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058:0160
Jianming Yang
Fall 2012
Chapter 2
1
Chapter 2: Pressure Distribution in a Fluid
1 Pressure and pressure gradient
In fluid statics, as well as in fluid dynamics, the forces acting
on a portion of fluid, control volume (C.V.), bounded by a
control surface (C.S.) are of two kinds: body forces and
surface forces.
Body Forces: act on the entire body of the fluid (force per
unit volume).
Surface Forces: act at the C.S. and are due to the surrounding
medium (force per unit area: stress).
In general the surface forces can be resolved into
two components: one normal and one tangential to the surface.
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y/00/19/77/26/19772652/1743-8454-6-12-1.jpg
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Considering a cubical fluid element, we see that
the stress in a moving fluid comprises a 2nd order tensor.
Since, by definition, a fluid cannot withstand a shear stress without moving
(deformation), a stationary fluid must necessarily be completely free of shear stress
(σij=0, i ≠ j).
The only non-zero stress is the normal stress, which is referred to as pressure:
𝑝 = −𝜎𝑥𝑥 = −𝜎𝑦𝑦 = −𝜎𝑧𝑧 ,
taken positive for compression by common convention.
Chapter 2
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(one value at a point, independent
of direction, p is a scalar)
n
p x = p y = p z = pn = p
i.e. normal stress (pressure) is isotropic. This can be easily seen
by considering the equilibrium of a wedge shaped fluid element
Force balance:
 Fx  0  pxbz  pnbs sin 
 Fz  0  pz bx  pnbs cos  12 gbxz
From geometry of the wedge
s sin   z
s cos  x
therefore
p x  pn
pz  pn  12 gz
In the limit as the fluid wedge shrinks to a
point z  0
px  pz  pn  p
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Note: For a fluid in motion, the normal stress is different on each face and not equal to p.
σxx ≠ σyy ≠ σzz ≠ -p
By convention p is defined as the average of the normal stresses
p
1
1
 xx   yy   zz     ii

3
3
The fluid element experiences a force on it as a result of the fluid pressure distribution if
it varies spatially. Consider the net force in the x direction due to p(x,t).
p 
p

dFx  p dy dz   p 
dx dy dz   dx dy dz
x 
x

The result will be similar for dFy and dFz; consequently, we conclude:
 p
p
p 
dFpress   
i
j
k dxdydz

x

y

z


Or: f press  p
is the force per unit volume due to p(x,t).
Note: if p=constant, f press  0 .
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2 Equilibrium of a fluid element
Consider now a fluid element which is acted upon by both surface forces and a body
force due to gravity
dFgrav  gdxdydz or fgrav  g (per unit volume)
Application of Newton’s law yields:
ma   F  dxdydza   f dxdydz
 a   f  fbody  fsurface (per unit volume)
f body  g   gk g   gk 
fsurface  f press  f visc  p    2S 
for incompressible Newtonian fluid flow, S is the stain rate tensor
a  g 
inertial
gravity
p
pressure gradient
   2S 
viscous
and
V
a
 V  V
t
This is called the Navier-Stokes equation and will be discussed further in Chapter 4.
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Consider solving the N-S equation for p when 𝐚 and 𝐕 are known.
p   g  a    2S  B(x, t )
This is simply a first order p.d.e. for p and can be solved readily. For the general case (𝐕
and p unknown), one must solve the N-S and continuity equations, which is a formidable
task since the N-S equations are a system of 2nd order nonlinear p.d.e.’s.
We now consider the following special cases :
1) Hydrostatics (𝐚 = 0 and 𝐕 = 0 ⇒ 𝐒 = 0)
1
2) Rigid body translation or rotation (no relative motion: 𝐒 = [∇𝐕 + (∇𝐕)T ] = 0)
2
3) Irrotational motion (   V  0 )
If viscous effects are neglected, the incompressible Navier-Stokes equation becomes
incompressible Euler equation:
 V

 V  V    g  p
 t


also,
  V  0  V  
and if  = constant,
2
 V  0     0
   2 p

 V



 V  V    g   p  

  gz   0


2

 t

 t

  2 p


  gz  const
t
2

(Bernoulli’s equation)
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3 Case (1): Hydrostatic Pressure Distribution
p  g   gk
p p
i.e. x  y  0
and
p
  g
z
or
2
2
1
1
dp   gdz
p  p    gdz   g   ( z )dz
2
1
r 
g  g0  0 
r
2
r
g  , constant near earth’s surface 𝑟
0
liquids  𝜌 = const. (for one liquid)
𝑝 = −𝜌𝑔𝑧 + constant
gases
 𝜌 = 𝜌(𝑝, 𝑡) which is known from the equation of state:
𝑝 = 𝜌𝑅𝑇 ⇒ 𝜌 = 𝑝/𝑅𝑇
dp
p
  g  
g
dz
RT

dp
g dz

p
R T (z )
which can be integrated if 𝑇 = 𝑇(𝑧) is known as it is for the atmosphere.
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Chapter 2
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4 Manometry
Manometers are devices that use liquid columns for measuring differences in pressure. A
general procedure may be followed in working all manometer problems:
1.) Start at one end (or a meniscus if the circuit is continuous) and write the pressure
there in an appropriate unit or symbol if it is unknown.
2.) Add to this the change in pressure (in the same unit) from one meniscus to the next
(plus if the next meniscus is lower, minus if higher).
3.) Continue until the other end of the gage (or starting meniscus) is reached and equate
the expression to the pressure at that point, known or unknown.
𝑝𝐴 − 𝑝𝐵
= (𝑝𝐴 − 𝑝1 )
+(𝑝1 − 𝑝2 )
+(𝑝2 − 𝑝3 )
+(𝑝3 − 𝑝𝐵 )
= −𝛾1 (𝑧𝐴 − 𝑧1 )
−𝛾2 (𝑧1 − 𝑧2 )
−𝛾3 (𝑧2 − 𝑧3 )
−𝛾4 (𝑧3 − 𝑧𝐵 )
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5 Hydrostatic forces on plane surfaces
The force on a body due to a pressure distribution is: F  A pn dA
where for a plane surface n = constant and we need only consider |F| noting that its
direction is always towards the surface: F  A p dA .
Consider a plane surface entirely submerged in a liquid such that the plane of the surface
intersects the free-surface with an angle α. The surface centroid is denoted ( x, y ).
dF  p dA   hdA   y sin  dA
F    y sin  dA
A
  sin   y dA
A
  sin  y A by definition , y 
1
y dA

A
A
 pA
F   sin  yA  pA
where p is the pressure at the centroid.
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To find the line of action of the force which we call the center of pressure (xcp, ycp) we
equate the moment of the resultant force to that of the distributed force about any
arbitrary axis.
ycp F   ydF   sin   y 2dA
A
Note: dF   y sin  dA
y 2 dA  I xx  y A  I xx
(parallel axis theorem)
A
A
2
𝐼𝑥𝑥 : moment of inertia about O-x
̅ : moment of inertia w.r.t. horizontal centroidal axis
𝐼𝑥𝑥

2
ycp F  ycp sin  y A   sin   y A  I xx 



ycp  y 
I xx
yA

xcp  x 
I xy
yA
and similarly for xcp

xcp F   x dF   sin   xydA   sin  x y A  I xy
A
A
where the product of inertia

A xydA  I xy  x yA  I xy (parallel axis theorem)
Note that the coordinate system in the textbook has its origin at the centroid and is related
to the one just used by:
xtextbook  x  x and ytextbook  ( y  y)
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To obtain equation (2.29) in the textbook:
𝑥CP,textbook = 𝑥𝑐𝑝 − 𝑥̅ = (𝑥̅ +
̅
𝐼𝑥𝑦
𝑦̅𝐴
) − 𝑥̅ =
𝑦CP,textbook = −(𝑦𝑐𝑝 − 𝑦̅) = − [(𝑦̅
̅
𝐼𝑥𝑦
𝑦̅𝐴
̅
𝐼𝑥𝑥
+ ̅𝐴 ) −
𝑦
=(
𝑦̅]
̅
𝐼𝑥𝑦
=
̅ sin 𝜃
𝐼𝑥𝑦
ℎ𝑐 ⁄sin 𝜃 )𝐴
ℎ𝑐 𝐴
̅
̅
𝐼𝑥𝑥
𝐼𝑥𝑥
= − ̅𝐴 = − ( ⁄
𝑦
ℎ𝑐 sin 𝜃 )𝐴
=−
̅ sin 𝜃
𝐼𝑥𝑥
ℎ𝑐 𝐴
Notice:
ℎ𝑐 here is the same ℎ𝐶𝐺 in the textbook: ℎ𝑐 = 𝑦̅ sin 𝜃
𝐼𝑥𝑥 is the moment of inertia w.r.t. horizontal centroidal axis, which is 𝐼𝑥𝑥 in the textbook
̅ is the product of inertia w.r.t. horizontal centroidal axis, which is −𝐼𝑥𝑦 in the textbook
𝐼𝑥𝑦
because the y axis has a different direction when using this definition: 𝐼𝑥𝑦 = ∫𝐴 𝑥𝑦𝑑𝐴
Therefore
𝑥CP,textbook = −
𝑦CP,textbook = −
𝐼𝑥𝑦 sin 𝜃
ℎ𝐶𝐺 𝐴
𝐼𝑥𝑥 sin 𝜃
ℎ𝐶𝐺 𝐴
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With the equations in the textbook, the origin
of the coordinates is defined at CG,
8ft
𝜃 = cos−1 10ft
ℎ𝐶𝐺 = 12ft;
𝐴 = width × length = 5ft × 10ft = 50ft 2
Therefore,
𝐹 = 𝛾ℎ𝐶𝐺 𝐴
The Gate is rectangular, from Fig. 2.13a
𝑏𝐿3
𝐼𝑥𝑦 = 0, 𝐼𝑥𝑥 =
12
From Eq. (2.29)
𝑥CP = −
𝑦CP = −
𝐼𝑥𝑦 sin 𝜃
ℎ𝐶𝐺 𝐴
𝐼𝑥𝑥 sin 𝜃
ℎ𝐶𝐺 𝐴
=0
=…
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6 Hydrostatic Forces on Curved Surfaces
F   pn dA
In general,
A
Horizontal Components:
Fx  F  i    pn  i dA
A
Fy   
Ay
p dAy
𝑑𝐴𝑥 = 𝐧 ∙ 𝐢𝑑𝐴 = projection of 𝐧𝑑𝐴
onto a plane perpendicular to x
direction
That is, the horizontal component of force acting on a curved surface is equal to the force
acting on a vertical projection of that surface (which includes both magnitude and line of
action) and can be determined by the methods developed for plane surfaces.
Fz   pn  k dA  
A
Az
p dAz    h dAz  
Az
Where h is the depth to any element area dA of the surface. That is, the vertical
component of force acting on a curved surface is equal to the net weight of the total
column of fluid directly above the curved surface and has a line of action through the
centroid of the fluid volume.
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Example Drum Gate
p  h  R1 cos  
n   sin  i  cos k
dA  lRd

F    pn dA    R1  cos   sin  i  cos  k  lRd



A
0  
p
dA
n

1



F  i  Fx   l R 2  1  cos   sin d   l R 2   cos  0  cos 2 0   2 l R 2
0
4


 
R2 Rl
p
 Same force as that on projection of gate onto vertical plane
A

Fz   l R 2
0

2


1  cos cosd   l R  sin    1 sin 2 
2 4

0
 R 2 
    Net weight of water above surface
 lR
 l 


2
 2 
1
1 



F1    2 R 2l  R 2l  / 2  lR 2  2    / 2
2
2 



Another approach:
1
1

F2    R 2l  F1 

F  F2  F1  R 2l
2
2

2

058:0160
Jianming Yang
Fall 2012
Chapter 2
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7 Hydrostatic Forces in Layered Fluids
Formulas for plane and curved surfaces apply separately to each layer: compute and sum
the separate layer forces and moments.
058:0160
Jianming Yang
Fall 2012
8 Buoyancy and Stability
8.1 Archimedes Principle
𝐹𝐵 = 𝐹𝑉(2) − 𝐹𝑉(1)
= fluid weight above 2ABC – fluid weight above 1ADC
= weight of fluid equivalent to body volume
In general,
𝐹𝐵 = 𝜌fluid 𝑔∀displaced
(∀displaced = displaced fluid volume).
The line of action is through the centroid of the displaced
volume, which is called the center of buoyancy.
Example: Oscillating floating block
Weight of the block 𝑊 = 𝜌𝑏 𝐿𝑏ℎ𝑔 = 𝑚𝑔 = 𝛾∀0 where
∀0 is displaced water volume by the block and 𝛾 is the
specific weight of the liquid, waterline area 𝐴𝑤𝑙 = 𝐿𝑏.
𝑊 = 𝐵 ⇒ 𝜌𝑏 𝐿𝑏ℎ𝑔 = 𝜌𝑤 𝐿𝑏𝑑𝑔
𝜌
⇒ 𝑑 = 𝑏⁄𝜌𝑤 ℎ
Instantaneous displaced water volume:
Chapter 2
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Jianming Yang
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Fall 2012
∀= ∀0 − 𝑦𝐴𝑤𝑙
∑ 𝐹𝑉 = 𝑚𝑦̈ = 𝐵 − 𝑊 = 𝛾∀ − 𝛾∀0 = −𝛾𝐴𝑤𝑙 𝑦
𝑚𝑦̈ + 𝛾𝐴𝑤𝑙 𝑦 = 0
𝑦̈ +
𝛾𝐴𝑤𝑙
𝑚
𝑦=0
Solution for this homogeneous linear 2nd-order ODE:
𝑦 = 𝐴cos𝜔𝑡 + 𝐵sin𝜔𝑡
Use initial condition (𝑡 = 0: 𝑦 = 𝑦0 , 𝑦̇ = 𝑦̇ 0 ) to determine 𝐴 and 𝐵:
𝑦 = 𝑦0 cos𝜔𝑡 + 𝑦𝜔̇ 0sin𝜔𝑡
Where the angular frequency
𝛾𝐴𝑤𝑙
𝜔=√
period
𝑚
𝑇=
2𝜋
𝜔
= 2𝜋√
𝑚
𝛾𝐴𝑤𝑙
Spar Buoy
We can increase period 𝑇 by increasing block mass 𝑚
and/or decreasing waterline area 𝐴𝑤𝑙 .
http://upload.wikimedia.org/wikipedia/com
mons/0/03/Lateral_view_of_spar-buoy.png
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8.2 Stability: Immersed Bodies
Stable
Neutral
Unstable
Condition for static equilibrium: (1) ∑Fv=0 and (2) ∑M=0
Condition (2) is met only when C and G coincide, otherwise we can have either a righting
moment (stable) or a heeling moment (unstable) when the body is heeled.
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Chapter 2
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8.3 Stability: Floating Bodies
For a floating body the situation is slightly more complicated since the center of
buoyancy will generally shift when the body is rotated, depending upon the shape of the
body and the position in which it is floating.
The center of buoyancy (centroid of the displaced volume) shifts laterally to the right for
the case shown because part of the original buoyant volume aOc is transferred to a new
buoyant volume bOd.
The point of intersection of the lines of action of the buoyant force before and after heel
is called the metacenter M and the distance GM is called the metacentric height.
If GM is positive, that is, if M is above G, then the ship is stable;
however, if GM is negative, then the ship is unstable.
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Consider a ship which has taken a small angle of heel 𝜃
1. evaluate the lateral displacement
of the center of buoyancy, 𝐵𝐵′
2. then from trigonometry, we can
solve for GM and evaluate the
stability of the ship
Recall that the center of buoyancy is
at the centroid of the displaced
volume of fluid (moment of volume
about y-axis – ship centerplane)
This can be evaluated conveniently as follows:
𝑥̅ ∀= ∫𝑐𝑂𝑑𝑒𝑎 𝑥𝑑∀ + ∫𝑂𝑏𝑑 𝑥𝑑∀ − ∫𝑐𝑂𝑎 𝑥𝑑∀ = 0 + ∫𝑂𝑏𝑑 𝑥 (𝐿𝑑𝐴) − ∫𝑐𝑂𝑎 𝑥(𝐿𝑑𝐴)
= 0 + ∫𝑂𝑏𝑑 𝑥𝐿(𝑥tan𝜃𝑑𝑥 ) − ∫𝑐𝑂𝑎 𝑥𝐿(−𝑥tan𝜃𝑑𝑥 )
= tan𝜃 ∫waterline 𝑥 2 𝑑𝐴waterline = 𝐼𝑂 tan𝜃
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∫𝑐𝑂𝑑𝑒𝑎 𝑥𝑑∀: moment of ∀ before heel (goes to zero due to symmetry of original buoyant
volume about centerplane)
𝑑𝐴 = 𝑦𝑑𝑥 = 𝑥tan𝜃𝑑𝑥
𝐼𝑂 = ∫𝑤𝑙 𝑥 2 𝑑𝐴𝑤𝑙 : area moment of inertia of ship waterline about its tilt axis 𝑂
𝑥̅ ∀= 𝐼𝑂 tan𝜃
𝐵𝐵′ = 𝑥̅ =
𝐵𝑀 =
𝑥̅
tan𝜃
=
𝐺𝑀 =
𝐼𝑂
∀
𝐼𝑂
∀
𝐼𝑂 tan𝜃
∀
= 𝐺𝑀 + 𝐵𝐺
− 𝐵𝐺
This equation is used to determine the
stability of floating bodies:
 If GM is positive, the body is stable
 If GM is negative, the body is unstable
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8.4 Roll
The rotation of a ship about the longitudinal
axis through the center of gravity.
Consider symmetrical ship heeled to a very
small angle θ. Solve for the subsequent
motion due only to hydrostatic and
gravitational forces.
𝐅𝐵 = (cos 𝜃 𝐣 − sin𝜃𝐢)𝜌𝑔∀
𝐌𝑔 = 𝐫 × 𝐅𝐵 = (−𝐺𝐵𝐣 + 𝐵𝐵′ 𝐢) × (cos 𝜃 𝐣 − sin𝜃𝐢)𝜌𝑔∀
= (−𝐺𝐵sin𝜃 + 𝐵𝐵′ cos 𝜃)𝜌𝑔∀𝐤 = (−𝐺𝐵sin𝜃 + 𝐵𝑀 tan 𝜃 cos 𝜃)𝜌𝑔∀𝐤
= (−𝐺𝐵 + 𝐵𝑀)sin𝜃𝜌𝑔∀𝐤 = 𝐺𝑀sin𝜃𝜌𝑔∀𝐤
Note: recall that 𝑀𝑂 = |𝐅|𝑑, where 𝑑 is the perpendicular distance from 𝑂 to the line of
action of 𝐅: 𝑀𝐺 = 𝐺𝑓𝜌𝑔∀= 𝐺𝑀sin𝜃𝜌𝑔∀
Angular momentum:
∑ 𝑀𝐺 = −𝐼𝜃̈
𝐼 = mass moment of inertia about long axis through 𝐺
𝜃̈= angular acceleration
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𝐼𝜃̈ + 𝐺𝑀sin𝜃𝜌𝑔∀= 0
̈ + 𝑚𝑔𝐺𝑀𝜃 = 0
For small 𝜃: 𝐼𝜃̈ + 𝐺𝑀𝜌𝑔∀𝜃 = 0 ⇒ 𝜃̈ + 𝐺𝑀𝜌𝑔∀
𝜃
=
0
⇒
𝜃
𝐼
𝐼
Definition of radius of gyration: 𝑘 = √𝐼⁄𝑚
𝑘 2 = 𝐼⁄𝑚 ⇒ 𝑚𝑘 2 = 𝐼 ⇒
𝑚𝑔𝐺𝑀
𝐼
=
𝑔𝐺𝑀
𝑘2
The solution to equation 𝜃̈ + 𝑔𝐺𝑀
𝜃 = 0 is,
𝑘2
𝜃(𝑡) = 𝜃0 cos 𝜔𝑡 +
𝜃̇0
sin 𝜔𝑡
𝜔
= 𝜃0 cos 𝜔𝑡
where 𝜃0 = the initial heel angle, 𝜃̇0 = 0 for no initial velocity, the natural frequency
𝑔𝐺𝑀
𝜔=√
𝑘2
=
√𝑔𝐺𝑀
𝑘
Simple (undamped) harmonic oscillation with period of the motion: 𝑇 =
2𝜋
𝜔
=
2𝜋𝑘
√𝑔𝐺𝑀
Note that large GM decreases the period of roll, which would make for an uncomfortable
boat ride (high frequency oscillation).
Earlier we found that GM should be positive if a ship is to have transverse stability and,
generally speaking, the stability is increased for larger positive GM. However, the
present example shows that one encounters a “design tradeoff” since large GM decreases
the period of roll, which makes for an uncomfortable ride.
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9 Case (2): Rigid Body Translation or Rotation
In rigid body motion, all particles are in combined translation and/or rotation and
there is no relative motion between particles; consequently, there are no strains or strain
rates and the viscous term drops out of the N-S equation.
∇𝑝 = 𝜌(𝐠 − 𝐚)
from which we see that ∇𝑝 acts in the direction of (𝐠 − 𝐚), and lines of constant pressure
must be perpendicular to this direction (by definition, ∇𝑓 is perpendicular to 𝑓 = const.).
For the general case of rigid body translation/rotation of fluid shown in the figure, if the
center of rotation is at 𝑂 where 𝐕 = 𝐕0 , the velocity of any arbitrary point 𝑃 is:
𝐕 = 𝐕0 + 𝛀 × 𝐫0
where 𝛀 = the angular velocity vector, and the acceleration is:
𝑑𝐕
𝑑𝑡
=𝐚=
First term =
Second term =
Third term =
𝑑𝐕0
𝑑𝑡
+ 𝛀 × (𝛀 × 𝐫0 ) +
𝑑𝛀
𝑑𝑡
× 𝐫0
acceleration of 𝑂
centripetal acceleration of 𝑃 relative to 𝑂
linear acceleration of 𝑃 due to 𝛀
Usually, all these terms are not present. In fact, fluids can rarely move in rigid body
motion unless restrained by confining walls for a long time.
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9.1 Uniform Linear Acceleration
∇𝑝 = 𝜌(𝐠 − 𝐚) = −𝜌[(𝑔 + 𝑎𝑧 )𝐤 + 𝑎𝑥 𝐢] = const.
𝜕𝑝
𝜕𝑥
1. 𝑎𝑥 < 0,
2. 𝑎𝑥 > 0,
= −𝜌𝑎𝑥
𝑝 increase in +𝑥
𝑝 decrease in +𝑥
𝜕𝑝
𝜕𝑧
= −𝜌(𝑔 + 𝑎𝑧 )
1. 𝑎𝑧 > 0,
𝑝 decrease in +𝑧
2. 𝑎𝑧 < 0 and |𝑎𝑧 | < 𝑔,
𝑝 decrease in +𝑧
3. 𝑎𝑧 < 0 and |𝑎𝑧 | > 𝑔,
𝑝 increase in +𝑧
Unit vector in the direction of ∇𝑝:
𝐬=|
∇𝑝
∇𝑝
=−
|
(𝑔+𝑎𝑧 )𝐤+𝑎𝑥 𝐢
1⁄2
[(𝑔+𝑎𝑧 )2 +𝑎𝑥2 ]
Lines of constant pressure are perpendicular to ∇𝑝.
Angle between the surface of constant pressure and the 𝑥 axes: 𝜃 = tan−1
𝑎𝑥
𝑔+𝑎𝑧
In general the rate of increase of pressure in the direction (𝐠 − 𝐚) is given by:
𝑑𝑝
𝑑𝑠
= ∇𝑝 ∙ 𝐬 = 𝜌[(𝑔 + 𝑎𝑧 )2 + 𝑎𝑥2 ]1⁄2 = 𝜌𝐺
𝑝 = 𝜌𝐺𝑠 + const. = 𝜌𝐺𝑠
gage pressure
.
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9.2 Rigid Body Rotation
Consider rotation of the fluid about the 𝑧 axis without any translation.
𝐚 = 𝛀 × (𝛀 × 𝐫0 ) = −𝑟Ω2 𝐢𝑟
∇𝑝 = 𝜌(𝐠 − 𝐚) = 𝜌(−𝑔𝒌 + 𝑟Ω2 𝐢𝑟 )
𝜕𝑝
𝜕𝑟
= 𝜌𝑟Ω2 ,
𝜕𝑝
𝜕𝑧
= −𝜌𝑔 = −𝛾
and
1
𝑝 = 𝜌𝑟 2 Ω2 + 𝑓(𝑧) + 𝑐
𝜕𝑝
2
= 𝑓 ′ = −𝛾 ⇒ 𝑝 = −𝛾𝑧 + 𝑓(𝑟) + 𝑐
𝜕𝑧
⇒ 𝑓(𝑧) = −𝛾𝑧
1 2 2
𝑝 = 𝜌𝑟 Ω − 𝛾𝑧 + const.
2
The constant is determined by specifying the pressure at one point; say,
𝑝 = 𝑝0 at (𝑟, 𝑧) = (0,0)
1
𝑝 = 𝑝0 − 𝛾𝑧 + 𝜌𝑟 2 Ω2 (Note: Pressure is linear in 𝑧 and parabolic in 𝑟)
2
Curves of constant pressure are given by:
𝑧=
𝑝0 −𝑝
𝛾
+
𝑟 2 Ω2
2𝑔
= 𝑎 + 𝑏𝑟 2
which are paraboloids of revolution, concave upward, with their minimum points on the
axis of rotation.
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The position of the free surface is found, as it is for linear acceleration, by conserving the
volume of fluid.
Unit vector in the direction of ∇𝑝:
𝐬=
∇𝑝
|∇𝑝|
=
−𝛾𝐤+𝜌𝑟Ω2 𝐢𝑟
− [ 2 ( 2)2 ]1⁄2
𝛾 + 𝜌𝑟Ω
𝑑𝑧
𝑔
Slope of 𝐬: tan 𝜃 =
𝑑𝑟
=
𝑟Ω2
.
(𝜃 is the angle between the surface of constant
pressure and the 𝑥 axis)
i.e.,
𝑟 = 𝐶1 exp (−
Ω2 𝑧
𝑔
)
is the equation of ∇𝑝 surfaces.
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10 Case (3): Pressure Distribution in Irrotational Flow
Potential flow solutions also solutions of NS under such conditions:
1. If viscous effects are neglected, Navier-Stokes equation becomes Euler equation:
𝜕𝐕
𝜌 ( + 𝐕 ∙ ∇𝐕) = 𝜌𝐠 − ∇𝑝
𝜕𝐕
𝜕𝑡
𝜕𝑡
1
+ ∇ ( 𝐕 ∙ 𝐕) − 𝐕 × (𝛁 × 𝐕) =
𝜌𝐠−∇𝑝
2
𝜌
1
= − ∇(𝑝 + 𝜌𝑔𝑧)
𝜌
1
Vector calculus identity: ∇ ( 𝐕 ∙ 𝐕) = 𝐕 ∙ ∇𝐕 + 𝐕 × (𝛁 × 𝐕)
2
2. If  = const.,
𝜕𝐕
𝜕𝑡
+ ∇(
3. Assume a steady flow:
𝑉2
2
𝜕
𝜕𝑡
𝑝
+ + 𝑔𝑧) = 𝐕 × 𝛚
𝜌
(𝑉 2 = 𝐕 ∙ 𝐕)
=0
∇(
𝑉2
2
𝑝
+ + 𝑔𝑧) = ∇𝐵 = 𝐕 × 𝛚
𝜌
Consider: ∇𝐵 perpendicular to 𝐵 = const., also 𝐕 × 𝛚 perpendicular to 𝐕 and 𝛚.
Stream lines 𝐞𝑠 : 𝐕 × 𝐞𝑠 = 0; vortex lines 𝐞𝑣 : 𝛚 × 𝐞𝑣 = 0
𝐞𝑠 ∙ ∇𝐵 =
𝜕𝐵
𝜕𝑠
= 𝐞𝑠 ∙ (𝐕 × 𝛚) = 𝛚 ∙ (𝐞𝑠 × 𝐕) = 0
𝐞𝑣 ∙ ∇𝐵 = 𝐞𝑣 ∙ (𝐕 × 𝛚) = 𝐕 ∙ (𝛚 × 𝐞𝑠 ) = 0
Therefore, 𝐵 =
𝑉2
2
𝑝
+ + 𝑔𝑧 = const. contains streamlines and vortex lines:
𝜌
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1. Assuming irrotational flow: 𝛚 = 0
∇𝐵 = 0
𝐵 = const. (everywhere same constant)
2. Unsteady irrotational flow
𝐕 = ∇𝜑
𝑉 2 = ∇𝜑 ∙ ∇𝜑
∇(
𝜕𝜑
𝜕𝑡
+
𝑉2
2
𝑝
+ + 𝑔𝑧) = 0
𝜌
𝜕𝜑 𝑉 2 𝑝
+
+ + 𝑔𝑧 = 𝐵(𝑡)
𝜕𝑡
2 𝜌
𝐵(𝑡) is a time-dependent constant.
Alternate derivation using streamline coordinates:
𝐕 = 𝑣𝑠 (𝑠, 𝑡)𝐞𝑠 + 𝑣𝑛 𝐞𝑛 = 𝑣𝑠 (𝑠, 𝑡)𝐞𝑠
𝜕
∇=
𝐚=
=[
𝐞𝑠 +
𝜕𝑠
𝐷𝐕
𝐷𝑡
𝜕𝑣𝑠
𝜕𝑡
=
𝜕
𝐞
𝜕𝑛 𝑛
𝜕𝐕
𝜕𝑡
+ 𝐕 ∙ ∇𝐕
𝐞𝑠 + 𝑣𝑠
𝜕𝐞𝑠
𝜕𝑡
] + 𝑣𝑠 [
𝜕𝑣𝑠
𝜕𝑠
𝐞𝑠 + 𝑣𝑠
𝜕𝐞𝑠
𝜕𝑠
]
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Time increment:
𝜕𝐞𝑠
Space increment:
𝐚=[
𝜕𝑣𝑠
𝜕𝑡
𝜕𝑣𝑛
−
𝜕𝑡
+ 𝑣𝑠
𝜕𝑠
𝜕𝑣𝑠
𝜕𝑠
=
𝜕𝜃
𝐞
𝜕𝑡 𝑛
𝜕𝜃
− 𝐞𝑛
𝜕𝑠
1
= − 𝐞𝑛
] 𝐞𝑠 + [−𝑣𝑠
𝑅
𝜕𝜃
𝜕𝑡
−
𝑣𝑠2
𝑅
] 𝐞𝑛
: local 𝑎𝑠 in the direction of flow
𝜕𝑡
𝑣𝑠
𝜕𝑣𝑠
𝜕𝑡
𝜕𝐞𝑠
=−
= −𝑣𝑠
𝜕𝑣𝑠
𝜕𝑠
𝑣𝑠2
𝑅
𝜕𝜃
𝜕𝑡
: local 𝑎𝑛 normal to the direction of flow
: convective 𝑎𝑠 due to convergence/divergence of streamlines
: normal 𝑎𝑛 due to streamline curvature
Euler Equation: 𝜌𝐚 = −∇(𝑝 + 𝜌𝑔𝑧)
𝜕𝑣
Steady flow 𝑠-direction equation: 𝜌𝑣𝑠 𝑠 = −
𝜕𝑠
𝜕
𝜕𝑠
(
𝑣𝑠2
2
𝜕
𝜕𝑠
(𝑝 + 𝜌𝑔𝑧)
𝑝
+ + 𝑔𝑧) = 0, i.e., B=const. along streamline
𝜌
Steady flow 𝑛-direction equation: −𝜌
−∫
𝑣𝑠2
𝑅
𝑝
𝑣𝑠2
𝑅
=−
𝜕
𝜕𝑛
(𝑝 + 𝜌𝑔𝑧)
𝑑𝑛 + + 𝑔𝑧 = const. across streamline
𝜌
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11 Flow Patterns: Streamlines, Streaklines, Pathlines
1.) A Streamline is a curve everywhere tangent to the local velocity vector at a given
instant. Instantaneous lines; convinent to compute mathematically.
2.) A Pathline is the actual path traveled by an individual fluid particle over some time
period. Generated as the passage of time; convinent to generate experimentally.
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3.) A Streakline is the locus of particles that have earlier passed through a prescribed
point. Generated as the passage of time; convinent to generate experimentally.
4.) A Timeline is a set of fluid particles that form a line at a given instant.
Instantaneous lines; convinent to generate experimentally.
Note:
1. Streamlines, pathlines, and streaklines are identical in a steady flow.
2. For unsteady flow, streamline pattern changes with time, whereas pathlines and
streaklines are generated as the passage of time.
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11.1
Streamline
By definition we must have 𝐕 × 𝑑𝐫 = 0 which upon expansion yields the equation of the
streamlines for a given time 𝑡 = 𝑡1
𝑑𝑥
𝑢
=
𝑑𝑦
𝑣
=
𝑑𝑧
𝑤
= 𝑑𝑠
𝑠 = integration parameter
So if (𝑢, 𝑣, 𝑤) is known, integrate with respect to 𝑠 for 𝑡 = 𝑡1 with initial condition
(𝑥0 , 𝑦0 , 𝑧0 , 𝑡0 ) at 𝑠 = 0 and then eliminate 𝑠.
11.2
Pathline
The pathline is defined by integration of the relationship between velocity and
displacement.
𝑑𝑥
𝑑𝑡
=𝑢
𝑑𝑦
𝑑𝑡
=𝑣
𝑑𝑧
𝑑𝑡
=𝑤
Integrate 𝑢, 𝑣, 𝑤 with respect to 𝑡 using initial condition (𝑥0 , 𝑦0 , 𝑧0 , 𝑡0 ), then eliminate 𝑡.
11.3
Streakline
To find the streakline, use the integrated result for the pathline retaining time as a
parameter. Now, find the integration constant which causes the pathline to pass through
(𝑥0 , 𝑦0 , 𝑧0 ) for a sequence of times 𝜏 < 𝑡. Then eliminate 𝜏.
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Example: an idealized velocity distribution is given by:
𝑢=
𝑥
1+𝑡
𝑣=
𝑦
𝑤=0
1+2𝑡
calculate and plot: 1) the streamlines 2) the pathlines 3) the streaklines which pass
through (𝑥0 , 𝑦0 , 𝑧0 ) at 𝑡 = 0.
1. First, note that since 𝑤 = 0 there is no motion in the 𝑧 direction and the flow is 2-D
𝑑𝑥
𝑥
𝑑𝑦
𝑦
=𝑢=
=𝑣=
𝑑𝑠
1+𝑡
𝑠
𝑑𝑠
𝑥 = 𝐶1 exp( )
1+𝑡
𝑠 = 0 at (𝑥0 , 𝑦0 ):
1+2𝑡
𝑠
𝑦 = 𝐶2 exp(
)
1+2𝑡
𝐶1 = 𝑥0 𝐶2 = 𝑦0
and eliminating 𝑠:
𝑠 = (1 + 𝑡)ln
𝑥
𝑥0
𝑛
𝑥
= (1 + 2𝑡)ln
𝑦 = 𝑦0 ( ) where 𝑛 =
𝑥0
𝑦
𝑦0
1+𝑡
1+2𝑡
This is the equation of the streamlines which pass through (𝑥0 , 𝑦0 ) for all times 𝑡.
𝑦
𝑡 = 0,
𝑡 = ∞,
𝑦0
𝑦
𝑦0
=
𝑥
𝑥0
𝑥 1⁄2
=( )
𝑥0
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2. To find the pathlines we integrate
𝑑𝑥
𝑑𝑡
=𝑢=
𝑥
𝑑𝑦
1+𝑡
𝑑𝑡
=𝑣=
𝑦
1+2𝑡
𝑐
𝑐
𝑥 = 𝐶1 (1 + 𝑡) 𝑦 = 𝐶2 (1 + 2𝑡)1⁄2 (∫
𝑑𝑥 = 𝑙𝑛|𝑎𝑥 + 𝑏| + 𝐶)
𝑎𝑥+𝑏
𝑎
𝑡 = 0 (𝑥, 𝑦) = (𝑥0 , 𝑦0 ): 𝐶1 = 𝑥0 𝐶2 = 𝑦0
now eliminate 𝑡 between the equations for (𝑥, 𝑦)
𝑦 = 𝑦0 [1 + 2 (
𝑥
𝑥0
− 1)]
1⁄2
This is the pathline through (𝑥0 , 𝑦0 ) at 𝑡 = 0 and does not coincide with the streamline
at 𝑡 = 0.
3. To find the streakline, we use the pathline equations to find the family of particles
that have passed through the point (𝑥0 , 𝑦0 ) for all times 𝜏 < 𝑡.
𝑥 = 𝐶1 (1 + 𝑡) 𝑦 = 𝐶2 (1 + 2𝑡)1⁄2
𝑥0 = 𝐶1 (1 + 𝜏) 𝑦0 = 𝐶2 (1 + 2𝜏)1⁄2
𝐶1 =
𝑥=
𝑥0
1+𝜏
𝑥0
1+𝜏
(1 + 𝑡)
𝜏 = (1 + 𝑡)
𝑥0
𝑥
𝑦0
1+2𝜏)1⁄2
𝑦
= ( 0)1⁄2 (1 + 2𝑡)1⁄2
1+2𝜏
1
𝑦0 2
𝐶2 = (
𝑦
− 1 = [(1 + 2𝑡) ( ) − 1]
2
𝑦
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𝑦0 2
( ) =
𝑦
𝑡 = 0:
1+2𝑡
𝑥
1+2[(1+𝑡)( 0 )−1]
𝑥
𝑦
𝑦0
= [1 + 2 (
𝑥0
𝑥
− 1)]
−1⁄2
The streakline does not coincide with either the equivalent streamline or pathline.
Physically, the streakline reflects the streamline behavior before the specified time 𝑡 =
0, while the pathline reflects the streamline behavior after 𝑡 = 0.
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11.4
Stream Function
The stream function is a powerful tool for 2D flows in which 𝐕 is obtained by
differentiation of a scalar 𝛹 which automatically satisfies the continuity equation.
Continuity equation:
say: 𝑢 =
𝜕
then:
𝐕=
𝜕𝑥
𝜕𝛹
𝜕𝑦
𝜕𝛹
(
𝜕𝛹
)+
𝜕𝑦
𝜕𝛹
𝜕𝑥
𝜕
𝜕𝑦
+
𝜕𝑣
=0
𝜕𝑥
𝜕𝑦
𝜕𝛹
𝑣=−
𝜕𝑦
𝐢−
𝜕𝑢
(−
𝜕𝑥
𝜕𝛹
𝜕𝑥
)=
𝜕2 𝛹
𝜕𝑥𝜕𝑦
−
𝜕2 𝛹
𝜕𝑥𝜕𝑦
=0
𝐣 ⇒ ∇ × 𝐕 = 𝛚 ⇒ 𝜔𝑧 = 𝜔 = −∇2 𝛹
2D vorticity transport equation:
𝜕𝜔
𝜕𝑡
+𝑢
𝜕𝜔
𝜕𝑥
+𝑣
𝜕𝜔
𝜕𝑦
= 𝜈∇2 𝜔
Replace 𝑢, 𝑣, 𝜔:
𝜕
𝜕𝑡
2
(∇ 𝛹) +
𝜕𝛹 𝜕
(∇ 𝛹) −
𝜕𝑦 𝜕𝑥
𝜕𝛹 𝜕
Steady flow:
2
𝜕𝑦 𝜕𝑥
𝜕𝛹 𝜕
4
(∇ 𝛹) = 𝜈∇ 𝛹
𝜕𝑥 𝜕𝑦
𝜕𝛹 𝜕
(∇2 𝛹) −
2
𝜕𝑥 𝜕𝑦
4
(∇ 𝛹 =
𝜕4 𝛹
𝜕𝑥 4
+2
𝜕4 𝛹
𝜕𝑥 2 𝜕𝑦 2
(∇2 𝛹) = 𝜈∇4 𝛹
It is a single fourth-order scalar equation, which requires 4 boundary conditions
At infinity: 𝑢 = 𝜕𝛹/𝜕𝑦 = 𝑈∞ 𝑣 = −𝜕𝛹/𝜕𝑥 = 0
On body: 𝑢 = 𝑣 = 0 = 𝜕𝛹/𝜕𝑦 = −𝜕𝛹/𝜕𝑥
+
𝜕4 𝛹
𝜕𝑦 4
)
058:0160
Jianming Yang
Chapter 2
38
Fall 2012
11.4.1 Irrotational Flow
∇×𝐕=0
⇒ ∇2 𝛹 = 0
Second-order linear Laplace equation
At infinity 𝑆∞ : 𝛹 = 𝑈∞ 𝑦 + const.
On body 𝑆𝐵 : 𝛹 = const.
11.4.2 Geometric Interpretation of 𝜳
Besides its importance mathematically 𝛹 also has
important geometric significance.
𝛹 = const. = streamline
Recall definition of a streamline:
𝐕 × 𝑑𝐫 = 0
𝑑𝑥
𝑢
=
𝑑𝑦
𝑣
⇒
𝑢𝑑𝑦 − 𝑣𝑑𝑥 = 0
Compare with 𝑑𝛹 =
𝜕𝛹
𝜕𝑥
𝑑𝑥 +
𝜕𝛹
𝜕𝑦
𝑑𝑦 = −𝑣𝑑𝑥 + 𝑢𝑑𝑦
i.e., 𝑑𝛹 = 0 along a streamline
Or 𝛹 = const. along a streamline and curves of constant 𝛹 are the flow streamlines. If
we know 𝛹(𝑥, 𝑦) then we can plot 𝛹 = const. curves to show streamlines.
058:0160
Jianming Yang
Chapter 2
39
Fall 2012
11.4.3 Physical Interpretation
𝑛𝑥 =
𝑑𝑦
𝑑𝑠
𝑛𝑦 = −
𝑑𝑥
𝑑𝑠
𝜕𝛹
𝜕𝛹
𝑑𝑦
𝑑𝑥
𝜕𝛹
𝜕𝛹
𝑑𝑄 = 𝐕 ∙ 𝐧𝑑𝐴 = (
𝐢−
𝐣) ∙ ( 𝐢 −
𝐣) 𝑑(𝑠 ∙ 1) =
𝑑𝑦 +
𝑑𝑥 = 𝑑𝛹
𝜕𝑦
𝜕𝑥
𝑑𝑠
𝑑𝑠
𝜕𝑦
𝜕𝑥
i.e. change in 𝑑𝛹 is the volume flux and along a streamline 𝑑𝑄 = 0.
Consider flow between two streamlines
𝐵
𝐵
𝑄𝐴𝐵 = ∫𝐴 𝐕 ∙ 𝐧𝑑𝐴 = ∫𝐴 𝑑𝛹 = 𝛹𝐵 − 𝛹𝐴
058:0160
Jianming Yang
Chapter 2
40
Fall 2012
11.4.4 Incompressible Plane Flow in Polar Coordinates
Continuity equation:
or
𝜕
𝜕𝑟
(𝑟𝑣𝑟 ) +
say: 𝑣𝑟 =
then:
𝜕
𝜕𝑟
𝜕𝑣𝜃
𝜕𝜃
1 𝜕𝛹
𝑟 𝜕𝜃
1 𝜕𝛹
(𝑟
𝑟 𝜕𝜃
1 𝜕
𝑟 𝜕𝑟
(𝑟𝑣𝑟 ) +
𝑟 𝜕𝜃
=0
=0
𝑣𝜃 = −
)+
1 𝜕𝑣𝜃
𝜕
𝜕𝜃
(−
𝜕𝛹
𝜕𝑟
𝜕𝛹
𝜕𝑟
)=0
As before 𝑑𝛹 = 0 along a streamline and
𝑑𝑄 = 𝑑𝛹
Volume flux = change in stream function
11.4.5 Incompressible Axisymmetric Flow
Continuity equation:
say: 𝑣𝑟 = −
then:
1 𝜕
𝑟 𝜕𝑟
(𝑟
1 𝜕𝛹
𝑟 𝜕𝑧
−1 𝜕𝛹
𝑟 𝜕𝑧
1 𝜕
𝑟 𝜕𝑟
𝑣𝑧 =
)+
(𝑟𝑣𝑟 ) +
𝜕𝑣𝑧
𝜕𝑧
=0
1 𝜕𝛹
𝑟 𝜕𝑟
𝜕 1 𝜕𝛹
(
𝜕𝑧 𝑟 𝜕𝑟
)=0
As before 𝑑𝛹 = 0 along a streamline and
𝑑𝑄 = 𝑑𝛹
Volume flux = change in stream function
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Jianming Yang
Chapter 2
41
Fall 2012
11.4.6 Generalization to Steady Plane Compressible Flow
In steady compressible flow, the continuity equation is
𝜕𝜌𝑢
𝜕𝑥
+
𝜕𝜌𝑣
𝜕𝑦
Define: 𝜌𝑢 =
=0
𝜕𝛹
𝜕𝑦
𝜌𝑣 = −
𝜕𝛹
𝜕𝑥
Streamline: 𝑢𝑑𝑦 − 𝑣𝑑𝑥 = 0
Compare with
𝑑𝛹 =
𝜕𝛹
𝜕𝑥
1 𝜕𝛹
𝜌 𝜕𝑦
𝑑𝑥 +
𝑑𝑦 +
𝜕𝛹
𝜕𝑦
1 𝜕𝛹
𝜌 𝜕𝑥
𝑑𝑦 ⇒
𝑑𝑥 = 0
1
𝜌
(𝑑𝛹) = 0 ,
i.e., 𝑑𝛹 = 0 and 𝛹 = const. is a streamline.
Now: 𝑑𝑚̇ = 𝜌(𝐕 ∙ 𝐧)𝑑𝐴 = 𝑑𝛹
𝐵
𝑑𝑚̇𝐴𝐵 = ∫𝐴 𝜌(𝐕 ∙ 𝐧)𝑑𝐴 = 𝛹𝐵 − 𝛹𝐴
Change in 𝛹 is equivalent to the mass flux.