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1
MA3006 Fluid MechanicsSemester-2 (13/14)
Dr. YEO Joon HockN3-02b-576790-5500
E-mail: mjhyeo@ntu.edu.sg
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Prof Yeo JH1. Momentum and Energy equation (7 hr) 2. Dimensional analysis and similitude (4 hr)
Prof Chan WK3. Internal flows and piping systems (8 hr)4. Principles and applications of fluid machines (7 hr)
Textbook BR Munson, DF Young and TH Okiishi, Fundamentals of Fluid Mechanics, 7th edition, John Wiley & Sons,
3
Or along a streamline.
constant2
1 2 gzVp
constant2
2
zg
V
g
p
RevisionsBernoulli’s equation•Applies to steady, incompressible (density ρ = constant), inviscid, and irrotational flow along a streamline.
where p = pressure, z = elevation and V = velocity
System vs. control volumeA system (sys) is a collection of fluid particles. It moves, flows and interacts with its surroundings by exerting a force.
4
5
A mass of air drawn into the compressor of a jet engine is considered a ‘system’. The air changes its shape, size and temperature as it is compressed and expelled through the outlet.
In the system, we (the observers) follow the fluid and observe its behavior as it moves.
6
A control volume (C.V.) is a volume (not mass) in space through which fluid may or may not flow.
C.V. is similar to the free-body diagram in dynamics.
(a) A fluid flowing through a pipe: (1) entrance, (2) exit
(b) A deflating balloon
7
A control surface (C.S.) is a surface of a control volume.
Sections (1) and (2) in (a) are C.S. with flow passing through. The pipe wall is a C.S. without flow passing through.Case (a) is a fixed C.V. but case (b) is a deforming C.V.
In C.V., we (the observers) are usually stationary and observe the fluid behavior.Governing equations of fluid motion are stated in terms of fluid systems, not control volumes.
8
Continuity: mass flow rate is conserved or
time rate of change of Msys (system mass) = 0,
0sysMDt
Dwhere V
sys
sys ρdM
In terms of C.V. and C.S.,
0)ˆ(.. ..
VC SC
dAnVVdt
9
If the flow is steady, then
&0..
VC
Vdt
0)ˆ(..
SC
dAnV
: sum over C.S.
dA: differential area A on C.S. : outward normal vector : component C.S.⊥
..SC
)ˆ( nV
n̂nV ˆ
V
10
11
5.2 Linear momentum equation
Newton’s 2nd law: )( Vmdt
dF
Fluids: continuity equation
.. ..
)ˆ(VC SC
sys dAnVVdt
MDt
D
Inserting into each term,V
.. ..
)ˆ()(VC SC
sys dAnVVVdVt
VMDt
D
(Pg 211)
12
.. ..
)ˆ(VC SC
dAnVVVdVt
F
If the flow is steady, then properties like ρ & are independent of time in C.V.
V
0..
VC
VdVt
Then, ..
)ˆ(SC
dAnVVF
in out
dAnVVdAnVVF )ˆ()ˆ(
13
If ρ & are constant at each inlet, thenV
inin
VAVdAnVV
)ˆ(
Similarly, at the outlet out
out
VAVdAnVV
)ˆ(
Therefore, inout VAVVAVF
inout VmVmF )()(
where is the mass flow rate through A.
AnVm )ˆ(
14
If there are many inlets and outlets, then
inout VmVmF )()(
Example 5.10 : A jet of water exits a nozzle (area A1) with a uniform speed V1. It strikes a vane and is deflected through an angle θ . Find the force needed to hold the vane stationary.
Ignore gravity.
Flow is steady.
15
Select C.V. and assign axes.
Continuity: mmm inout
16
Bernoulli’s equation: (1)→(2)
2
22
21
21
1 22gz
Vpgz
Vp
Ignore gravity: z1 = z2 p1 = p2 = patm
Outcome: V1 = V2 and A1 = A2
Force to hold vane:
Momentum eqn. in x-dir. inxoutxx VmVmF )()(
12 cos VmVmFAx
17
)cos1(1 VmFAx Momentum eqn. in z-dir.
inzoutzz VmVmF )()(
sinsin 12 VmVmFAz
Example 5.15 : A jet engine is mounted on a thrust stand. Velocity at inlet V1, at outlet V2 inlet area A1, outlet area A2 air density ρ , inlet pressure p1 (abs), outlet pressure p2 (abs). Find thrust Fth.
18
Steady flow:Select C.V. & assign axes.
mmAVm outin 11
Momentum eqn. in x-dir. inxoutxx VmVmF )()(
19
)()()( 122211 VVmAppAppF atmatmth )(
)(
12
212211
VVm
AApApApF atmth
Example 5.12 : Water enters a horizontal 180° pipe bend at speed V1. X-sectional area of bend is uniformly A1 = A2. Internal gauge pressures at (1)& (2) are p1 and p2. Find the force needed to hold the bend in place.Weight of nozzle: Wn
Weight of water in nozzle: Ww
20
Flow is steady.Select C.V. & assign axes. , ,
11AVmin
22 AVmout
As A1 = A2 = A, V1 = V2. Force: Inlet , outlet
Continuity: mmm inout outin AVAV )()(
21
Momentum eqn. in y-dir.
inyoutyy VmVmF )()(
)()( 1121 VmVmApApFF Ayy 121 2)( VmAppFAy
There is no flow in the x and z direction
Momentum eqn. in x-dir.
Momentum eqn. in z-dir.
inxoutxx VmVmF )()( 0AxF
inzoutzz VmVmF )()(
0 nwAz WWF
22
23
Example 5.11 Water at flow rate 0.6 L/s exits a vertical nozzle of mass m = 0.1 kg.
Inlet (1) diameter d1 = 16mm
Outlet (2) diameter d2 = 5mmz1 − z2 = 0.03mVn = 2.84×10−6m3
Find inlet pressure p1 and the force to hold nozzle.
24
25
Steady flow: Q = 0.6 ×10−3m3/s
Select C.V. & assign z −axis.
Continuity: outin mm m/s6.30/ 2112 AAVV
26
Bernoulli’s equation: (1)→(2)
2
22
21
21
1 22gz
Vpgz
Vp
m03.0 ,0 212 zzpp atm
)(2
)(21
21
22
1 zzgVV
p
So, p1 = 463.5 kPa (gauge)
Nozzle weight: Wn = 0.1(9.81) NWater weight: Ww = 1000(9.81)2.84×10−6 N
27
Pressure force at inletF1 = p1A1 = 93.3 NPressure force at outletF2 = p2A2 = 0
Momentum eqn. in z −dir.
N8.77
)()(
)()(
12
21
A
wnAz
inzoutzz
F
VmVm
FFWWFF
VmVmF
28
Example 5.16The width of gate is b. The water depth is H upstream of the gate. Find the force required to hold the gate when h is the waterdepth downstream the gate. Ignore friction.
29
Momentum equation:
inxoutxx VmVmF )()(
Section (1): hydrostatic force
HbAH
hAghF cc ,2
,1
Section (2): hydrostatic force
, ,
2
21
2
hbumHbum
hbh
gF
outin
30
If Rx is the force from the gate onto the water, then
HbuhbuFFR
Hbuhbu
umumVmVm
FFRF
x
inoutinxoutx
xx
21
2221
21
22
12
21
)()(
Continuity:
h
Huu
mm outin
12
31
Bernoulli’s:
Hbuhbh
HuFFR
hH
ghu
ghu
gHu
x2
1
2
121
2
1
22
21
2
22
Continuity equation & linear momentum equation for ‘moving’ C.V. in steady flow
.where CV
SC
SC
VVWdAnWWF
dAnW
)ˆ(
)ˆ(0
..
..
32
Example 5.17Water exits a nozzle (A = 5.6×10−4 m3) with a speed 31 m/s. It strikes a vane moving at speed 7 m/s and is turned through angle θ = 45°.Find the force exerted by the water on the vane. Ignore gravity. Ww = 2N
33
Steady flow: attach C.V. on the vane so
(1):
(2):
Force on C.V. from vane:Mass flow rate: AWm 11
34
Continuity:
..
12
122
)ˆ(
24
SC
dAnWWF
WW
mAWm
m/s
Assume uniform at inlet/outlet.)ˆ( nW
N
N
230sin
sin
)()(
95)cos1(
cos
)()(
11
22
11
1122
WmWR
WmWR
WmWmWR
WmR
WmWmR
WmWmR
wz
wz
inzoutzwz
x
x
inxoutxx
dir.z
dir.x
35
36
Angular momentum equation (Sect. 5.2)
....
)ˆ(SCVC
sys dAnVVdt
MDt
D
V
Continuity into Insert
....
....
)ˆ(
)ˆ()(
SCVC
SCVC
sys
dAnVVVdVt
F
dAnVVVdVt
VMDt
D
momentum linear get to
37
Vr
dAnVVrVdVrt
Fr
r
SCVC
velocity having particle fluidof vector position :
: momentum angular get to equation momentum linear into Inserting
....
)ˆ()()(
38
in
outSC
AnVVr
AnVVrdAnVVr
nVVr
)ˆ()(
)ˆ()()ˆ()(
)ˆ()(
..
then lets, inlets/out at uniform is If
inout
VrmVrmFr )()(
Or,
. to equal is
torque shaft the then involved, is shaft aIf
Fr
T
39
Example 5.7 Water enters a sprinkler from its base and leaves through 2 horizontal nozzles in the tangential direction. (1) inlet, (2) outlet
40
Find the speed of water leaving the nozzle (relative to nozzle) if (a) the arms are stationary, (b) each arm rotates at 600 rpm.
Note the direction of ω.
41
Rotating arms: C.V. (dashed line) attached to arms
..22
2
2
:
:
VCVWV
V
W
ground fixed to relative fluid of velocity
C.V. to relative fluid of velocity
42
..
....
....
0)ˆ(
0)ˆ(
VC
SCVC
SCVC
VVW
dAnWVdt
dAnVVdt
where
to changed is
Continuity
! of tindependen is m/s . , Here
:flow Steady
7.16)2/(
)2(
00)ˆ(
0/
222
2221
12
..
WAQW
WAmQm
m mdAnW
t
SC
43
Example 5.18 For the setting in Example 5.7, find (a) torque to hold sprinkle stationary (i.e. ω = 0), (b) torque when the sprinkle rotates at 500 rpm,(c) angular speed of sprinkle if the torque is zero.
(a) Stationary sprinkle:select a stationary C.V. (i.e. disk) and z−axis.
44
/sm36101000
)()(
ˆ
Q
VrmVrm
FrkT
inout
shaft
At the inlet (1), V1 = Q/A1. But A1 = ?
Velocity vector coincides with the z−axis. Therefore, r1, which is the distance between and z−axis, is zero, i.e. r1 = 0.
1V
1V
45
m/smm
(sec.2), outlet the At
. So,
2 67.16 ,30
2
0)(
22
22
11
VA
A
QV
VrmVrmin
46
(b) Rotating arms: select C.V. attached to arms
47
From Example 5.7,W2 = Q/(2A2) = 16.7m/s
Consider the direction of angular speed ω.
With its direction shown, the C.V. moves at
where U2 = r2ω is the speed of each arm-tip.
48
49
(c) Consider Tshaft = 0.
it is concluded that V2 = 0 because , r2 ≠ 0.
Therefore, W2 = V2−(−r2ω)
is reduced toW2 = 0−(−r2ω)
or
0m
rpm 7972
2 r
W
50
51
Energy equation:e: total energy per unit
in fluid particle : internal energy ( ) per unit
( : heat capacity, T: temperature)V2/2: kinetic energy per unit gz: potential energy ( ) per unit
It is actually “power” (instead) of energy that it dealt with here as
gzV
ue 2
2m
u Tcm p m
pc2/2Vm m
gzm m
.dtdmdAnVm /)ˆ(
52
to get momentum equation
....
)ˆ()(SCVC
sys dAnVVVdVt
VMDt
D
Insert e into continuity
0)ˆ(....
SCVC
dAnVVdt
VdDt
D
to get energy equation
Insert into continuityV
0)ˆ()(....
SCVC
sys dAnVVdt
MDt
D
53
: rate of heat transfer in/out of system : rate of work done (or transfer) into/out
of system
and are associated with the ‘+’ sign as they are into the system. and are associated with the ‘−’ sign as they are out of the system.
outinQ /
outinW /
inW
outW
inQ
outQ
sysoutinsysoutin
SCVCsys
WWQQ
dAnVeVdet
VdeDt
D
....
)ˆ(Equ 5.59Page 235
54
A common example of work done (or transfer) into the system through the rotating shaft in a pump is . )(shaftinW
: shaft torqueω : angular speed of shaft
shaftTshaftshaftin TW )(
55
An example of work transfer out of the system through the rotating shaft in a turbine is . shaftoutshaft TW
56
Another important example of is due to pressure:
inW
CS
pin dAnVpW )ˆ)(()(
Note that multiplying the force (−p)dA by velocity gives the power.
In MA3006, fluid flow is usually adiabatic i.e NO heat transfer, .
)ˆ( nV
sysoutsysin QQ 0
57
For Steady, adiabatic flow with , the energy equation is
)()( shaftinpinin WWW
sysoutinsysoutin
SCVC
WWQQ
dAnVeVdet
....
)ˆ(
is reduced to
outshaftin
CSSC
WWdAnVpdAnVe )(
..
)ˆ)(()ˆ(
Recall: ,gzV
ue 2
2
58
CSSC
dAnVpdAnVe )ˆ)(()ˆ(..
gz
Vpu
2
2
Let be uniform at inlets/outlets.
inin
outout
SC
dAnVgzVp
u
dAnVgzVp
u
dAnVgzVp
u
)ˆ(2
)ˆ(2
)ˆ(2
2
2
..
2
..
2
)ˆ(2SC
dAnVgzVp
u
59
out
out
in
in
mdAnVmdAnV
)ˆ()ˆ( ,
outshaftin
inout
WW
gzVp
umgzVp
um
)(
22
22
)(
22
22
shaftin
inout
W
gzVp
umgzVp
um
Energy Equation
60
Example 5.20Pump delivers water at Q = 0.02m3/s. Find pump power , when z1 = z2, D1 = 0.09m, D2 = 0.025 m, p1 = 124kPa, p2 = 414kPa, , ρ = 1000kg/m3
)(shaftinW
J/kg27912 uu
61)(
22
22
11
22
7.40/
1.3/
200
shaftin
inout
out
W
gzVp
umgzVp
um
AQV
AQV
QmW
m/sm/s
kg/s ,
outshaftin
inout
WW
gzVp
umgzVp
um
)(
22
22
62
)(
21
2212
12
27882
2
shaftinWLHS
VVppuumLHS
W
)(
22
22
shaftin
inout
W
gzVp
umgzVp
um
Energy Equation
63
Or
m
Wgz
Vp
uugzVp
shaftin
in
inout
out
)(2
2
2
2
: loss of energy per unit mass flow rate to overcome friction in
fluid flowHead loss:
Energy loss:
inout uu
guuh inoutL /)(
)( inoutL uumghm
m
64
outin
L
L
outin
shaftin
zg
V
g
pz
g
V
g
p
h
hzg
V
g
pz
g
V
g
p
W
22
0
22
0
22
22
)(
then ,If
then ,If
Bernoulli’s equation
65
66
Example 5.24A fan inputs 0.4 kW of power to the blades to produce an air stream of diameter D2 = 0.6m at 12m/s. Find hL and η efficiency of fan.
ρair = 1.23kg/m3
67
Using energy equation
m
WW
gzVp
uugzVp
outshaftin
in
inout
out
)(
22
2)(
2
kg/s
,
,
m/s ,
17.44
0
0
12/)(
22
222
121
21)(
2
DVAVm
Vppp
zzW
Vguuh
atm
shaftout
inoutL
68
%7575.0
44.2
17.4
)1000(4.081.9
2
)12(
4.0
)(
)(
2
)(
shaftin
Lshaftin
input
outputL
L
shaftin
W
ghmW
W
W ηh
h
W
m,
,
kW
25% of input power is lost to overcome friction in air flow or dissipated as heat.
69
Example 5.25A pump adds 7.5kW to deliver water from (B) to (A) with head loss hL = 4.6m. Find Q and power loss. Let zA − zB = 9.1m.
70
Using energy equation:
m
VVppp
hW
WW
guuhm
WW
gzVp
uugzVp
atmBA
Lshaftout
pumpshaftin
inoutLoutshaftin
in
inout
out
7500)6.41.9(81.9
00
6.40
5.7
/)(
2)(
2
21
)(
)(
)(
22
, ,
m. ,
kW
where
71
Power loss
Steam enters a turbine at speed 30 m/s with and leaves the turbine at speed 60 m/s with .
kJ/kg 3348)/( pu
kJ/kg2550)/( pu
mW shaftout /)(
For adiabatic flow with z1 = z2, find .
Example 5.22
72
Using energy equation
m
WW
gzVp
uugzVp
outshaftin
in
inout
out
)(
22
2)(
2
kJ/kg
kJ/kg
& With
35.12
3060
22
79833482550
0
2222
21)(
inout
inout
shaftin
VV
pu
pu
zzW
73
kJ/kg 65.79635.1798
22
)(
22)(
m
W
VVpu
pu
m
W
shaftout
inoutinout
shaftout
Example: Steady adiabatic flow through the horizontal connection without & .Find the power lost in this horizontal connection.ρ = 1000 kg/m3
)(shaftinW
outW
74
Continuity:
m/s4.162
332211
321
V
AVAVAV
mmm
Energy equation:
02
22
1
2
1
3
2
3
2
2
2
Vpum
Vpum
Vpum
(2 outlets & 1 inlet)
75
Power lost is
MW 16.1
222
)()()(
3
2
3
2
2
2
1
2
1
113322
Vpm
Vpm
Vpm
umumum
umumghm ininoutoutL
76
77
Flow rate meters (Sec. 8.6, Munson)
Various flow-measuring devices are frequently used to determine the flow rate Q in a pipe experimentally.1. Orifice meter :
Page 457
78
It is a plate with a circular hole of diameter D2 fitted in a pipe of diameter D1 > D2. As the flow is partially blocked by the plate, a pressure drop (p1 − p2 > 0) across the plate is expected. Two pressure taps are drilled upstream and downstream of the plate to measure the pressure drop (Δp = p1 − p2).
2. Nozzle meter: D → d
79
3. Venturi meter : D → d
Generally, a decrease in cross-sectional area A in apipe causes an increase in velocity V because Q=VA.And increasing V leads to a drop in pressure p at a given elevation z, according to Bernoulli’s equation:
constant2
2
zg
V
g
p
80
Consider an incompressible fluid of density ρ flowing through a horizontal pipe of diameter D inserted with a restriction of diameter d.
Let z1 = z2. If the loss of energy hL = 0, then
g
V
g
p
g
V
g
p
22
222
211
81
Cancelling gravity g and rearranging terms,
212
2
212
2
21
pp
V
VV
Continuity (V1A1 = V2A2):44
2
2
2
1
dV
DV
421
2
21422
2
2
1
1
)(2
)(21
ppV
ppV
V
V
D
d
Or,
And, . then , If
82
Define a theoretical or ideal flow rate as Qideal = A2V2.
From V2 derived, 421
2 1
)(2
pp
AQideal
In reality where ,0Lh
Lhg
V
g
p
g
V
g
p
22
222
211
with z1 = z2. In the absence of an accurate ‘theoretical’ formula for hL, the actual flow rate is taken to be Q = CnQideal
83
Cn : discharge coefficient (subscript n for nozzle) found from many experimental measurements
Cn = f (β ,Re) is plotted as a function of β & Re = ρDV /μ where V = V1, and μ is the dynamic viscosity of fluid.
84
Re is the Reynolds number which indicates if the pipe flow (either laminar or turbulent).
Procedure of finding Q when d, D, ρ, μ, Cn are given.1. Measure p1 − p2 = 71.5Pa2. Calculate β = d / D = 0.625,
A1 = πD2/4 = 0.005m3,A2 = πd2/4 = 0.002m3 and
3m022.0
1
)(2421
2
pp
AQideal
85
For an orifice meter of diameter d fitted in a pipe of diameter D > d, Co (subscript ‘o’ for nozzle) is the discharge coefficient in Q = CoQideal.
86
For a Venturi meter of diameter d fitted in a pipe of diameter D > d, Cv (subscript ‘v’ for Venturi) is the discharge coefficient in Q = CvQideal.
87
The orifice meter, although not as accurate as thenozzle meter and the Venturi meter, is widely used because of its simplicity in design and cost involved. Standards are available for values of Co, Cn and Cv for various specific geometric configurations.Example 8.15Alcohol flows through a pipe D = 0.06m at Q = 0.003m3/s. If p1 − p2 = 4kPa across the nozzle is measured, find nozzle d.Given: ρ = 789kg/m3. μ = 1.19×10−3 Ns/m2
88
4
23
421
2
2
221
1
11
21
1
1102.1
)1(
)(2
44
10?
42200Re06.14
dC
ppACQCQ
dApp
dD
d
DV
A
QV
DA
n
nidealn
kPa
, ,
m/s ,
89
Solve this equation for d with the Cn graph.
577.0/
0346.0 102.1
11123
4
Dd
dd
Cn
m
. and Assume
90
568.006.0/0341.0 0341.0
1102.1
577.0972.0
972.0
577.042200Re
4
23
m
. , Letgraph) (from
, & With
d
dC
C
C
n
n
n
With Re = 42200 & β = 0.568, estimate Cn ≈ 0.972 from graph. As this value is near the previous one, the calculation is stopped. Otherwise, repeat the calculation with d = 0.0341m until Cn does not change.
91
In practice, iterative calculations are carried out in a computer, if the lines in the graph are expressed as a function of β and Re.
.
over meters Venturi for74
15.115.42
1.4
10Re107 ,44.03.0
Re
1060033.000175.0
2262.099.0
nC