To determine the conservation of linear momentum of a fluid you will need to apply Newton’s second law of motion. For a system, Newton’s second law states the following. The time rate of change of the linear momentum of the system will equal sum of all external forces acting on the system. Due to the fact that momentum is mass times velocity, the momentum of a fluid particle is $vρdV$. As a result the momentum of the whole system will be $\int{_{sys}}vρdV$. In turn Newton’s second law will become the following equation.
(Eq 1) $\frac{D}{Dt}\int{_{sys}}~vρdV = \sum{F_{sys}}$
When relating the conservation of momentum of a system to a control volume at an instant in time, the forces acting on the control volume are instantaneously identical to the forces acting on the system. Hence,
(Eq 2) $\sum{F_{sys}} = \sum{F_{CV}}$
Conservation of Momentum for a Fixed, Non-Deforming Control Volume
Conservation of momentum can be defined using the Reynolds transport theorem.
(Eq 3) $\frac{DB_{sys}}{Dt}= \frac{∂}{∂t}\int{_{CV}}~ρbV +\int{_{CS}}~ρbv·\hat{n}dA$
Where “B” is the extensive property which represents the system momentum and “b” is the intensive property which represents the velocity. As a result, the Reynolds transport theorem will become the following equation for conservation of momentum.
(Eq 4) $\frac{D}{Dt}~\int{_{sys}}~vρdV= \frac{∂}{∂t}\int{_{CV}}~vρV +\int{_{CS}}~vρv·\hat{n}dA$
The Reynolds transport theorem states that the time rate of change of linear momentum for a system is the sum of two control volume properties. These are the following. The first part is linear momentum of the contents within the control volume. In addition the second part represents the net rate of linear momentum that is moving through the control volume. As a result as a mass particle moves in and out of the control volume through a control surface, linear momentum is carried in and out of the control volume. For a fixed, non-deforming control volume, the linear momentum equation will become the following.
(Eq 5) $\frac{∂}{∂t}\int{_{CV}}~vρV +\int{_{CS}}~vρv·\hat{n}dA = \sum{F_{CV}}$
$v$ = velocity
$ρ$ = density
$V$ = volume
$A$ = cross-sectional area
Considerations When Using Linear Momentum Equation
There are several things that you may have to consider when you use the linear momentum equation. These considerations are focused on fixed, non-deforming control volumes.
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A Uniformly Distributed Flow
If the flow is uniform over a section of the control surface, in relation to the fluid flowing into and out of the control volume, than the integral operations can be simplified. In turn one dimensional flows will be easier to work with. While flows that involve nonuniform velocity distribution will be more difficult.
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Linear Momentum is Directional
When you are solving a linear momentum problem, there can be as many as three orthogonal coordinate direction. In addition to this, in one of these three directions a fluid particle can be moving in a positive or negative direction. Hence it will be considered a negative or a positive quantity. However, what ever the case is, linear momentum is always directional.
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$v·\hat{n}$ Product and Flow Direction
When the fluid is flowing into the control volume, regardless of if it is a negative or positive linear momentum, the $v·\hat{n}$ product will be negative. On the other hand, when the fluid is flowing out of the control volume, the $v·\hat{n}$ will be positive.
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Time Rate of Change
If the control volume is non-deforming, and the flow is steady, than $\int{_{CV}}~vρdV$ will equal zero and can be removed from the equation.
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Perpendicular Control Surface
The forces on a control surface that is perpendicular to the flow will only be due to the pressure difference between the fluid inside and outside the control volume. To further this statement, if the flow is subsonic, atmospheric pressure will prevail where the subsonic flow exits the control volume.
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Atmospheric Pressure
Forces on the control surfaces caused by atmospheric pressure will cancel each other out. In turn the sum of the forces caused by atmospheric pressure will equal zero. In turn gage pressure can be used for any other pressure that do not equal atmospheric pressure.
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External Forces and Direction
The algebraic sign of all external forces will correspond to the chosen coordinate system. If the direction of a force is in the positive direction of the coordinate system, it will be a positive force. On the other hand, if it is in the negative direction of the coordinate system it will be a negative force.
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External Forces and Linear Momentum
When setting up the linear momentum equation, you will only need to consider the external forces that are acting on the contents of the control volume.
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Anchoring Force
An anchoring force is the force that will anchor the object that is being analyzed in place. As a result it must also be considered in the linear momentum equation.
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Moving and Deforming Control Volumes
Equation 5 currently can only be used for fixed, non-deforming control volume. Generally, this will normally be the type of problem that you will have to solve. However, there could be cases when the control volume is moving or deforming.
Let’s discuss a non-deforming, moving control volumes. First, the Reynolds transport theorem will become the following.
(Eq 6) $\frac{D}{Dt}~\int{_{sys}}~vρdV = \frac{∂}{∂t}~\int{_{CV}}~vρdV + \int{_{CS}}~vρW·\hat{n}dA$
$W$ = relative velocity
After equation 6 is combined with equation 1 and 2 the following equation will result.
(Eq 7) $\frac{∂}{∂t}~\int{_{CV}}~vρdV + \int{_{CS}}~vρW·\hat{n}dA = \sum{F_{CV}}$
Next, equation 7 needs to be related the absolute, relative, and control volume velocities. This is done by using the vector equation $v=W+v_{CV}$ where $v$ is the absolute velocity, $W$ is the relative velocity, and $v_{CV}$ is the velocity of the control volume. In turn this will result in the following equation.
(Eq 8) $\frac{∂}{∂t}~\int{_{CV}}~(W+v_{CV})ρdV + \int{_{CS}}~(W+v_{CV})ρW·\hat{n}dA = \sum{F_{CV}}$
If you have a case where the control volume velocity is constant and the flow is steady you would use the following equation.
(Eq 9) $\frac{∂}{∂t}~\int{_{CV}}~(W+v_{CV})ρdV = 0$
In addition, the following equation would be used for an inertial, non-deforming control volume.
(Eq 10) $\int{_{CS}}~(W+v_{cv})ρW·\hat{n}dA=\int{_{CS}}~WρW·\hat{n}dA+v_{CV}~\int{_{CS}}~ρW·\hat{n}dA$
Finally, the conservation of mass for steady flow of a moving control volume also needs to be considered.
(Eq 11) $\int_{_{CS}}~ρW·\hat{n}dA = 0$
Combining equations 8-11 will give us the linear momentum equation for an inertial, moving, non-deforming control volume with steady flow.
(Eq 12) $\int{_{CS}}~WρW·\hat{n}dA = \sum{F_{CV}}$