Homework set 3. Problems 7.30,58,72,82,122.
7.30 Repeat Prob. 7.16 if the fluid is water and the plate is smooth, T=20 deg C.
Look up \rho = 998 kg/m^3 and \mu = 0.001 kg/m s in Appendix A. Recall the problem was a thin plate 55 cm by 110 cm imersed in SAE 30 oil flowing at 6 m/s, find the frictional drag force if the stream is aligned with (a) the long side; or (b) the short side. In problem 7.16 the flow was laminar and the forces were (a) 181 N; and (b) 256 N. For water flow, we find that the boundary layer is turbulent:
(a) L = 110 cm, Re_L = 998(6)(1.1)/).001 = 6.59E6 (turbulent), C_D = 0.031/Re_L ^1/7 = 0.0039, so F_drag = C_D (\rho/2)U^2 b L (2sides) = 72 N
(b) L = 55 cm, Re_L = 3.29E6 (turbulent), C_D=0.00363, so F_drag = 79 N.
7.58 A long cylinder of rectangular cross section, 5 cm high and 30 cm long, is immersed in water at 20 deg 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 has (a) a flat face, or (b) a rounded nose.
Solution: For water at 20 deg C take \rho = 998 kg/m^3 and \mu=0.001 kg/m s. Assume a two dimensional flow, ie 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 (1 m) = 0.9(998/2)(12)^2(0.05) = 3200 N/m
(b) Round nose, Table 7.2: C_D=0.64, F=0.64/0.09 F_flat = 2300 N/m.
7.72 A settling tank for a municipal water supply is 2.5 m deep, and 20 deg C water flows through continuously at 35 cm/s. Estimate the minimum length of the tank which will ensure that all sediment (SG=2.55) will fall to the bottom for particle diameters greater than (a) 1 mm and (b) 100 micron.
Solution: For water at 20 deg C take \rho = 998 kg/m^3 and \mu=0.001 kg/m s. /tge oartuckes travek wutg tge strean fkiw Y = 35 cm/s (no horizontal drag) and fall at speed V_f with drag equal to their net weight in water:
W_net = (SG-1)\rho_water g (\pi/6) D^3 = Drag = C_D (\rho_water/2)V_f^2 (\pi/4) D^2, so V_f^2 = 4(SG-1)gD/3C_D,
where C_D = function (Re_D) fro Fig 7.16b. Then L=Uh/V_f where h=2.5 m.
(a) D=1 mm: V_f^2 = 4(SG-1)gD/3C_D, iterate Fig. 7.16b to C_D = 1.0, Re_D=140, V_f = 0.14 m/s, hence L=Uh/V_f = 0.35(2.5)/0.14 = 6.3 m
(b) D=100 micron: V_f^2 = 4(2.55-1)9.81(0.0001)/3C_D, iterate Fig. 7.16b to C_D=36, Re_D=0.75, V_f=0.0075 m/s, L=0.35(2.5)/).0075 = 120 m.
7.82 The average skydiver, with parachute unopened, weighs 175 lbf and has a drag area C_DA=9 ft^2 spread eagled and 1.2 ft^2 falling feet first (see Table 7.3). What are the minimum and maximum terminal speeds achieved by a skydiver at 5000 ft standard altitude?
Solution: At 5000 ft (1524 m) altitude, from Table A-6 with units conversion, \rho = 0.0025 slug/ft^3. With drag area known, we may solve the weight drag relation for V:
W=C_D (\rho/2) V^2 A, or V = (2W/\rho C_D A)^1/2 = (2(175)/0.00205 C_D A)^1/2 = 413/(C_D A)^1/2
(a) Minimum: V_min = 413/(9)^1/2 = 138 ft/s (b) Maximum: V_max = 413/(1.2)^1/2 = 377 ft/s.
7.122 A boat of mass 2500 kg has two hydrofoils, each of chord 30 cm and span 1.5 m, with C_L,max = 1.2 and C_D, inf = 0.08. Its engine can deliver 130 kW to the water. For seawater at 20 deg C, estimate (a) the minimum speed for which the foils support the boat, and (b) the maximum speed attainable.
Solution: For seawater at 20 deg C, take \rho = 1025 kg/m^3 and \mu = 0.00107 kg/m s. With two foils, total planform area is 2(0.3)(1.5) = 0.9 m^2. Thus the stall speed is
V_min = (2W/\rho C_L,max A)^1/2 = (2(2500)(9.81)/1025(1.2)(0.9))^1/2 = 6.66 m/s (13 knots)
Given AR = 1.5/0.3 = 5.0. At any speed during lifting operation (V>V_min), the lift and drag coefficients, from Eqs. 7.70 and 7.71 are
C_L = 2W/\rho A V^2 = 2(2500)(9.81)/1025(0.9)V^2 = 53.2/V^2
C_D = C_D,inf + C_L^2/\pi AR = 0.08 + (53.2/V^2)^2/\pi(5.0) = 0.08 + 180.0/V^4
Power = DV = (0.08 + 180/V^4)(1025/2)V^2(0.9)V = 130 hp x 745.7 = 96900 W
Clean up and rearrange: V^4 - 2627V + 2250 = 0, Solve V=13.5 m/s = 26 knots
Three other roots: 2 imaginary and V_4=0.86 m/s (impossible, below stall speed).