57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 1
Chapter 5 Finite Control Volume Analysis 5.1 Continuity Equation RTT can be used to obtain an integral relationship expressing conservation of mass by defining the extensive property B = M such that β = 1. B = M = mass β = dB/dM = 1 General Form of Continuity Equation
∫ ∫ ⋅ρ+ρ==CV CS
dAVVddtd0
dtdM
or
∫ρ−=⋅∫ρCVCS
VddtddAV
net rate of outflow rate of decrease of of mass across CS mass within CV Simplifications:
1. Steady flow: 0Vddtd
CV=∫ρ−
2. V = constant over discrete dA (flow sections):
∫ ∑ ⋅ρ=⋅ρCS CS
AVdAV
57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 2
3. Incompressible fluid (ρ = constant)
CS CV
dV dA dVdt
⋅ = −∫ ∫ conservation of volume
4. Steady One-Dimensional Flow in a Conduit:
∑ =⋅ρCS
0AV
−ρ1V1A1 + ρ2V2A2 = 0 for ρ = constant Q1 = Q2
Some useful definitions: Mass flux ∫ ⋅ρ=
AdAVm&
Volume flux ∫ ⋅=
AdAVQ
Average Velocity A/QV =
Average Density ∫ρ=ρ dAA1
Note: m Q≠ ρ& unless ρ = constant
57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 3
Example
*Steady flow
*V1,2,3 = 50 fps
*@ V varies linearly from zero at wall to Vmax at pipe center *find 4m& , Q4, Vmax
0 *water, ρw = 1.94 slug/ft3
∫ρ−=∫ =⋅ρCVCS
Vddtd0dAV
i.e., -ρ1V1A1 - ρ2V2A2 + ρ3V3A3 + ρ ∫4A
44dAV = 0
ρ = const. = 1.94 lb-s2 /ft4 = 1.94 slug/ft3
∫ρ= 444 dAVm& = ρV(A1 + A2 – A3) V1=V2=V3=V=50f/s
= ( )222 5.1214
50144
94.1 −+π××
= 1.45 slugs/s
4m&
57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 4
Q4 = 75.m4 =ρ& ft3/s = ∫
4A44dAV
velocity profile
Q4 = ∫ ∫ θ⎟⎟⎠
⎞⎜⎜⎝
⎛−
πor
0
2
0 omax drrd
rr1V
max
2o
2omax
r
0o
3r
0
2
max
r
0 o
2
max
r
0 omax
Vr31
31
21rV2
r3r
2rV2
drrrrV2
rdrrr1V2
o0
o
o
π=⎥⎦⎤
⎢⎣⎡ −π=
⎥⎥⎦
⎤
⎢⎢⎣
⎡−π=
∫ ⎥⎦
⎤⎢⎣
⎡−π=
∫ ⎟⎟⎠
⎞⎜⎜⎝
⎛−π=
Vmax = 86.2r
31
Q2
o
4 =π
fps
V4 ≠ V4(θ) dA4
2o
max2o
4r
Vr31
AQV
π
π==
= maxV31
57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 5
5.2 Momentum Equation Derivation of the Momentum Equation Newton’s second law of motion for a system is
time rate of change of the momentum of the system
= sum of external forces acting on the system
Since momentum is mass times velocity, the momentum of a small particle of mass is and the momentum of the entire system is . Thus,
Recall RTT:
With and ,
57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 6
Thus, the Newton’s second law becomes
where, is fluid velocity referenced to an inertial frame (non-
accelerating) is the velocity of CS referenced to the inertial frame is the relative velocity referenced to CV ∑ ∑ ∑ is vector sum of all forces acting on
the CV
is body force such as gravity that acts on the entire mass/volume of CV
is surface force such as normal (pressure and
viscous) and tangential (viscous) stresses acting on the CS
57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 7
Note that, when CS cuts through solids, may also include reaction force, (e.g., the reaction force required to hold nozzle or bend when CS cuts through the bolts that are holding the nozzle/bend in place)
∑ ∑ = resultant force on fluid in CV due to and , i.e. reaction force on fluid
Free body diagram
57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 8
Important Features (to be remembered) 1) Vector equation to get component in any direction must use
dot product x equation
∑ ∫ ⋅ρ+∫ρ=CS
RCV
x dAVuVuddtdF
y equation
∑ ∫ ⋅ρ+∫ρ=CS
RCV
y dAVvVvddtdF
z equation
∑ ∫ ⋅ρ+∫ρ=CS
RCV
z dAVwVwddtdF
2) Carefully define control volume and be sure to include all
external body and surface faces acting on it. For example,
(Rx,Ry) = reaction force on fluid
(Rx,Ry) = reaction force on nozzle
Carefully define coordinate system with forces positive in positive direction of coordinate axes
57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 9
3) Velocity V and Vs must be referenced to a non-accelerating
inertial reference frame. Sometimes it is advantageous to use a moving (at constant velocity) reference frame: relative inertial coordinate. Note VR = V – Vs is always relative to CS.
4) Steady vs. Unsteady Flow
Steady flow ⇒ ∫ =ρCV
0VdVdtd
5) Uniform vs. Nonuniform Flow
∫ ⋅ρ
CSR dAVV = change in flow of momentum across CS
= ΣVρVR⋅A uniform flow across A 6) Fpres = − ∫ dAnp ∫ ∫=∇
V SdsnfVfd
f = constant, ∇f = 0 = 0 for p = constant and for a closed surface i.e., always use gage pressure 7) Pressure condition at a jet exit
at an exit into the atmosphere jet pressure must be pa
57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 10
Applications of the Momentum Equation Initial Setup and Signs
1. Jet deflected by a plate or a vane 2. Flow through a nozzle 3. Forces on bends 4. Problems involving non-uniform velocity distribution 5. Motion of a rocket 6. Force on rectangular sluice gate 7. Water hammer 8. Steady and unsteady developing and fully developed pipe
flow 9. Empting and filling tanks 10. Forces on transitions 11. Hydraulic jump 12. Boundary layer and bluff body drag 13. Rocket or jet propulsion 14. Propeller
57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 11
1. Jet deflected by a plate or vane Consider a jet of water turned through a horizontal angle
x-equation: ∑ ∫ ∫ ⋅ρ+ρ==
CSxx dAVuVud
dtdFF
steady flow = )AV(V)AV(V 22x211x1 ρ+−ρ continuity equation: ρA1V1 = ρA2V2 = ρQ Fx = ρQ(V2x – V1x) y-equation: ∑ ∑ ⋅ρ==
CSyy AVvFF
Fy = ρV1y(– A1V1) + ρV2y(– A2V2) = ρQ(V2y – V1y) for above geometry only where: V1x = V1 V2x = -V2cosθ V2y = -V2sinθ V1y = 0 note: Fx and Fy are force on fluid - Fx and -Fy are force on vane due to fluid
∑ ⋅ρ=CS
x AVuF
CV and CS are for jet so that Fx and Fy are vane reactions forces on fluid
for A1 = A2 V1 = V2
57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 12
If the vane is moving with velocity Vv, then it is convenient to choose CV moving with the vane i.e., VR = V - Vv and V used for B also moving with vane x-equation: ∫ ⋅ρ=
CSRx dAVuF
Fx = ρV1x[-(V – Vv)1A1] + ρV2x[(V – Vv)2A2] Continuity: 0 = ∫ ⋅ρ dAVR
i.e., ρ(V-Vv)1A1 = ρ(V-Vv)2A2 = ρ(V-Vv)A Qrel Fx = ρ(V-Vv)A[V2x – V1x]
Qrel on fluid V2x = (V – Vv)2x V1x = (V – Vv)1x Power = -FxVv i.e., = 0 for Vv = 0 Fy = ρQrel(V2y – V1y)
For coordinate system moving with vane
57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 13
2. Flow through a nozzle Consider a nozzle at the end of a pipe (or hose). What force is required to hold the nozzle in place? Assume either the pipe velocity or pressure is known. Then, the unknown (velocity or pressure) and the exit velocity V2 can be obtained from combined use of the continuity and Bernoulli equations.
Bernoulli: 2222
2111 V
21zpV
21zp ρ+γ+=ρ+γ+ z1=z2
22
211 V
21V
21p ρ=ρ+
Continuity: A1V1 = A2V2 = Q
1
2
12
12 V
dDV
AAV ⎟
⎠⎞
⎜⎝⎛==
0dD1V
21p
42
11 =⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎠⎞
⎜⎝⎛−ρ+
Say p1 known: ( )
2/1
41
1
dD1
p2V⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
⎟⎠⎞⎜
⎝⎛ −ρ
−=
To obtain the reaction force Rx apply momentum equation in x-direction
CV = nozzle and fluid ∴ (Rx, Ry) = force required to hold nozzle in place
57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 14
∫ ⋅ρ+∫ ρ=∑CSCV
x dAVuVdudtdF
=∑ ⋅ρCS
AVu
Rx + p1A1 – p2A2 = ρV1(-V1A1) + ρV2(V2A2) = ρQ(V2 - V1) Rx = ρQ(V2 - V1) - p1A1 To obtain the reaction force Ry apply momentum equation in y-direction ∑ ∑ =⋅ρ=
CSy 0AVvF since no flow in y-direction
Ry – Wf − WN = 0 i.e., Ry = Wf + WN Numerical Example: Oil with S = .85 flows in pipe under pressure of 100 psi. Pipe diameter is 3” and nozzle tip diameter is 1”
V1 = 14.59 ft/s V2 = 131.3 ft/s Rx = 141.48 – 706.86 = −569 lbf Rz = 10 lbf This is force on nozzle
steady flow and uniform flow over CS
65.1g
S =γ=ρ
D/d = 3
Q = 2
2
V121
4⎟⎠⎞
⎜⎝⎛π
= .716 ft3/s
57:020 MeProfessor F
3. For Consiconsidsectionbend ideterm Contin x-mom p y-mom
echanics of FluFred Stern Fa
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uids and Transpall 2009
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: ∑ =Fy
port Processes
ough a bd uniformconcern
e reactionation of t
=⋅ρ AV= consta
∑ ⋅ρ VuxR =+θ
=
∑ ⋅ρ= Vv
bend in am across
is the fon forces the mom
ρ− 11AVant = 1AV
A (x1V −ρ=
= ( x2VQρ
⋅ A
a pipe. Tthe inle
orce requRx and R
mentum e
ρ+ 2AV21 AVA =
)11AV +−)x1x V−
Rx, R
The flowet and ouuired to hRy whichequation
2A 2A
(x2 VVρ+
Ry = reacti bend requir bend
Chapter 5 15
w is utlet hold theh can be n.
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5
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on e ld
57:020 MeProfessor F
p 4. ForThe fomome
∑ =Fx
G1 FF +
GWF =
=
GWF =
echanics of FluFred Stern Fa
22 sinAp
rce on a rorce on tentum eq
∑ρ= Vu
visGW F−
12 FF −=
= 2 y2y ⋅γ
( 22yb
21 γ=
uids and Transpall 2009
yRn −+θ
rectanguthe fluidquation:
⋅A
2sc F =−
( 2VQρ+
2 2yby γ−
)21
22 y +−
port Processes
f ww −− = ρ
ular sluicd due to t
(1 VV −ρ
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11 by
2+⋅
( 2VQ −ρ
1b Vw ρ=( y2VQ −ρ
ce gate the gate
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viscF+
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Chapter 5 16
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om the x-
ted
b
6
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-
57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 17
⎟⎟⎠
⎞⎜⎜⎝
⎛−ρ
12
2
y1
y1
bQ
5. Application of relative inertial coordinates for a moving but non-deforming control volume (CV) The CV moves at a constant velocity CSV with respect to the absolute inertial coordinates. If RV represents the velocity in the relative inertial coordinates that move together with the CV, then: R CSV V V= − Reynolds transport theorem for an arbitrary moving deforming CV:
SYS
RCV CS
dB d d V n dAdt dt
βρ βρ= ∀ + ⋅∫ ∫
For a non-deforming CV moving at constant velocity, RTT for incompressible flow:
syst
RCV CS
dBd V ndA
dt tβρ ρ β∂= ∀ + ⋅
∂∫ ∫
1) Conservation of mass systB M= , and 1β = :
RCS
dM V ndAdt
ρ= ⋅∫
For steady flow: 0R
CS
V ndA⋅ =∫
57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 18
2) Conservation of momentum
( )CSsyst RB M V V= + and syst R CSdB dM V Vβ = = +
( ) ( ) ( )[ ]CS CSR RCSR R
CV CS
d M V V V VF d V V V ndA
dt tρ ρ
+ ∂ += = ∀ + + ⋅
∂∑ ∫ ∫
For steady flow with the use of continuity:
( )CSR RCS
CSR R RCS CS
F V V V ndA
V V ndA V V ndA
ρ
ρ ρ
= + ⋅
= ⋅ + ⋅
∑ ∫
∫ ∫0
R RCS
F V V ndAρ= ⋅∑ ∫
Example (use relative inertial coordinates): Ex) A jet strikes a vane which moves to the right at constant velocity on a
frictionless cart. Compute (a) the force required to restrain the cart and (b) the power delivered to the cart. Also find the cart velocity for which (c) the force is a maximum and (d) the power is a maximum.
57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 19
Solution:
Assume relative inertial coordinates with non-deforming CV i.e. CV moves
at constant translational non-accelerating
then R CSV V V= − . Also assume steady flow with and neglect gravity effect. Continuity:
0RCS
V ndA⋅ =∫
Bernoulli without gravity:
1p0 2
1 212 RV pρ+ =
0 22
12 RVρ+
1 2R RV V=
Since 1 1 2 2R RV A V Aρ ρ=
1 2 jA A A= = Momentum: ∑
0 0
57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 20
cos 1 cos
1 cos
1 cos , 0 0 2 1 cos 2 1 cos
4 3 1 cos 0 3 4 0 4 16 126 4 26
3
3 23 1 cos 427 1 cos
57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 21
5.3 Energy Equation Derivation of the Energy Equation The First Law of Thermodynamics The difference between the heat added to a system and the work done by a system depends only on the initial and final states of the system; that is, depends only on the change in energy E: principle of conservation of energy ΔE = Q – W ΔE = change in energy Q = heat added to the system W = work done by the system E = Eu + Ek + Ep = total energy of the system potential energy kinetic energy The differential form of the first law of thermodynamics expresses the rate of change of E with respect to time
WQdtdE && −=
rate of work being done by system
rate of heat transfer to system
Internal energy due to molecular motion
57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 22
Energy Equation for Fluid Flow The energy equation for fluid flow is derived from Reynolds transport theorem with Bsystem = E = total energy of the system (extensive property) β = E/mass = e = energy per unit mass (intensive property) = u + ek + ep
∫ ⋅ρ+∫ ρ= CSCV dAVeVeddtd
dtdE
ˆ ˆ( ) ( )k p k pCV CS
dQ W u e e dV u e e V dAdt
ρ ρ− = + + + + + ⋅∫ ∫& & This can be put in a more useable form by noting the following:
2V
M2/MV
massVvelocitywithmassofKETotale
22
k =Δ
Δ==
gzVzV
ME
e pp =
ΔρΔγ=
Δ= (for Ep due to gravity only)
2 2
ˆ ˆ2 2CV Cs
d V VQ W gz u dV gz u V dAdt
ρ ρ⎛ ⎞ ⎛ ⎞− = + + + + + ⋅⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠∫ ∫& &
rate of work rate of change flux of energy done by system of energy in CV out of CV (ie, across CS) rate of heat transfer to sysem
VV2 =
57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 23
System at time t + Δt
System at time t
CS
Rate of Work Components: fs WWW &&& += For convenience of analysis, work is divided into shaft work Ws and flow work Wf Wf = net work done on the surroundings as a result of
normal and tangential stresses acting at the control surfaces
= Wf pressure + Wf shear Ws = any other work transferred to the surroundings
usually in the form of a shaft which either takes energy out of the system (turbine) or puts energy into the system (pump)
Flow work due to pressure forces Wf p (for system)
Work = force × distance
at 2 W2 = p2A2 × V2Δt rate of work⇒ 2222222 AVpVApW ⋅==&
at 1 W1 = −p1A1 × V1Δt 1111 AVpW ⋅=&
Note: here V uniform over A
(on surroundings)
neg. sign since pressure force on surrounding fluid acts in a direction opposite to the motion of the system boundary
CV
57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 24
In general, AVpWfp ⋅=& for more than one control surface and V not necessarily uniform over A:
∫ ⋅⎟⎠
⎞⎜⎝
⎛ρ
ρ=∫ ⋅= CSCSfp dAVpdAVpW&
fshearfpf WWW &&& += Basic form of energy equation
2 2
ˆ ˆ2 2
s fshear CS
CV CS
pQ W W V dA
d V Vgz u dV gz u V dAdt
ρρ
ρ ρ
⎛ ⎞− − − ⋅⎜ ⎟⎝ ⎠⎛ ⎞ ⎛ ⎞
= + + + + + ⋅⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
∫
∫ ∫
& & &
2
2
ˆ2
ˆ2
s fshear CV
CS
d VQ W W gz u dVdt
V pgz u V dA
ρ
ρρ
⎛ ⎞− − = + +⎜ ⎟
⎝ ⎠⎛ ⎞
+ + + + ⋅⎜ ⎟⎝ ⎠
∫
∫
& & &
h=enthalpy
Usually this term can be eliminated by proper choice of CV, i.e. CS normal to flow lines. Also, at fixed boundaries the velocity is zero (no slip condition) and no shear stress flow work is done. Not included or discussed in text!
57:020 MeProfessor F
Simp Energ Consi Energ
sQ W−& &
Q W−& &
*Althostreamconseqand thconsta
echanics of FluFred Stern Fa
plified F
gy Equati
der flow
gy Equati
s CSW ρ= ∫
1
2
s A
A
pWρ
⎛+ ⎜⎝⎛= ⎜⎝
∫
∫
ough thmlines quently, he pressant.
uids and Transpall 2009
Forms o
ion for S
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ion (stea2
2V gz⎛
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11
22
p gz
p gz
ρ
ρ
+ +
⎛ +⎝
he velocare asthere is
sure is h
port Processes
of the E
Steady O
h the pip
ady flow
ˆpz uρ
+ +
1 1 1
2 2
ˆ
ˆ
u V
u
ρ
ρ
⎞+ ⎟⎠⎞+ ⎟⎠
city varisumed
s no accehydrosta
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One-Dim
pe system
w)
u V dA⎞
⋅⎟⎠
11 1
2 2 2
AV A
V A
ρ
ρ
+
+
∫
∫
ies acroto be
elerationatically d
Equati
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2
31 1
1
32 2
2
2A
V dA
V
ρ
ρ∫
oss the e straign normaldistribute
ion
l Pipe Fl
wn
2dA
flow seght andl to the ed, i.e.,
Chapter 5 25
low
ections d paralstreamlip/ρ +gz
5
the llel; ines z =
57:020 MeProfessor F
*Furthconstaquanti
2
sQ W
pρ
−
⎛= ⎜⎝
& &
RecallSo tha
Define
i.e.,
Q W−& &
(1 Qm
−&&
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lamina
echanics of FluFred Stern Fa
hermoreant acroities can
1
2 ˆ
spW
gz u
ρ⎛+ +⎜⎝
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l that at
e:
=αA1
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) pWρ
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e that:
ar flow α
uids and Transpall 2009
, the inoss the f
then be
2
1 1
2
ˆ
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u Vρ
⎞+ +⎠
⎞⎟⎠∫
VQ ∫=dV∫ ⋅ρ
d2
VA
3
∫ρ
K.E. flu
∫ ⎟⎠⎞
⎜⎝⎛
A
3
VV
A1
1 1gz u+ +
11
p gzρ
+ +
α α > 1 i
α = 2
port Processes
nternal flow sectaken ou
11
2 2
AV dA
V dA
ρ
ρ
⎞⎟⎠
+
∫
VdAV =⋅AVdA ρ=
dA ρα=
ux K.3
dA = k
21
1 2V mα
⎞⎟+⎟⎠
1 1ˆ2Vu α+
= 1 if Vf V is no
turbule
energy ctions, iutside th
1
2
1
32
2
A
A
A
V dA
ρ
ρ
+ ∫
∫
AV mA &=
2AV3
α=ρ
.E. flux for
kinetic en
2pmρ
⎛⎜= +⎜⎝
&
21 2
2pVρ
= +
V is constonunifor
ent flow α
u can i.e. T =he integr
31
1
2
2V dA
A
m
m2
V2
&α
r V= V =co
nergy co
2 2ˆgz u+ +
2 2ˆgz u+ +
tant acrorm
α = 1.05
be con= constaral sign t
mass flow
onstant acro
orrection
22
2 2Vα
⎞+
⎠22
2 2 2Vα+
oss the fl
∼ 1 may
Chapter 5 26
nsidered ant. Thto yield
w rate
oss pipe
factor
m⎞⎟⎟⎠&
flow sect
be used
6
as hese
tion
57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 27
Shaft Work Shaft work is usually the result of a turbine or a pump in the flow system. When a fluid passes through a turbine, the fluid is doing shaft work on the surroundings; on the other hand, a pump does work on the fluid pts WWW &&& −= where tW& and pW& are
magnitudes of power ⎟⎠⎞
⎜⎝⎛
timework
Using this result in the energy equation and deviding by g results in
2 2
1 1 2 2 2 11 1 2 2
ˆ ˆ2 2
p tW Wp V p V u u Qz zmg mg g mg
α αγ γ
−+ + + = + + + + −& &&
& & &
mechanical part thermal part Note: each term has dimensions of length Define the following:
QW
QgW
gmW
h pppp γ
=ρ
==&&
&
&
gmWh t
t&
&=
2 1ˆ ˆ
Lu u Qh head lossg mg−= − =
&
&
57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 28
Head Loss In a general fluid system a certain amount of mechanical energy is converted to thermal energy due to viscous action. This effect results in an increase in the fluid internal energy. Also, some heat will be generated through energy dissipation and be lost (i.e. -Q& ). Therefore the term from 2nd law
2 1ˆ ˆ
0Lu u Qhg gm−= − >
&
&
Note that adding Q& to system will not make hL = 0 since this also increases Δu. It can be shown from 2nd law of thermodynamics that hL > 0. Drop ⎯ over V and understand that V in energy equation refers to average velocity. Using the above definitions in the energy equation results in (steady 1-D incompressible flow)
Lt2
22
22
p1
21
11 hhz
g2Vphz
g2Vp +++α+
γ=++α+
γ
form of energy equation used for this course!
represents a loss in mechanical energy due to viscous stresses
57:020 MeProfessor F
Comp Apply
Energ
•If hL conser •Therinterpreffects Summ The en B = E β = e =
echanics of FluFred Stern Fa
parison o
y energy
gy eq :
= 0 (i.e.rvation o
refore, enreted as s (hL) an
mary of th
nergy eq
= total e
= E/M =
uids and Transpall 2009
of Energy
equation
1
2Vp +
γ
., μ = 0) of mecha
nergy eqa modif
nd shaft w
he Energ
quation i
energy o
= energy
port Processes
y Equati
n to a str
1
21 zg
V =+
we get Banical en
quation ffied Bernwork (hp
gy Equa
is derive
of the sy
y per uni
Infi
ion and B
ream tub
222
g2Vp +
γ=
Bernoullnergy alo
for steadynoulli eqp or ht)
tion
d from R
stem
it mass
initesimal s
Bernoull
be witho
2
22 zg
++
li equationg a str
y 1-D piquation t
RTT wit
stream tub
li Equati
out any s
Lh
on and reamline
ipe flow o includ
th
be ⇒ α1=
Chapter 5 29
ion
haft wor
e
can be de viscou
=α2=1
9
rk
us
57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 30
heat add
Neglected in text presentation
= u + 2V21 +gz
internal KE PE
WQdAVeVeddtd
dtdE
CSCV
&& −=∫ ⋅ρ+∫ρ=
vps WWWW &&&& ++=
( )∫ ⋅ρρ∫ =⋅=CSCV
p dAVpdAVpW&
pts WWW &&& −=
( )∫ ⋅+ρ+∫ρ=+−CSCV
pt dAVepeVeddtdWWQ &&&
21ˆ2
e u V gz= + + For steady 1-D pipe flow (one inlet and one outlet): 1) Streamlines are straight and parallel
⇒ p/ρ +gz = constant across CS
work done
from 1st Law of Thermodynamics
shaft work done on or by system (pump or turbine)
pressure work done
on CS
Viscous stress work on CS
57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 31
mechanical energy Thermal energy
Note: each term has
units of length
V is average velocity (vector dropped) and corrected by α
2) T = constant ⇒ u = constant across CS
3) define ∫ ⎟⎠⎞
⎜⎝⎛=α
CS
3
dAVV
A1 = KE correction factor
⇒ ∫ α=ρα=ρ m2
VA2VdAV
2
233 &
Lt2
22
22
p1
21
11 hhz
g2Vphz
g2Vp +++α+
γ=++α+
γ
gmWh pp &&=
gmWh tt &&=
2 1ˆ ˆ
Lu u Qhg mg−= − =
&
& head loss
> 0 represents loss in mechanical energy due to viscosity
EGL1 =for hp =
57:020 MeProfessor F
Conc
1p α+γ
Define
HE
pressu
stagna
EGL1 EGL2
= EGL2 + hL= ht = 0
echanics of FluFred Stern Fa
cept of H
21
1 zg2
V +α
e HGL
EGL
HGL corEGL corr
ure tap:
ation tub
+ hp = E= EGL1
L
uids and Transpall 2009
abrupchange dto hp or
Hydrau
p1 hz =+
L = p +γ
L = zp +γ
rrespondresponds
hp2 =γ
be: p2 +γ
EGL2 + h + hp − h
port Processes
pt due r ht
ulic and
2p α+γ
=
z
g2Vz
2α+
ds to press to stag
g2V2
2 =α+
ht + hL ht − hL
E
DLf
d Energ
22
2 zg2
V +
2
ssure tapgnation tu
h
pagd
EGL = HGL
g2V2
gy Gra
t2 hz ++
p measurube mea
point-byapplicatigraphicadisplayed
h = h ta
if V = 0
hL = f
i.e., liV, an
de Line
Lh+
rement +asuremen
-point ion is ally d
eight of ap/tube
g2
2VDLf
inear variatiod f constant
f = ff = f
Chapter 5 32
es
+ z nt + z
f fluid in
on in L for D
friction factof(Re)
2
D,
or
1p+
γHGL
Lh =
57:020 MeProfessor F
Helpfu 1. EG
2.&3.
g2
21V
1z =++
L2 = EGL1 - h
g2
2V
= for ab
echanics of FluFred Stern Fa
ful hints
GL = HG
DLfhL =
downwa
2z2p+
γ=
hL
brupt expans
uids and Transpall 2009
for draw
GL + αV
g2V
DL 2
in
ard, exce
Lhg2
22V
++
sion
port Processes
wing HG
V2/2g = H
n pipe me
ept for a
L
L and E
HGL for
eans EG
abrupt ch
GL
V = 0
GL and H
hanges d
HGL will
due to ht
Chapter 5 33
l slope
or hp
3
57:020 MeProfessor F
4. p =
5. for
EG
6. for
i.e.
echanics of FluFred Stern Fa
0 ⇒ HG
DLfhL =
L/HGL
change i
. V1A
V1π
1DV
uids and Transpall 2009
GL = z
g2V
DL 2
=
slope do
in D ⇒
A1 = V2A
V4D2
1 =π
121 DVD =
port Processes
constant
ownward
change i
A2
4DV
22
2π
22D
⇒
t × L
d
in V
chHch
i.e., line
increasi
hange inHGL & Ehange du
early incr
ng L wit
n distanceEGL and ue to cha
Chapter 5 34
reased fo
th slope
e betweeslope
ange in h
4
for
g2V
Df 2
en
hL
57:020 MeProfessor F
7. If H
c
gage p
echanics of FluFred Stern Fa
HGL < z
condition
pressure
uids and Transpall 2009
z then p/
n for cav
vapp =
g,vap =
p g,va ≈γ
port Processes
/γ < 0
vitation:
a 2000=
aA pp −=
m10−≈
i.e., cav
2mN0
aatm p−≈
9810 N/
vitation p
atm 10−=
/m3
possible
mN000,00
Chapter 5 35
2mN
5
57:020 MeProfessor F
echanics of FluFred Stern Fa
uids and Transpall 2009
port Processes Chapter 5 36
6
57:020 MeProfessor F
echanics of FluFred Stern Fa
uids and Transpall 2009
port Processes Chapter 5 377
57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 38
Application of the Energy, Momentum, and Continuity Equations in Combination In general, when solving fluid mechanics problems, one should use all available equations in order to derive as much information as possible about the flow. For example, consistent with the approximation of the energy equation we can also apply the momentum and continuity equations Energy:
Lt2
22
22
p1
21
11 hhz
g2Vphz
g2Vp +++α+
γ=++α+
γ
Momentum: ( )∑ −ρ=ρ−ρ= 121
212
22s VVQAVAVF
Continuity: A1V1 = A2V2 = Q = constant
one inlet and one outlet ρ = constant
57:020 MeProfessor F
AbrupConsito kno
separa Apply
1 zp +γ
Mome ∑ =Fs
next d
echanics of FluFred Stern Fa
pt Expander the f
ow hL = h
ation, i.e
y Energy2
11 g2
Vz =+
entum eq
−= Ap 21
divide m
uids and Transpall 2009
nsion flow fromhL(V1,V
e, p1, an a
y Eq from2 zp +
γ=
q. For CV
−− Ap 22
W siomentum
LA2Δγ
port Processes
m a sma2). Anal
approxim
m 1-2 (α22
2 g2V ++
V shown
αsinW
in α m equati
LzΔ
all pipe tolytic solu
mate solu
α1 = α2 =
Lh+
n (shear
∑ρ= Vu
= (V1 −ρ= 2
2 AVρ
ion by γA
o a largeution to
extremto theflow turbuthe asthat thseparremaiconstvalue
ution for
= 1)
stress ne
⋅ AV
)AV 11 +−2
12 VA ρ−
A2
er pipe. exact prmely dife occurreseparati
ulence. Hssumptiohe pressration regins apprtant and e at the pr hL is po
eglected
V(V 22ρ+
1A
Chapter 5 39
Would lroblem isfficult duence of ons and Howeveron is masure in thgion oximateat the
point of ossible:
d)
)A22
9
like s ue
r, if ade he
ly
57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 40
÷ γA2 ( ) ⎟⎟⎠
⎞⎜⎜⎝
⎛−=−=−−
γ−
γ1
AA
AA
gV
AA
gV
gVzzpp
2
1
2
12
1
2
12
122
2121
from energy equation ⇓
2
12
122
L
21
22
AA
gV
gVh
g2V
g2V −=+−
⎟⎟⎠
⎞⎜⎜⎝
⎛−+=
2
12
122
L AA21
g2V
g2Vh
⎥⎦
⎤⎢⎣
⎡−+=
2
121
21
22L A
AV2VVg2
1h
−2V1V2
[ ]212L VV
g21h −=
If 2 1V V ,
2L 1
1h = V2g
continutity eq. V1A1 = V2A2
1
2
2
1
VV
AA =
57:020 MeProfessor F
Forces Examp
First a ∑ =Fx Fx + p Fx = ρ
f
echanics of FluFred Stern Fa
s on Tra
ple 7-6
apply mo
∑ρ= Vu
p1A1 − p2
ρQ(V2 −
force req
uids and Transpall 2009
ansitions
omentum
⋅A
2A2 = ρV
V1) − p1
quired to
port Processes
m theorem
V1(−V1A
1A1 + p2A
hold tra
m
A1) + ρV2
A2
ansition i
Q = .7
head lo
(empir Fluid p1 = 2D1 = 3D2 = 2Fx = ?
2(V2A2)
in place
707 m3/s
oss = V1.
rical equ
= water 250 kPa 30 cm 20 cm ?
Chapter 5 41
g2V2
2
uation)
1
57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 42
The only unknown in this equation is p2, which can be obtained from the energy equation.
L
222
211 h
g2Vp
g2Vp ++
γ=+
γ note: z1 = z2 and α = 1
⎥⎦
⎤⎢⎣
⎡+−γ−= L
21
22
12 hg2
Vg2
Vpp drop in pressure
⇒ ( ) 11L
21
22
1212x Aphg2
Vg2
VpAVVQF −⎥⎦
⎤⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛+−γ−+−ρ=
p2 In this equation, V1 = Q/A1 = 10 m/s V2 = Q/A2 = 22.5 m/s
m58.2g2
V1.h22
L ==
Fx = −8.15 kN is negative x direction to hold
transition in place
(note: if p2 = 0 same as nozzle)
continuity A1V1 = A2V2
12
12 V
AAV =
i.e. V2 > V1
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