Lecture Week 3

Chapter 7 White: Boundary-Layer Flows

%Introduction to boundary layers

%Karman's integral method review

%Extra problem discussion

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Purpose

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The purpose of boundary layer theory is to describe the flow external to a body immersed in a fluid stream. Close to the body the flow is affected by viscous forces. Turbulence may develop. Far from the body potential flow may develop. Matching such solutions is an important part of the process of solving boundary layer problems. Often no solution is possible because the flow separates and becomes turbulent. This is a major problem to be avoided in the design of wings, but is unavoidable for blunt objects. A large number of empirical formulas and drag coefficient plots are given in the Chapter to describe lift and drag coefficients determined experimentally for a wide variety of object shapes.

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Laminar versus turbulent boundary layers

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No satisfactory boundary layer theory exists for Reynolds numbers 1 to 1000. For larger Re values, the accepted expressions for boundary layer thickness \delta (99% = u/U) are:

** \delta / x = 5.0 / Re_x ^{1/2} laminar (Blasius profile)

** \delta / x = 0.16/ Re_x ^{1/7} turbulent

where Re_x = Ux/\nu is the local Reynolds number based on the distance x from the beginning of the boundary layer.

As shown in the table of formula values, the boundary layer is thin, never over 5% :

Re_x	10^4	10^5	10^6	10^7	10^8
\delta / x _ lam	0.050	0.016	0.005
\delta / x _ turb.			0.022	0.016	0.011

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Karman momentum integral

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Recall the derivation of drag D(x) = \rho b int [0,\delta] u(U-u) dy from Chapter 3, using the Reynolds transport theorem for mass and momentum for a control volume bounded by stream surfaces h from the wall upstream and \delta from the wall downstream, b wide and L long. In terms of the momentum thickness \theta:

** D(x) = \rho b U^2 \theta, where

** \theta = int [0,\delta] u(U-u)\U^2 dy .

Increasing \theta gives increasing drag. The drag is also the integrated shear stress

** D(x) = b int[0,x] \tau_w(x) dx, or

** dD/dx = b \tau_w.

But

** dD/dx = \rho b U^2 d\theta/dx,

so

** \tau_w = \rho U^2 d\theta/dx, valid for either laminar or turbulent flow.

Recall Karman assumed a parabolic velocity profile u(x,y) = U(2y/\delta - y^2/\delta^2) which gives \theta = 2/15 \delta. The stress at the wall \tau = 2 \mu U / \delta from Karman's profile, and \rho U^2 d\theta/dx. Differentiate \theta = 2/15 \delta and integrate to give

** \delta / x = 5.5 Rex^1/2 ; which is within 10% of the Blasius profile result.

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Extra problem discussion:

As a review for Exam 1, problems 6.23, 44, and 140 should be considered since they illustrate important aspects of viscous and turbulent flows in tubes and the use of so-called Bernoulli effect meters (eg. orifice meters).

6.23 Gycerin at 20 deg C is to be pumped through a horizontal smooth pipe at 3.1 m^3/hr (note there is a missprint in the text). It is desired that (1) the pressure drop be no more than 100 Pa/m and (2)the flow be laminar. What is the minimum pipe diameter allowable?

Solution: Look up \rho = 1260 kg/m^3 and \mu = 1.49 kg/m s in Appendix A. Compute d from each of the two constraints; that is, set delta p / L = 128 \mu Q/\pi d^4 <= 100 Pa/m and Re_d = 4\rhoQ/\pi \mu d <= 2000. The first constraint gives d >= 0.15 m and the second gives d >= 0.00046. Thus d = 0.15 will satisfy both constraints. Check out the case for Q=3.1 m^3/s. Where do the formulas come from?

6.44 Mercury at 20 deg C flows through 4 m of 7 mm diameter glass tubing at an average velocity of 3 m/s. Estimate the head loss in m and the pressure drop in kPa.

Solution: Look up \rho = 13550 kg/m^3 and \mu = 0.00156 kg/m s in Appendix A for mercury. Glass is considered smooth (this means the roughness is much smaller than the expected Kolmogorov scale). Compute the Reynolds number:

Re_d = 13550(3)(0.007)/0.00156 = 182000 (turbulent); Moody chart or Eq. 6.64 gives f = 0.0159.

h_f = f (L/d) (V^2/2g) = (0.0159)(4/0.007)(3^2/2x9.81) = 4.18 m

delta p = \rho g h_f = (13550)(9.81)(4.18) = 555000 Pa or 555 kPa. Check the formulas. What power is required to pump the fluid through the tubing?

6.140 Kerosine at 20 deg C flows at 18 m^3/h in a 5 cm diameter pipe. If a 2 cm diameter thin plate orifice with corner taps is installed, what will the measured drop be, in Pa?

Solution: Look up \rho = 804 kg/m^3 and \mu = 192E-3 kg/ m s in Appendix A for kerosine. The orifice plate ratio \beta = d/D = 2/5 = 0.4. The pipe velocity and Reynolds numbers are:

V=Q/A = (18/3600)/(\pi/4)(0.05^2) = 2.55 m/s; Re_d = 804(2.55)(0.05)/1.92E-3 = 53300.

Use Eq. 6.132 and 6.133a, or Fig 6.37 to estimate C_D=0.6030. Then the orifice pressure drop formula predicts Q= 18/3600 = 0.6030(\pi/4)(0.02^2)(2 delta p /(804)(1-(0.4^4))^1/2 , solve for delta p to give 273 kPa.

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