AA210 HOMEWORK 1 SOLUTION 2015-2016 Spring September 30, 2015 Problem 1.2 (10 points) Assumptions: steady, incompressible, 2D flow Under the assumptions, we can relate velocity to the streamline height from mass conservation. Using continuity equation, m = U A U h = constant By measuring the height we can determine the change in U: U2 h1 = U1 h2 The approximate values are: Location free stream A B C D U UF S Height (mm) 3 1.5 4 2.5 3.5 Percentage change 0 100 -25 20 -14 1 2 0.75 1.2 0.86 Problem 1.3 (20 points) Given: dy = g(x)y + f (x) dx The ODE can be written in the differential form: (g(x)y f (x))dx + dy = 0 By multiplying the integrating factor M = e e R g(x)dx It is a perfect differential if V y = U x , R g(x)dx to the differential equation, (g(x)y f (x))dx + e R g(x)dx dy = 0 where R g(x)dx R g(x)dx U = e V = e (g(x)y f (x)) It is easy to see that: R V U = = g(x)e g(x)dx y x 1 Therefore, the converted differential is a perfect differential. To find the stream function , we can make use of the fact that: R d = U = e g(x)dx dy R R d = V = e g(x)dx g(x)y e g(x)dx f (x) dx d dy R = e g(x)dx R = e g(x)dx y + h(x) R d = e g(x)dx g(x)y + h0 (x) dx R h0 (x) = e g(x)dx f (x) Z R h(x) = e g(x)dx f (x)dx Z R R g(x)dx = e y e g(x)dx f (x)dx If g = sin x, f = cos x, U = V = e cos x sin xe cos x y + e cos x cos x Z = e cos x y e cos x cos xdx 2 Problem 1.5 (20 points) Given: (sin x sin y)dx + (cos x cos y)dy = 0 Therefore, d = U = cos x cos y dy d = V = sin x sin y dx To check whether it is a perfect differential: V U = = sin x cos y y x The expression is a perfect differential. = V = sin x sin y x = cos x sin y + f (y) = y f 0 (y) = 0 f (y) = C = cos x sin y + C U = cos x cos y The acceleration fields are: DU Dt = = U U U +U +V t x y sin x cos x cos2 y sin x cos x sin2 y sin x cos x V V V = +U +V t x y 2 = cos x sin y cos y + sin2 x sin y cos y = DV Dt = sin y cos y 3 4 Problem 1.7 (10 points) The velocity in the fixed frame is U = (100, 60, 175)m/s. The velocity of the moving frame is X 0 = (25, 110, 90)m/s. Thus, the velocity in the moving frame is: U 0 = U X0 = (100, 60, 175)m/s (25, 110, 90)m/s = (75, 50, 85)m/s In the fixed frame, the kinetic energy is given by: k= 1 1 U U = (1002 + 602 + 1752 ) = 22112.5J/kg 2 2 In the moving frame, the kinetic energy is given by: k= 1 1 U 0 U 0 = (752 + 502 + 852 ) = 7675J/kg 2 2 Problem 1.8 (20 points) The given stream function is: = xy 1+x+y As stated in the problems, we assume the flow is steady, compressible and 2D. Therefore, we have: d y + y2 = V = dx (1 + x + y)2 x + x2 d = U = dy (1 + x + y)2 One of the plausible expression of density and velocity fields are: U V C (1 + x + y)2 1 = (x + x2 ) C 1 = (y + y 2 ) C = where C is a positive constant This satisfies the corner flow such that U is positive and V is negative in the first quadrant. Streamlines are also symmetric along the bisection line in the first quadrant with this expression of density and velocity. Using method of contradiction, first suppose a pressure field p(x, y) exists if the flow is assumed to be inviscid. The momentum equations of steady 2D inviscid flow are: U p U U + V + =0 x y x V V p U + V + =0 x y y 5 From the plausible expression of velocity fields: U x U y V x V y = 1 (1 + 2x) C = 0 = 0 = 1 (1 + 2y) C Therefore, 1 (1 + 2x)(x + x2 ) p = x C (1 + x + y)2 p 1 (1 + 2y)(y + y 2 ) = y C (1 + x + y)2 From the first equation, 2p 2(1 + 2x)(x + x2 ) = yx C(1 + x + y)3 From the second equation, 2p 2(1 + 2y)(y + y 2 ) = xy C(1 + x + y)3 This indicates the two equations are inconsistent since we have: 2p 2p 6= yx xy There is contradiction! This means a smooth p(x, y) does not exist if the flow is inviscid. Problem 1.9 (20 points) First part of the question asks us to find the particle paths given the unsteady velocity field: U= xt , t20 + t2 V = yt , t20 + t2 W = zt t20 + t2 Along the particle path, position is only a function of time. Starting from U, U (x, y, z) = U (x(t), y(t), z(t)) = dx xt = 2 dt t0 + t2 By separation and integration: Z x x1 dx Z = 2 0 t0 x0 x [ln x]x0 = \u0012 \u0013 x ln = x0 x = t \u0014 t dt + t2 \u0001 1 ln t2 + t20 2 1 t2 + t20 ln 2 t2 s t2 + t20 x0 t20 6 \u0015 Similarly, y z s t2 + t20 t20 s t2 + t20 t20 = y0 = z0 The second part asks us to find the substantial derivative of the density The problem can be simplified if we realize that: x2 + y 2 + z 2 = (x20 + y02 + z02 ) D Dt . t2 + t20 t20 Substituting the above equation into the equation of : = 0 t30 2 (t0 + t2 )3/2 \u0012 1 x20 + y02 + z02 2 Rinitial \u00133/2 The substantial derivative is: D Dt = = = + ui t xi t \u0012 \u00133/2 30 t30 t x20 + y02 + z02 1 2 Rinitial (t20 + t2 )5/2 7