Lecture 7
Integral budgets: mass and momentum
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Motivation
• Statics problems basically require only the density of
the fluid and knowledge of the free surface position
• Most flow problems require the analysis of an
arbitrary state of variable fluid motion defined by the
geometry, boundary conditions and the law of
mechanics
• Three basic approaches to the analysis of arbitrary
flow problems:
1. Control volume, or large-scale, analysis (Chap. 3).
2. Differential, or small-scale, analysis (Chap. 4).
3. Experimental, or dimensional, analysis (Chap. 5).
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Motivation
• In analyzing fluid motion, we might take one of
two paths:
• (1) seeking to describe the detailed flow pattern at
every point (x, y, z) in the field (“differential”
approach) .
• (2) working with a finite region, making a balance of
flow in versus flow out, and determining gross flow
effects such as the force or torque on a body or the
total energy exchange (“control volume” method).
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Aim: Provide tools to perform simple
integral calculations
• E.g. Compute the force exerted by a water jet on
a plate, as in the figure
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System versus control volumes
• All the laws of mechanics are written for a system
• System is defined as an arbitrary quantity of mass of fixed
identity
• Everything external to this system is denoted by the term
surroundings (“vizinhanças”), and the system is
separated from its surroundings by its boundaries
(”fronteiras”)
• The laws of mechanics then state what happens when
there is an interaction between the system and its
surroundings
• In order to convert a system analysis into a controlvolume analysis we must convert our mathematics to
apply to a specific region rather than to individual
masses (Reynolsd transport theorem).
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Laws of mechanics
• First, the system is a fixed quantity of mass, denoted by m.
Thus the mass of the system is conserved and does not
change (conservation of mass):
Msys=const or 𝑑𝑚 𝑑𝑡 = 0
• Second, if the surroundings exert a net force F on the system,
Newton’s second law states that the mass in the system will
begin to accelerate:
𝐹 = 𝑚𝑎 = 𝑚
𝑑𝑉
𝑑𝑡
=
𝑑
𝑑𝑡
𝑚𝑉
• Third, if the surroundings exert a net moment M about the
center of mass of the system, there will be a rotation effect:
𝑑𝐻
𝑀=
𝑑𝑡
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Laws of mechanics
• Fourth, if heat dQ is added to the system or work dW is done
by the system, the system energy dE must change according
to the energy relation, or first law of thermodynamics:
𝑑𝑄 𝑑𝑊 𝑑𝐸
𝑑𝑄 − 𝑑𝑊 = 𝑑𝐸 𝑜𝑟
−
=
𝑑𝑡
𝑑𝑡
𝑑𝑡
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System versus control volumes
• In analyzing a control volume, we convert the
system laws to apply to a specific region, which
the system may occupy for only an instant
• The basic laws are reformulated to apply to this
local region called a control volume
• All we need to know is the flow field in this
region, and often simple assumptions will be
accurate enough (such as uniform inlet and/or
outlet flows)
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Volume and Mass Rate of Flow
• All the control volume analyses involve evaluation of
the volume flow Q (“caudal volúmico”) or mass flow
𝑚 (“caudal mássico”) passing through a surface
(imaginary) defined in the flow.
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Volume and Mass Rate of Flow
• How much
in unit time?
volume
of
fluid
passes
through
S
• the amount of fluid swept through dA in time dt is the volume of the
slanted parallelepiped in (b)
𝑑𝑣𝑜𝑙 = 𝑉𝑑𝑡𝑑𝐴 cos 𝜃 = 𝑉. 𝑛 𝑑𝐴𝑑𝑡
• The integral of dVol/dt is the total volume rate of flow Q through the
surface S:
𝑄=
𝑆
𝑉. 𝑛 𝑑𝐴 =
𝑆
Volume flow can be multiplied
by density to obtain the mass
flow:
𝑚=
𝑠
𝜌 𝑉. 𝑛 𝑑𝐴
If =const𝑚 = 𝜌𝑄
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𝑉𝑛 𝑑𝐴
{the rate of accumulation of a property inside a control volume}
= {what flows in – what flows out} + {production – destruction}
• In case of mass there is no production or destruction,
and the rate of accumulation is due only to advective
transport (conservation of mass):
• Msys=const or 𝑑𝑚 𝑑𝑡 = 0
• If the fluid is incompressible, density is constant and
consequently it is not possible to accumulate mass
inside a rigid volume and outflow has to balance
inflow.
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What is production - destruction
• Depends on the property considered.
• In case of organisms it is growth minus
respiration/excretion
• In case of momentum it is the result of forces
• In case of kinetic energy it is the result of work
supplied
• ….
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Reynolds transport theorem
• The changing rate of a property inside a control
volume occupied by the fluid is equal to the changing
rate inside the material system located inside the
control plus what is flowing in, minus what is flowing
out.

t
d
  dVol  dt
CV
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 
  dVol    v .n dS
system
CS
Demonstration of Reynolds theorem
Let’s consider a conduct and 3
portions fluid (systems), SYS 1,
SYS2 and SYS 3 that are moving.
Time = t
SYS 1
CV
SYS 2
SYS 3
At time t+t the fluid SYS 2 has
moved slightly to the right and
fluid SYS 1 has also moved to the
right.
Time = t+∆t
SYS 1
CV SYS 2
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Let’s consider a space control
volume (not moving -fixed) that at
time “t” is completely filled by the
fluid SYS 2 (coincident with SYS 2).
SYS 3
Between time= t and time =(t+∆t)
inside the control volume
properties can change because
some fluid flew in (SYS1) and
other fluid flew out (SYS2) and
the properties of those systems
have changed in time.
Demonstration of Reynolds theorem
Rates of change
• Let B be any property of the fluid (energy, momentum, enthalpy,
𝑑𝐵
etc.) and and let 𝛽 =
be the intensive value, or the amount of B
𝑑𝑚
per unit mass in any small element of the fluid.
• In a material system:
•
𝐵𝑠𝑦𝑠 𝑡+∆𝑡 −𝐵𝑠𝑦𝑠 𝑡
∆𝑡
𝐵𝐶𝑉 𝑡+∆𝑡 −𝐵𝐶𝑉 𝑡
Inside the control volume:
∆𝑡
• SYS 2 was coincident with CV at time t:
𝐵𝐶𝑉 𝑡 = 𝐵𝑠𝑦𝑠2 𝑡
• At time t+t:
𝐵𝐶𝑉 𝑡 + ∆𝑡 = 𝐵𝑠𝑦𝑠2 𝑡 + ∆𝑡 +
+ 𝑖𝑛𝑓𝑙𝑜𝑤 − 𝑜𝑢𝑡𝑓𝑙𝑜𝑤
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SYS 1
SYS 1
CV
SYS 2
SYS 3
CV SYS 2
SYS 3
Demonstration of Reynolds theorem
• Computing the budget per unit of time and using the
specific property (per unit of volume)
𝐵𝐶𝑉 𝑡+∆𝑡 −𝐵𝐶𝑉 𝑡
∆𝑡
=
𝐵𝑠𝑦𝑠2 𝑡+∆𝑡 −𝐵𝑠𝑦𝑠2 𝑡
∆𝑡
− 𝑜𝑢𝑡𝑓𝑙𝑜𝑤
𝑑𝐵
𝛽=
𝑑𝑉
𝐵𝐶𝑉 𝑡 + ∆𝑡 − 𝐵𝐶𝑉 𝑡
=
∆𝑡
𝜕
=
𝛽𝑑𝑉
𝜕𝑡 𝐶𝑉
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𝐵=
𝐶𝑉
𝛽𝑑𝑉
+ 𝑖𝑛𝑓𝑙𝑜𝑤 −
𝛽𝑑𝑉
𝑡+∆𝑡
−
∆𝑡
𝐶𝑉
𝛽𝑑𝑉
𝑡
Demonstration of Reynolds theorem
• Identically for the material System
𝑑𝐵
𝛽=
𝐵=
𝛽𝑑𝑉
𝑑𝑉
𝐵𝑠𝑦𝑠2 𝑡 + ∆𝑡 − 𝐵𝑠𝑦𝑠2 𝑡
𝑑
=
∆𝑡
𝑑𝑡
𝑖𝑛𝑓𝑙𝑜𝑤 − 𝑜𝑢𝑡𝑓𝑙𝑜𝑤 = −
𝛽𝑑𝑉
𝑠𝑦𝑠2
𝛽 𝑣. 𝑛 𝑑𝐴
• If material is flowing in, the internal product is
negative and if is flowing out is positive
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Reynolds transport theorem – physical
interpretation

t
d
  dV  dt
cv
 
  v .n dA
  dV 
system
surface
Or:
d
dt

  dV  t
system
1
  dV 
cv
2
 
  v .n dA
surface
3
1-time rate of change of an arbitraryextensive parameter of a system
(rate of change of mass, momentum...)
2 –rate of change of B within the control volume as the fluid flows
through it
3- net flowrate of the parameter B across the entire control surface
(may be negative, zero or positive depending on the particular
situation)
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If the Volume is infinitesimal

t
  dV 
cv
d
dt
  dV 
system
 


v
 .n dA
surface
Becomes:
 
 

d
 V    V    v .n Aentrance   v .n Aexit
t
dt
But:
u k
d (  V )
d ( )
d ( V )
d ( )
 V

 V
  V
dt
dt
dt
dt
x k
And thus:
u

d  
 x 1x 2 x 3
  x 1x 2 x 3 k 
t
dt
x k
 x 2  x 3  v 1  x 1   x 2  x 3  v 1  x   x 
1
1
x 1x 2 x 3
x 1x 3  v 2 x 2  x 1x 3  v 2 x
x 1x 2  v 3 x 3  x 1x 2  v 3 x
Dividing by the volume (x1x2x3):
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2   21
3  x 3

Reynolds transport theorem
•
𝜕𝛽
𝜕𝑡
•
𝑑𝛽
𝑑𝑡
•
𝑑𝛽
𝑑𝑡
•
The Total derivative is the rate of change in a material system (Lagrangian
description) ;
The Partial derivative is the rate of change in a control volume (eulerian
description) ;
The advective derivative account for the transport by the velocity.
•
•
=
𝑑𝛽
𝑑𝑡
=
𝜕𝛽
𝜕𝑡
=
𝜕𝛽
𝜕𝑡
+
+
+
𝜕𝑣𝑘
𝛽
𝜕𝑥𝑘
𝜕𝛽𝑣𝑗
𝜕𝑥𝑗
−
−
𝜕
𝜕𝑥𝑗
𝛽𝑣𝑗
𝜕𝑣𝑘
𝛽
𝜕𝑥𝑘
𝜕𝛽
𝑣𝑗
𝜕𝑥𝑗
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Momentum conservation
{the rate of accumulation of a property inside a control volume}
= {what flows in – what flows out} + {production – destruction}

t
d
  dV  dt
cv
  dV 
system
 
  v .n dA
surface
Production is the result of the forces applied in the sense of the
flow and destruction of the forces in the opposite sense.
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Momentum conservation
• Forces applied to a volume of fluid (momentum
production/consumption)
 yx y dy
 u 

  
 y  y  dy
Forças superficie – pressão (força
normal) e o atrito (força tangencial)
Forças mássicas – peso e força
electromagnética
 xy y
 u 

  
 y  y
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Peso  g dxdydz
Integral momentum budget
• A taxa de variação da quantidade de movimento pode então
ser calculada através do balanço da quantidade de
movimento que entra/sai e das forças aplicadas ao volume de
controlo:

t
 u i dV

cv

p




u
u
n
dA


n
dA


 i j j
 ji j
  x
surface
surface

i

 g i dVol

• Se o escoamento for estacionário:
 p


0    u i u j n j dA    ji n j dA    
 g i dVol
 x i

surface
surface
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Integral momentum budget
• Se a velocidade for uniforme na entrada e saída do volume
(funções integradas podem sair dos integrais)e se o fluido for
incompressível (𝜕𝜌 𝜕𝑡 = 0):
 u i u j n j dA   U i Q   U i Q
Outlets
surface
Intlets
 g i dVol  Weight
 p
   x
i


dVol 

  pni dA
(Gradient Theorem)
surface
• If p is uniform along inlet and outlet:
  pni dA 
surface
  pni A j  FP i
inlet  outlet
• And finally we get:
 U i Q   U i Q 
Outlets
Intlets
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  pni A j  FPi  FVi  g Vol
In Outlet
i
Integral Momentum Budget
 U i Q   U i Q 
Outlets
Intlets
  pni A j  Fi  g Vol
In Outlet
i
• Where Fi is the summation of pressure forces other than
those acting at inlet and outlet plus the summation of
the friction forces.
• This is an algebraic equation applicable if:
• the flow is stationary and incompressible
• the velocity and pressure are uniform at each inlet and
outlet
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Example 1
• Calculate the force exerted by the fluid
over the deflector neglecting friction
V=2 m/s and jet radius is 2 cm and theta
is 45.
 U i Q   U i Q 
Outlets
•
•
•
•
Intlets
  pni A j  Fi  g Vol
We have a flow with an inlet and an
outlet.
Velocity has a component at the inlet
and two at the outlet
Pressure is atmospheric at inlet and
outlets and thus the velocity modulus
remains constant.
We have to compute budgets along
both directions x and y.
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i
In Outlet
D=
2Cm
A=
0.000315m2
V=
2m/s
Q=
0.000629m3/s
Theta 45º
1.26N
0.88971N
0.88971N
Fx=
-0.37N
Fy=
0.88971N
F=
0.963015N
Example 2
• The Jet is hitting the surface
perpendicularly (Vj=3m/s), but
the
surface
is
moving
(Vc=1m/s). D=10 cm. Calculate
the force and power supplied.
 U i Q   U i Q 
Outlets
Intlets
  pni A j  Fi  g Vol
i
In Outlet
 



u
u
 i .n dA   U i Q   U i Q
Outlets
surface
•
Intlets
If the control volume is moving, fluxes depend on the flow velocity relative to the control volume:
 

 
u j u.n  dA 
surface
 U j Q   U j Q 
Outlets
Intlets
 U j U j  U c A j   U j U j  U c A j
Outlets
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Intlets
Relative or absolute reference
 
 u i u.n dA 
surface
but
 U  UC A j
Outlets

 U i Q   U i Q
Outlets
Intlets
 U  UC A j
 QR
Intlets
So
 



u
u
 i .n dA   U i QR   U i QR
surface
Outlets

Intlets
 U R i QR   U R i QR   UC i QR   UC i QR
Outlets
Intlets
Outlets
Intlets
• Discharge must be computed using relative velocity.
• Transported velocity can be the relative or the absolute velocity.
Usually the relative velocity is more intuitive.
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Symmetrical
 U i Q   U i Q 
Outlets
Example 2
Intlets
Vc=
D=
  pni A j  FPi  FVi  g Vol
i
In Outlet
1m/s
10cm
Vj=
3m/s
A=
0.007864m2
VRj=
2
Q=
0.015728m3/s
Theta
90
31.46N
0N
0Symmetrical
Fx=
-31.46N
Fy=
0N
F=
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31.456N
Summary
• The integral (algebraic) momentum equation
describes the momentum conservation principle
(Newton law) assuming simplified solutions for
momentum flux calculations.
• It is useful when flow is stationary and
incompressible.
• Becomes more useful when associated to the
Bernoulli Energy Conservation Equation.
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