The Equation of the Motion

The continuity equation and the linear momentum equation was developed for finite systems and their were related to control volume. The equations give an average values of quantities or only components of resultant forces. They don't give any detail information of the flow everywhere inside the control volume. Here, we are trying to find out differential equations valid at any point of fluid.

Writing continuity equation in the form of differential equation, we consider an infinitesimal control volume in the shape of a rectangular parallelepiped fixed in xyz for a general flow v(x, y, z) measured relative to xyz.

Figure 2.4: Volume element

Figure 2.4: Volume element

The net efflux rate along the plane of control volume that are perpendicular to the x axis (x planes of the element) is, pvx |x+Ax AyAz — pvx \x AyAz (figure 2.4) If we write the similar equation for the axis in y and z direction and add them up we get the net efflux rate as following:

Net efflux rate = [pvx \x+Ax AyAz - pvx \x ayaz] + [pvy \y+Ay axaz - pVy \y AxAz] + [pVz |z+Az Ax Ay - pvz \z Ax Ay]

The rate of mass accumulation inside the control volume would be || ax ayaz. Equating the net efflux rate and the rate of decrease of mass inside the control volume and dividing all resulting equation by AxAyAz and taking the limit as Ax, Ay and Az approach zero, we get:

dx dy dz dt

This is called differentia continuity equation and one special cases of it would be case of steady flow which results in:

The other case of incompressible flow would be as following:

dvx dvy dvz

dx dy dz

Introducing the divergence operator which for a vector field v can be defined as:

Then we can write the general form of the continuity equation 2.11 in the following form:

Writing the linear momentum equation in the form of differential equation, we consider an infinitesimal system in the shape of a rectangular parallelepiped fixed in xyz (figure 2.5 ). The linear momentum equation is a form of Newton's second law which has an Eulerian viewpoint. Consider a volume element axayaz as shown in figure2.5. The surface stress on each side of

Figure 2.5: Rectangular Parallelpiped element

Figure 2.5: Rectangular Parallelpiped element a cubic element cab be written as Tj, i, j = x, y, z where i is an indication of the side, plane, in which the stress is located on and j is the direction of the stress. For example Txyis the components of the stress on a plane normal to x axis and directed toward y axis. It can be shown that for equilibrium we must have Tj = Tjj. As it was already mentioned equation of linear momentum is a vector equation (equation 2.9). Thus we can write for each components of it. Here, we start by considering the x direction(equation 2.10). The first term on the left hand side of the equation 2.10 is all of the surface forces acting on the element at the x direction. The resultant forces acting on the x planes along the x directions is Txx |x+Ax AyAz — Txx |x AyAz. The resultant of forces acting on the y plane along the x direction is Tyx |y+Ay AxAz — Tyx |y AxAz. Similarly, for the forces on the z plane along the x direction is Tzx |z+Az AxAz — Tzx |z Ax Ay. If we add all of the surface forces we will get:

The second term on the left hand side of the equation 2.10 is the body force and is:

pgx axayaz where gx is the component of the gravitational force along the x axis.

The first term on the right hand side of equation 2.10 is the net rate of efflux. the rate at which x component of momentum enters at x is vxpvx \x ayaz and the rate at which it leaves at x + Ax is vxpvx |x+Ax AyAz. The same can be written for the y and z planes. Thus if we add them up we get:

[vxpvx |x+Ax -vxpvx |x] ayaz + [vxpvy |y+Ay -vx Pvy \y] AxAz + [vxpvz |z+Az -vxpvz |z ] ax ay

The second term on the right hand side of the equation 2.10 is the accumulation of the momentum inside the element and is:

^ AxAyAz dt

If we substitute all the terms into the equation 2.10 from the above equations and dividing the entire resulting equation by AxAyAz and taking limit as Ax, Ay and Az approaches zero, we obtain the x-component of the equation of the motion as:

dTxx + dTXy + dTxz + _ d(vxpvx) + 3(vxpvy) + d(vxpvz) + 3(pvx) (2 12) dx dy dz P^x dx dy dz dt

Similarly, for the components in the xx and y direction it can be written as:

d Txy

1 dTyy

1 dTZy

d x

' dy

d z

d Txz

1 dTyZ

! dTZZ

d x

' dy

+ pgz = 0 (VZ PVX) + 8 (VZ PVy ) + 3 (VZ PVZ ) + d(PVz ) (214) z dx dy dz dt

The quantity pvx, pvy and pvz are the components of the pv ,mass velocity vector. The terms pvxvx, pvxvy, pvxvz, pvyvz, etc. are the nine com-ponents(six components because of symmetry) of the momentum flux pvv, which is the dyadic product of pv and v. Similarly, Txx, Txy, Txz, Tyx, etc. are the nine components( six components because of symmetry) of the T, known as the stress tensor. We can combine, the above three equations and write them in a vector format as followings:

There should be noticed that V.pvv and V.T are not simple divergence because of tensorial nature of the V.pvv and V.T. The above equation is the general form of the linear momentum equation.

If we use the equation of continuity to substitute it in the equation 2.12, we will get:

dx dy dz Dt

The same can be written for y and z components. When three components added together vectorially, we get:

, surface forces on element body forces on element mass per unit volume per unit volume per unit volume time acceleration

This is an statement of the Newton's second law which was developed for a volume of fluid element moving with acceleration because of the forces acting upon it. It can be seen that momentum balance is equivalent to Newton's second law of motion. The above equation is valid for any continuous medium. In equation 2.16 we can insert expressions for various stresses in forms of velocity gradient and pressure. Therefore, we need to find a relation between various stress and velocity gradient..

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