Homework set 3. Problems 7.1,6,8,15,16,25.

7.1 Air at 20 deg C and 1 atm flows at 20 m/s past a thin flat plate. Estimate the distance x from the leading edge at which the boundary thickness will be (a) 1 mm and (b) 10 cm. Assume Re_x, crit = 5 E 5.

Solution: for air at 20 deg C, take \rho = 1.2 kg/m^3 and \mu = 1.8 E-5 kg/m s. Guess whether the b. l. is laminar or turbulent: For (a), \delta = 1 mm, guess laminar. \delta / x laminar = 5.0 / Re_x ^1/2 , or x = \delta^2 / 25 \rho U / \mu (substitute values) = 0.0533 meters. Check Re_x = 71000. ok, laminar.

(b) For the thicker b. l., guess turbulent flow; \delta/x turbulent = 0.16 / Re_x ^1/7. Solve for x = (\delta/0.16)^7/6 (\rho U/\mu)^1/6 (substitute values) = 6.06 m. Check Re_x = 8.1 E6, ok, turbulent.

7.6 For the laminar parabolic boundary layer profile of Eq. 7.6, compute the shape factor "H" and compare with the exact Blasius theory result Eq. 7.1.

Solution: Given the profile approximation u/U = 2\eta - \eta^2, where \eta = y/\delta, compute

\theta = int[0,\delta]u/U(1-u/U)dy = \delta int[0,1](2\eta - \eta^2)(1 - 2\eta + \eta^2) d\eta = 2 \delta/15.

\delta* = int[1,1](1-u/U) dy = \delta int[0,1] (1 -2\eta + \eta^2) d\eta = \delta/3.

Hence H = \delta* / \theta = (\delta/3)/(2/\delta/15) = 2.5 (compared to 2.59 for the Blasius solution).

7.8 Repeat the flat-plate momentum integral analysis of section 7.2 by replacing the parabolic profile, Eq. 7.6 with the linear profile u/U = y/\delta. Computec_f, \theta/x, \delta/x, and H.

Solution: Introduce this linear profile into the Karman integral relation Eq. 7.5:

\tau_w = \mu du/dy|_w = \mu U/\delta = d/dx int[0,\delta] \rho (Uy/\delta)[1 - (Uy/\delta)] dy = \rho U^2/6 d\delta/dx,

or: \delta d\delta = 6 \nu /U dx , integrate to \delta ^2 = 12 \nu x / U , or \delta / x = 3.46 / (Re_x)^1/2 (about 35% low).

With \delta known, introduce \delta(x) int the difinitions of c_f, \theta, and \delta* to complete the work:

\theta = int[0,\delta] (y/\delta)(1-y\delta) dy = \delta/6 , hence \theta/x = 0.577/(Re_x)^1/2

\delta* = int[0,\delta] (1- y/\delta) dy = \delta/2 , hence \delta* / x = 1.732 / (Re_x)^1/2

Hence, H = \delta*/\theta = 3.0. Finally, c_f = \mu U/\delta / (1/2) \rho U^2 = 0.577/(Re_x)^1/2 (about 13% low)

7.15 Discuss whether fully developed laminar incompressible flow between parallel plates, Eq. 4.143 and Fig. 4.16b, represents an exact solution to the boundary layer euations 7.19 and the boundary conditions 7.20. In what sense, if any, are duct flows also boundary layer flows?

The analysis for flow between parallel plates leads to 4.143:

u = (dp/dx) (h^2/2\mu) (1 - y^2/h^2) , v = 0; dp/dx = constant < ) ; dp/dy = 0, u(+-h) = 0.

This is indeed a boundary layer, with v << u and dp/dy =0. The "freestreem" is the centerline velocity u_max = (-dp/dx)(h^2/2\mu). The b. l. does not grow because it is constrained by the two walls. The entire duct is filled with boundary layer.

7.16 A thin flat plate 55 by 110 cm is immersed in a 6 m/s stream of SAE 30 oil at 20 deg C. Compute the total friction drag if the stream is parallel to (a) the long side and (b) the short side.

Solution: For SAE 30 oil at 20 deg C, take \rho = 891 kg/m^3 and \mu = 0.29 kg/m s. Part (a):

L=110 cm, Re_L = 981(6.0)(1.1)/0.29 = 20300 (laminar), C_D = 1.328/(20300)^1/2 = 0.00933

Thus, F = C_D (\rho/2)U^2(2bL) = 0.00933(891/2)(6^2)[2(0.55)(1.1)] = 181 Newtons.

Part (b): L=55 cm, Re_L = 10140, C_D = 0.0132, F= 256 N (41% more).

7.25 Modify Prob. 7.24 to the following somewhat more difficult scenario. Let the known data be U = 15 m/s and h_1 = 8 mm of water. Use this information to determine (a) L, in cm, and (b) h_2, in mm.

Solution: For air at 20 deg C, take \rho = 1.2 kg/m^3 and \mu = 1.8 E-5 kg/m s. We can first use h_1 to find u_1 and then \eta_1 and finally x_1 (similar to prob. 7.2):

delta p_1,manometer = (\rho_w - \rho_air)gh_1 = (998-1.2)(9.81)(0.008) = 78.2 Pa

u_1 = (2 delta p_1 /\rho)^1/2 = [2(78.2)/1/2]^1/2 = 11.42 m/s, hence f' = 11.42/15 = 0.761

Table 7.1: read \eta_1 = 2.55 = y(U/\nux)^1/2, so x_1 = L = (15/1.5E-5)(0.002/2.55)^2 = 62 cm (ans a)

With L known, now we know x_2 = 2L and we can proceed to find \eta_2, u_2, and h_2:

\eta_2 = y(U/\nu x_2) = (0.002) (15/1.5E-5 (1.24))^1/2 = 1.80, Table 7.1, read f' = 0.575,

Thus, u_2 = 15(0.575) = 8.63 m/s , delta p_2 = 1.2/2 (8.63)^2 = 45 Pa = (998 - 1.2)(9.81)h_2 , or h_2 = 4.6 mm (ans b).