fluid_mechanics
298 Chapter 6 ■ Differential Analysis of Fluid Flow SOLUTION (a) The velocity is given by Eq. 6.101 as At point 122, u p2, and since this point is on the surface 1Eq. 6.1002 Thus, V 2 U 2 a1 2 b r b1p u2 r sin u cos u b2 r pb 2 (1) and it follows that V 2 147.4 mihr2 a 5280 ft mi 3600 shr b 69.5 ft s p 1 p 2 10.00238 slugs ft 3 2 3169.5 fts2 2 158.7 fts2 2 4 2 10.00238 slugsft 3 2132.2 fts 2 21100 ft 0 ft2 9.31 lbft 2 0.0647 psi (Ans) and the magnitude of the velocity at 122 for a 40 mihr approaching wind is (Ans) (b) The elevation at 122 above the plain is given by Eq. 1 as y 2 pb 2 Since the height of the hill approaches 200 ft and this height is equal to pb, it follows that 200 ft y 2 100 ft (Ans) 2 From the Bernoulli equation 1with the y axis the vertical axis2 COMMENTS This result indicates that the pressure on the hill at point 122 is slightly lower than the pressure on the plain at some distance from the base of the hill with a 0.0533 psi difference due to the elevation increase and a 0.0114 psi difference due to the velocity increase. By repeating the calculations for various values of the upstream wind speed, U, the results shown in Fig. E6.7b are obtained. Note that as the wind speed increases, the pressure difference increases from the calm conditions of p 1 p 2 0.0533 psi. The maximum velocity along the hill surface does not occur at point 122 but farther up the hill at u 63°. At this point V surface 1.26U. The minimum velocity 1V 02 and maximum pressure occur at point 132, the stagnation point. so that p 1 g V 2 1 2g y 1 p 2 g V 2 2 2g y 2 2 b F I G U R E E6.7b V 2 2 U 2 c 1 b2 1pb22 2 d U 2 a1 4 p 2b V 2 a1 4 12 p 2b 140 mihr2 47.4 mihr 0.14 0.12 0.10 p 1 – p 2 , psi 0.08 0.06 0.04 (40 mph, 0.0647 psi) p 1 p 2 r 2 1V 2 2 V 2 12 g1y 2 y 1 2 0.02 and with V 1 140 mihr2 a 5280 ft mi 3600 shr b 58.7 ft s 0 0 20 40 60 80 100 U, mph 6.6.2 Rankine Ovals The half-body described in the previous section is a body that is “open” at one end. To study the flow around a closed body, a source and a sink of equal strength can be combined with a uniform flow as shown in Fig. 6.25a. The stream function for this combination is c Ur sin u m 2p 1u 1 u 2 2 (6.103) and the velocity potential is f Ur cos u m 2p 1ln r 1 ln r 2 2 (6.104)
6.6 Superposition of Basic, Plane Potential Flows 299 U y Stagnation point ψ = 0 Stagnation point r 2 r r 1 Source a θ2 θ a θ 1 Sink x +m –m a a h h (a) (b) F I G U R E 6.25 The flow around a Rankine oval: (a) superposition of source–sink pair and a uniform flow; (b) replacement of streamline C 0 with solid boundary to form Rankine oval. Rankine ovals are formed by combining a source and sink with a uniform flow. Large Ua/m Small Ua/m As discussed in Section 6.5.4, the stream function for the source–sink pair can be expressed as in Eq. 6.93 and, therefore, Eq. 6.103 can also be written as or (6.105) The corresponding streamlines for this flow field are obtained by setting c constant. If several of these streamlines are plotted, it will be discovered that the streamline c 0 forms a closed body as is illustrated in Fig. 6.25b. We can think of this streamline as forming the surface of a body of length 2/ and width 2h placed in a uniform stream. The streamlines inside the body are of no practical interest and are not shown. Note that since the body is closed, all of the flow emanating from the source flows into the sink. These bodies have an oval shape and are termed Rankine ovals. Stagnation points occur at the upstream and downstream ends of the body as are indicated in Fig. 6.25b. These points can be located by determining where along the x axis the velocity is zero. The stagnation points correspond to the points where the uniform velocity, the source velocity, and the sink velocity all combine to give a zero velocity. The locations of the stagnation points depend on the value of a, m, and U. The body half-length, / 1the value of 0x 0 that gives V 0 when y 02, can be expressed as or (6.106) (6.107) The body half-width, h, can be obtained by determining the value of y where the y axis intersects the c 0 streamline. Thus, from Eq. 6.105 with c 0, x 0, and y h, it follows that or c Ur sin u m 2p tan1 a c Uy m 2ay 2p tan1 a x 2 y 2 a 2b / a ma 12 pU a2 b / a a m 12 pUa 1b h h2 a 2 tan 2pUh 2a m 2ar sin u r 2 a b 2 h a 1 2 cah a b 2 1 d tan c 2 a pUa m b h a d (6.108) (6.109) Equations 6.107 and 6.109 show that both /a and ha are functions of the dimensionless parameter, pUam. Although for a given value of Uam the corresponding value of /a can be determined directly from Eq. 6.107, ha must be determined by a trial and error solution of Eq. 6.109. A large variety of body shapes with different length to width ratios can be obtained by using different values of Uam, as shown by the figure in the margin. As this parameter becomes large, flow
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6.6 Superposition of Basic, Plane Potential Flows 299<br />
U<br />
y<br />
Stagnation<br />
point<br />
ψ = 0<br />
Stagnation<br />
point<br />
r 2<br />
r r 1<br />
Source<br />
a<br />
θ2<br />
θ<br />
a<br />
θ 1<br />
Sink<br />
x<br />
+m –m<br />
a<br />
a<br />
h<br />
h<br />
<br />
<br />
(a)<br />
(b)<br />
F I G U R E 6.25 The flow around a Rankine oval: (a) superposition of source–sink pair<br />
and a uniform flow; (b) replacement of streamline C 0 with solid boundary to form Rankine oval.<br />
Rankine ovals are<br />
formed by combining<br />
a source and<br />
sink with a uniform<br />
flow.<br />
Large Ua/m<br />
Small Ua/m<br />
As discussed in Section 6.5.4, the stream function for the source–sink pair can be expressed as in<br />
Eq. 6.93 and, therefore, Eq. 6.103 can also be written as<br />
or<br />
(6.105)<br />
The corresponding streamlines for this flow field are obtained by setting c constant. If several<br />
of these streamlines are plotted, it will be discovered that the streamline c 0 forms a closed body<br />
as is illustrated in Fig. 6.25b. We can think of this streamline as forming the surface of a body of<br />
length 2/ and width 2h placed in a uniform stream. The streamlines inside the body are of no practical<br />
interest and are not shown. Note that since the body is closed, all of the flow emanating from the<br />
source flows into the sink. These bodies have an oval shape and are termed Rankine ovals.<br />
Stagnation points occur at the upstream and downstream ends of the body as are indicated<br />
in Fig. 6.25b. These points can be located by determining where along the x axis the velocity is<br />
zero. The stagnation points correspond to the points where the uniform velocity, the source velocity,<br />
and the sink velocity all combine to give a zero velocity. The locations of the stagnation points<br />
depend on the value of a, m, and U. The body half-length, / 1the value of 0x 0 that gives V 0<br />
when y 02, can be expressed as<br />
or<br />
(6.106)<br />
(6.107)<br />
The body half-width, h, can be obtained by determining the value of y where the y axis intersects<br />
the c 0 streamline. Thus, from Eq. 6.105 with c 0, x 0, and y h, it follows that<br />
or<br />
c Ur sin u m 2p tan1 a<br />
c Uy m 2ay<br />
2p tan1 a<br />
x 2 y 2 a 2b<br />
/ a ma 12<br />
pU a2 b<br />
/<br />
a a m 12<br />
pUa 1b<br />
h h2 a 2<br />
tan 2pUh<br />
2a m<br />
2ar sin u<br />
r<br />
2 a b 2<br />
h<br />
a 1 2 cah a b 2<br />
1 d tan c 2 a pUa<br />
m b h<br />
a d<br />
(6.108)<br />
(6.109)<br />
Equations 6.107 and 6.109 show that both /a and ha are functions of the dimensionless parameter,<br />
pUam. Although for a given value of Uam the corresponding value of /a can be determined<br />
directly from Eq. 6.107, ha must be determined by a trial and error solution of Eq. 6.109.<br />
A large variety of body shapes with different length to width ratios can be obtained by using<br />
different values of Uam, as shown by the figure in the margin. As this parameter becomes large, flow