Integral Momentum Theorem
We can learn a great deal about the overall behavior of propulsion systems using the
integral form of the momentum equation. The equation is the same as that used in fluid
mechanics and in air-breathing propulsion.
The momentum theorem is an expression of Newton’s 2nd Law:


 F  dt mV 
d
1
We want to look at this expression in a form that is most relevant to propulsion studies.
To do so, consider the two coordinate systems shown in the figure below:
(1) Inertial Coordinate System (labeled
with subscript ‘I’ in the figure.
(2) Fixed to the Vehicle (labeled with
subscript ‘V’ in the figure). This
coordinate system moves with a
velocity uo relative to the inertial
coordinate system. All velocities
relative to the vehicle-fixed coordinate
frame are denoted by u.
ZV
uo
ZI
XV
XI
YV
YI
Newton’s 2nd Law for a control volume of fixed mass can be expressed as:



Du o  u 
 F    Dt dV
V
2
We can also write this as:

F
=

ao  dV
+
V
All external forces
on control volume
(pressure forces,
shear forces, body
forces)
Force due to change
in inertia for
accelerating vehicle

Du
  Dt dV
V
Change in
momentum of mass
in control volume
3

 Fo
We can explain this equation further, by considering a simple example of a falling block
and examining the application of two different control volumes. The falling block on the
left has a control volume fixed to it. In this case, the first term on the right hand side of
the equation is nonzero since the control volume is accelerating relative to an intertial
1
reference frame. The second term is zero because the block is not accelerating relative to
a coordinate system fixed to the control volume. The opposite is true for the falling block
on the left, which is falling within a fixed control volume. The first term of the above
equation is zero in this case because the control volume is not accelerating relative to an
inertial reference frame. The second term is nonzero because the block is moving relative
to a coordinate system fixed to the control volume.

 mao  0

 ma

 
Du
F

a

dV


dV

o

Dt
V
V

Du
 0m
Dt

 ma
As expected, the result is the same for both. The integral momentum equation reduces to
the familiar form, F=ma.
Equation 3 is exact and completely valid for propulsion system applications. What we
want to do now, and this is what is often lost in many derivations, is to make it easier to
use for propulsion system applications. To do so, we’ll rearrange some of the terms in
equation 3 and put them in a useful form. What follows is a lot of vector algebra and
elementary vector calculus. Please keep in mind as you go through this what the goal is:
simplify the equation for utility to propulsion applications. It is still F=ma, just in a
different form, and knowing what the expressions of this alternate form are important.
To continue, equation 3 can be written as follows:

 

 u 
 F  Fo     t  u  u dV
V


     
 u
  
  u   u  u
 u dV

t

t
t 

V


   
 u 
 
  u   u  u dV
t
t 
V 
2
3.a
3.b
3.c
Why did we do all this? In this form we can apply conservation of mass, in differential
form. The general form of conservation of mass (for incompressible, compressible, nonconstant density flows) can be written as:


    u   0
t
  

u
 u    u   0
t
4.a
4.b
In equation 4.b, we just multiplied the mass conservation equation by u to match the form
of equation 3.c. We can now substitute to arrive at:

 



 u  
 F  Fo    t  u  u    u  u dV
V
5
Remember that this is a vector equation. Most propulsion problems that we will
encounter only need to consider one-dimension (along say the rocket’s primary line of
travel), so examine the x-component of equation 5:
F
x


 u x 

 Fox   
 u x  u    u   u x dV
t

V 
  u x  
u y
u
u
ˆj  u x y
 
  u x x iˆ  u x
t
x
y
y

V 
6
 
u
u
u
kˆ    u x x iˆ  u y x ˆj  u z x
x
y
z
 

kˆ dV

 u x   
u y
u y
u
u  
u  
u  
 
   u x x  u x x iˆ   u x
 u y x  ˆj   u x
 u z x kˆ dV
x
x  
y
y  
y
z  
V 

 t

 u x 
 
   u x u dV
t

V 
Now, one last step and we obtain at the most common and useful form for propulsion and
fluid mechanics. This last step is the application of the divergence theorem.


   AdV   nˆ AdS
V
7
S
Finally we have the form of the momentum equation that we want.
F
x
 Fox
Sum of forces acting
on control volume
=
 u 
V  t x dV
Change in
momentum of mass
contained in control
volume over time
3
+

 u u  nˆdS
x
S
Change in
momentum flux
across surface of
control volume
8
One of the most common simplifications is the case of steady flow, with no acceleration:

 F   u u  nˆ dS
x
x
9
S
Below equation 8 we noted what the terms in these equations mean. Here we list exactly
what these terms are and some important items to remember when solving problems
using equations 8 and 9.
u x is the velocity relative to an inertial frame.

The quantity u  nˆ  is the velocity relative to the control surface.
We could also write the momentum equation for a moving control volume as:

d

  
ˆ



u
dV


u
u

u

n
dS

F

cs
S
dt V
8.a
Again, note the significance of the two velocity terms in the momentum flux across the

control surface. If we have a moving control surface, u , is the velocity measured in the
 
inertial frame, and the entire quantity u  ucs   nˆ is the velocity relative to the control

surface. If ucs = 0, then equation 8.a reduces to equation 9.
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