From the previous chapter, we saw that the continuity equation is a conservation of mass equation with which mass transfers across boundaries and mass storage within control volumes can be accounted for.
Now we can derive a conservation
of linear and angular momentum equations using the same technique.
Linear Momentum Equation
Using the general conservation
equation below:
(S = System, CV = Control volume, CS = Control surface)
Let N be momentum mV, and n, which is N per unit mass, becomes V. Thus, the general conservation equation becomes:
From Newton’s law for the system, we have:
Combining these equations with
Equation 1, we get a conservation of linear momentum equation for the three
principal directions:
Two velocity terms appear in the
integrals of the last terms on the right-hand sides of equations 2a, 2b and 2c.
One of the velocities refers to a principal flow direction; for example, Vx.
The other velocity, Vn, is normal to the control surface where mass
crosses the boundary. These two velocities are not always equal. Because force
and velocity have magnitude and direction, they are vectors.
Thus, Equation 2 can be written
in vector form as:
This equation is a conservation
of linear momentum equation.
As required by Newton’s law, the
term ΣF represents all
forces applied externally to the control volume. These include forces due to
gravity, electric and magnetic fields, surface tension effects, pressure
forces, and viscous forces (friction).
The first term on the right-hand
side of equation 3 represents the rate of storage of linear momentum in the
control volume. The last term is a net rate out (out minus in) of linear
momentum from the control volume. For steady one-dimensional flow, Equation 3
becomes, for any direction i,
For one fluid stream entering and
one leaving the control volume, we have the following for one-dimensional flow:
The momentum equation can be used
to set up a general equation for frictional effects existing within a pipe, for
a certain type of phenomenon in open channel flow, and for other kinds of
applications-oriented problems.
Angular Momentum Equation
For some fluid mechanics problems,
it is essential to be able to evaluate moments exerted by moving fluid volumes
especially for rotating machinery like turbines and pumps. The equation
applicable to these instances is known as angular momentum equation and is
derived using the linear momentum equation.
Let us consider a control volume
located in the xy plane as shown below.
This control volume is located a
distance, r, from the origin; and as fluid passes through the control volume, a
force is exerted. The force can be resolved into two components: one normal and
one tangential to r. The torque exerted by the force equals the product
of force and moment arm. In differential form, we have:
where: dT0 is the
differential torque exerted about the origin
dFt is the tangential
component of the differential force perpendicular to r
Because the force is regarded as
being caused by fluid motion through the control volume, we can use the linear
momentum equation (Equation 2) applied in the tangential direction to obtain the
following:
The differential torque is then,
Integrating the above expression
over all control volumes that contribute to the total torque, we get:
The left-hand side of equation 5
represents the sum of all externally applied torques on the control volume. On
the right-hand side, the first term represents a storage of angular
momentum in the control volume; the last term is the net rate out
(out minus in) of angular momentum from the control volume.
As seen in xy graph above,
V sin θ = Vt.
Equation 5 can then be expressed as:
If we use vector notation, the
cross product is defined as:
r x V = rV
sinθ
Now we can express equation 6 in
vector form:
Equation 7 is applicable to both
2-dimensional and 3-dimensional cases.
