Homework set5. Problems 8.5,8,10,11,14,23

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.

8.8 For the velocity distribution of Prob. 8.5, u=-By, v=+By, evaluate the circulation \Gamma around the rectangular closed curve defined by (x,y) = (1,1), (3,1), 3,2), and (1,2).

Solution: Given \Gamma = int[counter-clockwise]v \dot ds around the curve, divide the rectangle into (a,b,c,d) pieces between (1,1), (3,1), 3,2), and (1,2), respectively. Thus:

\Gamma = int[a]uds + int[b]vds + int[c]uds + int[d]vds = (-B)(2) + (3B)(1) + (2B)(2) + (-B)(1) = 4B

From 8.5 |curl v| = 2B. Also \Gamma = |curl v| A = 2B2=4B (check).

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.