1. Fluid Flow Across a Vertical Surface
Group #10
Andrea Watson
Rena Armstrong
Oluseyi Harris
Crystal Casey

2. Fluid flows across a surface with uniform velocity v
Fluid flows across a surface with uniform velocity v. The surface has height H and width W as shown. Give and expression for the mass flow rate m across the surface is ρ is not constant. Simplify as much as possible.

3. Mass flow rate (m) : m =    (vn) dA m =    [v] cos dA
The velocity is uniform, so m = [v] cos -W/2W/2 -H/2H/2  dydz
We need  as a function of y and z to evaluate.

4. Volumetric flow rate (Q):
Q=   (vn) dA
Q=   (vn) dydz
Q=  [v] [n] cos y dz
Q= [v] [1] cos y z
Q=H W [v] cos

5. [v] cos = v  n = 9m/s(½3 i+½ j)  I [v] cos = 9m/s(½3) = 7.79 m/s
For a constant density of 55 lbm/ft3 and v=9m/s(½3 i+½ j), what is the numerical value of Q?
[v] cos = v  n = 9m/s(½3 i+½ j)  I
[v] cos = 9m/s(½3) = 7.79 m/s
Thus, Q = (3.5m)(2.0m)(7.79m/s)
Q = 54.5m3/s

6. What is the value of the x-momentum flux across the surface in metric units?
Px =   vx(vn) dA
Px = vx [v] cos HW
vx is positive, so we can drop absolute value.
Px = vx2 HW
 = 55lbm/ft3 (999kg/m3/62.4 lbm/ft3) = 881 kg/m3
Px=(881kg/m3)(7.79m/s)2(3.5m)(2.0m)
Px = 374,000 kg m/s2

7. Find the value of the y-momentum flux across the surface.
Py =    vy(vn) dA
Py =  vy [v] cos HW
Py =  vy vx HW
vy = [v] sin = 4.50 m/s
Py=(881kg/m3)(4.50m/s)(7.79m/s) (3.50 m)(2.0m)
Py= 216,000 kg m/s2

8. What is the z-momentum flux across the surface?
There is no z-component to the velocity.
Pz = 0

9. The End