Potential flow:

10 CHAPTER 1. POTENTIAL FLOW:=( 2πR 3 ρ3= M bdU idt) dUidt(1.21)Here, the outward unit normal to the sphere is given by n i = (x i /r), and wehave used the identity∫dSx i x j = 4πR4 δ ij(1.22)S 3sphereto obtain the third step in equation 1.22. Not surprisingly, the force requried toaccelerate the particle through the fluid is the product of the added mass andthe acceleration.1.2.2 General three-dimensional potential flows:Some general results can be obtained for the form of the velocity potential for anobject of arbitrary shape translating through a fluid in potential flow. The solutionfor the velocity potential is, in general, more complicated than the solution1.6 for the motion of a sphere. However, the leading contribution to the potentialis still the dipole contribution which decays proportional to (1/r 2 ) in threedimensions. There is no source term proportional to (1/r) due to the incompressibilitycondition and the constant volume of the object. However, higherorder terms which decay faster than (1/r 2 ) could be present in the solution.Equation 1.18 for the pressure is valid for an object of arbitrary shape, sinceno assumption was made regarding the specific form of the velocity potentialwhile deriving the pressure.It is possible to show that the force required for the steady motion of anobject of arbitrary shape through the fluid,∫F i = dSpn iS object∫( )u2j= ρ dSn i S 2 − U ju j (1.1)objectis equal to zero. Consider the volume integral∫VdV∂∂x i(u2j2 − U ju j)(1.2)evaluated over the volume of the fluid surrounding the fluid. Using the divergencetheorem, this volume integral can be written as,∫ ( )dV∂ u2 ∫ ( )ju2 ∫( )V ∂x i 2 − U ju2ju j = dSn iS ∞2 − U jju j − dSn i S 2 − U ju jobject