Dr. MohsinSiddique Assistant Professor · 2018. 7. 11. · to mean velocity of flow (V) as long as...
Transcript of Dr. MohsinSiddique Assistant Professor · 2018. 7. 11. · to mean velocity of flow (V) as long as...
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Dynamics of Viscous Fluid Flow in
Closed Pipe: Darcy-Weisbach equation for flow in pipes. Major and minor losses in pipe
lines.
Dr. Mohsin Siddique
Assistant Professor
1
FLUID MECHANICS
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Steady Flow Through Pipes
2
� Laminar Flow:
flow in layers
Re4000 (pipe flow)
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Steady Flow Through Pipes
3
� Reynold’s Number(R or Re): It is ratio of inertial forces (Fi) to viscous forces (Fv) of flowing fluid
� For laminar flow: Re
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Steady Flow Through Pipes
4
� Hydraulic Radius (Rh) or Hydraulic Diameter: It is the ratio of area of flow to wetted perimeter of a channel or pipe
P
A
perimeterwetted
AreaRh ==
For Circular Pipe
( )( )
h
h
RD
D
D
D
P
AR
4
4
4/ 2
=
===π
π
For Rectangular pipe
DB
BD
P
ARh
2+==
B
D
Note: hydraulic Radius gives us indication for most economical section. More the Rh more economical will be the section.
ννh
h
VRVDR
4==
By replacing D with Rh, Reynolds’ number formulae can be used for non-circular sections as well.
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Head Loss in Pipes
5
� Total Head Loss=Major Losses+ Minor Losses
� Major Loss: Due to pipe friction
� Minor Loss: Due to pipe fittings, bents and valves etc
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Head Loss in Pipes due to Friction
6
The head loss due to friction in a given length of pipe is proportional to mean velocity of flow (V) as long as the flow in laminar. i.e.,
But with increasing velocity, as the flow become turbulent the head loss also varies and become proportion to Vn
Where n ranges from 1.75 to 2
Log-log plot for flow in uniform pipe (n=2.0 for rough wall pipe; n=1.75 for smooth wall pipe
VH f ∝
n
f VH ∝
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Frictional Head Loss in Conduits of Constant
Cross-Section
7
� Consider stead flow in a conduit of uniform cross-section A. The pressure at section 1 & 2 are P1 & P2 respectively. The distance between the section is L. For equilibrium in stead flow,
∑ == 0maF
Figure: Schematic diagram of conduit
0sin 21 =−−− APPLWAP oτα
P= perimeter of conduit= Avg. shear stress
between pipe boundary and liquid
oτ
01221 =−
−−− PL
L
zzALAPAP oτγ
αsin12 =
−
L
zz
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Frictional Head Loss in Conduits of Constant
cross-section
8
01221 =−
−−− PL
L
zzALAPAP oτγ
Dividing the equation by Aγ
( ) 01221 =−−−−A
PLzz
PP o
γ
τ
γγ
fLo hhA
PLPz
Pz ===
+−
+
γ
τ
γγ2
21
1
Therefore, head loss due to friction hf can be written as
h
oof
R
L
A
PLh
γ
τ
γ
τ==
Lhg
vz
P
g
vz
P+++=++
22
2
22
2
2
11
1
γγ
Remember !! For pipe flow
For stead flow in pipe of uniform diameter v1=v2
LhzP
zP
=
+−
+ 2
21
1
γγ
This is general equation and can be applied to any shape conduit having either Laminar or turbulent flow.
P
ARh =Q
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Determining Shear Stress
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� For smooth-walled pipes/conduits, the average shear stress at the wall is
� Using Rayleigh's Theorem of dimensional analysis, the above relation can be written as;
� Rewriting above equation in terms of dimension (FLT), we get
( ),,,, VRf ho ρµτ =
( )
=
ncb
a
T
L
L
FT
L
FTLK
L
F4
2
22
( )ncbaho VRk ... ρµτ =
LlengthR
L
Fareaforce
h
o
==
==2
/τ
( )22
4233
/.
////
/
−==
===
=
FTLmsN
LFTLaFLM
TLV
µ
ρ( ) ( ) ( ) ( )( )ncba TLLFTFTLLKFL /4222 −−− =
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Determining Shear Stress
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� According to dimensional homogeneity, the dimension must be equal on either side of the equation, i.e.,
( ) ( ) ( ) ( )( )ncba TLLFTFTLLKFL /4222 −−− =
)(20:
)(422:
)(1:
iiincbT
iincbaL
icbF
→−+=
→+−−=−
→+=Solving three equations, we get
1;2;2 −=−=−= ncnbna
� Substituting values back in above equation
( ) ( ) 22
122 ...... VVR
kVRkVRk
n
hnnnn
h
ncba
ho ρµ
ρρµρµτ
−
−−−
===
( ) 22 VRk neo ρτ−
=
� Setting = we get
2
2V
C fo ρτ = Where, Cf is coefficient of friction
( ) 2−neRk 2/fC
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Determining Shear Stress
11
� Now substituting the equation of avg. shear stress in equation of head loss,
� For circular pipe flows, Rh=D/4
� Where, f is a friction factor. i.e.,
� The above equation is known as pipe friction equation and as Darcy-Weisbach equation.
� It is used for calculation of pipe-friction head loss for circular pipes flowing full (laminar or turbulent)
2/2VC fo ρτ =
h
of
R
Lh
γ
τ=
h
f
h
f
fgR
LVC
R
LVCh
22
22
==γ
ρ
g
V
D
Lf
g
V
D
LC
Dg
LVCh f
f
f22
442
4 222
===
( )Re4 fCf f ==
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Friction Factor for Laminar and
Turbulent Flows in Circular Pipes
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� Smooth and Rough Pipe
� Mathematically;
� Smooth Pipe
� Rough Pipe
� Transitional mode
Turbulent flow near boundary
=
=
v
e
δ
Roughness height
Thickness of viscous sub-layer
Smooth pipe
Rough pipe
f
D
fVv
Re
14.1414.14==
νδ
ve δ<
ve δ>
ve δ<
ve δ14>
vv e δδ 14≤≤
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Friction Factor for Laminar and Turbulent Flows in
Circular Pipes
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� For laminar flow
� For turbulent flow
Re
64=f2000Re <
51.2
Relog2
1 f
f=
9.6
Relog8.1
1=
f
25.0Re
316.0=f
7/1
max
=
or
y
u
u
Def /
7.3log2
1=
+−=
f
De
f Re
51.2
7.3
/log2
1
+
−=
Re
9.6
7.3
/log8.1
111.1
De
f
Colebrook Eq. for turbulent flow in all pipes
Halaand Eq. For turbulent flow in all pipes
Von-karman Eq. for fully rough flow
Blacius Eq. for smooth pipe flow
Seventh-root law
510Re3000 ≤≤
From Nikuradse experiments
Colebrook Eq. for smooth pipe flow
for smooth pipe flow
4000Re >
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Friction Factor for Laminar and Turbulent Flows
in Circular Pipes
14
� The Moody chart or Moody diagram is a graph in non-dimensional form that relates the Darcy-Weisbach friction factor, Reynoldsnumber and relative roughness for fully developed flow in a circular pipe.
� The Moody chart is universally valid for all steady, fully developed,incompressible pipe flows.
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Friction Factor for Laminar and Turbulent Flows
in Circular Pipes
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� For laminar flow For non-laminar flow
eRf
64=
+−=
f
De
f Re
51.2
7.3
/log2
1 Colebrook eq.
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Friction Factor for Laminar and
Turbulent Flows in Circular Pipes
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� The friction factor can be determined by its Reynolds number (Re) and the Relative roughness (e/D) of the Pipe.( where: e = absolute roughness and D = diameter of pipe)
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Problem Types
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� Type 1: Determine f and hf,
� Type 2: Determine Q
� Type 3: Determine D
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Problem
19
� Find friction factor for the following pipe
� e=0.002 ft
� D=1ft
� Kinematic Viscosity, ν=14.1x10-6ft2/s
� Velocity of flow, V=0.141ft/s
� Solution:
� e/D=0.002/1=0.002
� R=VD/ ν =1x0.141/(14.1x10-6)=10000
� From Moody’s Diagram; f=0.034___________
Re
51.2
7.3
/log2
1
=
+−=
f
f
De
f
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Problem-Type 1
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� Pipe dia= 3 inch & L=100m
� Re=50,000 ʋ=1.059x10-5ft2/s
� (a): Laminar flow:
� f=64/Re=64/50,000=0.00128
ftgD
fLVH Lf 0357.0
)12/3)(2.32(2
)12.2)(100(00128.0
2
22
===
sftVVVD
/12.210059.1
)12/3(50000Re
5=⇒
×=⇒=
−ν
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Problem-Type 1
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� Pipe dia= 3 inch & L=100m
� Re=50,000 ʋ=1.059x10-5ft2/s
� (b): Turbulent flow in smooth pipe: i.e.: e=0
0209.0
50000
51.2
7.3
0log2
Re
51.2
7.3
/log2
1
=
+−=
+−=
f
ff
De
f
ftgD
fLVH Lf 582.0
)12/3)(2.32(2
)12.2)(100(0209.0
2
22
===
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Problem-Type 1
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� Pipe dia= 3 inch & L=100m
� Re=50,000 ʋ=1.059x10-5ft2/s
� (c): Turbulent flow in rough pipe: i.e.: e/D=0.05
0720.0
50000
51.2
7.3
05.0log2
Re
51.2
7.3
/log2
1
=
+−=
+−=
f
ff
De
f
ftgD
fLVH Lf 01.2
)12/3)(2.32(2
)12.2)(100(0720.0
2
22
===
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Problem-Type 1
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hL=?
memDmL 0005.0;25.0;1000 ===
smsmQ /10306.1;/051.0 263 −×== ν
( ) ( )002.025.0/0005.0/
10210306.1/25.0039.1/ 56
==
×=××== −
De
VDR ν
0.0245f
Diagram sMoody' From
=
mgD
fLVhL 39.5
2
2
==
smAQV /039.1/ ==Q
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Problem-Type 2
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gDfLVhL 2/2=
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Problem-Type 2
25
� For laminar flow For non-laminar flow
eRf
64=
+−=
f
De
f Re
51.2
7.3
/log2
1 Colebrook eq.
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Problem-Type 3
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Problem
27
gD
flVh
f
De
f
Lf2
Re
51.2
7.3
/log2
1
2
=
+−=
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Problem
28
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Problem
29
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MINOR LOSSES
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� Each type of loss can be quantified using a loss coefficient (K). Losses are proportional to velocity of flow and geometry of device.
� Where, Hm is minor loss and K is minor loss coefficient. The value of K is typically provided for various types/devices
� NOTE: If L > 1000D minor losses become significantly less than that of major losses and hence can be neglected.
g
VKHm
2
2
=
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Minor Losses
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� These can be categorized as
� 1. Head loss due to contraction in pipe
� 1.1 Sudden Contraction
� 1.2 Gradual Contraction
� 2. Entrance loss
� 3. Head loss due to enlargement of pipe
� 3.1 Sudden Enlargement
� 3.2 Gradual Enlargement
� 4. Exit loss
� 5. Head loss due to pipe fittings
� 6. Head loss due to bends and elbows
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Minor Losses
32
� Head loss due to contraction of pipe (Sudden contraction)
� A sudden contraction (Figure) in pipe usually causes a marked drop in pressure in the pipe because of both the increase in velocity and the loss of energy of turbulence.
g
VKH cm
2
2
2=
Head loss due to sudden contraction is
Where, kc is sudden contraction coefficient and it value depends up ratio of D2/D1 and velocity (V2) in smaller pipe
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Minor Losses
33
� Head loss due to enlargement of pipe (Gradual Contraction)
� Head loss from pipe contraction may be greatly reduced by introducing a gradual pipe transition known as a confusor as shown Figure.
g
VKH cm
2'
2
2=
Head loss due to gradual contraction is
Where, kc’ is gradual contraction
coefficient and it value depends up ratio of D2/D1 and velocity (V2) in smaller pipe
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Minor Losses
34
� Entrance loss
� The general equation for an entrance head loss is also expressed in terms of velocity head of the pipe:
� The approximate values for the entrance loss coefficient (Ke) for different entrance conditions are given below
g
VKH em
2
2
=
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Minor Losses
35
� head loss due to enlargement of pipe (Sudden Enlargement)
� The behavior of the energy grade line and the hydraulic grade line in the vicinity of a sudden pipe expansion is shown in Figure
The magnitude of the head loss may be expressed as
( )g
VVHm
2
2
21 −=
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Minor Losses
36
� head loss due to enlargement of pipe (Gradual Enlargement)
� The head loss resulting from pipe expansions may be greatly reduced by introducing a gradual pipe transition known as a diffusor
The magnitude of the head loss may be expressed as
( )g
VVKH em
2
2
21 −=
The values of Ke’ vary with the diffuser angle (α).
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Minor Losses
37
� Exit Loss
� A submerged pipe discharging into a large reservoir (Figure ) is a special case of head loss from expansion.
( )g
VKH dm
2
2
=
Exit (discharge) head loss is expressed as
where the exit (discharge) loss coefficient Kd=1.0.
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Minor Losses
38
� Head loss due to fittings valves
� Fittings are installed in pipelines to control flow. As with other losses in pipes, the head loss through fittings may also be expressed in terms of velocity head in the pipe:
g
VKH fm
2
2
=
-
Minor Losses
39
� Head loss due to bends
� The head loss produced at a bend was found to be dependent of the ratio the radius of curvature of the bend (R) to the diameter of the pipe (D). The loss of head due to a bend may be expressed in terms of the velocity head as
� For smooth pipe bend of 900, the values of Kb for various values of R/D are listed in following table.
g
VKH bm
2
2
=
-
Minor Losses
40
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Numerical Problems
41
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Numerical Problems
42
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Thank you
� Questions….
� Feel free to contact:
43
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Fluid Dynamics:
(ii) Hydrodynamics: Different forms of energy in a flowing
liquid, head, Bernoulli's equation and its application, Energy
line and Hydraulic Gradient Line, and Energy Equation
Dr. Mohsin Siddique
Assistant Professor
1
Fluid Mechanics
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Forms of Energy
2
� (1). Kinetic Energy: Energy due to motion of body. A body of mass, m, when moving with velocity, V, posses kinetic energy,
� (2). Potential Energy: Energy due to elevation of body above an arbitrary datum
� (3). Pressure Energy: Energy due to pressure above datum, most usually its pressure above atmospheric
2
2
1mVKE =
mgZPE =
m andV are mass and velocity of body
Z is elevation of body from arbitrary datumm is the mass of body
hγ=PrE !!!
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Forms of Energy
3
� (4). Internal Energy: It is the energy that is associated with the molecular, or internal state of matter; it may be stored in many forms, including thermal, nuclear, chemical and electrostatic.
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HEAD
4
� Head: Energy per unit weight is called head
� Kinetic head: Kinetic energy per unit weight
� Potential head: Potential energy per unit weigh
� Pressure head: Pressure energy per unit weight
g
VmgmV
Weight
KE
2/
2
1head Kinetic
22 =
== mgWeight =Q
( ) ZmgmgZWeight
PE=== /head Potential
γ
P
Weight==
PrEhead Pressure
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TOTAL HEAD
5
� TOTAL HEAD
= Kinetic Head + Potential Head + Pressure Head
g
VPZ
2HHead Total
2
++==γ
g
V
2
2
γ
PZ
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Bernoulli’s Equation
6
� It states that the sum of kinetic, potential and pressure heads of a fluid particle is constant along a streamline during steady flow when compressibility and frictional effects are negligible.
� i.e. , For an ideal fluid, Total head of fluid particle remains constant during a steady-incompressible flow.
� Or total head along a streamline is constant during steady flow when compressibility and frictional effects are negligible.
21
22
22
12
11
2
22
2Head Total
HH
g
VPZ
g
VPZ
consttg
VPZ
=
++=++
=++=
γγ
γ
1
2
Pipe
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Derivation of Bernoulli’s Equation
7
� Consider motion of flow fluid particle in steady flow field as shown in fig.
� Applying Newton’s 2nd Law in s-direction on a particle moving along a streamline give
� Where F is resultant force in s-direction, m is the mass and as is the acceleration along s-direction.
ss maF =
Assumption:Fluid is ideal and incompressibleFlow is steadyFlow is along streamlineVelocity is uniform across the section and is equal to mean velocityOnly gravity and pressure forces are acting
ds
dVV
dtds
dsdV
dsdt
dsdV
dt
dVas ====
Eq(1)
Eq(2)
Fig. Forces acting on particle along streamline
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Derivation of Bernoulli’s Equation
8
W=weight of fluidWsin( )= component acting along s-directiondA= Area of flowds=length between sections along pipe
θ
( ) θsinWdAdpPPdAFs −+−=
Substituting values from Eq(2) and Eq(3) to Eq(1)
Eq(3)
( )ds
dVmVWdAdpPPdA =−+− θsin
ds
dVdAdsV
ds
dzgdAdsdpdA ρρ =−−
( )gdAdsmgW ρ==
ds
dz=θsin
Cancelling dA and simplifying
VdVgdzdp ρρ =−−
Note that 2
2
1dVVdV =
2
2
1dVgdzdp ρρ =−−
Eq(4)
Eq(5)
Fig. Forces acting on particle along streamline
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Derivation of Bernoulli’s Equation
9
� Dividing eq (5) by
� Integrating
� Assuming incompressible and steady flow
� Dividing each equation by g
ρ
02
1 2 =++ dVgdzdp
ρ
conttdVgdzdp
=
++∫
2
2
1
ρ
conttVgzP
=++ 2
2
1
ρ
conttg
Vz
g
P=++
2
2
ρ
� Hence Eq (9) for stead-incompressible fluid assuming no frictional losses can be written as
Eq (6)
Eq (7)
Eq (8)
Eq (9)
( ) ( )21
22
22
12
11
Head TotalHead Total
22
=
++=++g
VPZ
g
VPZ
γγ
Above Eq(10) is general form of Bernoulli’s Equation
Eq (10)
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Energy Line and Hydraulic Grade line
10
� Static Pressure :
� Dynamic pressure :
� Hydrostatic Pressure:
� Stagnation Pressure: Static pressure + dynamic Pressure
Hg
Vz
P=++
2
2
γ
Head TotalheadVelocity headElevation head Pressure =++
P
gZρ
2/2Vρ
conttV
gzP =++2
2
ρρ
Multiplying with unit weight,γ,
stagPV
P =+2
2
ρ
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Energy Line and Hydraulic Grade line
11
� Measurement of Heads
� Piezometer: It measures pressure head ( ).
� Pitot tube: It measures sum of pressure and velocity heads i.e.,
g
VP
2
2
+γ
γ/P
What about measurement of elevation head !!
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Energy Line and Hydraulic Grade line
12
� Energy line: It is line joining the total heads along a pipe line.
� HGL: It is line joining pressure head along a pipe line.
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Energy Line and Hydraulic Grade line
13
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Energy Equation for steady flow of any fluid
14
� Let’s consider the energy of system (Es) and energy of control volume(Ecv) defined within a stream tube as shown in figure. Therefore,
� Because the flow is steady, conditions within the control volume does not change so
� Hence
in
CV
out
CVCVs EEEE ∆−∆+∆=∆
in
CV
out
CVs EEE ∆−∆=∆
0=∆ CVE
Eq(1)
Eq(2)
Figure: Forces/energies in fluid flowing in streamt ube
-
Energy Equation for steady flow of any
fluid
15
� Now, let’s apply the first law of thermodynamics to the fluid system which states ” For steady flow, the external work done on any system plus the thermal energy transferred into or out of the system is equal to the change of energy of system”
( )
( ) inCVout
CV
s
EEshaftworklowwork
Eshaftworklowwork
∆−∆=++
∆=++
=+
ferredheat transf
ferredheat transf
energy of changeferredheat transdone work External
� Flow work: When the pressure forces acting on the boundaries move, in present case when p1A1 and p2A2 at the end sections move
through ∆s1 and ∆s2, external work is done. It is referred to as flow work.
msAsApp
mg
sAp
sAp
sApsAp
∆=∆=∆
−∆=
∆−∆=∆−∆=
222111
2
2
1
1
222
2
2111
1
1111111
workFlow
workFlow
ρργγ
γγ
γγ
Q Steady flow
Eq(3)
Eq(4)
Eq(5)
-
Energy Equation for steady flow of any
fluid
16
� Shaft work: Work done by machine, if any, between section 1 and 2
( ) ( ) mm
m
hmghsA
thdt
dsAtime
weight
energy
time
weight
∆=∆=
∆
==
111
111
Shaft work
Shaft work
γ
γ
� Where, hm is the energy added to the flow by the machine per unit weight of flowing fluid. Note: if the machine is pump, which adds energy to the fluid, hm is positive and if the machine is turbine, which remove energy from fluid, hm is -ve
� HeatTransferred: The heat transferred from an external source
into the fluid system over time interval ∆t is
( ) ( ) HH
H
QmgQsA
tQdt
dsA
∆=∆=
∆
=
111
111
ferredHeat trans
ferredHeat trans
γ
γ
� Where, QH is the amount of energy put into the flow by the external heat source per unit weight of flowing fluid. If the heat flow is out of the fluid, the value QH is –ve and vice versa
Eq(6)
Eq(7)
-
Energy Equation for steady flow of any
fluid
17
� Change in Energy: For steady flow during time interval ∆t, the weight of fluid entering the control volume at section 1 and leaving at section 2 are both equal to g∆m . Thus the energy (Potential+Kinetic+Internal) carried by g∆m is;
( )
( )
++∆=∆
++=∆
++
=∆
++∆=∆
++=∆
++
=∆
2
2
22
2
2
222222
2
22
222
1
2
11
1
2
111111
2
11
111
2
22
2
22
Ig
VzmgE
Ig
VzdsAtI
g
Vz
dt
dsAE
Ig
VzmgE
Ig
VzdsAtI
g
Vz
dt
dsAE
out
CV
out
CV
in
CV
in
CV
α
αγαγ
α
αγαγ
α is kinetic energy correction factor and ~ 1
Eq(8)
Eq(9)
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Energy Equation for steady flow of any
fluid
18
� Substituting all values from Eqs. (5),(6), (7), (8), & (9) in Eq(4)
( ) inCVout
CV EEshaftworklowwork ∆−∆=++ ferredheat transf
( ) ( )
++∆−
++∆=∆+∆+
−∆ 1
2
112
2
22
2
2
1
1
22I
g
VzmgI
g
VzmgQmghmg
ppmg Hm αα
γγ
++−
++=++
− 1
2
112
2
22
2
2
1
1
22I
g
VzI
g
VzQh
ppHm αα
γγ
+++=++
++− 2
2
22
2
21
2
11
1
1
22I
g
Vz
pQhI
g
Vz
pHm α
γα
γ
This is general form of energy equation, which applies to liquids, gases, vapors and to ideal fluids as well as real fluids with friction, both incompressible and compressible. The only restriction is that its for steady flow.
Eq(10)
-
Energy Equation for steady flow of
incompressible fluid
19
� For incompressible fluids
� Substituting in Eq(10), we get
( )122
22
2
2
11
1
22II
g
Vz
pQh
g
Vz
pHm −+
++=++
+−
γγ
γγγ == 21
( ) Hm QIIg
Vz
ph
g
Vz
p−−+
++=+
+− 12
2
22
2
2
11
1
22 γγ
Lm hg
Vz
ph
g
Vz
p+
++=+
+−
22
2
22
2
2
11
1
γγ Eq(11)
( ) HL QIIh −−= 12Q
� Where hL=(I2-I1)-QH= head loss. It equal to is gain in internal energy minus any heat added by external source.
� Hm is head removed/added by machines. It can also be referred to head loss due to pipe fitting, contraction, expansion and bends etc in pipes.
-
Energy Equation for steady flow of
incompressible fluid
20
� In the absence of machine, pipe fitting etc, Eq(11) can be written as
� When the head loss is caused only by wall or pipe friction, hLbecomes hf, where hf is head loss due to friction
Lhg
Vz
p
g
Vz
p+
++=
++
22
22
22
21
11
γγEq(12)
-
Power
21
� Rate of work done is termed as power
Power=Energy/time
Power=(Energy/weight)(weight/time)
� If H is total head=total energy/weight and γQ is the weight flow ratethen above equation can be written as
Power=(H)(γQ)= γQH
In BG:
Power in (horsepower)=(H)(γQ)/550
In SI:
Power in (Kilowatts)=(H)(γQ)/1000
1 horsepower=550ft.lb/s
-
Reading Assignment
22
� Kinetic energy correction factor
� Limitation of Bernoulli’s Equation
� Application of hydraulic grade line and energy line
-
NUMERICALS
23
� 5.2.1
-
24
� 5.2.3
-
25
� 5.3.2
-
26
� 5.3.4
-
27
� 5.3.6
-
28
� 5.9.6
-
Momentum and Forces in Fluid Flow
29
� We have all seen moving fluids exerting forces. The lift force on an aircraft is exerted by the air moving over the wing. A jet of water from a hose exerts a force on whatever it hits.
� In fluid mechanics the analysis of motion is performed in the same way as in solid mechanics - by use of Newton’s laws of motion.
� i.e., F = ma which is used in the analysis of solid mechanics to relate applied force to acceleration.
� In fluid mechanics it is not clear what mass of moving fluid we should use so we use a different form of the equation.
( )dt
mdma s
VF ==∑
-
Momentum and Forces in Fluid Flow
30
� Newton’s 2nd Law can be written:
� The Rate of change of momentum of a body is equal to the resultant force acting on the body, and takes place in the direction of the force.
� The symbols F and V represent vectors and so the change in momentum must be in the same direction as force.
It is also termed as impulse momentum principle
( )dt
md sVF =∑
=
=∑
mV
F Sum of all external forces on a body of fluid or system s
Momentum of fluid body in direction s
( )smddt VF =∑
-
Impact of a Jet on a Plane
31
-
Impact of a Jet on a Plane
32
-
Thank you
� Questions….
� Feel free to contact:
33
Unit 4 Fluid Dyanamics.pdf (p.1-43)unit 4 part 2fluiddynamics.pdf (p.44-76)