Homework hints, first problem set: 7.1,3,8,15,16,25,35,59,66,69,71,93

7.1. Air at 20 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 layer thickness will be a. 1 mm and b. 10 cm. Assume Rex,crit = 5E5.

Look up \rho for air to be 1.2 kg/m^3, and \mu = 1.8E-5 kg/m s. For a., guess laminar:

\delta lam / x = 5.0 / Re_x^{1/2}. Solve for x = 0.0533 m. Check Rex = 71000, ok.

b. guess turbulent flow

\delta turb. / x = 0.16 / Re_x^{1/7}. Solve for x = 6.06 m. Check Rex = 8.1E6, ok.

7.3. Equation 7.1b [\delta turb. / x = 0.16 / Rex^{1/7}] assumes that the b. l. on the plate is turbulent from the leading edge onward. Devise a scheme for determining the b. l. thickness more accurately when the flow is laminar up to a point Rex,crit and turbulent therreafter. Apply this scheme to computation of the b. l. thickness at x = 1.5 m in 40 m/s flow of air at 20 C and 1 atm past a flat plate. Compare with 7.1b. Assume Re_x,crit 1.2E6.

Given xcrit, Re_xcrit, calculate the l. b. l. thickness \delta c / xc = 5.0 / Rec^{1/2}. Then find the apparent distance upstream Lc which gives the same turbulent b. l. thickness \delta c / Lc = 0.16 / Rec^{1/7}. Then begin xeffective at this apparent origin and calculate the remainder of the turbulent b. l. as \delta /xeff = 0.16 / Rexeff^{1/7}. Illustrate with the numerical example requested. Look up \rho for air to be 1.2 kg/m^3, and \mu = 1.8E-5 kg/m s.

From Rex,crit 1.2E6 find xc = 0.45 m, then \delta c = 0.00205 m from 5.0 / Rec^{1/2}. Compute Lc = 0.0731 m from \delta c / Lc = 0.16 / Rec^{1/7}.

Now, at x=1.5m, xeff = x + Lc - xc = 1.5 + 0.0731 - 0.45 = 1.123 m, Reeff = 2.995 E6.

\delta 1.5m = 0.16 xeff / Reeff ^{1/7} = 0.0213 m.

Compare with 7.1b: Rex = 4.0E6, whence \delta = 0.027 m, 25% higher.

7.8 Repeat the flat-plate momentum integral analysis of 7.2 by replacing the parabolic profile with a linear profile u/U = y/\delta. Compute cf, \theta/x, \delta*/x, and H.

Substitute the linear velocity profile into the Karman integral relation 7.5:

\tau w = \mu du/dy wall = \mu U/\delta = d/dx int[0,\delta] \rho U y/\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 = 3.4/Rex^{1/2}, 35% low.

With \delta known, introduce \delta(x) into the definitions of cf, \theta, and \delta*:

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

\delta* = int[0,\delta] [1 - y/\delta] dy = \delta/2, hence \delta*/x 1.732/Rex^{1/2}

H = \delta*/\theta = 3.0. cf = 0.577/Rex^{1/2} , 13% low.

7.15 Discuss whether fully developed laminar incomp. flow between paralel plates is a b. l. flow.

Previously we solved this laminar flow problem to obtain u = [ - dp/dx ] [h^2/2\mu] [1-y^2/h^2],

v=0, dp/dx < 0. This is a b. l. flow, with v << u and dp/dy = 0. The freestream flow is the centerline velocity. The flow can't grow because of the wall constraint.

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

For SAE 30 oil take \rho = 891 kg/m^3 and \mu = 0.28 kg/m s. For a., L=110cm, ReL=20300 (laminar), CD=1.328/ReL^1/2 =0.0093, F=CD(\rho/2)U^2(2bL)=181N.

For b., L-55cm, ReL=10140, CD=0.0132, F=256N, 41% more.

**7.24 Air at 20 C and 1 atm flow past a flat plate. Two pitot tubes are 2 mm from the wall. The manometer fluid is water at 20 C. If U = 15 m/s and L=50cm, determine the values of the manometer readings h1 and h2 in cm. Assume laminar b.l. flow.

Look up \rho for air to be 1.2 kg/m^3, and \mu = 1.8E-5 kg/m s. The velocities u at each pitot inlet can be estimated from the Blasius Solution in Table 7.1.

1. \eta1 = y[U/\nu x1]^{1/2} = 2.83, read f' = 0.816 from 7.1, then u1 = Uf' = 12.25 m/s.

2. \eta2 = y[U/\nu x2]^{1/2} = 2.0 read f' = 0.63 from 7.1, then u2 = Uf' = 9.45m/s.

Assume constant stream pressure, then the manometers are a measureof the local velocity u at each position of the pitot inlet, so we can find \delta p across each manometer:

\delta p1 = \rho u1^2/2 = 90 Pa = \delta \rho g h1 ; h1 = 9.2 mm.

\delta p2 = \rho u2^2/2 = 54 Pa = \delta \rho g h2 ; h2 = 5.5 mm.

7.25 Modify 7.24 to the following somewhat more difficult scenario. Let the known data be U=15m/s and hz=8mm of water. Use this information to determine a. L in cm and b. h2 in mm.

Look up \rho for air to be 1.2 kg/m^3, and \mu = 1.8E-5 kg/m s. Use h1 to find u1, then \eta 1,

and then x1. \delta p1 manom. = \delta rho g h1 = 78.2 Pa. u1 = (2\delta p/\rho)^{1/2} = 11.42 m/s. Hence f' = 11.42/15 = 0.761 . From 7.1 read \eta 1 = 2.55 = y(U/\nu x)^{1/2}. Thus x1=L=62 cm.

With known L, since x2 = 2L, we find \eta 2, u2, and h2. \eta 2 = y(U/\nu x2)^{1/2}= 1.8, so f'=0.575 from 7.1. Thus u2=Uf'=8.63 m/s, \delta p2=45 Pa = \delta \rho g h2, so h2= 4.6mm.

**7.34 A thin equilateral-triangle plate 2m per side is immersed parallel to a 12 m/s stream of water at 20 C. Assuming Retrans = 5E5, estimate the drag of this plate.

Use a strip dx long and L-x wide parallel tothe leading edge of the plate. Let the side length be a: Strip dA = (L-x)dx, where L = a sin 60 deg, and a = 2 m.

Laminar part: dF_lam = \tau_w dA = 0.332(\rho \mu/x)^{1/2}U^{3/2}(L-x)dx (2sides)

Integrate from x=0 to xcrit.: F_lam = 0.6664(\rho \mu)^{1/2}U^{3/2}(2L xcrit^{1/2} - 3/2 xcrit^{3/2})

Turbulent part: dF_turb = \tau_w dA = 0.027(\rho U^2 /2)(\nu/Ux)^{1/7}(L-x) dx (2sides)

Integrate from x_cr to L: F_turb = 0.027 \rho \nu^{1/7} U^{13/7} [7/6 (L^{13/7} - x_cr^{13/7})]

The total force is F_lam + F_turb. L=1.732 m. For water at 20 C take \rho = 998 kg/m^3 and \mu = 0.001 kg/m s. Evaluate x_crit and then F:

Re_cr = 5E5 = \rho V x_crit / \mu , so x_crit = 0.042 m. F_lam = 19.4 N. F_turb. = 609.2 N. F_total = 629 N.

7.35 Develop, using strip theory, an analytical formula for the drag of the triangular plate of 7.34. Let the stream velocity be U, side length a, assume complete turbulence.

Use the expression in prob. 7.34. F = 0.027 \rho ^{6/7} \mu ^{1/7} U^{13/7} (49/78) L^{13/7}, where L = a sine 60 deg.

**7.58 A long cylinder of rectangular cross section, 5 cm high and 30 cm long, isimmersed in water at 20 C flowing at 12 m/s parallel to the long side of the rectangle. Estimate the drag force on the cylinder, per unit length, if the rectangle a. has a flat face, or b. has a rounded nose.

For water at 20 C take \rho = 998 kg/m^3 and \mu = 0.001 kg/m s. Assume a 2D flow. Use table 7.2. If the nose is flat, L/H = 6, then C_D = 0.9:

a. Flat nose: F = C_D \rho/2 U^2 H (1m) = 3200 N/m

b. Round nose: C_D = 0.64, F = 0.64/0.9 F_flat = 2300 N/m

7.59 Repeat 7.58 if the water flow is parallel to the short side.

We reverse L and H (now=30 cm), look in 7.2 for L/H=0.15 for a flat nose and L/H = 0.67 for a rounded nose:

a. Flat nose: C_D = 2.0, F = 43000 N/m ; b. Round nose: C_D = 1.16, F = 1.16/2 F_flat = 25000 N/m.

7.66 A sphere of density \rho_s and diameter D is dropped from rest in a fluid of density \rho and viscosity \mu. Assuming a constant drag coefficient C_do, derive a differential equation for the fall velocity V(t) and show that the solution is

V = [4gD(S-1)/3C_do]^{1/2} tanh Ct

C = [3gC_do(S-1)/4S^2D]^{1/2}, where S = \rho_s/\rho is the specific gravity of the sphere material.

Newton's law gives W-B-C_D \rhoV^2A/2 = W/g dV/dt, where A=\piD^2/4 and W-B=\rho_s(S-1)g\piD^3/6. Rearrange to dV/dt=\beta-\alphaV^2, \beta=g(1-1/S) and \alpha=\rho g C_D A /2W. Separate the variables and integrate from rest, V=0 at t=0: int dt = int dV/(\beta-\alphaV^2), or

V= (\beta/\alpha)^{1.2} tanh (t(\alpha \beta)^{1/2}) = V_final tanh(Ct), where V_final = [4gD(S-1)/3C_do]^{1/2} and C = [3gC_do(S-1)/4S^2D]^{1/2} , S=\rho_s/\rho.

7.69 Two baseballs from Prob. 7.68 are connected to a rod 7 mm in diameter and 56 cm long. What power, in watts, is required to keep the system spinning at 400 r/min? Include the drag of the rod, and assume sea level standard air.

Look up \rho for air to be 1.225 kg/m^3, and \mu = 1.78E-5 kg/m s. Assume a laminar drag coefficient C_D = 0.47 from table 7.3. Convert \Omega to 41.9 rad/s. V_b = 13.3 m/s. Re=67000, check C_D. F_b = 0.215 N. V_rod = 5.86 m/s, Re = 2800, C_D=1.2, F_rod = 0.0674, Power= 2F_bV_b + 2F_rodV_rod = 6.5 W.

7.71 A football weighs 0.91 lbf and approximates an ellipsoid 6 in in diameter and 12 in long (table 7.3). It is thrown upward at a 45 deg angle with an initial velocity of 80 ft/s. Neglect spin and lift. Assuming turbulent flow, estimate the horizontal distance traveled, a. neglecting drag and b. accounting for drag with a numerical model.

For sea level air, take \rho = 0.00238 slug/ft^3 and \mu = 3.71E-7 slug/ft s. For a 2:1 ellipsoid, 7.3, assume C_D = 0.13 (turbulent flow).

a. For zero drag, L = 2(V_o cos 45 deg)^2 /g = 199 ft

b. With drag of an ellipsoid (C_D = 0.13) included, we need to use Newton's law in both the x and z directions. The equations of motion of the ball with constant drag coefficient are:

(0.91/32.2)dV_x/dt = -Fcos\theta, where F=C_D(\rho/2)(V_x^2 + V_z^2)(\pi D^2 /4)

(0.91/32.2)dV_z/dt = -Fsin\theta - W, where \theta = tan_{-1} (V_z / V_x).

Integrate numerically for x(t) and z(t) until the ball comes down to z=0. Thus, x_max = L = 171 ft at t=3.4 s (z_max = 46 ft.

7.93 A hot-film probe is mounted on a cone and rod system in a sea level airstream ofr 45 m/s. Estimate the maximum cone vertex angle allowable if the flow induced bending moment at the root of the rod is not to exceed 3 N cm.

Assume bending moment is 30 N cm instead of 3 (otherwise the rod alone exceeds the small limit given).

F_rod = 1.49 N; M_base,rod = 14.9 N cm. M_base,cone = 30 - 14.9 = 15.1 n cm = C_D_cone (\rho/2)V^2A . Solve for C_D = 0.8. Table 7.3 read \theta_cone = 60 deg.