Reynolds Transport Theorem
Steven A. Jones
Biomedical Engineering
January 8, 2008
Louisiana Tech University
Ruston, LA 71272
Things to File Away
• Divergence Theorem
• If the integral of some differential entity
over an arbitrary sample volume is zero,
then the differential entity itself is zero.
Louisiana Tech University
Ruston, LA 71272
Conservation Laws
• Conservation of mass:
Increase of mass = mass generated + mass flux
• Conservation of momentum
Increase of momentum = momentum generated +
momentum flux
• Conservation of energy
Increase of energy = energy generated + energy flux
If more mass goes in
than comes out, mass
accumulates (unless it
is destroyed).
If we take in more
calories than we use,
we get fat.
Louisiana Tech University
Ruston, LA 71272
Conservation Laws: Mathematically
All three conservation laws can be expressed mathematically
as follows:
d d
   dV    v  ndA
CS
dt
dt CV
Production of the
entity (e.g. mass,
momentum,
energy)


Increase of
“entity per unit
volume”
Flux of “entity per unit
volume” out of the surface
of the volume
(n is the outward normal)
is some entity. It could be mass, energy or momentum.
is some property per unit volume. It could be density, or
specific energy, or momentum per unit volume.
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Ruston, LA 71272
The Bowling Ball
If you are on a skateboard, traveling west
and someone throws a bowling ball to you
from the south, what happens to your
westward velocity component?
myou v you  mball  0    myou  mball  v you and ball
myou
myou  mball

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vyou and ball
vyou
 vyou and ball  vyou
(You slow down).
Reynolds Transport Theorem:
Mass
If we are concerned with the entity “mass,” then the
“property” is mass per unit volume, i.e. density.
dm d
   dV    v  ndA
CS
dt dt CV
Production of
mass within
the volume
Effect of
increased mass
on density within
the volume.
Flux of mass through
the surface of the
volume
Mass can be produced by:
1. Nuclear reactions.
2. Considering a certain species (e.g. production of ATP).
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Ruston, LA 71272
Reynolds Transport Theorem:
Momentum
If we are concerned with the entity “mass,” then the
“property” is mass per unit volume, i.e. density.
d  mvx 
dt
Production of
momentum
within the
volume
d
    vx  dV     vx  v  ndA
CS
dt CV
Increase of
momentum within
the volume.
Momentum can be produced by:
External Forces.
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Ruston, LA 71272
Flux of momentum
through the surface of
the volume
Mass Conservation in an Alveolus
dm d
   dV    v  ndA
CS
dt dt CV
Density remains constant, but mass increases
because the control volume (the alveolus)
increases in size. Thus, the limits of the
integration change with time.
Term 1: There is no production of mass.
Control Volume (CV)
Term 2: Density is constant, but the control
volume is growing in time, so this term is positive.
Control Surface CS
Term 3: Flow of air is into the alveolus at the inlet,
so this term is negative and cancels Term 2.
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Ruston, LA 71272
Mass Conservation in an Alveolus
dm d
   dV    v  ndA
CS
dt dt CV
N2, O2,
CO2 and
others.
O2
Can look separately at O2
and CO2.
CO2
Third term is different:
(Inflow of O2 from the bronchiole) – (Outflow of O2 into the capillary system)
Louisiana Tech University
Ruston, LA 71272
Heating of a Closed Alveolus
dm d
   dV    v  ndA
CS
dt dt CV
Density can be “destroyed”
through energy influx, but
the transport theorem still
holds.
Heat
Term 1 is zero. No mass is created inside the control volume.
Term 2 is zero. The decrease in density is cancelled by the
increase in volume.
Term 3 is zero. There is no flux of mass through the walls.
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Ruston, LA 71272
Air Compressed into a Rigid Vessel
dm d
   dV    v  ndA
CS
dt dt CV
Density increases so mass increases while the
control volume (vessel) remains constant.
Term 1: There is no production of mass in the
container.
Term 2: There is an increase in the total mass
of air in the container.
Region R(m)
Surface S(m)
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Term 3: There is flow of air into the alveolus at
the inlet.
Differential Form
dx
vy
vy
x  x
dy
vz
vx
vx
vz
dz
Along the 2 faces shown, vy and vz do
not contribute to changes in the mass
x
within the cube. Only vx contributes.
dm d
   dV    v  ndA
CS
dt dt CV
The left hand term is production of mass. The first term on the right is an
increase in density within the cube, and the second term on the right is the
outward flux of fluid. If the control volume is stationary, then:
d
d
 dV  
dV

CV
CS
dt
dt
Because mass is not being created or destroyed, the left hand term is 0.
Louisiana Tech University
Ruston, LA 71272
Differential Form – Conservation of
Mass
dx
vy
vy
dy
vz
x  x
vx
vz
dm d
   dV    v  ndA
CS
dt dt CV
vx
We can get a differential form if we
convert the last integral to a volume
integral. The divergence theorem says:
dz

CS
 v  ndA      v  dV
CV
dm d
so
   dV       v  dV
CV
dt dt CV
Louisiana Tech University
Ruston, LA 71272
Differential Form
0
   x  dx v x  x  dx     x v x  x 

t
dx

  y  dy v y  y  dy     y v y  y 

dy
 z  dz v z z  dz    z v z z 

dz
  v x v y v z 
  0
 


t  x
y
z 
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vy
vy
x  x
xv
vz
x
vz
Continuity Equation,
Differential Form
vx
Divergence
Conservation of mass reduces to:

   v  0
t
If density is constant then   v  0.
When is density constant?
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Ruston, LA 71272
Constant Density
• Generally density is taken as constant when
the Mach number Mv/c is much less than 1
(where c is the speed of sound).
• For biological and chemical applications,
this condition is almost always true.
• For design of aircraft, changes in density
cannot necessarily be ignored.
• In acoustics (but nobody pays any attention
when I say this).
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Ruston, LA 71272
RTT Applied to Momentum
d
d
    dV    v  ndA
CS
dt
dt CV
Production of the
entity (e.g. mass,
momentum,
energy)
Increase of
“entity per unit
volume”
Flux of “entity per unit
volume” out of the surface
of the volume
(n is the outward normal)


is now momentum.
  mv
  v
is momentum per unit volume.
Louisiana Tech University
Ruston, LA 71272
RTT Applied to Momentum
d
d
    dV    v  ndA
CS
dt
dt CV
Production of the
entity momentum
Increase of
“momentum per
unit volume”
Flux of “momentum per unit
volume” out of the surface
of the volume
(n is the outward normal)


is now momentum.
  mv
  v
is momentum per unit volume.
Louisiana Tech University
Ruston, LA 71272
RTT Applied to Momentum
d  mv syst
dt
d
   vdV     v  v  ndA
CS
dt CV
This v is part of the property being transported.
This v transports the property.
Louisiana Tech University
Ruston, LA 71272
RTT Applied to Momentum
d  mv syst
dt
d
   vdV     v  v  ndA
CS
dt CV
Momentum has three components. Therefore, this is really 3 equations.
Momentum is “produced” by external forces. Therefore, the first term
represents the forces on the control volume.
d
 F  dt CV  vdV  CS   v  v  ndA
Louisiana Tech University
Ruston, LA 71272
Example 3.7
F
What resultant force is
required to hold the
section of tubing in place?
v2
d
 F  dt CV  vdV  CS   v  v  ndA
Steady state
v1
 F     v  v  ndA    A v  v  n     A v  v  n 
White reduces to:  F    A v V    A v V   m  v  v 
2
CS
2
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2
2
2
2
2
2
1 1 1
2
1 1 1
1
1
2
1
1
Example 3.7
F
v2
 v1  n1   V1
 v2  n2   V2
Fx   2 A2V2 V2   1 AV
1 1 cos  V1 
v1
Fy    2 A2V2 sin 0 V2   1 AV
1 1 sin  V1 
   1 AV
1 1 sin  V1 
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Momentum and Pressure
Pout
Pin
twall
d  mv syst
dt
d
   vdV     v  v  ndA
CS
dt CV
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Ruston, LA 71272
CS
Example 3.1 from White
2
Find the rate of change of
energy in the control volume.
CV
1
3
Section
1
2
Type
Inlet
Inlet
 (kg/m2)
3
Outlet
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800
800
V (m/s)
5
8
A (m2)
2
3
e (J/kg)
300
100
800
17
2
150
Example 3.1 Continued
d
 dE 

 
 dt  syst dt

CV

e dV  e3m3  e1m1  e2 m2
If the system is in steady state (i.e. there is no change with
time of the energy within the control volume), then the
integral is zero. Thus, the loss of energy through the control
surface must be balanced by a “production” of energy.
This example is a bit misleading because “production” may
be considered to be a flux of energy through the control
surface. However, production could also be caused by, for
example, a chemical reaction.
Louisiana Tech University
Ruston, LA 71272
Example 3.1 Continued
 dE 
1 1  e2  2 A2V2

  e3 3 A3V3  e1 1 AV
 dt  syst
4.08 MW
-1.92 MW
-2.4 MW
Louisiana Tech University
Ruston, LA 71272
-2.4 + -1.92 = -4.32 (>4.08),
so there is more energy
coming in than going out.
Therefore, the “box” must
“destroy” the energy (e.g. by
doing work).