fluid_mechanics
268 Chapter 6 ■ Differential Analysis of Fluid Flow The three components, v x , v y , and v z can be combined to give the rotation vector, , in the form v x î v y ĵ v z kˆ (6.15) An examination of this result reveals that is equal to one-half the curl of the velocity vector. That is, 1 2 curl V 1 2 V (6.16) Vorticity in a flow field is related to fluid particle rotation. Wing since by definition of the vector operator V î 1 2 V 1 2 ∞ 0 0x u ĵ 0 0y v kˆ 0 ∞ 0z w 1 2 a 0w 0y 0v 0z b î 1 2 a 0u 0z 0w 0x b ĵ 1 2 a 0v 0x 0u 0y b kˆ The vorticity, Z, is defined as a vector that is twice the rotation vector; that is, Z 2 V (6.17) The use of the vorticity to describe the rotational characteristics of the fluid simply eliminates the 1 1 22 factor associated with the rotation vector. The figure in the margin shows vorticity contours of the wing tip vortex flow shortly after an aircraft has passed. The lighter colors indicate stronger vorticity. (See also Fig. 4.3.) We observe from Eq. 6.12 that the fluid element will rotate about the z axis as an undeformed block 1i.e., v OA v OB 2 only when 0u0y 0v0x. Otherwise the rotation will be associated with an angular deformation. We also note from Eq. 6.12 that when 0u0y 0v0x the rotation around the z axis is zero. More generally if V 0, then the rotation 1and the vorticity2 are zero, and flow fields for which this condition applies are termed irrotational. We will find in Section 6.4 that the condition of irrotationality often greatly simplifies the analysis of complex flow fields. However, it is probably not immediately obvious why some flow fields would be irrotational, and we will need to examine this concept more fully in Section 6.4. E XAMPLE 6.1 Vorticity GIVEN For a certain two-dimensional flow field the velocity is given by the equation V 1x 2 y 2 2î 2xyĵ FIND Is this flow irrotational? SOLUTION For an irrotational flow the rotation vector, , having the components given by Eqs. 6.12, 6.13, and 6.14 must be zero. For the prescribed velocity field and therefore u x 2 y 2 v 2xy w 0 v x 1 2 a 0w 0y 0v 0z b 0 v y 1 2 a 0u 0z 0w 0x b 0 v z 1 2 a 0v 0x 0u 0y b 1 312y2 12y24 0 2 Thus, the flow is irrotational. (Ans) zero, since by definition of two-dimensional flow u and v are not functions of z, and w is zero. In this instance the condition for irrotationality simply becomes v z 0 or 0v0x 0u0y. The streamlines for the steady, two-dimensional flow of this example are shown in Fig. E6.1. (Information about how to calculate y COMMENTS It is to be noted that for a two-dimensional flow field 1where the flow is in the x–y plane2 and will always be v x v y F I G U R E E6.1 x
6.2 Conservation of Mass 269 streamlines for a given velocity field is given in Sections 4.1.4 and 6.2.3.) It is noted that all of the streamlines (except for the one through the origin) are curved. However, because the flow is irrotational, there is no rotation of the fluid elements. That is, lines OA and OB of Fig. 6.4 rotate with the same speed but in opposite directions. As shown by Eq. 6.17, the condition of irrotationality is equivalent to the fact that the vorticity, Z, is zero or the curl of the velocity is zero. In addition to the rotation associated with the derivatives 0u0y and 0v0x, it is observed from Fig. 6.4b that these derivatives can cause the fluid element to undergo an angular deformation, which results in a change in shape of the element. The change in the original right angle formed by the lines OA and OB is termed the shearing strain, dg, and from Fig. 6.4b where dg is considered to be positive if the original right angle is decreasing. The rate of change of dg is called the rate of shearing strain or the rate of angular deformation and is commonly denoted with the symbol g. The angles da and db are related to the velocity gradients through Eqs. 6.10 and 6.11 so that and, therefore, dg g lim dtS0 dt lim dtS0 dg da db c 10v 0x2 dt 10u0y2 dt d dt g 0v 0x 0u 0y (6.18) As we will learn in Section 6.8, the rate of angular deformation is related to a corresponding shearing stress which causes the fluid element to change in shape. From Eq. 6.18 we note that if 0u0y 0v0x, the rate of angular deformation is zero, and this condition corresponds to the case in which the element is simply rotating as an undeformed block 1Eq. 6.122. In the remainder of this chapter we will see how the various kinematical relationships developed in this section play an important role in the development and subsequent analysis of the differential equations that govern fluid motion. 6.2 Conservation of Mass Conservation of mass requires that the mass of a system remain constant. As is discussed in Section 5.1, conservation of mass requires that the mass, M, of a system remain constant as the system moves through the flow field. In equation form this principle is expressed as DM sys Dt We found it convenient to use the control volume approach for fluid flow problems, with the control volume representation of the conservation of mass written as 0 0t cv r dV cs r V nˆ dA 0 (6.19) where the equation 1commonly called the continuity equation2 can be applied to a finite control volume 1cv2, which is bounded by a control surface 1cs2. The first integral on the left side of Eq. 6.19 represents the rate at which the mass within the control volume is changing, and the second integral represents the net rate at which mass is flowing out through the control surface 1rate of mass outflow rate of mass inflow2. To obtain the differential form of the continuity equation, Eq. 6.19 is applied to an infinitesimal control volume. 0 6.2.1 Differential Form of Continuity Equation We will take as our control volume the small, stationary cubical element shown in Fig. 6.5a. At the center of the element the fluid density is r and the velocity has components u, v, and w. Since the element is small, the volume integral in Eq. 6.19 can be expressed as 0 (6.20) 0t r dV 0r cv 0t dx dy dz
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6.2 Conservation of Mass 269<br />
streamlines for a given velocity field is given in Sections 4.1.4<br />
and 6.2.3.) It is noted that all of the streamlines (except for the one<br />
through the origin) are curved. However, because the flow is irrotational,<br />
there is no rotation of the <strong>fluid</strong> elements. That is, lines<br />
OA and OB of Fig. 6.4 rotate with the same speed but in opposite<br />
directions.<br />
As shown by Eq. 6.17, the condition of irrotationality is equivalent<br />
to the fact that the vorticity, Z,<br />
is zero or the curl of the velocity is zero.<br />
In addition to the rotation associated with the derivatives 0u0y and 0v0x, it is observed<br />
from Fig. 6.4b that these derivatives can cause the <strong>fluid</strong> element to undergo an angular<br />
deformation, which results in a change in shape of the element. The change in the original right<br />
angle formed by the lines OA and OB is termed the shearing strain, dg, and from Fig. 6.4b<br />
where dg is considered to be positive if the original right angle is decreasing. The rate of change<br />
of dg is called the rate of shearing strain or the rate of angular deformation and is commonly<br />
denoted with the symbol g. The angles da and db are related to the velocity gradients through Eqs.<br />
6.10 and 6.11 so that<br />
and, therefore,<br />
dg<br />
g lim<br />
dtS0 dt lim<br />
dtS0<br />
dg da db<br />
c 10v 0x2 dt 10u0y2 dt<br />
d<br />
dt<br />
g 0v<br />
0x 0u<br />
0y<br />
(6.18)<br />
As we will learn in Section 6.8, the rate of angular deformation is related to a corresponding shearing<br />
stress which causes the <strong>fluid</strong> element to change in shape. From Eq. 6.18 we note that if<br />
0u0y 0v0x, the rate of angular deformation is zero, and this condition corresponds to the case<br />
in which the element is simply rotating as an undeformed block 1Eq. 6.122. In the remainder of this<br />
chapter we will see how the various kinematical relationships developed in this section play an important<br />
role in the development and subsequent analysis of the differential equations that govern <strong>fluid</strong> motion.<br />
6.2 Conservation of Mass<br />
Conservation of<br />
mass requires that<br />
the mass of a<br />
system remain<br />
constant.<br />
As is discussed in Section 5.1, conservation of mass requires that the mass, M, of a system remain<br />
constant as the system moves through the flow field. In equation form this principle is expressed as<br />
DM sys<br />
Dt<br />
We found it convenient to use the control volume approach for <strong>fluid</strong> flow problems, with the control<br />
volume representation of the conservation of mass written as<br />
0<br />
0t cv<br />
r dV cs<br />
r V nˆ dA 0<br />
(6.19)<br />
where the equation 1commonly called the continuity equation2 can be applied to a finite control<br />
volume 1cv2, which is bounded by a control surface 1cs2. The first integral on the left side of Eq. 6.19<br />
represents the rate at which the mass within the control volume is changing, and the second integral<br />
represents the net rate at which mass is flowing out through the control surface 1rate of mass<br />
outflow rate of mass inflow2. To obtain the differential form of the continuity equation, Eq. 6.19<br />
is applied to an infinitesimal control volume.<br />
0<br />
6.2.1 Differential Form of Continuity Equation<br />
We will take as our control volume the small, stationary cubical element shown in Fig. 6.5a. At<br />
the center of the element the <strong>fluid</strong> density is r and the velocity has components u, v, and w. Since<br />
the element is small, the volume integral in Eq. 6.19 can be expressed as<br />
0<br />
(6.20)<br />
0t r dV 0r<br />
cv 0t dx dy dz