同一流線上
@ (5,-)), (4, 1)
④求流體穎粒軌跡y(t)=?
⑤兩點之間需要時間七
-x² - y² = x²+ y² = c **
h 1
每一條流線有一個(值
15²+ 2 = 33
不同
4+ (-1) = 15
3
③ 5²+(-2):17
4² + 13 = 17
V
dy
同
dt je x x x y
dt = dy
好複雜。
-2X
S
dy
-> (17-y³) =

ページ23:

Timeline 時間線
.每秒放出一排泡泡
外域
無滑動邊界
Motions of fluid element
1. rate of translation 移動率
u = dx
dt
I. rate of deformation 線變形率
x方向長度變化
Ⅲ
Exx
(0(x)
Ju
ot
8x
2X 變形率
rate of angular information 角變形率
Ju
Exy
=
+
x
sy

ページ24:

- ) 逆時針為正
IV. rate of rotation
Ju
+) R₂ = = (dx - Jy )
V.
vorticity
旋度
52 = 2R = 2 < Rx, Ry, Rz>
<52x, szy, 528>
1x (curl)
irrotational flow 52=0
非旋流
->
·0/5
3

ページ25:

Ex. 5
順旋流
逆旋流
野
Ex.6
< u, v, W> = < x² + x,- (axy+y); 0>
07
今
r =
下
d
ry
: -2y- (2y) : 0
=
x²-1+x-(2xy+y) 0
irotational

ページ26:

Reynolds Transport Theorem RTT 雷諾傳輸定理
= dB sys d
=
RTT - JBuy ISIS pedu of ev· JA
dt
C.V.
+
CS.
流場在系統
流場在女
穿過C 表面
的變化
的變化
的流場淨通量
system
rate of change net outflow rate
out - in
(+) (-)
人,裡面的東西一樣
(Lagrangian)
0
② contral volume C.V.
可以移動的空間
。
裡面的東西可以進去出來
③ contral surface C.S.
Cv. 的表面

ページ27:

extensive
properties
B
和所有關
e.g.
F = ma
M = m v
③intensive properties B.
qo m zx l. u. P. T
Continuity Equation CE
When
3 = 1, B = m
ß
Jmsys
dť
0
質量守恒
(1) steady flow JB = 0.
(2)
incompensi ble fluid (= const.
A,
=
C.S.
A₂
Q=A. V. = A₂ V₂

ページ28:

Ex.1
V = 0.05 m²
P = 800 kPa
T= 15°C
2
-5 2
A = 65mm = 6.5×10³ m²
V = 3111
1 = 6.13 £9
m³
I mays 0-seat
dt
=
=
+
§ evdÃ
.C.V.
v不變但又變
0
de
dt
CS.
(十) (-)
出去一進來
(SSC dt) + (VA-0
C.V.
0 = 11 v + eva
de
-eva
dt.
=
- 6.13 x 311 × (65 × 10 5 ).
0.05
2.478 (7))

ページ29:

Ex. 2
Q₁ = 1.2 cms
dt
(1 = 0.013 £23
3
m
=
-0.015.
la
某一瞬間
d
σ = 8 m³
O.V.
CA = 0.02
dma
dt
kg
=
-45
day
-45 kg
24x60
Q2. 0.8 m³ (cms)
12. CA
JCA - ?
kg
dt
m³
myslet & evañ
dt
dma
dt
-45
24x602
-45
24x602
=
=
dt
+
=
2-
C. V.
C. S.
d
dt
e SSS dt + P₂ Q₂- li O,
et +
CAQ₂ - CQ
dt
C.V.
2
(l+ t + (vt) + 0.02x0.8 -0.013 × 12
Chain rule'
-11 (48-002 × (-0015)) -0.014
-45
24x602
de
dt
=
=
-7.76x103
kg
m's
x

ページ30:

Divergence theorem of Gauss
I mays
dt
=
d
dt
Sssed
C.V.
+
Surface 淨通量
BedÅ = 0
C. S.
||
SSS Fev st
C.V.
d
dt
=
=
ev =
d
ay
>
<en, ev, ew>
=
8114)
JX
+
dev) + allw
ry.
volume 裡的改變量
1410
· ( (44 33 +4477) + (4+1 +244 + 1)
Sssedutel
·C.V.
SSS
C.V.
DP
Dt
+
7
t
t
dz
V
V
ry
Ju
dx 7
by + am) + (u de + very + wal) st
() -
t
=
0
CE in differential form
=
For I const. fluid du
(im compensible)
JW
+
8x
t
dy
= 0
dz
=
Exx + Eyj + Ezz 0 →
Moohi

ページ31:

Ex. 2D
irrotational Water flow
V
(xy plane) <u, v> =
V = f(x, y) = ?
=
• For const. fluid
(im compensible)
For
imrotational
15221 = 0
siz
H7-24)
dy.
↑ Ĵ ĥ
Ju
8x
< x-2y, σ> at (0,0)
Ju
8x
X-zy
dx
+
t
+
JV
ay
+ 0 = 0
ay.
= -2
-
0
今
54
|-
u
V
W
=
x-2y
d
d
:ry:
v = -2x-y+c
8x
d
0
前言
πt' (x, y, v)
= (0, 0, 0)
→ C = 0.