By Roel Snieder

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This means that the flux of the flow through any closed surface S must be zero. ”) This means that the divergence of the flow is zero, except possibly near the source at r = 0: (∇ · v) = 0 f or r =0. 5) In addition we know that due to the symmetry of the problem the flow is directed in the radial direction and depends on the radius r only: v(r) = f (r)r. 6) Problem a: Show this. This is enough information to determine the flow field. 5) because we have not yet derived an expression for the divergence in cylinder coordinates.

2: Definition of the geometric variables for the computation of the divergence in cylinder coordinates. 2). Let us first consider the flux of v through the surface elements perpendicular to ˆ r. The size of this surface is rdϕdz and (r + dr)dϕdz respectively at r and r + dr. The normal components of v through these surfaces are vr (r, ϕ, z) and vr (r + dr, ϕ, z) respectively. Hence the total flux through these two surface is given by vr (r + dr, ϕ, z)(r + dr)dϕdz − vr (r, ϕ, z)(r)dϕdz. ∂ (rvr ) drdϕdz.

THE DIVERGENCE OF A VECTOR FIELD The constant A is yet to be determined. Let at the source r = 0 a volume V per unit time be injected. Problem d: Show that V = v·dS (where the integration is over an arbitrary surface around the source at r = 0). By choosing a suitable surface derive that v(r) = V ˆ r . 9) From this simple example of a single source at r = 0 more complex examples can be obtained. Suppose we have a source at r+ = (L, 0) where a volume V is injected per unit time and a sink at r− = (−L, 0) where a volume −V is removed per unit time.