Mech 280: Frigaard
Lecture 1: Kinematics, ideal mechanical systems and Bernoulli’s equation
Should be able to:
Understand some terms used in flow visualization and
kinematics
Understand Lagrangian and Eulerian frames of reference
Understand the material (or total) derivative
Derive Bernoulli’s theorem using momentum balance along a
streamline
Solve simple problems using the Bernoulli theorem
Mech 280: Frigaard
Streamlines, Pathlines, Streaklines
A streamline is everywhere tangent to the velocity field In steady 2D flow can integrate
A pathline is the actual path travelled by a fluid particle over some time interval E.g. with a tracer particle
A streakline is locus of fluid particles that pass through a given point in the flow Common experimentally: e.g. inject
dye in water, smoke in air
uv
dxdy
Vds
wdz
vdy
udx
=
===
Mech 280: Frigaard
Natural ways to describe flows?
Lagrangian description A fluid is comprised of a large number of fluid
particles having mass, momentum, internal
energy, and other properties. Mathematical
laws can then be written for each fluid particle.
Eulerian description In the Eulerian description of fluid motion, we
consider how flow properties change at a fluid
element that is fixed in space and time (x,y,z,t),
rather than following individual fluid particles.
Mech 280: Frigaard
Eulerian description = most common
Eulerian description of fluid flow: a flow domain or control volumeis defined by which fluid flows in and out
We define field variables which are functions of space and time Pressure field: P=P(x,y,z,t) Velocity field: V=V(x,y,z,t) = u(x,y,z,t)i + v(x,y,z,t)j + w(x,y,z,t)k Acceleration field, a=a(x,y,z,t) = ax(x,y,z,t)i + ay(x,y,z,t)j + az(x,y,z,t)k These (and other) field variables define the flow field.
Suitable for formulation of initial boundary-value problems (PDE's). Leonhard Euler (1707-1783).
Particle’s velocity at a point is same as fluid velocity
Time derivative of particle position is fluid velocity
What about acceleration?
( ) ( ) ( )tzdtdty
dtdtx
dtd
particleparticleparticleparticle ++== VV
Acceleration in Eulerian coordinates?
Acceleration is the time derivative of the particle's velocity:
To take the time derivative of the velocity, chain rule must be used
Advective acceleration term: acceleration of fluid even in steady flow
Advective acceleration is nonlinear: source of many phenomena and main
challenge in solving fluid flow equations
( ) ( ) ( )tzdtd
zty
dtd
ytx
dtd
xt particleparticleparticleparticle ∂∂
+∂∂
+∂∂
+∂∂
=VVVVa
onaccelerati Advectiveonaccelerati Local
zw
yv
xu
tparticle ∂∂
+∂∂
+∂∂
+∂∂
=VVVVa
particleparticle dtd Va =
( ) ( ) ( )( )ttztytx particleparticleparticleparticle ,,,VV =
Material derivative & acceleration:
Total derivative operator is called material derivative
Other names for the material derivative include: total, particle, Lagrangian, Eulerian, and substantial derivative
Lagrangian:
Eulerian:
Consider advective acceleration tangential & normalto steady streamline:
∇⋅+∂∂
=∂∂
+∂∂
+∂∂
+∂∂
= Vtz
wy
vx
utDt
D
∇⋅+∂∂
= VtDt
Ddtd
Mech 280: Frigaard
( )
sVV
tVa
dsdAmgdsdAW
WdAdPPPdAFma
s
ss
∂∂
+∂∂
=
==
−+−== ∑
ρρ
θsin
Bernoulli’s equation
Apply Newton’s second law along a streamline Assume no viscous effects
Assume no heat transfer
Acceleration = material derivative of velocity V along a streamline s
Steady incompressible flow, divide by m and take ds0
Along a streamline: constant=++
=+∂∂
+∂∂
gzPVdsdzg
sP
sVV
ρ
ρ
2
01
2
Bernoulli’s equation
Mech 280: Frigaard
Frictionless flow with no energy transfer
3 terms in Bernoulli equation Potential energy
Flow energy
Kinetic energy
Alternatively, dividing by g: Static head
hydrostatic head (pressure head)
Dynamic head
Unsteady, variable density version of Bernoulli
Using the Bernoulli equation
P1ρ
+V1
2
2+ g z1 =
P2ρ
+V2
2
2+ g z2 = const
P2 − P1ρ
+V2
2 − V12
2+ g z2 − z1( )= 0
( ) ( ) 021
122
12
2
2
1
2
1
=−+−++∂∂
∫∫ zzgVVdpdstV
ρ
22V
Pgz
ρ
gV
gPz
22
ρ
Bernoulli’s equation is an energy equation for an ideal mechanical system
Mech 280: Frigaard
Example 1
Air flows past a ball, as shown. It is determined that the velocity along the x-axis for x<-a is given by:
Determine the pressure variation along the x-axis and the stagnation pressure at x = -a
+= 3
3
0 1xaVV
Mech 280: Frigaard
Example 2Water flows from a garden hose. A child places his thumb to cover most of the outlet. A thin stream of water jets upwards. a) If the gage pressure in the hose just upstream of his
thumb is 400kPa what is the maximum height the jet can achieve, (assuming no energy losses at position 1)?
b) Assuming that soon above position 1 the pressure in the jet is Patm , derive an expression for the velocity VJ as a function of height z above 1
Mech 280: Frigaard
( )
R
2
,cos
Va
dndAmgdndAWWdAdPPPdAFma
n
nn
nnnnn
=
==
−+−== ∑ρρ
θ
F=ma: Normal to a streamline?
Newton’s second law normal to streamline Assume no viscous effects
Assume no heat transfer
Steady incompressible flow, divide by m and take dn 0 n = in direction of curvature
Circular streamlines at same height implies radial increase in pressure
R
21 Vdndzg
nP
=−∂∂
−ρ
Mech 280: Frigaard
Example 3:
Flows a & b have circular streamlines:
Sketch the pressure distributions, assuming p(r0) = p0
Mech 280: Frigaard
Fluids in the news:
Mech 280: Frigaard
Lecture 2: Applications and limitations of Bernoulli’s equation
Bernoulli states that total pressure is constant
Stagnation pressure is the pressure at any point where the fluid is brought to rest isentropically
This can be used with a Pitot tube and piezometer to measure fluid velocity
gzVPPtotal ρρ++=
2
2
2
2VPPstagρ
+=
Mech 280: Frigaard
Example 4:
A piezometer and a Pitot tube are tapped into a horizontal water pipe as shown. Determine the velocity at the center of the pipe?
Mech 280: Frigaard
Example 5:a) What is the pressure at the nose of a torpedo moving in salt water at
100ft/s at a depth of 30ft?b) If the pressure at a point on the side of the torpedo, at the same depth
as the nose, is 10.0psi gage, what is the fluid velocity at that point?
Mech 280: Frigaard
Free streams and jets:
Example 6: Water flows along an undulating path, as shown. What is the pressure variation: a) between points 1 & 2?b) between points 3 & 4?
Mech 280: Frigaard
Example 7: A stream of liquid flows from a hole of diameter 0.01m near the bottom of a reservoir of diameter 0.2m, with height as shown. The liquid is replenished continuously.
Find the flow rate Q that keeps the level constant
Mech 280: Frigaard
Example 8: Water flows under the sluice gate as shown.
Estimate the flow rate per unit width Q
Mech 280: Frigaard
Bernoulli: frictionless flow with no energy transfer Energy equation for an ideal
mechanical system
Track these quantities along a flow path EGL = total head
HGL = static + pressure head
Measure with series of: Pitot tubes (EGL)
Piezometric taps (HGL)
Energy & Hydraulic Grade Lines
constant 2
headDynamic
2
headPressure
headStatic
=++g
Vg
Pzρ
HGL and EGL Plot of EGL and HGL generally
decays along flow path
Variation in EGL along real
flow paths is a measure of
energy losses/gains (see later)
2
2P VEGL zg gρ
= + +
PHGL zgρ
= +
Frictionless flow: HGL indicates positive & negative pressure
Limitations on using the Bernoulli Equation
Steady flow: d/dt = 0
Frictionless flow
No shaft work: wpump=wturbine=0
Incompressible flow: ρ = constant
No heat transfer: qnet,in=0
Applied along a streamline
Mech 280: Frigaard
Fluids in the news:
Mech 280: Frigaard
What we covered
Eulerian and Lagrangian frames of reference
Converting time derivatives between the two systems
Material derivative and acceleration
Bernoulli’s equation
Derivation from momentum balance
Interpretation as an energy balance
Examples of applicationconstant=++
=+∂∂
+∂∂
gzPVdsdzg
sP
sVV
ρ
ρ
2
01
2
Mech 280: Frigaard
Lecture 3: Integral analyses of fluid flow
Systems and control volumes
Reynolds transport theorem
General principals of conservation laws for a control volume
Conservation of mass for a control volume
Systems and control volumes
Generally in mechanics we apply conservation laws to estimate quantities of interest
A (closed) system is a region in space consisting of a fixed quantity of matter No mass crosses system
boundary A control volume, is any
properly selected region in space Analogous to the Free
body diagram in Solid mechanics / dynamics
Mass can cross control surfaces
26
Mech 280: Frigaard
The fundamental conservation laws (conservation of mass, energy &momentum) apply directly to systems
However, in most fluid mechanics problems, control volume analysis is preferred over system Control volumes are chosen more relevant to the problem at hand
Therefore, we need to transform the conservation laws from a system to a control volume. This is accomplished with the Reynolds transport theorem (RTT)
At a time t, we define a system to be the mass in the control volume (CV) bounded by the control surface (CS)
Reynolds Transport Theorem (RTT)
nReynolds transport theorem: Rate of change of ‘B’ in the system is equal to the rate of change of ‘B’ in the CV plus the net flow of ‘B’ out of the CV, across the CS
Mech 280: Frigaard
Mathematical statement
What is B? Any extensive property Associated intensive property is β β is the amount of B per unit mass
∫=V
sys dvB ρβ
dmdB
=β
Mass Momentum Energy Angular momentum
B, Extensive properties m mV E mr×V
β, Intensive properties 1 V e r×V
( )∫∫ ⋅+
=
CSr
CVsys dAdv
dtdB
dtd nVρβρβ
Reynolds transport theorem: Rate of change of ‘B’ in the system is equal to the rate of change of ‘B’ in the CV plus the net flow of ‘B’ out of the CV, across the CS
Mech 280: Frigaard
Moving / deforming control volume
Reynolds transport theorem
Is valid for fixed, moving and/or deforming control volumes,
The velocity Vr in the second term is the relative velocity:
Vr = V -VCS
V is the fluid velocity, (typically given in an inertial system)
VCS is velocity of the moving/deforming CS, (in the same
system as V)
( )∫∫ ⋅+
=
CSr
CVsys dAdv
dtdB
dtd nVρβρβ
Mech 280: Frigaard
Conservation of Mass
Conservation of mass is one of the most fundamental principles in nature.
Mass, like energy, is a conserved property, and it cannot be created or destroyed during a process.
For closed systems mass conservation is implicit since the mass of the system remains constant during a process.
For control volumes, mass can cross the boundaries which means that we must keep track of the amount of mass entering and leaving the control volume.
Mech 280: Frigaard
Mass Conversvation
B =m = Massn
( )∫∫ ⋅+
=
CSr
CVsys dAdv
dtdB
dtd nVρβρβ
( )∫∫ ⋅+
=
CSr
CV
dAdvdtd nVρρ0
1==dmdmβ
0 =sysBdtd :conserved Mass
Mech 280: Frigaard
Mass and Volume Flow Rates
Consider a fixed CV and CS, so that Vr = V
Consider flow through a control surface, with area Ac . Flow rates obtained by integration
The amount of mass flowing through a control surface per unit time is called the mass flow rate and is denoted
The dot over a symbol is used to indicate time rate of change.
Volume flow rate through Ac
Often density doesn’t change much
c c
n cA A
m m V dAδ ρ= =∫ ∫
V
( ) cavgA
cnA
c AVdAVdAQcc
==⋅= ∫∫ nV
1
c
avg n cc A
V V dAA
= ∫
avg cm V Aρ=
Conservation of Mass Principle For fixed CV of arbitrary shape,
rate of change of mass within the CV
net mass flow rate
Therefore, general conservation of mass for a fixed CV is:
Same expression for deforming or moving CV, except net mass flow rate in/out are relative to the CS
= ∫
CVCV dv
dtdm
dtd ρ
( )∫∫ −=⋅==CS
inoutCS
net mmdAmm nVρδ
CVin out
dmm mdt
− =
Steady Flow Processes For steady flow, total amount of
mass contained in CV is constant. Total amount of mass entering
must be equal to total amount of mass leaving
For incompressible flows,in out
m m=∑ ∑
n n n nin out
V A V A=∑ ∑
Mech 280: Frigaard
Example 1:
If the average water velocity in a in a 300mm pipe is 0.5m/s, what is the
average velocity in a 75mm water jet coming from the pipe?
Mech 280: Frigaard
Example 2:An airplane moves forward at speed 971km/hr. The frontal intake area of the jet
engine is 0.8m2 and the entering air density is 0.736kg/m3. A stationary observer
determines that relative to the Earth, the jet engine exhaust gases move away
from the engine with a speed 1050km/hr. The engine exhaust area is 0.558m2,
and the exhaust gas density is 0.515kg/m3.
Estimate the mass flowrate of fuel into the engine in kg/h
Mech 280: Frigaard
Example 3:
1m
2m
For an observer standing at the c.v. inlet (point 1)V1 = VJ – Uc = 7 – 2 = 5 m/s
= ρ1 V1 A1 = 998 kg/m3*5 m/s*π*.042/4 = 6.271 kg/sNote: The inlet velocity used to specify the mass flow rate is again measured relative to the inlet boundary, but now is given by VJ – Uc . Exit:
= 6.271 kg/s, Again, since ρ and A are cons., V2 = 5 m/sAgain, the exit flow is most easily specified by conservation of mass concepts.
VJ
A jet of water leaves the nozzle at mean speed 7 m/sAnd strikes the turning vane as shown. a) Compute the mean velocity exiting the CV if the
vane is stationaryb) Repeat the calculation if the vane moves to the
right at a steady velocity of 2 m/s.
Mech 280: Frigaard
Example 4:
A room contains dust at a uniform concentration, C=ρdust/ρ. The room is to be cleaned by flushing air through the room. Find an expression for time rate of change of dust mass in the room.
Vout
Mech 280: Frigaard
What we covered Derived the general Reynolds Transport Theorem (RTT)
Velocity is relative to moving / deforming control volumes
Used the RTT to derive conservation of mass for a CV
Mass flow rates:
Mass conservation can be written as: CVin out
dmm mdt
− =
( )∫∫ ⋅+
=
CSr
CVsys dAdv
dtdB
dtd nVρβρβ
( )∫∫ ⋅+
=
CSr
CV
dAdvdtd nVρρ0
( )∫ ∫∫ =⋅==CS CS
nCS
net dAVdAmm ρρδ nV
Mech 280: Frigaard
Lecture 4: Conservation of momentum
Should be able to:
Use Reynolds’ transport theorem to derive the conservation of
linear momentum relation for a CV
Examples for fixed & moving CV
Examples coupled with mass conservation CV
Last lecture we derived Reynolds’ transport theorem
Applied this to give conservation of mass relation in a CV
In steady flow conservation of mass is relation between flow rates
What about computing forces?
Mech 280: Frigaard
Newton’s Laws
Newton’s laws are relations between motions of bodies
and the forces acting on them
First law: a body at rest remains at rest, and a body in motion
remains in motion at the same velocity in a straight path when
the net force acting on it is zero
Second law: the acceleration of a body is proportional to the net
force acting on it and is inversely proportional to its mass
( )VVaF mdtd
dtdmm ===
Mech 280: Frigaard
Momentum Conservation for a CV
n
B =mV = Momentumn
( )∫∫ ⋅+
=
CSr
CVsys dAdv
dtdB
dtd nVρβρβ
( )∫∫∑ ⋅+
=
CSr
CVCVdAdv
dtd nVVVF ρρ
( ) VV == mdmdβ
∑= FsysBdtd :conserved Momentum
Mech 280: Frigaard
Orientation of CS’s and the Momentum Flux Correction Factor
Momentum is a vector
quantity
Flux term in momentum
balance is more complex
than for mass conservation
Nearly always we orient CS’s
to be normal to the flow
direction Does not affect mass balance
Simplifies flux term
Note that at inlet/outlet, Vavg
often only has component in
normal direction:
In an incompressible flow,
we have
β>1 always and for turbulent
flow β in range 1.01 – 1.05( ) ∫∫ ==⋅cc A
avgcnA
c mdAVdA VVnVV βρρ
∫
∫
∫
∫==
c
c
c
c
Ac
Ac
avg
Acn
Acn
avg dA
dA
dAV
dAV VV
VV
ρ
ρβ
∫
=
cAc
avgn
n
c
dAVV
A
2
,
1β
Choosing a Control Volume
CV is chosen by fluid dynamicist and there is a lot of freedom. However, selection of CV can either simplify or complicate analysis Clearly define all boundaries. Analysis is often simplified
if CS is normal to flow direction Clearly identify all fluxes crossing the CS Clearly identify forces and torques of interest acting on
the CV and CS Fixed, moving, and deforming control volumes:
For moving CV, use relative velocity,Vr = V - VCS = V -VCV
For deforming CV, use velocity relative to each deforming control surfaces
Vr = V –VCS Note that for momentum balance, as given, Newton’s
laws require an inertial frame of reference For non-inertial frames, add fictitious forces – addressed
later in course
Forces Acting on a CV
Forces acting on CV consist of body forces that act throughout the
entire body of the CV (such as gravity, electric, and magnetic forces) and
surface forces that act on the control surface (such as pressure and
viscous forces, and reaction forces at points of contact).
• Body forces act on each volumetric portion dv of the CV.
• Surface forces act on each portion dA of the CS.
n
dFdFsurface
45
Body Forces
The most common body force is
gravity, which exerts a downward
force on every differential element
of the CV
dFbody = dFgravity = ρgdV
Summing over all differential
volume elements, gives total body
force acting on CVi
dFbody = dFgravity = ρgdVk
j
g
ggF CVCV
body mdV == ∫∑ ρ
Surface Forces
Surface forces include both normal and
tangential components
Diagonal components σxx, σyy, σzz are called
normal stresses and are due to pressure and viscous stresses
Off-diagonal components σxy, σxz, etc., are
called shear stresses and are due solely to
viscous stresses
Force/area acting on CS with normal n:
Total surface force acting on CS
∫∑ ⋅=CS
surface dSσnF
( ) ∑=
=⋅321 ,,i
iijj nσσn
Body and Surface Forces
Surface integrals are cumbersome.
Careful selection of CV allows
expression of total force in terms of
more readily available quantities like
weight, pressure, and reaction forces.
Goal is to choose CV to expose only
the forces to be determined and a
minimum number of other forces.
Surface
otherviscouspressure
Body
gravity ∑∑∑∑∑ +++= FFFFF
∑∑∑ −=In
avgOut
avg mm VVF ββ
Simplification for steady flows:
Mech 280: Frigaard
Example 1: For the pipe-flow reducing section shown, D1 = 8cm, D2 = 5cm and all fluids are at 20°C. If V1 = 5m/s and the manometer reading is h = 58cm, estimate the total horizontal force on the flange bolts.
Mech 280: Frigaard
Example 2: Force on an elbow
Water flows steadily through a 90° elbow, exiting to atmosphere at position 2. If D = 5cm & P1 = 3.43 kPa, what anchoring force is needed to keep the elbow in place?
Mech 280: Frigaard
Example 3: Moving control volumeA water jet 4 cm in diameter with a velocity 7 m/s is directed towards a turning vane (θ = 40˚) that is moving with velocity, Uc = 2 m/s. Determine the force F necessary to hold the vane in steady motion.
iieeb VmVmF −=−
V1 = VJ – Uc = 7 – 2 = 5 m/s and V2 = 5 m/s inclined 40˚ .
-Fb = 6.271 kg/s * 5 m/s *cos 40˚ -6.271 kg/s * 5 m/s
and -Fb = - 7.34 kg m/s2 or Fb = 7.34 N ← ans.
Mech 280: Frigaard
Example 4: Non-uniform pressureA sluice gate across a channel of width b operates in both open and close positions, as shown. Is the anchoring force required to hold the gate in place larger when the gate is open or closed?
Mech 280: Frigaard
Fluids in the news
Mech 280: Frigaard
What we covered
Derived linear momentum equation
Momentum flux correction factor
Simplified forms for momentum balance
Sum of forces include: Pressure, body, shear and mechanical
Computed examples of: Stationary control volume
Moving, control volume in an inertial frame
( )∫∫∑ ⋅+
=
CSr
CVCVdAdv
dtd nVVVF ρρ
Mech 280: Frigaard
Lecture 5: Momentum II
Examples using the momentum balance to generate thrust
Derive the conservation of momentum for a control volume
moving in an accelerating frame of reference.
Examples of accelerating frames of reference
Mech 280: Frigaard
Example 1:A liquid jet of velocity Vj and area Aj strikes a single 180˚ bucket on a turbine wheel rotating at angular velocity Ω. a) Find an expression for the power deliveredb) At what Ω is the power a maximum
Mech 280: Frigaard
Example 2:A static thrust stand as sketched is to bedesigned for testing a jet engine. The following conditions are known for a typical test: • Intake air velocity: 200m/s • Exhaust gas velocity: 500m/s • Intake cross-sectional area: 1m2
• Intake static gage pressure: -22.5kPa• Intake static temperature: 268K• Exhaust static gage pressure: 0kPaEstimate the nominal anchoring force for design
Mech 280: Frigaard
Non-inertial frames of reference
Absolute Acceleration = Non-inertial Acceleration +
Relative Acceleration
Consider fluid particle, with position:
ri =
Velocity:
Vi =
V = velocity in non-inertial frame
How about acceleration?
X
Y
Z
y
x
z
Mech 280: Frigaard
Relative acceleration
X
Y
Z
y
x
z
reli dtd aVa +=
Mech 280: Frigaard
Non-inertial frames of reference
imaF = Newton’s 2nd Law
dtdmm
dtdmm relreli
VaFFaVa =−⇒=
+= ∑∑
∫∫ ∫∑ ∫ −+∂∂
=−ie A
icv A
ecvcv
cv mdmdVdt
md VVVaF ρ
Effect on control volume analysis is to add fictitious forces to the net force balance
In general, situations can be very complex. In many mechanical systems we restrict motion to specific degrees of freedom.
Mech 280: Frigaard
Example 1: Accelerating control volume
A water jet 4cm in diameter with a velocity of 7m/s is directed to a turning vane moving with velocity Uc(t), and with angle θ = 40˚ as shown. Determine the acceleration of the vane as a function of time, assuming initially stationary.
∫∫ ∫∑ ∫ −+∂∂
=−ie A
icv A
ecvcv
cv mdmdVdt
md VVVaF ρ
Mech 280: Frigaard
Example 2: Rocket acceleration
Calculate the rocket’s angular velocity, given the rocket’s initial mass, the propellant exit velocity and the exit mass flow rate (both assumed constant)?
atmeee ppVm =,,
ω
∫∫ ∫∑ ∫ −+∂∂
=−ie A
icv A
ecvcv
cv mdmdVdt
md VVVaF ρ
Mech 280: Frigaard
What we covered:
Examples of using the linear momentum balance to
calculate thrust, in inertial frames
We derived the fictional forces that we experience
in a non-inertial frame of reference.
Note that the fictional force (acceleration) is negative.
Examples
∫∫ ∫∑ ∫ −+∂∂
=−ie A
icv A
ecvcv
cv mdmdVdt
md VVVaF ρ
https://www.youtube.com/watch?v=X_UDguQ420IDon’t do this at home
Mech 280: Frigaard
Lectures 7 & 8: Energy balance for CV’s
Should be able to: Use the Reynolds Transport Theorem (RTT) to derive the
conservation of energy for a control volume
Understand forms of energy and how work is done on
system
Understand efficiency
Relationship with Bernoulli equation for single stream
systems
Application examples
Mech 280: Frigaard
General Energy Equation
One of the most fundamental laws in nature is the 1st law of
thermodynamics, which is also known as the conservation of
energy principle.
It states that energy can be neither created nor destroyed during a
process; it can only change forms
• Falling rock, picks up speed as PE is converted to KE.
• If air resistance is neglected, PE + KE = constant
• Enthalpy?
General Energy Equation
Energy in a system is in 3 forms Kinetic energy: 0.5mV2
Potential energy: mgz Enthalpy: H
The energy content of a closed system can be changed by two mechanisms: heat transfer Q and work transfer W
Conservation of energy for a closed system can be expressed in rate form as
Net heat transfer to the system:
Net power input to the system:
In open system (CV) work comes from Pressure work Viscous work Shaft work (pump / turbine)
Mech 280: Frigaard
Energy Conservation for a CV (from RTT)
B = Esys
β = e = u+0.5V2+gz = (internal + kinetic + potential) specific energies
General Energy Equation:
( )∫∫ ⋅+
=
CSr
CVsys dAdv
dtdB
dtd nVρβρβ n
𝑄𝑛𝑛𝑛𝑛𝑛𝑛,𝑖𝑖𝑛𝑛 + 𝑊𝑛𝑛𝑛𝑛𝑛𝑛,𝑖𝑖𝑛𝑛 =𝑑𝑑𝐸𝐸𝑠𝑠𝑠𝑠𝑠𝑠𝑑𝑑𝑑𝑑
=𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑
( )∫∫ ⋅+
==+
CSr
CV
sysinnetinnet dAedve
dtd
dtdE
WQ nVρρ,,
Pressure work
Where does expression for pressure work come from? When piston moves down ds under the influence of F=PA,
the work done on the system is dWboundary=PAds. If we divide both sides by dt, we have
For generalized control volumes:
Note sign conventions: n is outward pointing normal vector Negative sign ensures that work done is positive when is
done on the system.
∫ ⋅−=cs r
p dAPdt
dW)( nV
n
V
𝛿𝛿𝑊𝑝𝑝𝑝𝑝𝑛𝑛𝑠𝑠𝑠𝑠𝑝𝑝𝑝𝑝𝑛𝑛 = 𝛿𝛿𝑊𝑏𝑏𝑏𝑏𝑝𝑝𝑛𝑛𝑏𝑏𝑏𝑏𝑝𝑝𝑠𝑠 = 𝑃𝑃𝑃𝑃𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑
= 𝑃𝑃𝑃𝑃𝑉𝑉𝑝𝑝𝑖𝑖𝑠𝑠𝑛𝑛𝑏𝑏𝑛𝑛
𝛿𝛿𝑊𝑏𝑏𝑏𝑏𝑝𝑝𝑛𝑛𝑏𝑏𝑏𝑏𝑝𝑝𝑠𝑠 = −𝑃𝑃𝑑𝑑𝑃𝑃𝑉𝑉𝑛𝑛 = −𝑃𝑃𝑑𝑑𝑃𝑃(𝑽𝑽𝒓𝒓.n)
Mech 280: Frigaard
Viscous work
Viscous work is the work done on the control volume surface due
to shear force
Can be a real problem because we don't know velocity
distribution, therefore don't know shear stress.
Solution! Draw control volume carefully, e.g.
Perpendicular to flow
Where the velocity is zero, e.g., at a solid boundary
Or where the shear stress is zero, (or small, or perpendicular to the velocity)
∫ ⋅⋅−=cs ij dA
dtdW nV τν
Mech 280: Frigaard
Energy Analysis of Steady Flows
Simplifications: Kinetic energy correction factor set to 1 (see later)
Rate of change of energy content = 0
Well-chosen CV means rate of viscous work is zero
Pressure work rate is included in flux terms
• Note, could use enthalpy: h=u+P/ρ, but confusing later when h= “head”.
Single inflow/outflow: specific quantities
+++−
+++=+ ∑∑ gzVuPmgzVuPmWQ
inoutinnetshaftinnet 22
22
,,, ρρ
Fixed control volume
In
Out
2
1
innetshaftinnet WQ ,,, +
+++ 1
21
11
1
2gzVuPm
ρ
+++ 2
22
22
2
2gzVuPm
ρ
[ ]
lossmechturbinepump
e
innetinnetshaft
innetshaftinnetinnetshaftinnet
ewgzVPgzVPw
quugzVPgzVPw
gzVuPm
Wm
Qwq
lossmech
,2
22
2
21
21
1
1
,122
22
2
21
21
1
1,,
2
1
2,,,
,,,
22
22
2
,
++++=+++
−−+++=+++
+++=+=+
ρρ
ρρ
ρ
Mech 280: Frigaard
Example 1:A pump delivers water at a steady rate of 300 gal/min as shown. Upstream [sect. (1)]
pipe diameter is 3.5 in. & the pressure is 18 psi. Downstream [sect. (2)] the pipe diameter
is 1 in. & the pressure is 60 psi. The change in elevation is zero. The rise in internal
energy of water, u2-u1, associated with a temperature rise across the pump is 93 ft lb/lbm.
Q. Determine the power that the pump requires.
Mech 280: Frigaard
Example 2:A steam turbine generator unit is used to produce electricity. The steam enters the turbine with
a velocity of 30 m/s and enthalpy, h1, of 3348 kJ/kg. The steam leaves the turbine as a mixture
of vapor and liquid having a velocity of 60 m/s and an enthalpy h2 of 2550 kJ/kg. The flow
through the turbine is adiabatic, and changes in elevation are negligible.
Q. Determine the work output involved, per unit mass of steam through-flow
Mech 280: Frigaard
Head form of energy equation (single stream CV)
Divide by g to express each
term in units of length
Magnitude of each term is
expressed as an equivalent
height of fluid, i.e. a head
Note, with no shaft work or
irreversible head losses,
similar to Bernoulli’s equation
2 21 1 2 2
1 21 22 2pump turbine LP V P Vz h z h h
g g g gρ ρ+ + + = + + + +
Example 3:
Given the initial exit velocity what is the pump
power & the time taken to empty the tank?
2 2
2 2in in out out
in out f p tV P V Pz z h h h
g g g gρ ρ+ + = + + + − +
Mech 280: Frigaard
Example 4:The pump shown in the figure adds 10 horsepower to the water as it pumps water
from the lower lake to the upper lake. The elevation difference between the lake
surfaces is 30 ft and the head loss is 15 ft.
Q. Determine the flow rate and power loss associated with this flow
Mech 280: Frigaard
Efficiency
Transfer of emech is usually accomplished by a rotating shaft: shaft work
Pump, fan, propulsion: receives shaft work (e.g., from an electric motor) and transfers it to the fluid as mechanical energy
Turbine: converts emech of a fluid to shaft work. In absence of irreversible losses (e.g., friction), mechanical
efficiency of a device or process can be defined as
If ηmech < 100%, losses have occurred during conversion.
, ,
, ,
1mech out mech lossmech
mech in mech in
E EE E
η = = −
Pump and Turbine Efficiencies
In fluid systems, we are usually interested in increasing the pressure, velocity, and/or elevation of a fluid.
In these cases, efficiency is better defined as the ratio of (supplied or extracted work) vs. rate of increase in mechanical energy
Overall efficiency must include motor or generator efficiency.
Mech 280: Frigaard
Example 5:The pump of a water distribution system is powered by a 15-kW electric motor (efficiency 90%). The water flow rate is 50l/s. The inlet/outlet pipe diameters are the same and there is negligible increase in elevation across the pump. Determine a) the mechanical efficiency of the pump; b) the temperature rise of the water as it flows through the pump, due to inefficiency.
Mech 280: Frigaard
Example 6:In a hydroelectric power plant, 100m3/s of water flows from an elevation of 120m to a turbine. The total irreversible head loss is determined to be 35m. If the overall efficiency of the turbine generator is 80%, estimate the electric power output.
Mech 280: Frigaard
Example 7:An axial-flow ventilating fan driven by a motor thatdelivers 0.4 kW of power to the fan blades produces a 0.6-m diameter axial stream of air having a speed of 12 m/s. The flow upstream of the fan involves negligible speed.Q. Determine how much of the work to the air actually produces useful effects, that is, fluid motion and a rise in available energy. Estimate the mechanical efficiency of this fan.
Mech 280: Frigaard
Up to this point, the velocity used in the kinetic energy term of the energy equation has been the mass average velocity obtained from the definition of flow rate:𝑚 = 𝜌𝜌𝑃𝑃 𝑉𝑉
Kinetic Energy Correction Factor
3 3av
1 1u d A V A2 2
ρ ρ β=∫
3
av
1 u=A V
d Aβ
∫
However, since V varies over the flow area and the kinetic energy term varies with the square of the velocity, using this definition of Vmay not result in an accurate evaluation of the kinetic energy term for the flow. This problem can be corrected through the use of the kinetic energy correction factor, defined for incompressible flow from
Solving for 𝛽𝛽 we obtain
For fully developed laminar flow, β= 2 and for turbulent pipe flow, a value from 1.04 to 1.11 is common
Mech 280: Frigaard
Example 8:The small fan shown moves air at a mass flowrate of 0.1 kg/min. Upstream the pipe diameter is 60 mm, the flow is laminar and the kinetic energy coefficient, is 2.0. Downstream of the fan, the pipe diameter is 30 mm, the flow is turbulent and the kinetic energy coefficient, is 1.08. The rise in static pressure across the fan is 0.1 kPa, and the fan motor draws 0.14 W.Q. Compare the value of losses calculated: (a) assuming uniform velocity distributions, (b) considering actual velocity distributions.
Mech 280: Frigaard
What we covered
Derived conservation of energy for a CV
Converted to ‘head’ form for steady, single stream CV
Note: Bernoulli’s equation as a frictionless flow with no
energy in / out.
Efficiency of pumps & turbines
Examples using energy equation in different formats
2 2
2 2in in out out
in out f p tV P V Pz z h h h
g g g gρ ρ+ + = + + + − +
P1ρ
+V1
2
2+ g z1 =
P2ρ
+V2
2
2+ g z2 = const
Mech 280: Frigaard
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