Fluids Midterm Exam 2 Solutions

Midterm Exam II Solution, ME 342 Fluid Mechanics (Spring 2013) 1. Consider a steady, incompressible, y=+h b viscous flow of viscosity  due to y a pressure gradient x inside a channel of two fixed u=u(y) plates at a distance of 2h, as shown in the figure. Neglect a gravity effect. The channel width y=h b is W and it is very long compared to the channel length L (i.e., W>>L) so that it is still valid to assume that the channel flow is a two-dimensional plane flow ( ) and an axial flow (v = w = 0). In contrast to conventional no-slip boundary conditions, consider a slip boundary condition as shown in the figure, where us is defined as a slip velocity at the wall and b as a slip length (distance where the no-slip boundary condition will virtually be satisfactory by linear extrapolation). The relationship between the slip velocity and slip length is defined as ( ) . (a) Solve the momentum (Navier-Stokes) equation to find out u(y) and a maximum velocity umax (i.e., u at y=0) with the considered slip boundary condition. Use the definition of a slip velocity, ( ) , so that your solutions for the velocities should include the slip length b in the equations. (20 pt.)

u 

  2u  2u   u u  p  v   g x    2  2   x y  x y   x   

Solving Navier-Stokes Eq 10 points

  2u  p 0     2   y  x  

1 p y 2 u( y)   C1 y  C2  L 2

Applying BCs (for no slip condition) at y  h , u  0 or BCs 6 points at ,

C1  0 and C2  
For slip condition uslip  b p h L

p h 2 L 2

2 points

2 points

Midterm Exam II Solution, ME 342 Fluid Mechanics (Spring 2013) (b) Compute the volume flow rate (Q) and the average velocity (Vav). The parallel plates can be considered as a square channel so that the cross-sectional area of the channel can be estimated to be 2hW. Your solutions should include the slip length b in the equations. (20 pt.)
Q   udA ,
A  2hW

4 points

Q

2Wh3 p 2h2W p  b 3 L  L
QPoiseuille Qslip

6 points

Vave 

Q 1 