Homework hints, second problem set: 8.5,10,11,16,23,26,31,39,43,64,70
8.5 Consider the two-dimensional velocity distribution u = -By, v = +Bx, where B is a constant. If this flow possesses a stream function, find its form. If it has a velocity potential, find that also. Compute the local angular velocity of the flow, if any, and describe what the flow might represent.
The flow is 2D, steady, constant density, and the velocity satisfies the continuity equation when u and v are substituted into u,x + u,y = 0. u,x = partial -By / partial x = 0. v,y = partial Bx / partial y = 0. Now use u = \psi , y = -By. Integrate to get \psi = -By^2/2 + f(x). Substitute this into v = - \psi ,x = -df/dx = Bx. Integrate to get f = -Bx^2/2 + C. Combine to get \psi = -(B/2)(x^2 + y^2) + C. ans
The flow does not have a velocity potential because the vorticity is not zero. \omega _z = v,x - u,y = B - (-B) = 2B not equal zero.
The flow represents a solid body rotation at uniform clockwise angular velocity B.
8.10 How does Stoke's Theorem simplify for irrotational flow, and how des the resulting line integral relate to velocity potential?
For irrotational flow, v=grad \phi. Substitute into Stoke's Theorem that int[line, counterclockwise, closed]v dot dl= int[area within closed line]curl v dot dA. But:
int[line, counterclockwise, closed]v dot dl = int[line, counterclockwise, closed] d \phi = 0, and int[area within closed line]curl grad \phi dA = 0.
Note that for rotational flows, Stokes Theorem shows the circulation per unit area \Gamma / A = area averaged vorticity perpendicular to A = (1/A)int[area within closed line]curl v dot dA
8.11 By conservation of mass the radial velocity V_r = Q/2 \pi r b , where r is any radius, Q is the volume flow rate, and b is the length of the power plant discharge pipe. V_r = partial \phi / partial r , so \phi = m ln r with m = Q/2 \pi b , where b = 8 m and Q = sum[1,25000] Q_1 hole.
**8.14 A tornado may be modeled as the circulating flow shown in the figure P8.14 with zero radial and vertical velocity and v_@ = \Omega r, r <= R; v_@ = \Omega R^2 /r, r >= R. Determine whether this flow is rotational or irrotational in either the inner or outer region. Using the r-momentum equation E5 determine the pressure distribution p(r) inthe tornado, assuming p = p_infinity as r goes to infinity. Find the location and magnitude of the lowest pressure.
From Appendix E, the vorticity component \omega_z = 1/r d/dr r v_@ - 1/r d/d@ v_r ; but v_r = 0, so \omega_z = \Omega 1/r d/dr r ^2 = 2 \Omega , which is not zero for the inner region.
For the outer region v_r is still zero, so \omega_z = 1/r d/dr r v_@ = 1/r d/dr \Omega R ^2 = 0.
The pressure is found by integrating the r-momentum equation E5 in the Appendix (dropping the time, nonlinear, viscous and gravitational terms) to give dp/dr = \rho v_@^2/2.
In the outer region, p_outer = integral \rho/r (\Omega R^2 / 2 )^2 dr = - \rho \Omega^2 R^4 / 2 r^2 + constant.
The constant is p_infinity.
For the inner region, p_inner = int( \rho / r )(\Omega r)^2 dr = \rho \Omega^2 r^2/2 + const. Match to p(R) = p_infinity - \rho \Omega^2 R^2 / 2 gives p_inner = p_infinity - \rho \Omega^2 R^2 / 2 + \rho \Omega^2 r^2/2 .
The minimum pressure occurs at r=0. p_min = p_infinity - \rho \Omega^2 R^2 / 2
8.16 Evaluate Prob. 8.14 for the case of a large scale hurricane, R = 8 km, v_@max = 45 m/s, with sea level conditions at r = infinity. Plot p(r) out to r = 30 km.
Compute \Omega and then p(r): \Omega = v_max / R = 45/8000 = 0.005625 rad/s, p_outer = 101000 - 1.225(0.005625)^2(8000)^4 / 2r^2 . or p_outer = 101000 - 7.9E10/r^2 , p_inner = 98519 + 1.94E-5 r^2. Plot.
8.23 Find the resultant velocity vector at point A due to the combination of uniform stream, vortex and line source.
Add vectorially, to get v = 11.3 m/s at @ = 44.2 degrees.
8.26 Find the resultant vector velocity at A. v= 9.4 m/s at @ = -46.8 degrees.
8.31 A Rankine half body is formed by uniform flow of 7 m/s and a source at 0,0 , with dividing streamline at 0,3 m. Find a. the source strength m in m^2/s; b. the distance a of the stagnation point; c. the distance h of the dividing streamline at point 4,0; d. the total velocity at A = 4,h .
The equation for the body is r_body = m(\pi - r)/Usin@ . Thus, for @ = \pi / 2 and y=r_body = 3 m we find m = 13.4 m^2 / s. Also a = 1.91 m. At x=4 m, r_body = m(\pi - r)/Usin@ = 13.4(\pi - @)/7 sin@ = 4/cos@ . Solve for @= 47.8 degrees.
r_A = 4.0/cos47.8 = 5.95 m; h = r sin@ = 4.41 m. The velocity at A is computed from 8.18:
V_A = U(1 + a^2/r^2 + 2a/r cos @_A )^{1/2} = 8.7 m/s
8.39 Sketch the streamlines of a uniform stream U_infinit pas a line source sink pair aligned vertically with the source at +a and the sink at -a on the y axis. Does a closed body shape appear?
No close body shape appears.
8.43 Consider water at 20 C flowing past a 1 m diameter cylinder. What doublet strength in m^2/s is required to simulate this flow? If the stream pressure is 200 kPa use inviscid theory to estimate the surface pressure at a. 180 deg; b. 135 deg; and c. 90 deg.
The doublet strength \lambda = U_infinity a^2 = 6 0.5^2 = 1.5 m^3/s. Surface pressures from Bernoulli's equation, with V_surface = 2 U_infinity sin@ a. at 180 deg p_surf = 218000 ; b. 182000 ; c. 146000 Pa.
8.64 Determine qualitatively from boundary layer theory whether separation will occur.
In 8.61 for \psi = A r^n sin n@ the velocity component along the x axis with @=0 and r=x v_r = u = const x^{n-1}. Thus, for a. n=3, so U = const. x^2 ; b. n=2, U = const. x^1 ; c. n= 3/2, U = const. x^1/2 ; all give favorable pressure gradients, so there should be no separation.
8.70 Show that the complex potential f(z) = U[z + a/4 coth \pi z / a ] represents flow past an oval shape placed midway between two parallel walls y = a/2. What is the practical application?
The stream function is \psi = U[ y - ( a/4 sin 2 \pi y / a ) / ( cosh 2 \pi x / a - cot 2 \pi y / a )]
The streamlines are trapped between the wall and an oval body shape, with width a/2 and height 0.51 a. An application is the estimate of wall blockage effects when a body in a wind tunnel is trapped between walls.