Question

PIU 15 a 2.8) Olive oil (u = 3 X 10-4 slug/ftos, v = 1.8 x 10-4 ft2/s) flows over a wooden plank with a velocity profile given by u = 53y - 1215y2 ft/s where y is the distance from the plank in feet. What is the shear stress at the surface of the plank, and where is it a maxi- mum? fin mn chowe air at 150 flows

Answer 1

For all the calculation sake, I solve this on paper.

Given DATA: Dlive oil > = 3x10 " sug/it's *= 1.8X104 ft Velocity Poppile u= 53y - 1215 y? ft/s p. u= 534-1215 y 2 du = 53-2430 y / ŠO, T: (3x10+) / 53-2430 y ) at the As Olive oil is a Newtonian fluid, it must follow this egna sheas stress a rate of straen t - udr Edt • Now, when fluid moves parallely, Hic for the small time at Air can be written as y = dy at Hot and for small sy (change in length) ? surface of plank, y=0 T = (3x10-9) X 53 :- 159x10-9 sugltt? Por maximum shear force, die -2430 (-ve sign show maximum value) do 53 -2430 y = 0 4.53 = 0.021847 av de 243 0 Therefore, our equation become . t = le du Kohere, de lite dynamic viscosity = 3x10" slugift's y = ft., distance Therefore, maximum shear force T = (3x104) ( 53 - 2430X0.0218 ) - 13x10-4 0.026) 2 7.8x10-6 sengitt - Maximum Slicas stress is at surface, e. 159x1o-stay