Conservation of Momentum using Control Volumes
Conservation of Linear Momentum
Recall the conservation of linear momentum law for
In order to convert this for
use in a control volume, use RTT with B = mV, beta
NOTE: Recall that at any instant of time t, the system
& CV occupy the SAME physical space.
So, the forces of the system are the same at the
forces of the control volume at a given instant.
Now lets return to the left side of the CLM equation
(the Force term). This term represents the sum of all the forces acting
on the control volume. There are several types of forces that can act on
our control volume.
- First, let us assume that we have one dimensional inlets
and outlets. This implies that our velocity vector V is parallel
to our normal to the surface vector n. We also assume that the velocity
is constant across the inlet or outlet surface.
Assuming density is constant,
we can rewrite the last term in our CLM equation:
This holds true because V dot n = V for outlets and -V for
inlets. We can make a further simplification if we notice that the definition
for mass flow rate is (density)(velocity)(area). Therefore our equation
for the flux term becomes:
- If the flow is steady we can drop the (d/dt) term.
Most of the problems we deal with having steady flow also have one dimensional
inlets and outlets with constant density. Our equation for conservation
of linear momentum now becomes:
Notice that this is a vector equation. Therefore we can break this up into
Since the velocity vector = (u,v,w) and the force vector = (Fx,Fy,Fz), our
equation can be rewritten into three equations:
Let us revisit the pressure force. Recall that
P is always defined as positive inward and n is always defined
as positive outward.. These two directions are opposite. Therefore
it turns out that
- Body Forces:
Gravity is a body force that acts in the negative z-direction.
- Surface Forces: These forces include:
- Pressure-contributes only normal forces.
- Viscous-viscous and frictional forces contribute normal
and shear forces.
- Other-forces due to control volume cutting through bolts
You can use either absolute or gage pressure in this equation as
long as you are consistent everywhere! To prove this, consider an arbitrary
control volume with a certain pressure field. Suppose there is a high pressure
at two locations and atmospheric pressure everywhere else. Realizing the
definition of gage pressure, Pg = Pabsolute - Patmospheric, we can subtract
atmospheric pressure from everything. We realize that the sum of the forces
due to pressure is the same in both cases. The net pressure force
will not change. This holds true because any shape in a uniform
pressure field has zero net pressure force.
Now let us return to the viscous forces.
This force is difficult to calculate because it has a normal and a tangential
component acting in an arbitrary direction. In order to solve for this
term you would have to integrate the shear stress over the entire control
surface. Fortunately we usually do not have to integrate for this term.
We can often choose a "wise" control volume to eliminate this
difficult term. This is done by picking a control volume so that no part
of the control surface has viscous forces acting on it. We could also keep
the viscous term as are unknown term and solve for it by computing our
Finally let us consider other surface forces.
These forces include whatever is left. In other words, any force acting
on the control volume that is not accounted for. Some examples of these
types of forces include struts, bolts, cables, ropes in tension, forces
holding a control volume in place, etc.
The final form of the momentum equation for steady flow with one
dimensional inlets and outlets is:
Here are some rules when using the momentum equation.
- An incompressible jet has an exit pressure equal to
the ambient pressure. The ambient pressure is usually atmospheric.
Therefore, when you draw your control volume, it is wise to slice through
the jet at the exit plane. This will cause the pressure to be ambient everywhere.
- The pressure is approximately ambient in any slice
through an incompressible jet. As you move down the stream, the pressure
is still equal to the ambient pressure.
- The mass flow rate and momentum flow rate equal zero
across a streamline. Therefore it is often advantageous to pick a control
surface so as to run along a streamline.
When you do this V dot
n is equal to zero because they are perpendicular. This will cause
the flux terms(mass, momentum, and energy) to be zero along the
- Mass flow rate and momentum flow rate equal zero along
a solid wall. This is true since a solid wall is always a streamline.
No mass can go through a solid wall.
Note that the pressure and viscous force
terms may not be zero along streamlines, however sometimes it is "wise"
to pick a control volume along a solid wall.
Given: A water jet of velocity Vj and thickness Dj impinges on a turning block which is held in place by a force Fx, as shown in the sketch.
As the water leaves the block, the round jet flattens out and slows down due to friction along the wall. The water turns a full 180 degrees and flattens into a rectangle shape of thickness Dj/6 and width 10Dj in cross section. The flow is steady.
a) Find: Ve, the exit velocity of the jet.
- First, pick a control volume which cuts through the inlet and exit, but does not include the turning block.
- Use conservation of mass for steady 1-D flow:
b) Find: Fx, the force required to hold block in place.
- Choose a new control volume because friction along the wall would be very hard to evaluate with the previous C.V. Pick a C.V. which cuts through the inlet and outlet and that includes the whole turning block. The wisest choice of control volume is one which cuts through where the force Fx acts on the turning block, since Fx is our unknown.
- Write momentum equation for steady 1-D inlet/outlet in the x-direction:
Examine each term on the left: (no grav in x-dir), (p = pa everywhere), (wise choice of C.V), and
- Thus, the momentum equation reduces to
- Now, recall, at an inlet or exit. Here at the inlet,
- So, solve for :
Or, finally, , which is our final answer.
- Note: This is not merely an academic exercise. A testing facility at NASA Langley Research Center uses this principle to propel a test rig along a track to test airplane landing gear, etc. The experimental facility, called the Aircraft Landing Dynamics Facility, uses a high pressure water jet that hits a turning bucket on the test cart, much like in the above problem. The cart is propelled from zero to 250 miles per hour in two seconds flat!
Momentum Flux Correction Factor
- Again, just like in the conservation of mass equation, even if we don't really have one-dimensional (uniform) inlets and outlets, we would still like to use the simplified version of the conservation of momentum equation. I.e. we would like to use Vav instead of V in the equation.
- The equivalent profile and actual profile thus have identical mass flow rates, .
- What about momentum flow rate, or momentum flux, MF? Do actual and equivalent profiles have the same momentum flux? No! It turns out, after integration, that:
- So, we can't just substitute Vav for V in the momentum equation or we will get the wrong answer. So, instead, let's introduce a momentum flux correction factor, :
It turns out that: ,
1) 1-D uniform flow, by definition
2) fully developed laminar pipe flow,
3) fully developed turbulent pipe flow,
- Now, we can use the one-dimensional form of the momentum equation, but with these momentum flux correction factors thrown in:
For a fixed control volume with steady flow,
- Fortunately, most problems in real life applications are turbulent, not laminar, and we can usually neglect the momentum flux correction factors since is close to 1.0.
Most Useful Form of the Momentum Equation
- For steady flow with a fixed control volume, the most useful form of the momentum equation is thus:
, where momentum flux correction factor, and .
Given: Consider incompressible flow in the entrance of a circular tube.
The inlet flow is uniform u1 = U0. The flow at section 2 is developed laminar pipe flow.U0, p1, p2, R, and , are also known. At section 2, .
Find: Total friction force on the fluid from 1 to 2.
- First draw a C.V. Again cut through the inlet and outlet 1 and 2. But now, we want to cut along the walls since Ffriction is our unknown.
- Draw all the forces acting on the C.V. in the x-direction.
- Simplifications: Assume steady, incompressible, uniform one-dimensional inlet, but not a uniform one-dimensional outlet.
- So, let's use the modified momentum equation for non-1-D inlets/outlets in the x-direction:
- Now look at each term:
(no grav. in x-dir)
(this is our unknown)
(no struts, bolts, etc.)
(inlet is 1-D so )
: This term is the hardest. First of all, what is uav? I.e. what is the equivalent 1-D velocity? Well, since the area at the outlet is the same as that at the inlet, uav at the outlet has to equal U0.
- What is at the outlet? By definition, . You can plug the equation in and integrate [Try it on your own]. You will get (the value for laminar pipe flow).
- So, the outlet momentum flux term becomes
- Finally, then sum everything and solve for the unknown:
- Notice, if is ignored, i.e. (set ), this last term would cancel completely, and the answer would be wrong.
More Example Problems
Problem # 3.58 in the text:
Given: Cart with water
jet, deflector, as shown
Known in this problem are the jet area A, the average velocity
Vav, the jet deflection angle, ,
and the momentum. flux correction factor of the jet, .
Also, frictionless wheels are assumed.
Find: Tension in cable
at time t=0.
- As always, the first step in any control volume
problem is to pick and draw a control volume. The forces and
coordinate system have been labeled on the sketch. Here it is
appropriate to slice through the jet and slice through the unknown
force T (tension in cable), as shown in the sketch.
Incompressible? Yes (water is the fluid, which is approximately
an incompressible liquid)
Steady? - No, not really, since the water level in the tank is
falling. But all we are asked for is T at time t=0, so think
of this as a pseudo-steady problem in order to find T.
Note: The falling water level will have no effect on the x-momentum
equation anyway, since its velocity is vertical.
- Now, use the x-component of the momentum equation
in it steady form:
There is only one outlet, at which
The quantity uav in the momentum flux outlet term in
the above equation needs to be considered carefully. This is not
Vav, but rather is the x-component of Vav.
From a little trig one can see that
Thus the outlet term on the right hand side of the momentum equation
and the final form of the x-momentum equation is (solving for
- Plug in the numbers last: The density of water
at room temperature is 998. kg/m3, the jet velocity,
Vav is 8 m/s, the cross-sectional area A is that of
a circle of diameter 0.04 m, the jet angle is given as 60o,
and the jet momentum flux correction factor is 1.0 for a uniform
jet. This yields:
Or T = 40.1 N.
- Question: At what
angle, is the tension a maximum?
Answer: When the jet deflection angle is zero, i.e. the
jet simply exits horizontally into the atmosphere. This is clearly
seen in the above equation since the cosine of zero is unity.
Example A water-mounted fire pump (an old exam
Given: A pump is anchored
to the ground as shown, with Vj=35.0 m/s and dj=3.00
cm. Assume the jet has a fully developed turbulent pipe flow profile
at its exit.
Find: Horizontal force
required to hold platform in place.
- The first step in any control volume problem
is to pick and draw a control volume. As shown, the control volume
should slice through jet exit and through the bolts or whatever
is holding the platform solidly to the ground. The inlet to the
control volume is most easily taken at the surface of the water,
where the pressure and velocity are known (p = pa and
V = 0 at the surface). Note: When this problem was given as
an exam question, lots of students took their control volume inlet
at the pipe inlet. This makes the problem more difficult. Remember
the first rule about selecting a control volume - Be Lazy!
- Apply the x-momentum equation:
The gravity term on the left hand side is zero because gravity
does not act in the x direction. Likewise, the pressure term is
zero because everywhere on the control surface above the water,
the pressure is atmospheric (including the portion of the control
surface that slices through the jet). Below the water, the pressure
is hydrostatic, and whatever pressure force is exerted on the
left side of the control surface is exactly balanced by that on
the right side of the control surface. In the control volume
selected, there is no net viscous force acting on the control
volume. Thus the first three terms on the left hand side of the
x-momentum equation are zero. The only "other" force
acting on the control surface is the force of the ground acting
on the platform, as shown in the sketch. The direction of this
force is assumed to act to the right; if this is wrong, the result
will be negative.
On the right hand side (the momentum flux terms), there is no
x-component of velocity at any inlet, so the last term on the
right is zero. There is only one outlet, with uav
equal to the x-component of the velocity vector of the jet. Thus,
the x-momentum equation reduces to
This is the answer in variable form.
- Lastly, plug in the numbers, noting that for
a fully developed turbulent pipe flow the momentum flux correction
factor is around 1.02:
or finally, the horizontal force required to hold platform
in place is 799. N.
Problem # 3.51 in the text:
Given: A turbine wheel,
powered by a water jet, as shown in the sketch (at time t = 0)
The turbine is spinning at a constant rotational speed.
- 1-D inlet and outlet
- neglect friction in the turning bucket
(a) Find: The force of the turning bucket
on the turbine wheel at this instant of time.
- Pick a control volume. Here we will pick a
moving control volume, which makes things a little tricky.
When the control volume is moving, one must use the relative
velocity, i.e. relative to the moving C.V.
- At the inlet, then, the relative velocity at
the inlet is obtained by subtracting the velocity of the C.V.
from the absolute velocity at the inlet, i.e.
where Vj is the magnitude of the absolute velocity at the inlet,
and its direction is in the positive x-direction. The second
term on the right represents the velocity of the control volume
in an absolute reference frame, which is subtracted as shown.
The x-component of this relative inlet velocity is simply the
magnitude of the above velocity vector, since the vector acts
only in the positive x-direction, i.e.
- Now apply conservation of mass to this moving
control volume. Note: Here, relative velocities must be
used, since they represent the velocities actually entering and
leaving the control volume as it moves along.
Note that the areas on either side of the equation are identical
by the original assumption of negligible viscous effects. Also
the incompressible assumption causes the density to drop out.
At time t = 0, when the turning bucket is on the top of the turbine
wheel as shown, the velocity of the inlet is in the positive x-direction,
while that of the outlet is in the negative x-direction. Thus,
the x-component of the relative outlet velocity is simply the
negative of that of the relative inlet velocity, i.e.
In other words, the turning bucket changes the direction of the
water jet by 180 degrees, but it does not change the magnitude
of the jet velocity. (If friction along the walls of the turning
vane was taken into account, the jet velocity magnitude would
change as well.)
- Now apply conservation of x-momentum
for our moving C.V.
Notice that the relative x-velocity components are used
in the momentum flux terms on the right hand side since this is
a moving control volume. The gravity term on the left hand side
is zero because gravity does not act in the x direction. Likewise,
the pressure term is zero because everywhere on the control surface,
the pressure is atmospheric (including the portions of the control
surface that slice through the jets). In the control volume selected,
there is no net viscous force acting on the control volume. (This
would be true even if viscous forces along the turning vane were
not ignored, because the control surface does not pass along the
turning vane wall.) Thus the first three terms on the left hand
side of the x-momentum equation are zero. The only "other"
force acting on the control surface is the force of the turbine
wheel acting on the turning bucket, which is equal and opposite
to the force of the turning bucket acting on the turbine wheel,
as shown in the sketch. The direction of the force of the bucket
on the wheel is assumed to act to the right; if this is wrong,
the result will be negative.
On the right hand side (the momentum flux terms), there is only
one inlet and one outlet. At both the inlet and the outlet, the
x-component of the relative velocity vector of the jet is known
from the conservation of mass analysis above. Since friction
is being neglected, it is assumed that the momentum flux correction
factors are unity. Thus, the x-momentum equation reduces to
This is the force of the turning bucket on the turbine wheel.
(b) Find: The power, P, delivered to the
wheel at this instant of time (t = 0).
- Power on a rotating wheel is defined as the torque
times the angular velocity of the wheel. The torque due to this
turning bucket is the force just derived times the radius fo the
wheel, R. Thus,
(c) Find: The angular velocity which provides
the maximum power to the wheel.
- To find the maximum power, take the derivative
of power with respect to angular velocity.
- The values of angular velocity at which this
derivative is zero then represent conditions where the power is
either a minimum or a maximum. There are two such roots, i.e.
- The first of these represents the case where
the absolute jet speed and the speed of the turning bucket are
the same. Under such conditions, no power is transferred to the
turbine wheel at all. (This can be verified easily, i.e. P =
0 when the first root is plugged into the result of part (b) above.)
In other words, the first root is a minimum. The second root
is therefore the desired one; i.e. the angular velocity which
supplies the maximum power to the turbine wheel is that in which
the turbine wheel spins three times slower at its rim than the
jet speed, i.e.