Question 4.6: Calculation of Kinematic Properties in a Two-Dimensional Flo...
Calculation of Kinematic Properties in a Two-Dimensional Flow
Consider the steady, two-dimensional velocity field of Example 4–1:
\vec{V} = (u , \upsilon ) = (0.5 + 0.8x)\vec{i} + (1.5 – 0.8y)\vec{j} (1)
where lengths are in units of m, time in s, and velocities in m/s. There is a stagnation point at (−0.625, 1.875) as shown in Fig. 4–42. Streamlines of the flow are also plotted in Fig. 4–42. Calculate the various kinematic properties, namely, the rate of translation, rate of rotation, linear strain rate, shear strain rate, and volumetric strain rate. Verify that this flow is incompressible.
We are to calculate several kinematic properties of a given velocity field and verify that the flow is incompressible.
Assumptions 1 The flow is steady. 2 The flow is two-dimensional, implying no z-component of velocity and no variation of u or 𝜐 with z.
Analysis By Eq. 4–19, the rate of translation is simply the velocity vector itself, given by Eq. 1. Thus,
\vec{V} = u\vec{i} + \upsilon \vec{j} + w\vec{k} (4.19)
Rate of translation: u = 0.5 + 0.8x 𝜐 = 1.5 − 0.8y w = 0 (2)
The rate of rotation is found from Eq. 4–21. In this case, since w = 0 everywhere, and since neither u nor 𝜐 vary with z, the only nonzero component of rotation rate is in the z-direction. Thus,
\vec{\omega } = \frac{1}{2}\left(\frac{∂w}{∂y} – \frac{∂\upsilon }{∂z} \right) \vec{i} + \frac{1}{2}\left(\frac{∂u}{∂z} – \frac{∂w}{∂x} \right)\vec{j} + \frac{1}{2}\left(\frac{∂\upsilon }{∂x} – \frac{∂u}{∂y} \right)\vec{k} (4.21)
Rate of rotation: \vec{\omega } = \frac{1}{2}\left(\frac{∂\upsilon }{∂x} – \frac{∂u}{∂y} \right) \vec{k} = \frac{1}{2}(0 – 0)\vec{k} = 0 (3)
In this case, we see that there is no net rotation of fluid particles as they move about. (This is a significant piece of information, to be discussed in more detail later in this chapter and also in Chap. 10.)
Linear strain rates can be calculated in any arbitrary direction using Eq. 4–23. In the x-, y-, and z-directions, the linear strain rates are
\epsilon _{xx} = \frac{∂u}{∂x} \epsilon _{yy} = \frac{∂\upsilon}{∂y} \epsilon _{zz} = \frac{∂w}{∂z} (4.23)
\epsilon _{xx}=\frac{∂u}{∂x} = 0.8 s^{-1} \epsilon _{yy}=\frac{∂\upsilon}{∂y} = -0.8 s^{-1} \epsilon _{zz}=0 (4)
Thus, we predict that fluid particles stretch in the x-direction (positive linear strain rate) and shrink in the y-direction (negative linear strain rate). This is illustrated in Fig. 4–43, where we have marked an initially square parcel of fluid centered at (0.25, 4.25). By integrating Eqs. 2 with time, we calculate the location of the four corners of the marked fluid after an elapsed time of 1.5 s.
Indeed this fluid parcel has stretched in the x-direction and has shrunk in the y-direction as predicted.
Shear strain rate is determined from Eq. 4–26. Because of the two- dimensionality, nonzero shear strain rates can occur only in the xy-plane. Using lines parallel to the x- and y-axes as our initially perpendicular lines, we calculate \epsilon_{xy},
\epsilon _{xy} = \frac{1}{2}\left(\frac{∂u}{∂y} + \frac{∂\upsilon }{∂x}\right) \epsilon _{zx} = \frac{1}{2}\left(\frac{∂w}{∂x} + \frac{∂u }{∂z}\right) \epsilon _{yz} = \frac{1}{2}\left(\frac{∂\upsilon}{∂z} + \frac{∂w }{∂y}\right) (4.26)
\epsilon _{xy} = \frac{1}{2}\left(\frac{∂u}{∂y} + \frac{∂\upsilon }{∂x} \right) = \frac{1}{2}(0 + 0)= 0 (5)
Thus, there is no shear strain in this flow, as also indicated by Fig. 4–43. Although the sample fluid particle deforms, it remains rectangular; its initially 90° corner angles remain at 90° throughout the time period of the calculation.
Finally, the volumetric strain rate is calculated from Eq. 4–24:
\frac{1}{V} \frac{DV}{Dt} = \frac{1}{V}\frac{dV}{dt} = \epsilon _{xx} +\epsilon _{yy} + \epsilon _{zz} = \frac{∂u}{∂x} + \frac{∂\upsilon }{∂y} + \frac{∂w}{∂z} (4.24)
\frac{1}{V}\frac{DV}{Dt}= \epsilon _{xx} + \epsilon _{yy} + \epsilon _{zz} = (0.8 – 0.8 + 0)s^{-1}=0 (6)
Since the volumetric strain rate is zero everywhere, we can say definitively that fluid particles are neither dilating (expanding) nor shrinking (compressing) in volume. Thus, we verify that this flow is indeed incompressible. In Fig. 4–43, the area of the shaded fluid particle (and thus its volume since it is a 2-D flow) remains constant as it moves and deforms in the flow field.
Discussion In this example it turns out that the linear strain rates ( \epsilon _{xx} and \epsilon _{yy}) are nonzero, while the shear strain rates ( \epsilon _{xy} and its symmetric partner \epsilon _{yx}) are zero.
This means that the x- and y-axes of this flow field are the principal axes. The ( two-dimensional) strain rate tensor in this orientation is thus
\epsilon _{ij} = \left(\begin{matrix} \epsilon _{xx} & \epsilon _{xy} \\ \epsilon _{yx} & \epsilon _{yy} \end{matrix} \right) = \left(\begin{matrix} 0.8 & 0 \\ 0 & -0.8 \end{matrix} \right)s^{-1} (7)
If we were to rotate the axes by some arbitrary angle, the new axes would not be principal axes, and all four elements of the strain rate tensor would be nonzero.
You may recall rotating axes in your engineering mechanics classes through use of Mohr’s circles to determine principal axes, maximum shear strains, etc. Similar analyses are performed in fluid mechanics.
