Height Differences Using Flat-Plate Boundary Layer Problem 7.24* Pamela Christian BIEN 301 Individual Project February 15, 2007

Problem Diagram

Problem Givens  Air flows past the flat plate shown in Figure p7.24 under laminar conditions at T=20° and P=1 atm.  The two equally stagnant tubes are each placed y=2mm from the wall.  The manometer fluid is a) water at 20° and b) glycerin at 20°.  The velocity (U) is constant at 15 m/s and length (L) is 50 cm.

Problem Objectives  Find the values of the manometer reading h 1 and h 2, in mm for –a) water –b) glycerin

Assumptions  Steady, velocity is constant.  Laminar boundary-layer flow  Constant pressure across the manometers  Incompressible Fluid

Equations and Constants  For air at 20°, v = 1.5e-5 m 2 /s, ρ = 1.2 kg/m 3  For water at 20°, ρ = 998 kg/m 3,  For glycerin at 20°, μ = 1.49 kg/(m*s), ρ = 1260 kg/m 3  h = Δp/(Δρ*g) –p = pressure –ρ = Density –g = gravity  u/U = f’(η) –u = local velocity –U = Velocity = 15 m/s –f’(η) = first derivative of the dimensionless velocity profile with respect to η. –η = (y)[U/(vL)] 1/2

Solution  First, find the velocities at each manometer opening  For the fluids (water and glycerin) in the manometer: –η 1 = (y)[U/(vL)] 1/2 = (.002m)[15m/s(1.5e-5 m 2 /s *.5 m)] 1/2 –η 1 = –η 2 = (y)[U/(vL)] 1/2 = (.002m)[15m/s(1.5e-5 m 2 /s * 1 m)] 1/2 –η 2 = 2.0 From Table 7.1 in White’s, for η 1 = ~ 2.8, f’(η) = and for η 2 = 2, f’(η) = and for η 2 = 2, f’(η) = Therefore, using u = U*f’(η) u 1 = m/s u 2 = 9.45 m/s

Solution  Using Bernoulli’s equation for steady, incompressible flow, the equilibrium equation becomes: Δp/ρ air + ½(Δu 2 ) + g(Δz) = 0  With the assumption that the height between the two openings of the manometer is negligible to the pressure difference of the water/glycerin, the Bernoulli's equation becomes: Δ p = ½(ρ air )(u 2 ) Δ p = ½(ρ air )(u 2 ) –Therefore: Δ p 1 = ½(1.2 kg/m 3 )(12.17 m/s) 2 = 88.9 Pa Δ p 2 = ½(1.2 kg/m 3 )(9.45 m/s) 2 = 53.6 Pa

Solution  Finally, using the equation for Δ p in a manometer, the change in height h 1 and h 2 can be found for each the water and glycerin manometers.  Δ p = Δ ρ*g* Δ h  For the water manometer: –Δh = h 1 = 88.9 Pa/[(998 kg/m kg/m 3 ) * 9.81 m/s 2 ] = –h 1 = 9.09 mm –Δh = h 2 = 53.6 Pa/[(998 kg/m kg/m 3 ) * 9.81 m/s 2 ] = – h 2 = 5.48 mm

Solution  For the glycerin manometer: –Δh = h1 = 88.9 Pa/[(1260 kg/m^ kg/m^3)*9.81 m/s^2] = –h1 = 7.2 mm –Δh = h2 = 53.6 Pa/[(1260 kg/m^ kg/m^3)*9.81 m/s^2] = –h2 = 4.34 mm

Biomedical Application  When running tests using different types of fluids in the body, this method is helpful in finding the effects due to a difference in density of water or blood on that particular fluid inside the body.  Therefore, a more accurate prediction can be determined from the information collected by the fluid, and factors such as gravity are taken into account.