# Integral Momentum Theorem:

Electronics - Devices

Oct 10, 2013 (4 years and 9 months ago)

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1

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 2
nd

Law:

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 system
s

show
n

in the figure below:

(1)

Inertial Coordin
ate System (labeled
with subscript ‘I’ in the figure.

⠲(

Fixed to the Vehicle (labeled with
subscript ‘V’ in the figure
)
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c潯牤楮o瑥⁳y獴敭⁭潶os⁷楴栠愠

u
o

relative to the inertial
coordinate system. All velocities
relative to the vehicle
-
fixed
coordinate
frame are denoted by
u
.

Newton’s 2
nd

Law for a control volume of fixed mass can be expressed as:

2

We can also write this as:

=

+

3

All external forces
on control volume
(pressure forces,
shear forces, body
forces)

Force due to change
in inertia for
accelerating vehicle

Change in
momentum of mass
in control volume

We can explain thi
s 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
th
e equation is nonzero since the control volume is accelerating relative to an intertial
X
I

Y
V

Y
I

Z
I

X
V

Z
V

u
o

2

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.

As expected, the result is the same for both. The integral momentum equation reduces to
the familiar form,
F
=m
a
.

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
ep in mind as you g
o through this what the goal is:

s
implify the equation for utility to propulsion applications. It is still
F
=m
a
, just in a
different form
,

and knowing what the expressio
ns of this alternate form are

important.

To continue, equation 3 ca
n be written as follows:

3.a

3.b

3.c

3

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 in
compressible, compressible, non
-
constant density flows) can be written as:

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 sub
stitute to arrive at:

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 examin
e the x
-
component of e
quation 5:

6

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.

7

Finally we have the form of the
momentum equation that we want.

=

+

8

Sum of forces acting
on control volume

Change in
momen
tum of mass
contained in control
volume over time

Change in
momentum flux
across surface of
control volume

4

One of the most common simplifications is the case of steady flow, with no acceleration:

9

Below equation 8 we noted w
hat 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.

is the velocity
relative

to an inertial frame.

The quantity

is the
velocity relative to the control surface.

We could also write the momentum equation for a moving control volume as:

8.a

Again, note the significance of the two velocity terms in the momentum flux across t
he
control surface. If we have a moving control surface,

, is the velocity measured in the
inertial

frame, and the
entire
quantity

is the velocity
relative

to the control
surface. If
= 0
, then equation 8.a reduces to equation 9.