Solution Manual For Fundamentals of Aerodynamics, 6th Edition

Fundamentals of Aerodynamics 6th Edition Anderson Solutions Manual Full Download: http://testbanklive.com/download/fundamentals-of-aerodynamics-6th-edition-anderson-solutions-manual/ CHAPTER2 2.1 If p = constant = Poo However, the integral of the surface vector over a closed surface is zero, i.e., Hence, combining Eqs. (1) and (2), we have 2.2 t/f’,Per W11 I/ /LL.L/tL fQ. I / 7 7 7 7 7 7 i.OU1er Wa I I 7’t! 7 f.t {‘X ) 7 7 / 19 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education Full download all chapters instantly please go to Solutions Manual, Test Bank site: testbanklive.com Denote the pressure distributions on the upper and lower walls by pu(x) and p e (x) respectively. The walls are close enough to the model such that Pu and p 1 are not necessarily equal to Poo· Assume that faces ai and bh are far enough upstream and downstream of the model such that and v = 0 and ai and bh. Take they-component ofEq. (2.66) # (p v.dS) v – H(p dS)y L =- S abhi The first integral= 0 over all surfaces, either because V · ds = 0 or because v = 0. Hence -> fJ (pdS)y L’ = – b J Pu dx – J p dx] = – [ •bM h l; • 1 Minus sign because y-component is in downward Direction. Note: In the above, the integrals over ia and bh cancel because p = Poo on both faces. Hence h 2.3 b L’ = J pf dx – J Pu dx dy v cy/(x 2 +y 2 ) u 2 dx =-= dy dx y x 2 ex I ( x + y ) = y x -=- The streamlines are straight lines emanating from the origin. streamline pattern for a source, to be discussed in Chapter 3.) 2.4 dy v x dx u y -=-== (This is the velocity field and y dy = – x dx 20 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education y2 = -x2 + const x2 + y2 = const. The streamlines are concentric with their centers at the origin. (This is the velocity field and streamline pattern for a vortex, to be discussed in Chapter 3.) 2.5 From inspection, since there is no radial component of velocity, the streamlines must be circular, with centers at the origin. To show this more precisely, u = – v e sin = – er rr = – cy x v = Ve cos 8 = er – = ex r dy v x = = dx u y &2 + x2 = const.J This is the equation of a circle with the center at the origin. (This velocity field corresponds to solid body rotation.) 2.6 dy v y = = dx u x ~. )) ~. ~ dy dx = y x ~ fn y = X £n X + C1 ~ (/’ y = C2/x x f The streamlines are hyperbolas. 21 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education __, 1o 1 OV cr roe 8 In polar coordinates: V · V = — (r Vr) + – – – r Transformation: x = r cos 8 y = r sine Vr = u cos 8 + v sin 8 ve = – u sin e + v cos 8 22 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education u= v= ex er cosB c cosB (x2 + y2) ? c r cy (x2 + y2) er sine r1 = c sine r . 2 c v r = -c cos2e + -c sm e=r r r Ve= – ~ cose sine+ ~ cose sine= 0 r r . -> 18 1 8(0) V V = — (c) + – – = 0 r r !JB a: (b) From Eq. (2.23) V x V = ez [O + 0 – O] = ~ The flowfield is irrotational. 2.8 u= c sine cy er sine = = ? (x2 +y2) r c v= -ex er cose c cose – = ? (x2+y2) r c Yr= ~ cose sine – ~ cose sine= 0 r r ·2 c 2 c v e = – -c sm e – – cos e = – r (a) v· r v L~a: (O)+ = r r 18(-c/r) =O+O=~ r 8() 23 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education (b) V’x V= e–+ [O'(-c/r) _~_!b'(O)] z V’ x a r2 r oe V=~except at the origin, where r = 0. The flowfield is singular at the origin. 2.9 Ve=cr V’x V= e__,, [o(c/r) +er_..!_ o(O)J z a –7 r roe –7 ez (c + c – 0) = 2c ez The vorticity is finite. The flow is not irrotational; it is rotational. 2.10 c b a.. 24 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education Mass flow between streamlines = Ll f;t Lllf/ =pV Lln Ll f;t = (-p Ve) M + p Yr (r8) Let cd approach ab (1) Also, since f;t = f;t (r,9), from calculus – 3,;,. 3,;,. a 8e d If/ = _ ‘ f ‘ dr + _ ‘ f ‘ de (2) Comparing Eqs. (1) and (2) 8f;t -pVe=- a and 3 f;t P rV r = 8B or: 8f;t pVe=- – 2.11 u = ex = a 8 1// : ljf = cxy + f(x) (1) 07 v = – cy = – 8 lf : lJf = cxy + f(y) & (2) 25 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education Comparing Eqs. (1) and (2), f(x) and f(y) =constant ljf = c x y + const. I (3) u = ex = Olf/ : ~ = cx2 + f(y) & (4) v = – cy = Olf/ : ~ = – cy2 + f(x) (5) 0’ Comparing Eqs. (4) and (5), f(y) = – cy2 and f(x) = cx 2 (6) Differentiating Eq. (3) with respect to x, holding jf = const. 0 =ex dy + cy dx or, dy) =-y/x (dx lf=const (7) Differentiating Eq. (6) with respect to x, holding~= const. dy 0=2cx-2cy dx or, ( dy) = x/y dx ¢=const (8) Comparing Eqs. (7) and (8), we see that Hence, lines of constant jf are perpendicular to lines of constant~· 26 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education 2.12. The geometry of the pipe is shown below. t{ =/CJC> ..-w /:s-ec.,..(I As the flow goes through the U-shape bend and is turned, it exerts a net force Ron the internal surface of the pipe. From the symmetric geometry, R is in the horizontal direction, as shown, acting to the right. The equal and opposite force, -R, exerted by the pipe on the flow is the mechanism that reverses the flow velocity. The cross-sectional area of the pipe inlet is nd2/4 where dis the inside pipe diameter. Hence, A= nd2/4 = n(0.5)214 = 0.196m2 . The mass flow entering the pipe is • m = P1 A V1=(1.23)(0.196)(100)=24.11 kg/sec. Applying the momentum equation, Eq. (2.64) to this geometry, we obtain a result similar to Eq. (2.75), namely R =- # (p V · dS) V (1) Where the pressure term in Eq. (2.75) is zero because the pressure at the inlet and exit are the same values. In Eq. (1), the product (p V · dS) is negative at the inlet (V and dS are in opposite directions), and is positive at the exit (V and dS) are in the same direction). The magnitude of p • V · dS is simply the mass flow, m. Finally, at the inlet V 1 is to the right, hence it is in the positive x-direction. At the exit, V2 is to the left, hence it is in the negative x-direction. Thus, V2 = – V1. With this, Eq. (1) is written as . . ~ R = – [- m V 1 + m V2] = m (V 1 – V2) • • = m [V1 -(-V1)] = m (2V1) R = (24.11)(2)(100) = &822 NI 27 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education 2.13 From Example 2.1, we have u = V∞ (2.35) v = – V∞ h (2.36) and Thus, = u = V∞ (2.35a) Integrating (2.35a) with respect to x, we have = V∞ x = V∞ x + f(y) + f(y) (2.35b) From (2.36) = v = – V∞ h (2.36a) Integrating (2.36a) with respect to y, we have Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education = V∞ h + f(x) = + f(x) (2.36b) Comparing (2.35b) and (2.36b), which represent the same function for , we see in (2.36b) that f(x) = V∞ x. So the velocity potential for the compressible subsonic flow over a wavy well is: ________________________________________________________________ 2.14 The equation of a streamline can be found from Eq. (2.118) = For the flow over the wavy wall in Example 2.1, = As y → ∞, then → 0. Thus, → =0 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education Fundamentals of Aerodynamics 6th Edition Anderson Solutions Manual Full Download: http://testbanklive.com/download/fundamentals-of-aerodynamics-6th-edition-anderson-solutions-manual/ The slope is zero. Hence, the streamline at y → ∞ is straight. Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education Full download all chapters instantly please go to Solutions Manual, Test Bank site: testbanklive.com