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

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.

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.

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

[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

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

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

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

The End