Fluid Flow Across a Vertical Surface Group 10

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. 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) : l l m = (v n) d. A m = [v] cos d. A The velocity is uniform, so m = [v] cos -W/2 -H/2 dydz We need as a function of y and z to evaluate.

Volumetric flow rate (Q): l l l Q= (v n) d. A Q= (v n) dydz Q= [v] [n] cos y dz Q= [v] [1] cos y z Q=H W [v] cos

For a constant density of 55 lbm/ft 3 and v=9 m/s(½ 3 i+½ j), what is the numerical value of Q? l [v] cos = v n = 9 m/s(½ 3 i+½ j) I l [v] cos = 9 m/s(½ 3) = 7. 79 m/s l Thus, Q = (3. 5 m)(2. 0 m)(7. 79 m/s) l Q = 54. 5 m 3/s

What is the value of the x-momentum flux across the surface in metric units? l l l l Px = vx(v n) d. A Px = vx [v] cos HW vx is positive, so we can drop absolute value. Px = vx 2 HW = 55 lbm/ft 3 (999 kg/m 3/62. 4 lbm/ft 3) = 881 kg/m 3 Px=(881 kg/m 3)(7. 79 m/s)2(3. 5 m)(2. 0 m) Px = 374, 000 kg m/s 2

Find the value of the y-momentum flux across the surface. l l l Py = vy(v n) d. A Py = vy [v] cos HW Py = vy vx HW vy = [v] sin = 4. 50 m/s Py=(881 kg/m 3)(4. 50 m/s)(7. 79 m/s) (3. 50 m)(2. 0 m) Py= 216, 000 kg m/s 2

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

The End