Question 11.4: Consider an infinitesimal CV filled with fluid as shown in Figu...
Consider an infinitesimal CV filled with fluid as shown in Figure 11.3A. Apply a momentum balance in the x direction to this CV to derive the x component of the momentum equation in Cartesian coordinates.
We are asked to derive the x component of the momentum equation in Cartesian coordinates for a specified volume of fluid. Figure 11.3 serves as the sketch for this system. Recalling the procedure used to perform a mass balance for this CV in Example 11.2, we will use a Taylor series expansion to relate the values of density, velocity, and stress on each face to the values at the center of this cube. To apply a momentum balance, we write (Eq. 7.18):
\int_{CV}\frac{\partial}{\partial t}(\rho u)dV+\int_{CS}(ρu)(u • n)dS=\int_{CV}^{}{\rho \mathsf{f} dV}+\int_{CS}\sum{dS}
As discussed earlier, we will write the surface force in terms of the stress tensor rather than the stress vector, using Eq. 4.32, F_S =\int_{S}^{}{} (n • \pmb{\sigma} )dS. The resulting momentum balance is
\int_{CV}\frac{\partial}{\partial t}(\rho u)dV+\int_{CS}(ρu)(u • n)dS=\int_{CV}^{}{\rho \mathsf{f} dV}+\int_{CS}(n • \pmb{\sigma} )dS
The x component of this equation is \int_{CV}(∂/∂t)(ρu)dV+\int_{CS}^{}{} (ρu)(u • n)dS= \int_{CV}ρ f _x dV+\int_{CS}^{}{}(n • \pmb{\sigma} )_xdS, where the integrand of the stress integral, (n • \pmb{\sigma})x, gives the stresses that act on the faces in the x direction as shown in Figure 11.3B. Notice in this figure that a first order Taylor series expansion has been used to relate the value of a stress on a face to the value at the center of the cube.
Each of the integrals in the x component momentum balance has a constant integrand. We can therefore write the volume integrals in terms of the values of the integrand at the center of the cube multiplied by the volume of the cube (dx dy dz), to obtain
\int_{CV}\frac{\partial}{\partial t}(\rho u)dV=\frac{\partial}{\partial t}(\rho u)dx \ dy \ dz=\left(\rho \frac{\partial u}{\partial t}+u\frac{\partial \rho }{\partial t} \right)dx \ dy \ dz (A)
and
\int_{CV}\rho f_xdV=\rho f_x dx \ dy \ dz (B)
The stress integral \int_{CV} (n • \pmb{\sigma} )x dS = (n • \pmb{\sigma})x A is evaluated by using the stress values shown in Figure 11.3B and considering each pair of faces in turn. For example, on the near and far faces we find, respectively, (n •\pmb{\sigma} )x A = [σxx + (∂σxx/∂x)(dx/2)]dy dz and (n •\pmb{\sigma} )x A = − [σxx − (∂σxx/∂x)(dx/2)]dy dz. The net surface force on this pair of faces is therefore (n • \pmb{\sigma})x A = (∂σxx/∂x)dx dy dz. The contribution from the remaining two pairs of faces is found to be (∂σyx/∂y)dx dy dz + (∂σzx/∂z)dx dy dz, thus the total surface force on the cube is to first order
\left(\frac{\partial \sigma _{xx}}{\partial x}+\frac{\partial \sigma _{yx}}{\partial y}+\frac{\partial \sigma _{zx}}{\partial z}\right)dx \ dy \ dz (C)
The momentum flux integral is of the form \int_{CS} (ρu)(u • n)dS = (ρu)(u • n)A and may be evaluated by using a Taylor series expansion (see Example 11.2) to define the appropriate values of ρ, u, (u • n) and the area A on the six faces. For example, the momentum flux on the near face, where ρnear = ρ + (∂ρ/∂x)(dx/2) and (u • n) is given by + unear = u + (∂u/∂x)(dx/2), takes the form
= ρuu dy dz + \left(\rho u\frac{\partial u}{\partial x}+\frac{1}{2}uu\frac{\partial \rho }{\partial x} \right)dx \ dy \ dz to first order
where we have neglected higher order terms as usual. On the far face, where (u • n) takes the value − ufar = − u + (∂u/∂x)(dx/2), we find
= – ρuu dy dz + \left(\rho u\frac{\partial u}{\partial x}+\frac{1}{2}uu\frac{\partial \rho }{\partial x} \right)dx \ dy \ dz to first order
The sum of these two terms is
\left(2\rho u\frac{\partial u}{\partial x}+uu\frac{\partial \rho }{\partial x} \right)dx \ dy \ dz
or equivalently
\left[\rho u\frac{\partial u}{\partial x}+u\left(\rho \frac{\partial u}{\partial x}+u\frac{\partial \rho }{\partial x} \right)\right]dx \ dy \ dz
The two remaining pairs of faces contribute fluxes of
\left[\rho v\frac{\partial u}{\partial y}+u\left(\rho \frac{\partial v}{\partial y}+v\frac{\partial \rho }{\partial y} \right)\right]dx \ dy \ dz and \left[\rho w\frac{\partial u}{\partial z}+u\left(\rho \frac{\partial w}{\partial z}+w\frac{\partial \rho }{\partial z} \right)\right]dx \ dy \ dz
Thus, after some rearrangement, the total momentum flux is to first order
\left\{\rho \left(u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}+w\frac{\partial u}{\partial z} \right)+u \left[\left(u\frac{\partial \rho }{\partial x}+v\frac{\partial \rho }{\partial y}+w\frac{\partial \rho }{\partial z}\right)+\rho \left(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}+\frac{\partial w}{\partial z} \right)\right] \right\} dx \ dy \ dz (D)
Gathering terms A–D, rearranging, and dividing by the common factor dx dy dz yields
\rho \left(\frac{\partial u}{\partial t}+ u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}+w\frac{\partial u}{\partial z} \right)+u \left[\left(\frac{\partial \rho }{\partial t}+ u\frac{\partial \rho }{\partial x}+v\frac{\partial \rho }{\partial y}+w\frac{\partial \rho }{\partial z}\right)+\rho \left(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}+\frac{\partial w}{\partial z} \right)\right]=\rho f_x +\left(\frac{\partial \sigma _{xx}}{\partial x}+\frac{\partial \sigma _{yx}}{\partial y}+\frac{\partial \sigma _{zx}}{\partial z}\right)
The final step is to realize that since the term in square brackets is the continuity equation in the form of Eq. 11.2a, it has a value of zero. Therefore the final result is
\left(\frac{\partial \rho }{\partial t}+ u\frac{\partial \rho }{\partial x}+v\frac{\partial \rho }{\partial y}+w\frac{\partial \rho }{\partial z}\right)+\rho \left(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}+\frac{\partial w}{\partial z}\right)=0 (11.2a)
\rho \left(\frac{\partial u}{\partial t}+ u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}+w\frac{\partial u}{\partial z} \right) =\rho f_x +\left(\frac{\partial \sigma _{xx}}{\partial x}+\frac{\partial \sigma _{yx}}{\partial y}+\frac{\partial \sigma _{zx}}{\partial z}\right)
which is identical to Eq. 11.5a, as expected.
\rho \left(\frac{\partial u}{\partial t}+ u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}+w\frac{\partial u}{\partial z} \right) =\rho f_x +\left(\frac{\partial \sigma _{xx}}{\partial x}+\frac{\partial \sigma _{yx}}{\partial y}+\frac{\partial \sigma _{zx}}{\partial z}\right) (11.5a)