Fluids – Lecture 15 Notes - MIT

1 Fluids Lecture 15 Notes1. Uniform flow, Sources, Sinks, DoubletsReading: Anderson FlowDefinitionAuniform flowconsists of a velocity field where~V=u +v is a constant. In 2-D, thisvelocity field is specified either by the freestream velocitycomponentsu ,v , or by thefreestream speedV and flow angle .u=u =V cos v=v =V sin Note also thatV2 =u2 +v2 . The corresponding potential and stream functions are (x,y) =u x+v y=V (xcos +ysin ) (x,y) =u y v x=V (ycos xsin )Vvu Zero DivergenceA uniform flow is easily shown to havezero divergence ~V= u x+ v y= 0since bothu andv are constants. The equivalent statement is that (x,y) satisfiesLaplace s equation. 2 = 2(u x+v y) x2+ 2(u x+v y) y2= 0 Therefore, the uniform flow satisfies mass CurlA uniform flow is also easily shown to be irrotational, or to havezero vorticity.

2 ~V ~ =( v x u y) k= 01 The equivalent irrotationality condition is that (x,y) satisfies Laplace s equation. 2 = 2(u y v x) x2+ 2(u y v x) y2= 0 Source and SinkDefinitionA 2-Dsourceis most clearly specified in polar coordinates. The radial and tangential velocitycomponents are defined to beVr= 2 r,V = 0where is a scaling constant called thesource strength. The volume flow rate per unit span V across a circle of radiusris computed as follows. V = 2 0~V ndA= 2 0 Vrrd = 2 0 2 rrd = Hence we see that the source strength specifies the rate of volume flow issuing outwardfrom the source. If is negative, the flow is inward, and the flow is called VrV xyrCartesian representationThe cartesian velocity components of the source or sink areu(x,y) = 2 xx2+y2v(x,y) = 2 yx2+y2and the corresponding potential and stream functions are asfollows.

3 (x,y) = 2 ln x2+y2= 2 lnr (x,y) = 2 arctan(y/x) = 2 2It is easily verified that apart from the origin location (x,y) = (0,0), these functions satisfy 2 = 0 and 2 = 0, and hence represent physically-possible incompressible, origin location (0,0) is called asingular pointof the source flow. As we approach thispoint, the magnitude of the radial velocity tends to infinityasVr 1rHence the flow at the singular point is not physical, althoughthis does not prevent us fromusing the source to represent actual flows. We will simply need to ensure that the singularpoint is located outside the flow region of Flow with SourceTwo or more incompressible, irrotational flows can be combined bysuperposition, simply byadding their velocity fields or their potential or stream function fields.

4 Superposition of auniform flow in thex-direction and a source at the origin therefore hasu(x,y) = 2 xx2+y2+V v(x,y) = 2 yx2+y2or (x,y) = 2 ln x2+y2+V x= 2 lnr+V rcos or (x,y) = 2 arctan(y/x) +V y= 2 +V rsin The figure shows the streamlines of the two basic flows, and also the combined bullet-shaped heavy line on the combined flow corresponds to thedividing streamline,which separates the fluid coming from the freestream and the fluid coming from the we replace the dividing streamline by a solid semi-infinite body of the same shape, theflow about this body will be the same as the flow outside the dividing streamline in thesuperimposed Flow with Source and SinkWe now superimpose a uniform flow in thex-direction, with a source located at ( /2,0),and a sink of equal and opposite strength located at (+ /2,0), plus a freestream.

5 = 2 ( 1 2) +V rsin xy 2 1r1r2l The figure on the right shows the streamlines of the combined flow. The heavy line againindicates the dividing streamline, which traces out aRankine oval. All the streamlines insidethe oval originate at the source on the left, and flow into the sink on the right. The netvolume outflow from the oval is zero. Again, the dividing streamline could be replaced by asolid oval body of the same shape. The flow outside the oval then corresponds to the flowabout this a source-sink pair with strengths , located at ( /2,0). Now let the separationdistance approach zero, while simultaneously increasing the sourceand sink strengths suchthat the product remains constant.

6 The resulting flow is adoubletwith strength . = lim 0 =const. 2 = 2 sin rxyrl xy 4A similar limiting process can be used to produce the doublet s potential function. = 2 cos rThe streamline shapes of the doublet are obtained by setting = 2 sin r=c= constantr=dsin whered= 2 cIn polar coordinates this is the equation for circles of diameterd, centered onx,y= (0, d/2).Nonlifting Flow over Circular CylinderFlowfield definitionWe now superimpose a uniform flow with a doublet. =V rsin 2 sin r=V rsin (1 2 V r2)or =V rsin (1 R2r2)whereR2 /(2 V )This corresponds to the flow about a circular cylinder of radial and tangential velocities can be obtained by differentiating the stream functionas =V cos (1 R2r2)V = r= V sin (1 +R2r2)5 Surface velocities and pressuresOn the surface of the cylinder wherer=R, we haveVr= 0V = 2V sin The maximum surface speed of 2V occurs at = 90.

7 The surface pressure is then obtained using the Bernoulli equationp( ) =po 12 (V2r+V2 )SubstitutingVr= 0 andV ( ), and using the freestream value for the total pressure,po=p +12 V2 gives the following surface pressure ( ) =p +12 V2 (1 4 sin2 )The corresponding pressure coefficient is also readily ( ) p( ) p 12 V2 = 1 4 sin2 6