Stream function in polar coordinate 𝐹𝑜𝑟 𝑐𝑎𝑟𝑑𝑖𝑧𝑖𝑎𝑛 𝑐𝑜𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒 𝜓=𝑓(𝑥,𝑦) With 𝑢=𝑉 𝑥 = 𝜕𝜓 𝜕𝑦 𝑎𝑛𝑑 𝑣=𝑉 𝑦 =− 𝜕𝜓 𝜕𝑥 d𝜓= 𝜕𝜓 𝜕𝑥 𝑑𝑥+ 𝜕𝜓 𝜕𝑦 𝑑y = -v dx + u dy 𝜓= 𝜕𝜓 𝜕𝑥 𝑑𝑥 + 𝜕𝜓 𝜕𝑦 𝑑y + C = −𝑣 𝑑𝑥 + u dy +𝐶 𝐹𝑜𝑟 polar coordinate 𝜓=𝑓(𝑟,𝜃) With 𝑉 𝑟 = 𝜕𝜓 𝑟𝜕𝜃 𝑎𝑛𝑑 𝑉 𝑡 =− 𝜕𝜓 𝜕𝑟 d𝜓= 𝜕𝜓 𝑟𝜕𝜃 𝑟𝑑𝜃+ 𝜕𝜓 𝜕𝑟 𝑑𝑟= 𝑉 𝑟 𝑟𝑑𝜃 - 𝑉 𝑡 dr 𝜓= 𝜕𝜓 𝑟𝜕𝜃 𝑟𝑑𝜃 + 𝜕𝜓 𝜕𝑟 𝑑𝑟 + C= 𝑉 𝑟 𝑟𝑑𝜃 - 𝑉 𝑡 dr + C

Basic Flowfields 2- Source and Sink 2-1 Source 𝝍= 𝒒 𝟒𝝅 Source For the following flowfield 𝑣 𝑟 = 𝜕𝜓 𝑟𝜕𝜃 = 𝑞 2π𝑟 𝑎𝑛𝑑 𝑉 𝑡 =0 𝝍= 𝒒 𝟐 𝝍= 𝟎 𝜓= 𝜕𝜓 𝑟𝜕𝜃 𝑟𝑑𝜃 + 𝜕𝜓 𝜕𝑟 𝑑𝑟 + C = 𝑉 𝑟 𝑟𝑑𝜃 - 𝑉 𝑡 dr + C = ( 𝑞 2π𝑟 ) 𝑟𝑑𝜃 - (0) dr + C = 𝑞 2π 𝜃 + C Let 𝜓=0 at 𝜃 = 0 ∴ 0 = 𝑞 2π 0 + C C = 0 which means that 𝝍= 𝒒 𝟐𝝅 𝜽 𝝍= 𝟑𝒒 𝟒 𝝍= - 𝒒 𝟒𝝅 Sink 2-2 Sink In this case the stream function will be the inverse of source function ∴ 𝝍= - 𝒒 𝟐𝝅 𝜽 𝝍=- 𝒒 𝟐 𝝍= 𝟎 𝝍= −𝟑𝒒 𝟒

2- Source and Sink of Equal Strength Some Useful Combined Flowfields 2- Source and Sink of Equal Strength F𝑜𝑟 𝑠𝑜𝑢𝑟𝑐𝑒 𝝍 𝑨 = 𝒒 𝟐𝝅 𝜽 𝟏 For sink 𝝍 𝑩 = - 𝒒 𝟐𝝅 𝜽 𝟐 For combined flow 𝝍=𝝍 𝑨 + 𝝍 𝑩 or 𝜓 = 𝒒 𝟐𝝅 𝜽 𝟏 − 𝒒 𝟐𝝅 𝜽 𝟐 = 𝒒 𝟐𝝅 ( 𝜽 𝟏 − 𝜽 𝟐 ) For any point P , note that for the angle (𝜶) between r1 and r2 is 𝜶= 𝜽 𝟐 − 𝜽 𝟏 𝝍 =- 𝒒 𝟐𝝅 𝜶 For constant 𝝍 at any stream line, the 𝜶 is also a constant . That means all the stream line is circles through the source and sink 𝑭𝒐𝒓 𝒄𝒂𝒓𝒅𝒊𝒛𝒊𝒂𝒏 𝒄𝒐𝒐𝒓𝒅𝒊𝒏𝒂𝒕𝒆 𝜽 𝟏 = 𝒕𝒂𝒏 −𝟏 𝒚 𝒙+𝒂 and 𝜽 𝟐 = 𝒕𝒂𝒏 −𝟏 𝒚 𝒙−𝒂 ∴ 𝝍 = 𝒒 𝟐𝝅 𝒕𝒂𝒏 −𝟏 𝒚 𝒙+𝒂 − 𝒕𝒂𝒏 −𝟏 𝒚 𝒙−𝒂

3- Source and Sink of Equal Strength in Rectilinear Flow Some Useful Combined Flowfields 3- Source and Sink of Equal Strength in Rectilinear Flow F𝑜𝑟 Source and Sink of Equal Strength 𝝍 𝑨 = 𝒒 𝟐𝝅 𝒕𝒂𝒏 −𝟏 𝒚 𝒙+𝒂 − 𝒕𝒂𝒏 −𝟏 𝒚 𝒙−𝒂 For Rectilinear Flow 𝝍 𝑩 = U y So, For combined flow 𝝍=𝝍 𝑨 + 𝝍 𝑩 or 𝝍 = 𝒒 𝟐𝝅 𝒕𝒂𝒏 −𝟏 𝒚 𝒙+𝒂 − 𝒕𝒂𝒏 −𝟏 𝒚 𝒙−𝒂 + U y 𝑢=𝑉 𝑥 = 𝜕𝜓 𝜕𝑦 = 𝒒 𝟐𝝅 𝟏 𝒙+𝒂 𝟏+ 𝒚 𝒙+𝒂 𝟐 − 𝟏 𝒙−𝒂 𝟏+ 𝒚 𝒙−𝒂 𝟐 + U At point sL (-L/2 , 0) the velocity u = 0 ∴ 0 = 𝒒 𝟐𝝅 𝟏 ( 𝑳 𝟐 +𝒂) − 𝟏 ( 𝑳 𝟐 −𝒂) + U Or 𝑳 𝟐 =𝒂 𝟏+ 𝒒 𝒂𝝅𝑼 ( prove it ) at this point sL (-L/2 , 0) 𝝍 = 𝒒 𝟐𝝅 𝒕𝒂𝒏 −𝟏 𝟎 − 𝒕𝒂𝒏 −𝟏 𝟎 + U (0) = 𝒒 𝟐𝝅 (𝝅) = 𝒒 𝝅

Some Useful Combined Flowfields For the equation of the body 𝒒 𝝅 = 𝒒 𝟐𝝅 𝒕𝒂𝒏 −𝟏 𝒚 𝒙+𝒂 − 𝒕𝒂𝒏 −𝟏 𝒚 𝒙−𝒂 + U y Or 0 = 𝒒 𝟐𝝅 𝒕𝒂𝒏 −𝟏 𝒚 𝒙+𝒂 − 𝒕𝒂𝒏 −𝟏 𝒚 𝒙−𝒂 + U y - 𝒒 𝟐𝝅 The distance b could be find from the point (0, b/2) as follows From 𝟎 = 𝒒 𝟐𝝅 𝒕𝒂𝒏 −𝟏 𝒚 𝒙+𝒂 − 𝒕𝒂𝒏 −𝟏 𝒚 𝒙−𝒂 + U y - 𝒒 𝝅 𝟎 = 𝒒 𝟐𝝅 𝒕𝒂𝒏 −𝟏 𝒃/𝟐 𝟎+𝒂 − 𝒕𝒂𝒏 −𝟏 𝒃/𝟐 𝟎−𝒂 + U 𝒃 𝟐 - 𝒒 𝝅 𝟎 = 𝒒 𝟐𝝅 𝒕𝒂𝒏 −𝟏 𝒃 𝟐𝒂 − 𝒕𝒂𝒏 −𝟏 𝒃 −𝟐𝒂 + U 𝒃 𝟐 - 𝒒 𝝅 𝟎 = 𝒒 𝟐𝝅 𝒕𝒂𝒏 −𝟏 𝒃 𝟐𝒂 + 𝒕𝒂𝒏 −𝟏 𝒃 𝟐𝒂 + U 𝒃 𝟐 - 𝒒 𝝅 𝟎 = 𝒒 𝟐𝝅 𝟐 𝒕𝒂𝒏 −𝟏 𝒃 𝟐𝒂 + U 𝒃 𝟐 - 𝒒 𝝅 𝟎 = 𝒒 𝝅 𝒕𝒂𝒏 −𝟏 𝒃 𝟐𝒂 + U 𝒃 𝟐 - 𝒒 𝝅 Solve by trail and error to find b/2 Note that ∅=𝑼𝒓 𝒄𝒐𝒔𝜽 − 𝒒 𝟐𝝅 (𝒍𝒏 𝒓 𝟏 −𝒍𝒏 𝒓 𝟐 )

Some Useful Combined Flowfields Example 3 : Plot the body contour formed by a source at the origin of 40 π m3/(s.m) in a uniform horizontal stream (from left to right ) of velocity 10 m/s. 1- calculate the velocity at the body contour for the value of θ = π , (5/6) π, (4/6) π , π /2 and (2/6) π . 2- calculate the velocity at the radial distance r = 2, 3 and 4 for θ = π .