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ENVIRONMENTAL FLUID MECHANICS
–
Mass, Momentum & Energy Balanceswith some examples
Benoit Cushman-RoisinThayer School of Engineering
Dartmouth College
To extract useful quantitative information from a system, it is necessary to apply the laws of physics. The most important physical laws are:
Conservation of mass → mass of carrying fluid (air or water)→ mass of carried contaminant (ex. Mercury)→ mass of a relevant quantity (ex. dissolved oxygen)
Conservation of momentum (Newton’s Second Law)→ mass x acceleration = Σ forces
(Forces act as momentum sources.)
Conservation of energy → total energy (mechanical + thermal)→ usually thermal energy dominates,
and the energy budget devolves into a heat budget.
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And, there are two broad approaches:
1. The usual method is to establish the budgets for a small volume
and then shrink that volume to zero.
The result is a set of partial differential equations.
2. An alternative, which is often practical but overlooked,
is to perform the budgets on a finite volume of the fluid.
The result gives a coarse-grain picture of reality, but it can be extremely useful if done judiciously.
Inside:Chemical reactionsBiological processes
Physical settling
Inflow Outflow
In the environment, we are not dealing only with local processes, such as chemical reactions; we are mostly dealing with interconnected systems, implying that any sub-system is connected to other sub-systems through fluxes in and out.
We must therefore reckon with material balances in the presence of open systems.
Exchangewith theatmosphere
Waterexample
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Finite-volume budget methodology
Step 1: Choose a control volumeThis can be the entire system or only a well-chosen piece of it.
Step 2: Select the quantity for which the budget is to be madeThis can be mass of fluid, mass of a contaminant, or the fluid’s momentum or energy.
Step 3: Consider all contributions, positive and negativeAdd imports and subtract exports through boundaries.Add sources and subtract sinks within the control volume.
Step 4: Extract useful information from the resulting equationSolve the equation for the salient unknown in the budget equation.
Examples of practical control volumes
urban airshed
entire lake
section of a jet
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Sometimes it is difficult to know where to place the boundaries of the control volume…
(Phoenix, Arizona)
The choice of a control volume is more an art than a science, for there are countless possibilities. A good level of intuition is required, which can only be developed with practice.
Notwithstanding this, there are a couple of basic rules that apply:
1. A control volume ought to be practical, in the sense that it should yield budget equations with the minimum number of unknowns of the problem.
2. Its boundaries need to be clearly defined so that there is no ambiguity about what is inside and what is outside and what are the fluxes in and out of the domain.
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Let’s do a budget for a generic quantity, c.
Think of it as the concentration of a contaminant:
cVmV
mc
fluid carrying of volume
substance of mass
In words, the budget is:
Accumulation = imports – exports
+ sources – sinks
Accumulation = imports – exports
+ sources – sinks
Now, we can fill in the details of the budget.
dt
dcV
dt
tcVdttcVdt
tmdttmdt
tdtt
)()(
)()(
duration
)(at amount )(at amount onAccumulati
and construct each term on a “per-time” basis.
The term on the left is then expressed as:
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Notion of flux
The control volume is almost never bounded on all sides by impermeable boundaries. Thus, fluid enters the volume across some of its boundaries and leaves through some others, carrying with it mass, momentum, energy and any dissolved or suspended substance.
The quantity that describes such exchange between the control volume and its surroundings is the flux.
The flux of any quantity (mass, momentum, energy, dissolved or suspended substance) is defined as the amount of that quantity that crosses a boundary per unit area and per unit time:
uc
q
velocityionconcentrat
duration timeareaboundary
volume
volume
boundary thecrossingquantity
duration timeareaboundary
boundary thecrossingquantity flux
nucqnuu ˆˆ
For oblique flow:where is the perpendicular velocity component
is the 3D velocity vector
and is the outward normal unit vector.
u
n̂
u
Because the exports through the boundaries can be counted as negative imports, a single summation suffices for all imports and exports.
Per unit time, this sum is:
where the summation over the index i covers all the inlets and outlets.
iiii
iii
ii
Anuc
Auc
Aq
)ˆ(
boundaries through exports Imports
The determination of the sources and sinks inside the control volume requires the specification of the quantity for which the budget is performed (mass, momentum, energy, etc.) and a knowledge of the mechanisms by which this quantity may be generated or depleted. For the moment, let us subsume all sources and sinks in a single term
S volumecontrol e within thsinks Sources
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Now with all pieces together, the budget becomes:
SAucdt
dcV iii
In cases when system properties vary continuously inside the control volume and along its boundaries, the total amount of the quantity is to be expressed as a volume integral of its concentration, and the summation over all inlets and outlets must be replaced by an integration over all open portions of the surface enclosing the control volume:
dVsdAnucdVcdt
d)ˆ(
where dV is an elementary piece of volume, dA an element of boundary surface with unit outward normal vector through which passes vector velocity , and s the source/sink per volume.
n̂ u
Mass conservation
To state mass conservation, we use
And we state that there are no sources or sinks.
The budget becomes:
densityvolume
massc
dAnudVdt
d
Audt
dV iii
)ˆ(
or
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Air and water in the environmental systems behave as incompressible fluids. Thus,
constant i
And the mass budget reduces to a volume budget:
outlets
outinlets
inoutlets
outoutinlets
inin QQAuAu
Naturally !
0)ˆ(
0
dAnu
Au ii
More practically, when inlets and outlets can be counted and characterized each by a single volumetric flux Q:
Example of mass conservation
m/yr 7.8km1500
/yrkm )7487(2
3
evap
u
Consider Lake Nasser behind the Aswan High Dam in Egypt, which is fed by the Nile River.
Annual averages of the upstream and downstream river flow rates are 87 km3/yr and 74 km3/yr, respectively, indicating a loss of water from the lake. Assuming that there is no ground seepage, the loss can only be caused by evaporation at the surface. Knowing that the lake surface area is 1500 km2, we can determine the rate of evaporation by performing a water budget for the lake:
River inflow = River outflow + Evaporation
87 km3/yr = 74 km3/yr + uevap A
It follows:
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Momentum budget
Now, for the quantity c , we use the momentum per volume:
And forces act as sources (if accelerating) or sinks (if decelerating).
Forces are:
uV
um
pressure
gravity
friction
Momentum budget is then:
f
g
p
F
gVF
AnpF
ˆ (acting on external surfaces only)
(acting within the volume)
(left unspecified at this stage)
fFgVAnp
AuuAuuudt
dV
surfaces
outoutoutlets
outoutinininlets
inin
ˆ
Again, we invoke the assumption of incompressibility (= o = constant):
o
f
oo
FgVAn
pAuuAuu
dt
udV
surfacesoutoutout
outletsininin
inlets
ˆ
Except in the gravitational term !
Example: Turbulent discharge jet
Statistically steady state:
Uniform outside pressure → pressure term is self-cancelling
No friction on outside surfaces and Internal friction self-cancelling:
Horizontal jet → No gravitational term
0dt
ud
0fF
The momentum budget reduces to:
outoutoutlets
ininininlets
AuuAuu out
entering momentum flux = exiting momentum flux
One-dimensional flow further simplifies the equation
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12
22
22
21
212
221
21 u
R
RuRuRuAuAu
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So, velocity down a turbulent jet decreases inversely to distance.
There seems to be a problem, however: How can this be compatible with mass conservation?
Indeed, conservation of mass implies conservation of volume, and velocity should be inversely proportional to the cross-sectional area and thus to the square of radius and the square of distance, shouldn’t it?
oo uxR
Rxuu
R
Ru
)()(1
2
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Laboratory experiments show that the radius of a turbulent jet increases proportionally to distance:
005
)(
5)(
ux
Rxu
xxR
x
R
factor determinedexperimentally
The answer lies in the fact that the jet entrains ambient fluid.
We can calculate this entrainment by doing a mass (volume) budget:
A consequence of this is dilution. Should the discharge contain some concentration co of a contaminant, this concentration falls off downstream.
Performing a material balance for the contaminant yields:
)0()()( 222 xEuRxuxR oo E rate of entrainment ≠
0
20
2
ambient0022
5
)(5
5
)0()()()(
cx
Rc
cuRxcx
uRx
cxEcuRxcxuxR
o
oooo
o
=0
We find that the concentration decreases inversely with distance and, surprisingly, is independent of the exit velocity.
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Second Example: Uniform river flow down a sloping channel
Assumptions:
• Statistically steady state
• Uniform flow→ momentum in = momentum out
• Atmospheric pressure all along the way
This leaves only two forces that need to balance each other:
Downslope gravity and friction along the bottom.
hf
f
RSgP
ASg
S
dxPgdxA
)()sin)((0
withh
P
AR
S
h
radius hydraulic
slopesin
Take quadratic form for bottom stress: SRC
SRgUUC h
D
hDf .const2
(Chézy formula)
Bernoulli Principle
An interesting and very important corollary of the momentum budget is the so-called Bernoulli principle, which is obtained when the momentum budget is applied to an infinitesimal cylinder aligned with a streamline in a steady-state flow.
Mass conservation:
dsUdUR
UdsRURdUUR
n
n
2
)2()())(( 22
ds
dURUn 2
(1)
Momentum budget:
sin)(
))(()(
)2()())((
2
22
2222
gdsR
dppRpR
UUdsRURdUUR nooo
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constant2
1
0
sin2
2
zgpU
ds
dzg
ds
dp
ds
dUU
dsgdpdUUdUU
o
o
oo
After mathematical simplifications and elimination of Un with Equation (1), we get
ds
dz slopesindivide by ds and
gather terms on left side
integrate with
respect to s
Kinetic energy “Pressure” energy Potential energy
(all per unit volume)
The expression is called the Bernoulli functionzgpUo 2
2
1
Assuming = constantalong the streamline!
Application of mass, momentum and energy budgets to windmill efficiency: The Betz limit
Is there a limit to the power that we can extract from the wind?
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Consider that the wind cannot be brought to a complete stop because air needs to be evacuated to make room for new air to flow. The best we can expect is to slow down the wind some (0 < U2 < U1).
Obviously, the windmill exerts a resistance to the air flow. Therefore, the pressure on its front
surface (pfront) is higher than atmospheric (patm). Upon approaching, the wind of strength U1
faces an adverse pressure gradient and slows down to an intermediate value U.
Having extracted energy from the wind, the air pressure drops in the rear (down to pback),
which is lower than atmospheric. The flow decelerates further, to a speed U2, as it continues
to face an adverse pressure gradient before recovering atmospheric pressure (patm).
back2
atm22
front2
atm2
1
2211
2
1
2
12
1
2
1
pUpU
pUpU
UAUAUA
pUUUU
ppUU
))((2
1
)(2
1
2121
backfront22
21
pAUUAUUA 111222 )()(
UA UA
pAUUUA )( 21
2))((
2
1)( 21
212121
UUUUUUUUUU
Mass conservation:
Bernoulli in frontup to rotor:
Bernoulli in rearpast rotor:
Across the rotor itself: Downstream momentum = Upstream momentum – braking pressure force
(1)
(2)
Eliminate p between Equations (1) and (2)
Mass conservation
subtract
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Having established this relation between the velocities, let us now get to the power extracted from the wind.
1 with )1)(1(2
1
2
))(()()(2
2availablePower
extractedPower
Efficiency ePerformanc oft Coefficien
2
1
2
1
rate massmass
rateenergy kinetic availablePower
time
ntdisplaceme force
time
workextractedPower
1
2231
22
2121
31
212
122
212
1
31
312
1
311
21
U
U
U
UUUU
U
UUUU
U
Up
UA
UAp
C
UAUAU
KE
UAp
p
pressure difference x area
velocity
Note: U1 in this expression, because it is the strength of the local wind
Vary to optimize this expression and find:3
1for %3.59
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16max
pC
)(2
1
)(2
1
21
22
21
UU
UU
The optimal windmill, giving 59% efficiency:
How actual windmill designs perform
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Hydrostatic balance
If the fluid is at rest, there is no momentum and no friction, leaving an equation with only the pressure and gravitational forces:
gVAnp
surfaces
ˆ
This simply says that the various pressures on the sides need to support the weight of the fluid within the control volume.
Removal of the hydrostatic balance
The part of the pressure caused by the main o contribution of the density
can be eliminated leaving a smaller (dynamic) pressure in the pressure force and a density anomaly in the gravitational term
gVAnp o
surfaces
chydrostati ˆ
o
f
oo
FgVAn
pAuuAuu
dt
udV
0
surfaces
dynamicoutoutout
outletsininin
inlets
ˆ
where pdynamic = actual p – phydrostatic
Energy conservation
There are various forms of energy: kinetic, potential, “pressure”, and thermal.
In environmental fluid mechanics, the thermal energy dwarfs the other three.
For example: Heating 1 kg of water by 1oC takes 4184 J;Heating 1 kg of air by 1oC takes 1005 J (at constant pressure);1 kg of air or water moving at 10 m/s has 50 J of kinetic energy;1 kg of air or water falling down 50 m releases 490 J of potential energy.
The result is that the energy budget reduces in good first approximation to a heat budget:
Accumulation of heat = Heat entering – Heat exiting + Heat sources – Heat sinks
Furthermore, sources and sinks are rarely internal but occur most often at boundaries. Thus,
Accumulation of heat = Heat entering – Heat exiting
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Heat content = Thermal energy = Internal energy
TCm v
Per unit volume: TCvo
Budget:
)()()()()( outoutoutlets
outinininlets
in AuTCAuTCTCdt
dV vovovo
This means that temperature is conserved along the flow (no source and sinks).
Finally since temperature and density anomaly are related by the equation of state:
,
this means that density anomaly is conserved along the flow.
(This justifies bringing the density anomaly inside the streamwise derivative in establishing the Bernoulli principle, several slides ago.)
)(anomaly oo TT
Conclusions and word of caution
Finite-volume budget equations do not provide a detailed picture of a system, only some basic relations and properties of it.
These relations and properties, however, can be extremely useful.
The usefulness of the information obtained from budget analysis is only as good as the choice of the control volume…
… and only as good as the attending dynamical simplifications that can be made (such as considering only averaged velocities through cross-sections).
In other words, the tool is only as useful as you are smart.