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Consider a twodimensional body in a flow, as sketched in Figure A. A control volume is drawn around this body, as given in the dashed lines in Figure A. The control volume is bounded by:
The upper and lower streamlines far above and below the body (ab and hi, respectively.)
Lines perpendicular to the flow velocity far ahead of and behind the body (ai and bh, respectively).
A cut that surrounds and wraps the surface of the body (cdefg).
The entire control volume is abcdefhia. The width of the control volume in the z direction (perpendicular to the page) is unity. Stations 1 and 2 are inflow and outflow stations, respectively.
Assume that the contour abhi is far enough from the body such that the pressure is everywhere the same on abhi and equal to the freestream pressure p = p(. Also, assume that the inflow velocity u1 is uniform across ai (as it would be in a freestream, or a test section of a wind tunnel.) The outflow velocity u2 is not uniform across bh, because the presence of the body has created a wake at the outflow station. However, assume that both u1 and u2 are in the x direction; hence, u1 = constant and u2 = f(y).
Consider the surface forces on the control volume shown in Figure A. They stem from two contributions:
The pressure distribution over the surface, abhi,
EMBED Equation.2 p dS
2. The surface force on def created by the presence of the body
Figure A
The surface shear stress on ab and hi has been neglected. Also, note that in Figure A the cuts cd and fg are taken adjacent to each other; hence any shear stress or pressure distribution on one is equal and opposite to that on the other; i.e., the surface forces on cd and fg cancel each other. Also, note that the surface on def is the equal and opposite reaction to the shear stress and pressure distribution created by the flow over the surface of the body. To see this more clearly, examine Figure B. On the left is shown the flow over the body. The moving fluid exerts pressure and shear stress distributions over the body surface which create a resultant aerodynamic force per unit span R( on the body. In turn, by Newtons third law, the body exerts equal and opposite pressure and shear stress distributions on the flow, i.e., on the part of the control surface bounded by def. Hence, the body exerts a force R( on the control surface, as shown on the right of Figure B. With the above in mind, the total surface force on the entire control volume is
Surface force = EMBED Equation.2 p dS R( (1)
Moreover, this is the total force on the control volume shown in Figure A because the volumetric body force is negligible.
Consider the integral form of the momentum equation as given by Equation (2.11) in the text. The righthand side of this equation is physically the force on the fluid moving through the control volume. For the control volume in Figure A, this force is simply the expression given by Equation (1). Hence, using Equation (2.11), with the righthand side given by Equation (1), we have
EMBED Equation.2 EMBED Equation.2 r V EMBED Equation.2 d( + EMBED Equation.2 (rV . dS) V = EMBED Equation.2 p DS  R( (2)
Figure B
Assuming steady flow, Equation (2) becomes
R( =  EMBED Equation.2 (rV . dS) V EMBED Equation.2 p dS (3)
Equation (3) is a vector equation. Consider again the control volume in Figure A. Take the x component of Equation (3), nothing that the inflow and outflow velocities u1 and u2 are in the x direction and the x component of R( is the aerodynamic drag per unit span D(:
D( =  EMBED Equation.2 (rV . dS) u EMBED Equation.2 (p dS)x (4)
In Equation (4), the last term is the component of the pressure force in the x direction. [The expression (p dS)x is the x component of the pressure force exerted on the elemental area dS of the control surface.] Recall that the boundaries of the control volume abhi are chosen far enough from the body such that p is constant along these boundaries. For a constant pressure.
EMBED Equation.2 (p dS)x = 0 (5)
because, looking along the x direction in Figure A, the pressure force on abhi pushing toward the right exactly balances the pressure force pushing toward the left. This is true no matter what the shape of abhi is, as long as p is constant along the surface. Therefore, substituting Equation (5) into (4), we obtain
D( =  EMBED Equation.2 (rV . dS) u (6)
Evaluating the surface integral in Equation (6), we note from Figure A that:
The sections ab, hi and def are streamlines of the flow. Since by definition V is parallel to the streamlines and dS is perpendicular to the control surface, along these sections V and dS are perpendicular vectors, and hence V . dS = 0. As a result, the contributions of ab, hi and def to the integral in Equation (6) are zero.
The cuts cd and fg are adjacent to each other. The mass flux out of one is identically the mass flux into the other. Hence, the contributions of cd and fg to the integral in Equation (6) cancel each other.
As a result, the only contributions to the integral in Equation (6) come from sections ai and bh. These sections are oriented in the y direction. Also, the control volume has unit depth in the z direction (perpendicular to the page). Hence, for these sections, dS = dy(1). The integral in Equation (6) becomes
EMBED Equation.2 (rV . dS) u =  EMBED Equation.2 riu21dy + EMBED Equation.2 r2u22dy (7)
Note that the minus in front of the first term on the righthand side of Equation (7) is due to V and dS being in opposite directions along ai (station 1 is an inflow boundary); in contrast, V and dS are in the same direction over hb (station 2 is an outflow boundary), and hence the second term has a positive sign.
Before going further with Equation (7), consider the integral form of the continuity equation for steady flow. Applied to the control volume in Figure A, this becomes
 EMBED Equation.2 r1u1 dy + EMBED Equation.2 r2u2 dy = 0
or,
EMBED Equation.2 r1u1 dy = EMBED Equation.2 r2u2 dy (8)
Multiplying Equation (8) by u1, which is a constant, we obtain
EMBED Equation.2 r1u21dy = EMBED Equation.2 r2u2u1 dy (9)
Substituting Equation (9) into Equation (7), we have
EMBED Equation.2 (rV . dS) u =  EMBED Equation.2 r2u2u1 dy + EMBED Equation.2 r2u22dy
or
EMBED Equation.2 (rV . dS) u =  EMBED Equation.2 r2u2 (u1 u2) dy (10)
Substituting Equation (10) into Equation (6) yields
D( = EMBED Equation.2 r2u2 (u1 u2) dy (11)
Equation (11) is the desired result of this section; it expresses the drag of a body in terms of the known freestream velocity u1 and the flowfield properties r2 and u2, across a vertical station downstream of the body. These downstream properties can be measured in a wind tunnel, and the drag per unit span of the body D( can be obtained by evaluating the integral in Equation (11) numerically, using the measured data for r2 and u2 as a function of y.
Examine Equation (11) more closely. The quantity u1 u2 is the velocity decrement at a given y location. That is, because of the drag on the body, there is a wake that trails downstream of the body. In this wake, there is a loss in flow velocity u1 u2. The quantity r2u2 is simply the mass flux; when multiplied by u1 u2, it gives the decrement in momentum. Therefore, the integral in Equation (11) is physically the decrement in momentum flow that exists across the wake, and from Equation (11), this wake momentum decrement is equal to the drag on the body.
For incompressible flow, r = constant and is known. For this case, Equation (11) becomes
D( = r EMBED Equation.2 u2 (u1 u2) dy (12)
Equation (12) is the answer to the questions posed at the beginning of this section. It shows how a measurement of the velocity distribution across the wake of a body can yield the drag. These velocity distributions are conventionally measured with a Pitot rake.
2.2
Denote the pressure distributions on the upper and lower walls by pu(x) and p EMBED Equation.2 (x) respectively. The walls are close enough to the model such that pu and p EMBED Equation.2 are not necessarily equal to p(. Assume that faces ai and bh are far enough upstream and downstream of the model such that
p = p( and v = 0 and ai and bh.
Take the ycomponent of Eq. (2.11) in the text:
L =  EMBED Equation.2 . EMBED Equation.2 v  EMBED Equation.2
The first integral = 0 over all surfaces, either because EMBED Equation.2 . EMBED Equation.2 = 0 or because v = 0. Hence
L( =  EMBED Equation.2 =  [ EMBED Equation.2 pu dx  EMBED Equation.2 p EMBED Equation.2 dx]
Minus sign because ycomponent is in downward
direction.
Note: In the above, the integrals over ia and bh cancel because p = p( on both faces. Hence
L( = EMBED Equation.2 p EMBED Equation.2 dx  EMBED Equation.2 pu dx
PAGE
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