Height Differences Height Differences Using Flat-Plate Boundary LayerUsing Flat-Plate Boundary Layer

Problem 7.24*Problem 7.24*

Pamela ChristianPamela ChristianBIEN 301BIEN 301

Individual ProjectIndividual Project

February 15, 2007February 15, 2007

Problem DiagramProblem Diagram

Problem GivensProblem Givens

Air flows past the flat plate shown in Figure p7.24 Air flows past the flat plate shown in Figure p7.24 under laminar conditions at T=20under laminar conditions at T=20° and P=1 atm. ° and P=1 atm.

The two equally stagnant tubes are each placed The two equally stagnant tubes are each placed y=2mm from the wall. y=2mm from the wall.

The manometer fluid is a) water at 20° and b) The manometer fluid is a) water at 20° and b) glycerin at 20°.glycerin at 20°.

The velocity (U) is constant at 15 m/s and length The velocity (U) is constant at 15 m/s and length (L) is 50 cm. (L) is 50 cm.

Problem ObjectivesProblem Objectives

Find the values of the manometer Find the values of the manometer reading hreading h11 and h and h22, in mm for , in mm for – a) water a) water – b) glycerinb) glycerin

AssumptionsAssumptions

Steady, velocity is constant. Steady, velocity is constant. Laminar boundary-layer flowLaminar boundary-layer flow Constant pressure across the manometersConstant pressure across the manometers Incompressible FluidIncompressible Fluid

Equations and Constants Equations and Constants

For air at 20For air at 20°, v = 1.5e-5 m°, v = 1.5e-5 m22/s, /s, ρρ = 1.2 kg/m = 1.2 kg/m33

For water at 20°, For water at 20°, ρρ = 998 kg/m = 998 kg/m33, , For glycerin at 20°, For glycerin at 20°, μμ = 1.49 kg/(m*s), = 1.49 kg/(m*s), ρρ = 1260 kg/m = 1260 kg/m33

h = h = ΔΔp/(p/(ΔρΔρ*g)*g)– p = pressurep = pressure– ρρ = Density = Density– g = gravityg = gravity

u/U = f’(u/U = f’(ηη))– u = local velocityu = local velocity– U = Velocity = 15 m/sU = Velocity = 15 m/s– f’(f’(ηη) = first derivative of the dimensionless velocity profile with respect to ) = first derivative of the dimensionless velocity profile with respect to ηη. .

– ηη = (y)[U/(vL)] = (y)[U/(vL)]1/21/2

SolutionSolution

First, find the velocities at each manometer openingFirst, find the velocities at each manometer opening For the fluids (water and glycerin) in the For the fluids (water and glycerin) in the

manometer:manometer:– ηη11 = (y)[U/(vL)] = (y)[U/(vL)]1/21/2 = (.002m)[15m/s(1.5e-5 m = (.002m)[15m/s(1.5e-5 m22/s *.5 m)]/s *.5 m)]1/21/2

– ηη11 = 2.828 = 2.828

– ηη22 = (y)[U/(vL)] = (y)[U/(vL)]1/21/2 = (.002m)[15m/s(1.5e-5 m = (.002m)[15m/s(1.5e-5 m22/s * 1 m)]/s * 1 m)]1/21/2

– ηη22 = 2.0 = 2.0

From Table 7.1 in White’s, for From Table 7.1 in White’s, for ηη11 = 2.828 ~ 2.8, f’( = 2.828 ~ 2.8, f’(ηη) = .81152) = .81152

and for and for ηη22 = 2, f’( = 2, f’(ηη) = .62977) = .62977

Therefore, using u = U*f’(Therefore, using u = U*f’(ηη))

uu11 = 12.17 m/s = 12.17 m/s

uu22 = 9.45 m/s = 9.45 m/s

SolutionSolution

Using Bernoulli’s equation for steady, incompressible Using Bernoulli’s equation for steady, incompressible flow, the equilibrium equation becomes:flow, the equilibrium equation becomes:

ΔΔp/p/ρρairair + ½( + ½(ΔΔuu22) + g() + g(ΔΔz) = 0z) = 0 With the assumption that the height between the two With the assumption that the height between the two

openings of the manometer is negligible to the openings of the manometer is negligible to the pressure difference of the water/glycerin, the pressure difference of the water/glycerin, the Bernoulli's equation becomes:Bernoulli's equation becomes:

ΔΔ p = ½( p = ½(ρρairair)(u)(u22))

– Therefore: Therefore: ΔΔ p p11 = ½(1.2 kg/m = ½(1.2 kg/m33)(12.17 m/s))(12.17 m/s)22 = 88.9 Pa = 88.9 Pa ΔΔ p p22 = ½(1.2 kg/m = ½(1.2 kg/m33)(9.45 m/s))(9.45 m/s)22 = 53.6 = 53.6

PaPa

SolutionSolution

Finally, using the equation for Finally, using the equation for ΔΔp in a manometer, p in a manometer, the change in height hthe change in height h11 and h and h22 can be found for can be found for each the water and glycerin manometers. each the water and glycerin manometers.

ΔΔp = p = ΔΔρρ*g* *g* ΔΔh h For the water manometer:For the water manometer:

– ΔΔh = hh = h11 = 88.9 Pa/[( = 88.9 Pa/[(998 kg/m998 kg/m33 - - 1.2 kg/m1.2 kg/m33) * 9.81 m/s) * 9.81 m/s22] =] =

– hh11 = = 9.09 mm 9.09 mm

– ΔΔh = hh = h22 = 53.6 Pa/[( = 53.6 Pa/[(998 kg/m998 kg/m33 - - 1.2 kg/m1.2 kg/m33) * 9.81 m/s) * 9.81 m/s22] =] =

– hh22 = = 5.48 mm 5.48 mm

SolutionSolution

For the glycerin manometer:For the glycerin manometer:– ΔΔh = h1 = 88.9 Pa/[(h = h1 = 88.9 Pa/[(1260 kg/m^3 - 1260 kg/m^3 - 1.2 kg/m^31.2 kg/m^3)*9.81 )*9.81

m/s^2] = m/s^2] =

– h1h1 = 7.2 mm = 7.2 mm

– ΔΔh = h2 = 53.6 Pa/[(h = h2 = 53.6 Pa/[(1260 kg/m^3 - 1260 kg/m^3 - 1.2 kg/m^31.2 kg/m^3)*9.81 )*9.81 m/s^2] = m/s^2] =

– h2h2 = 4.34 mm = 4.34 mm

Biomedical ApplicationBiomedical Application

When running tests using different When running tests using different types of fluids in the body, this method types of fluids in the body, this method is helpful in finding the effects due to a is helpful in finding the effects due to a difference in density of water or blood difference in density of water or blood on that particular fluid inside the body.on that particular fluid inside the body.

Therefore, a more accurate prediction Therefore, a more accurate prediction can be determined from the can be determined from the information collected by the fluid, and information collected by the fluid, and factors such as gravity are taken into factors such as gravity are taken into account. account.