elementary mechanics of fluids ce 319 f daene mckinney momentum equation
TRANSCRIPT
Elementary Mechanics of Fluids
CE 319 F
Daene McKinney
Momentum
Equation
Momentum Equation
• Reynolds Transport Theorem
CSCV
sys bdbdtd
dt
dBAV
• b = velocity; Bsys = system momentum
CSCV
ddt
dAVVVF
FM
dt
d
dt
dB syssys
• Vector equation -- 3 components, e.g., x
CSCV
x ududtd
F AV
kjiVwvu
Ex (6.3)
• Given: Figure
• Find: (a) Force acting on bottom of the tank and (b) the force acting on the stop block.
• Solution: d=30 mm
v=20 m/s
Tankm=20 kgVo=20 L
T=15 oC
N
AVF
VAV
uF
ududtd
F
o
o
o
CS
CSCVx
97
70cos)20)(4/03.0*(*)999(
70cos
)(70cos
22
2
AV
AV
W
N
F
jiVvu
V
oVu 70cos
oVv 70sinA
Ex (6.3)
N
AVWN
VAV
vWN
vdvdtd
F
o
o
o
CS
CSCVy
658
70sin5.28281.9)999*02.020(
70sin
)(70sin2
AV
AV
W
N
F
jiVvu
V
oVu 70cos
oVv 70sinA
EX (6.5)• Given: Figure
• Find: Horizontal force required to hold plate in position
• Solution:
kNQVF
VAV
uF
ududtd
F
CS
CSCVx
9.43.12*4.0*999
)(
AV
AV
smp
V
gVp
gV
zp
gV
zp
AB
BA
BB
BAA
A
/3.12999/75000*22
2
222
22
T=15 oC
Q=0.4 m3/s
pA=75 kPa
F
B
iVV
HW (6.12)
Ex (6.17)• Given: Figure
• Find: External reactions in x and y directions needed to hold fixed vane.
• Solution:
)left theto(25.9
)30cos2728(2.0*1000*9.0
)30cos(
)(30cos)(
)()30cos()(
21
222111
22211
kNF
VVQ
AVVAVVF
AVVAVV
uF
ududtd
F
x
o
o
o
o
CS
CSCVx
AV
AV
V1=28 m/s
V2=27 m/s
Q=0.20 m3/s
Fx
Fy
Ex (6.17)
)d(43.2
)30sin27(2.0*1000*9.0
)30sin(
)(30sin
)()30sin(
2
222
222
ownkNF
VQ
AVVF
AVV
vF
vdvdtd
F
y
o
o
o
o
CSy
CSCVy
AV
AV
V1=28 m/s
V2=27 m/s
Q=0.20 m3/s
Fx
Fy
Ex (6.34)
• Given: Figure
• Find: Force applied to flanges to hold pipe in place
• Solution: Continuity equation
• Momentum
D=30 cm
Q=0.60 m3/s
P=100 kPa, gage
Vol=0.10 m3
W=500 N
NF
F
AVVAVVApApF
ududtd
F
x
x
x
CSCVx
325,24
6.0*1000*49.8*2)4/3.0*)(000,100(2
)()(2
2221112211
AV p1A1
p2A2
V1
V2
Fx
Fy
Wb+Wf
x
y
smAQV
AVAVQ
/49.8)4/3.0*/(6.0/ 2
2211
NF
WWF
vdvdtd
F
y
fby
CSCVy
14819810*1.0500
0
AV
HW (6.37)
Ex (6.72)• Given: Water jet, 6 cm diameter, with
velocity 20 m/s hits vane moving at 7 m/s.
• Find: Find force on vane by water.
• Solution: Select CV moving with the vane at constant velocity. The magnitude of the velocity along the vane is constant
Fx
FyV2
NF
AVV
AVVVVAVVVVF
ududt
dF
x
o
ov
vo
vvvx
CSCVx
7.815
)45cos1)(4/06.0*)(1000()720(
)45cos1()(
])[(45cos)(])([)(
22
2
21
AV
vVVVV 21
NF
AVV
AVVVVF
vdvdt
dF
y
o
ov
vo
vy
CSCVy
9.337
)45)(sin4/06.0*)(1000()720(
45sin)(
])[(45sin)(
22
2
2
AV
AVVAVAV v )(2211
HW (6.80)
HW (6.98)
HW (6.101)
Sluice Gate
• Find: Force due to pressure on face of gate
• Solution:
Assume: v1 and v2 are uniform (so pressure is hydrostatic)
)(2
)(
)()2
()2
(
)(
][][
22
2121
1222
11
122211
222111
yyb
vvQF
vvQFbyy
byy
vvQFApAp
AvvAvvF
ududt
dF
G
G
G
x
CSCVx
AV