Question 4.3: Material Acceleration of a Steady Velocity Field Consider th...

For the given velocity field, the material acceleration vector is to be calculated at a particular point and plotted at an array of locations in the flow field.
Assumptions   1  The flow is steady and incompressible. 2  The flow is twodimensional, implying no z-component of velocity and no variation of u or 𝜐 with z.
Analysis   (a) Using the velocity field of Eq. 1 of Example 4–1 and the equation for material acceleration components in Cartesian coordinates (Eq. 4–11), we write expressions for the two nonzero components of the acceleration vector:

                                  a_x = \frac{∂u}{∂t} + u\frac{∂u}{∂x} + \upsilon \frac{∂u}{∂y} + w\frac{∂u}{∂z}
Cartesian coordinates:                    a_y = \frac{∂\upsilon}{∂t} + u\frac{∂\upsilon}{∂x} + \upsilon \frac{∂u}{∂y} + w\frac{∂\upsilon}{∂z}          (4–11)                                                                          

 \begin{matrix}a_x = \frac{∂u}{∂t}     +  u\frac{∂u}{∂x} &                          +  \upsilon \frac{∂u}{∂y}  +  w\frac{∂\upsilon}{∂z}\end{matrix}

 = 0 + \overbrace{(0.5 + 0.8x)(0.8)}^{}  +  \overbrace{(1.5  –  0.8 y)(0)}^{ } + 0 = (0.4 + 0.64x)  m/s^2

and

a_y = \frac{∂\upsilon}{∂t}   +   u\frac{∂\upsilon}{∂x}                       +  \upsilon \frac{∂\upsilon}{∂y}    + w\frac{∂\upsilon}{∂z}

= 0 + \overbrace{(0.5 + 0.8x)(0)} + \overbrace{(1.5 – 0.8 y)(-0.8)} + 0 = (-1.2 + 0.64y)  m/s^2

At the point (x = 2 m, y = 3 m), a_x = 1.68 m/s²  and  a_y = 0.720 m/s². (b) The equations in part (a) are applied to an array of x- and y-values in the flow domain within the given limits, and the acceleration vectors are plotted in Fig. 4–14.
Discussion   The acceleration field is nonzero, even though the flow is steady. Above the stagnation point (above y = 1.875 m), the acceleration vectors plotted in Fig. 4–14 point upward, increasing in magnitude away from the stagnation point. To the right of the stagnation point (to the right of x = −0.625 m), the acceleration vectors point to the right, again increasing in magnitude away from the stagnation point. This agrees qualitatively with the velocity vectors of Fig. 4–4 and the streamlines sketched in Fig. 4–14; namely, in the upper-right portion of the flow field, fluid particles are accelerated in the upper-right direction and therefore veer in the counterclockwise direction due to centripetal acceleration toward the upper right. The flow below y = 1.875 m is a mirror image of the flow above this symmetry line, and the flow to the left of x = −0.625 m is a mirror image of the flow to the right of this symmetry line.