Homework 5 Due date: Thursday, Mar. 3 hapter 7 Problems 1. 7.88. 7.9 except assume the parachute has diameter of 3.5 meters and calculate how long it takes to stop. (Must solve differential equation) 3. 7.115 4. 7.116 hapter 4 Problems 1. 4.47. 4.48 3. 4.54 4. 4.56 FD Problem 1. Use Fluent code to calculate the flow in a channel of 0.5m x 0.1m for an inlet velocity of 1 cm/s. Approximately evaluate the thickness of the boundary layer. Hand in a velocity vector plot of the flow field and a contour plot of the pressure field.
hapter 7 Flow Past Immersed Bodies 57 7.88 A pickup truck has a clean drag-area D A of 35 ft. Estimate the horsepower required to drive the truck at 55 mi/h (a) clean and (b) with the 3- by 6-ft sign in Fig. P7.88 installed if the rolling resistance is 150 lbf at sea level. Solution: For sea-level air, take ρ = 0.0038 slug/ft 3 and µ = 3.7E 7 slug/ft s. onvert V = 55 mi/h = 80.7 ft/s. alculate the drag without the sign: ρ F = Frolling + DA V = 150 + 35(0.0038/)(80.7) 41 lbf Fig. P7.88 Horsepower = (41)(80.7) 550 6 hp ( clean) (a) With a sign added, b/h =.0, read D 1.19 from Table 7.3. Then 0.0038 F = 41clean + 1.19 (80.7) (6)(3) 587 lbf, Power = FV 86 hp (b) 7.89 The new AMTRAK high-speed Acela train can reach 150 mi/h, which presently it seldom does, because of the curvy coastline tracks in New England. If 75% of the power expanded at this speed is due to air drag, estimate the total horsepower required by the Acela. Solution: For sea-level air, take ρ = 1. kg/m 3. From Table 7.3, the drag-area D A of a streamlined train is approximately 8.5 m. onvert 150 mi/h to 67.1 m/s. Then 3 1. kg/m 3 ρ 0.75 Ptrain = ( DA) V V = (8.5 m ) (67.1 m/s) = 1.56E6 watts Solve for P =.08E6 W = 800 hp train
7.9 A 1500-kg automobile uses its drag-area, D A = 0.4 m, plus brakes and a parachute, to slow down from 50 m/s. Its brakes apply 5000 N of resistance. Assume sealevel standard air. If the automobile must stop in 8 seconds, what diameter parachute is appropriate?
530 Solutions Manual Fluid Mechanics, Fifth Edition Solution: For sea-level air take ρ = 1.5 kg/m 3. From Table 7.3 for a parachute, read Dp 1.. The force balance during deceleration is, with V o = 50 m/s, 1.5 π dv F = Froll Fdrag = 5000 0.4 + 1. Dp V = ( ma) car = 1500 4 dt Note that, if drag = 0, the car slows down linearly and stops in 50(1500)/(5000) = 15 s, not fast enough so we definitely need the drag to cut it down to 8 seconds. The firstorder differential equation above has the form dv dt D p 0.4 1. /4 1.5 + π 5000 = b av, where a= and b= 1500 1500 Separate the variables and integrate, with V = V o = 50 m/s at t = 0: 0 t dv Vo 0 1 1 a = dt, Solve: t = tan Vo = 8 s? b+ av ab b The unknown is D p, which lies within a! Iteration is needed an ideal job for EES! Well, anyway, you will find that D p = 3 m is too small (t 9.33 s) and D p = 4 m is too large (t 7.86 s). We may interpolate (or EES will quickly report): Dparachute(t=8 s) 3.9 m
548 Solutions Manual Fluid Mechanics, Fifth Edition For our particular data, evaluate (9810 sin 5.71 70) m Vf = = 46.0 ; 1.5(0.7) s (9810 sin 5.71 70)(0.7)(1.5/) = = 0.0197 s 1000 We need to know when the car reaches x = (0 m)/sin(5.71 ) 01 m. The above expression for V(t) may be readily integrated: t Vf 46.0 x V dt ln[ cosh(t)] ln[ cosh(0.0197t)] 01 m if t = = = = 1.4 s 0.0197 0 m km Then, at x = 01 m, z = 0 m, V = 46.0 tanh[0.0197(1.4)] 18.3 = 65.9 s h 1 7.115 The essna itation executive jet weighs 67 kn and has a wing area of 3 m. It cruises at 10 km standard altitude with a lift coefficient of 0.1 and a drag coefficient of 0.015. Estimate (a) the cruise speed in mi/h; and (b) the horsepower required to maintain cruise velocity. Solution: At 10 km standard altitude (Table A-6) the air density is 0.415 kg/m 3. (a) The cruise speed is found by setting lift equal to weight: 3 0.415 kg/m wing ρ Lift = 67000 N = L V A = 0.1 V (3 m ), m mi Solve V = 0 = 49 (a) s h (b) With speed known, the power is found from the drag: ρ 0.415 Power = FdragV = D V A V = 0.015 (0) (3) (0) = 1.05 MW =1410 hp (b) 7.116 An airplane weighs 180 kn and has a wing area of 160 m and a mean chord of 4 m. The airfoil properties are given by Fig. 7.5. If the plane is designed to land at V o = 1.V stall, using a split flap set at 60, (a) What is the proper landing speed in mi/h? (b) What power is required for takeoff at the same speed?
hapter 7 Flow Past Immersed Bodies 549 Solution: For air at sea level, ρ 1.5 kg/m 3. From Fig. 7.4 with the flap, L,max 1.75 at α 6. ompute the stall velocity: V stall W (180000 N) m = = = 3.4 (1.5 kg/m 3 )(1.75)(160 m ρ A ) s L, max p m Then Vlanding = 1. Vstall = 38.9 (a) s L L,max 1.75 = = = 1. ( V /V ) (1.) land stall For take-off at the same speed of 38.9 m/s, we need a drag estimate. From Fig. 7.5 with a split flap, D 0.04. We don t have a theory for induced drag with a split flap, so we just go along with the usual finite wing theory, Eq. (7.71). The aspect ratio is b/c = (40 m)/(4 m) =10. F D drag L (1.) = D + = 0.04 + = 0.087, πar π(10) 1.5 = (0.087) (38.9) (160) = 1900 N Power required = FV = (1900 N)(38.9 m/s) = 501000 W = 67 hp (b) 7.117 Suppose the airplane of Prob. 7.116 takes off at sea level without benefit of flaps and with constant lift coefficient and take-off speed of 100 mi/h. (a) Estimate the take-off distance if the thrust is 10 kn. (b) How much thrust is needed to make the take-off distance 150 m? Solution: For air at sea level, ρ = 1.5 kg/m 3. onvert V = 100 mi/h = 44.7 m/s. From Fig. 7.5, with no flap, read D 0.006. ompute the lift and drag coefficients: L W (180000 N) = = = 0.919 (assumed constant) ρv A (1.5 kg/m )(44.7 m/s) (160 m ) 3 p D L (0.919) = D + = 0.006 + = 0.039 πar π(10) The take-off drag is D o = D (ρ/)v A p = (0.039)(1.5/)(44.7) (160) = 6440 N. From Ex. 7.8, m T So = ln, K D( ρ/) Ap (0.039)(1.5/)(160) 3. kg/m K = = = T D o (180000/9.81) kg 10000 N Then So = ln = 940 m (a) (3. kg/m) 10000 N 6440 N
80 Solutions Manual Fluid Mechanics, Fifth Edition 4.46 Fluid from a large reservoir at temperature T o flows into a circular pipe of radius R. The pipe walls are wound with an electric-resistance coil which delivers heat to the fluid at a rate q w (energy per unit wall area). If we wish to analyze this problem by using the full continuity, Navier-Stokes, and energy equations, what are the proper boundary conditions for the analysis? Solution: Letting z = 0 be the pipe entrance, we can state inlet conditions: typically u z (r, 0) = U (a uniform inlet profile), u r (r, 0) = 0, and T(r, 0) = T o, also uniform. At the wall, r = R, the no-slip and known-heat-flux conditions hold: u z (R, z) = u r (R, z) = 0 and k( T/ r) = q w at (R, z) (assuming that q w is positive for heat flow in). At the exit, z = L, we would probably assume free outflow : u z / z = T/ z = 0. Finally, we would need to know the pressure at one point, probably the inlet, z = 0. 4.47 Given the incompressible flow V = 3yi + xj. Does this flow satisfy continuity? If so, find the stream function ψ(x, y) and plot a few streamlines, with arrows. Solution: With u = 3y and v = x, we may check u/ x + v/ y = 0 + 0 = 0, OK. Find the streamlines from u = ψ/ y = 3y and v = ψ/ x = x. Integrate to find 3 ψ = y x Set ψ = 0, ±1, ±, etc. and plot some streamlines at right: flow around corners of half-angles 39 and 51. y 51 39 Fig. P4.47 x 4.48 onsider the following two-dimensional incompressible flow, which clearly satisfies continuity: u = U o = constant, v = V o = constant Find the stream function ψ(r, θ) of this flow, that is, using polar coordinates. Solution: In cartesian coordinates the stream function is quite easy: u = ψ/ y = U o and v = ψ/ x = V o or: ψ = U o y V o x + constant But, in polar coordinates, y = rsinθ and x = rcosθ. Therefore the desired result is ψ(r, θ) = U o r sinθ V o r cosθ + constant
hapter 4 Differential Relations for a Fluid Particle 83 4.54 An incompressible stream function is defined by U 3 ψ ( x, y) = (3 x y y ) L where U and L are (positive) constants. Where in this chapter are the streamlines of this flow plotted? Use this stream function to find the volume flow Q passing through the Fig. E4.7 rectangular surface whose corners are defined by (x, y, z) = (L, 0, 0), (L, 0, b), (0, L, b), and (0, L, 0). Show the direction of Q. Solution: This flow, with velocities u = ψ/ y = 3U/L (x y ), and v = ψ/ x = 6xyU/L, is identical to Example 4.7 of the text, with a = 3U/L. The streamlines are plotted in Fig. E4.7. The volume flow per unit width between the points (L, 0) and (0, L) is U U 3 Q/b = ψ(l, 0) ψ(0, L) = (0 0) [3(0) L L ] = UL, or: Q = ULb L L Since ψ at the lower point (L, 0) is larger than at the upper point (0, L), the flow through this diagonal plane is to the left, as per Fig. 4.9 of the text. 4.55 In spherical polar coordinates, as in Fig. P4.1, the flow is called axisymmetric if υ θ 0 and / φ 0, so that υ r = υ r (r, θ ) and υ θ = υ θ (r, θ ). Show that a stream function ψ(r, θ ) exists for this case and is given by 1 ψ 1 ψ υr = υ θ r sinθ θ = rsinθ r This is called the Stokes stream function [5, p. 04]. Solution: From Prob. 4.1 with zero velocity υ φ, the continuity equation is 1 1 (r υ r) + ( υθ sin θ) = 0, or: (r υrsin θ) + (rυθ sin θ) = 0 r r rsinθ θ r θ ψ ψ ompare this to a stream function cross-differentiated form + = 0 r θ θ r 1 ψ 1 ψ It follows that: υr(stokes) = ; υ (Stokes) θ = rsinθ θ r sinθ r
84 Solutions Manual Fluid Mechanics, Fifth Edition 4.56 Investigate the velocity potential φ = Kxy, K = constant. Sketch the potential lines in the full xy plane, find any stagnation points, and sketch in by eye the orthogonal streamlines. What could the flow represent? Solution: The potential lines, φ = constant, are hyperbolas, as shown. The streamlines, Fig. P4.56 sketched in as normal to the φ lines, are also hyperbolas. The pattern represents plane stagnation flow (Prob. 4.48) turned at 45. 4.57 A two-dimensional incompressible flow field is defined by the velocity components x y y u= V v= V L L L where V and L are constants. If they exist, find the stream function and velocity potential. Solution: First check continuity and irrotationality: u v V V + = = 0 ψ exists; x y L L v u V xv= k = 0+ 0 x y k φ does not exist L To find the stream function ψ, use the definitions of u and v and integrate: ψ x y u V, V xy y = = ψ = + f( x) y L L L L ψ Vy df Vy Evaluate = + = v = x L dx L df xy y Thus = 0 and ψ = V dx + const L L 4.58 Show that the incompressible velocity potential in plane polar coordinates φ(r,θ) is such that φ 1 φ υr = υθ = r r θ