HW5_Fall2023_Soln

pdf

School

University of Illinois, Urbana Champaign *

*We aren’t endorsed by this school

Course

311

Subject

Mechanical Engineering

Date

Dec 6, 2023

Type

pdf

Pages

Uploaded by MajorRose12050

1. First we need to relate spin rate to lift. We can use the Kutta Jukowski theorem for this. L ′ = ρU inf Γ Γ = v θ 2 πR v θ = ωR ω = 2 πs 60 Where s is the spin rate in rpms. Plugging in the values given, the lift per unit depth L ‘ = mg 2 R = 19 . 86 N/m, which is generated by a spin rate of 447.83 rpms. To plot the streamlines, we will build up the flow field using known velocity fields from a doublet, vortex and uniform flow. import numpy as np import matplotlib.pyplot as plt from matplotlib.colors import BoundaryNorm from matplotlib.ticker import MaxNLocator def get_velocity_doublet(k, xd, yd, X, Y): """ Returns the velocity field generated by a doublet. Parameters ---------- k: float Strength of the doublet. xd: float x-coordinate of the doublet. yd: float y-coordinate of the doublet. X: 2D Numpy array of floats x-coordinate of the mesh points. Y: 2D Numpy array of floats y-coordinate of the mesh points. Returns ------- u: 2D Numpy array of floats x-component of the velocity vector field. v: 2D Numpy array of floats y-component of the velocity vector field. """ u = (-k / (2 * np.pi) * ((X - xd)**2 - (Y - yd)**2) / ((X - xd)**2 + (Y - yd)**2)**2) v = (-k / (2 * np.pi) * 2 * (X - xd) * (Y - yd) / ((X - xd)**2 + (Y - yd)**2)**2) return u, v 1

def get_velocity_vortex(gamma, xv, yv, X, Y): """ Returns the velocity field generated by a vortex. Parameters ---------- gamma: float Strength of the vortex. xv: float x-coordinate of the vortex. yv: float y-coordinate of the vortex. X: 2D Numpy array of floats x-coordinate of the mesh points. Y: 2D Numpy array of floats y-coordinate of the mesh points. Returns ------- u: 2D Numpy array of floats x-component of the velocity vector field. v: 2D Numpy array of floats y-component of the velocity vector field. """ u = +gamma / (2 * np.pi) * (Y - yv) / ((X - xv)**2 + (Y - yv)**2) v = -gamma / (2 * np.pi) * (X - xv) / ((X - xv)**2 + (Y - yv)**2) return u, v def get_velocity_freestream(U_inf, theta, x, y, dim=2): """ Returns the velocity field generated by a vortex. Parameters ---------- U_inf: float freestream velocity theta: float angle of freestream flow X: 2D Numpy array of floats x-coordinate of the mesh points. Y: 2D Numpy array of floats y-coordinate of the mesh points. Returns ------- u: 2D Numpy array of floats x-component of the velocity vector field. v: 2D Numpy array of floats y-component of the velocity vector field. """ if dim == 2: 2

u = U_inf * np.cos(theta) * np.ones((x.shape[0], y.shape[0])) v = U_inf * np.sin(theta) * np.ones((x.shape[0], y.shape[0])) elif dim == 1: u = U_inf * np.cos(theta) * np.ones(x.shape[0]) v = U_inf * np.sin(theta) * np.ones(x.shape[0]) return u, v We then define all of the necessary parameters. U_inf = 40.2 # m/s x_center = 0 # m y_center = 0 # m R = 0.037 # m P = 101325 # Pa F_g = 9.8 * 0.15 # N F_g_per_depth = F_g / (2*R) # N/m rho = 1.225 # kg/m^3 #calculate the doublet strength k = 2*np.pi*U_inf*R**2 gamma = F_g_per_depth / (rho*U_inf) v_theta = gamma/ (2 * np.pi * R) omega = v_theta / R s = 60*omega / (2*np.pi) Next we build up the velocity field using the three elementary flows. We can add together the velocity contributions from each of them. x = np.linspace(-0.2,0.2,100) y = x xx,yy = np.meshgrid(x,y) theta = np.linspace(0,2*np.pi,100) ball_shape_x = R*np.cos(theta) ball_shape_y = R*np.sin(theta) u_freestream, v_freestream = get_velocity_freestream(U_inf,0,xx,yy, dim=2) u_vortex, v_vortex = get_velocity_vortex(gamma,x_center,y_center,xx,yy) u_doublet, v_doublet = get_velocity_doublet(k,x_center,y_center,xx,yy) plt.figure(2, figsize= (4,4)) u = u_freestream + u_doublet + u_vortex v = v_freestream + v_doublet + v_vortex plt.streamplot(xx, yy, u, v,density=.9, linewidth=1, arrowsize=1.5, arrowstyle=’->’, broken_streaml plt.plot(ball_shape_x, ball_shape_y) plt.xlabel(’x (m)’) plt.ylabel(’y (m)’) plt.show() 3

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

As a sanity check, you can verify that v r = 0 on the surface of the baseball. v r = xu + yv p x 2 + y 2 u_freestream, v_freestream = get_velocity_freestream(U_inf,0,ball_shape_x, ball_shape_y, dim=1) u_vortex, v_vortex = get_velocity_vortex(gamma, x_center, y_center, ball_shape_x, ball_shape_y) u_doublet, v_doublet = get_velocity_doublet(k, x_center, y_center, ball_shape_x, ball_shape_y) u = u_freestream + u_doublet + u_vortex v = v_freestream + v_doublet + v_vortex v_r = (ball_shape_x * u + ball_shape_y*v) / (np.sqrt(ball_shape_x**2 + ball_shape_y**2)) v_theta = (ball_shape_x*v - ball_shape_y*u) / np.sqrt(ball_shape_x**2 + ball_shape_y**2) plt.figure(0) plt.plot(theta, v_r) plt.ylim([-1,1]) plt.xlabel(’Angle (radians)’) plt.ylabel(’Raidal Velocity’) plt.savefig(’5_1_vr’) plt.show() 4

Now to calculate the coefficient of pressure use the following equation: C p = 1 − ( U U inf ) 2 Because v r = 0 on the surface of the ball, U = v θ , which was calculated in the previous code block. 5

2. Here we will solve the problem numerically. We will build off of the code for the previous question, specifically using the same functions for the elementary flow fields. U_inf = 40.2 # m/s x_center = 0 y_center = 0 R = 0.037 P_inf = 101325 F_g = 9.8 * 0.15 F_g_per_depth = F_g / (2*R) # N/m L_freestream_per_depth = F_g_per_depth rho = 1.225 #calculate the doublet strength k = 2*np.pi*U_inf*R**2 #calculate vortex strength gamma = L_freestream_per_depth / (0.02*rho*U_inf) Next we will iterate over many different heights and calculate the lift force per unit depth if the ball is located at that height. We will start with small heights, and check to see if the lift satisfies our conditions. As soon as we calculate a lift that is with 1% of the freestream L ′ , that is the numerical solution for the smallest height we can use. If you smaller increments between each h that you test, then you will find a more accurate answer. As part of this, we also numerically integrate the pressure distribution over the surface by calculating the pressure at a discrete number of points, and adding together the vertical contribution from each of them. L_list = [] h_list = np.linspace(0.05, 0.5, 10000) theta, dtheta = np.linspace(0,2*np.pi,100, endpoint=False, retstep=True) ball_shape_x = R*np.cos(theta) ball_shape_y = R*np.sin(theta) found_min = 0 for h in h_list: x = np.linspace(-3*h,3*h,101) y = np.linspace(-3*h,3*h,101) xx,yy = np.meshgrid(x,y) # (0,0) is at the center of the ball, meaning the wall is at y = -h # and the image is at (0, -2h) lower_wall = -h lower_image_y = -2*h #calculate velocity along the surface of the ball #freestream velocity u_freestream, v_freestream = get_velocity_freestream(U_inf,0,ball_shape_x,ball_shape_y, dim=1) # velocity due to the vortex and doublet representing the ball u_vortex, v_vortex = get_velocity_vortex(gamma,x_center,y_center,ball_shape_x,ball_shape_y) u_doublet, v_doublet = get_velocity_doublet(k,x_center,y_center,ball_shape_x,ball_shape_y) # velocity due to the vortex and doublet representing the image 6

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

u_vortex_lower, v_vortex_lower = get_velocity_vortex(-gamma,x_center,lower_image_y,ball_shape_x u_doublet_lower, v_doublet_lower = get_velocity_doublet(k,x_center,lower_image_y,ball_shape_x,b # add all velocity components together and calculate the magnitude u = u_freestream + u_doublet + u_vortex + u_vortex_lower + u_doublet_lower v = v_freestream + v_doublet + v_vortex + v_vortex_lower + v_doublet_lower V_mag = np.sqrt(u**2 + v**2) # calculate the pressure coefficient C_p = 1 - (V_mag/U_inf)**2 # calculate the pressure P = C_p * 1/2 * rho * U_inf**2 + P_inf # calculate the lift per depth and store the result L = np.sum(-P*np.sin(theta) * R * dtheta) L_list.append(L) # check if this Lift is within 1% of the free stream lift if np.abs(L-L_freestream_per_depth)/L_freestream_per_depth < 0.01 and not found_min: print(f’Minimum height is {h:0.4f} m’) found_min = True min_h = h # sanity check to make sure there’s no flow through the wall y_lower = np.ones(101)*lower_wall _, v_freestream = get_velocity_freestream(U_inf,0,x,y_lower, dim=1) _, v_vortex = get_velocity_vortex(gamma,x_center,y_center,x,y_lower) _, v_doublet = get_velocity_doublet(k,x_center,y_center,x,y_lower) _, v_vortex_lower = get_velocity_vortex(-gamma,x_center,lower_image_y,x,y_lower) _, v_doublet_lower = get_velocity_doublet(k,x_center,lower_image_y,x,y_lower) v_wall = v_freestream + v_doublet + v_vortex + v_vortex_lower + v_doublet_lower # raise an error if there is flow through the wall eps = 1.e-10 if (np.abs(v_wall) > eps).any(): print(’Flow through wall’) print(v_wall) Using this code we calculate a minimum height of approximately 0.21 m. Next we will plot the relationship between height and lift. I’ve added an orange line to represent 1% less than the freestream lift. plt.figure(0) plt.plot(h_list, L_list) plt.plot([h_list.min(), h_list.max()], [L_freestream_per_depth*0.99, L_freestream_per_depth*0.99]) 7

plt.xlabel(’Height (m)’) plt.ylabel("L’ (N/m)") plt.savefig(’5_2_lift’) plt.show() Lastly, we will plot the streamlines of the flow at this minimum height. I’ve added a black line to represent the wall, and an orange circle to represent the ball. h = min_h x = np.linspace(-3*h,3*h,101) y = np.linspace(-3*h,3*h,101) xx,yy = np.meshgrid(x,y) lower_wall = -h lower_image_y = -2*h #calculate velocity along the surface of the ball u_freestream, v_freestream = get_velocity_freestream(U_inf,0,xx,yy, dim=2) u_vortex, v_vortex = get_velocity_vortex(gamma,x_center,y_center,xx,yy) u_doublet, v_doublet = get_velocity_doublet(k,x_center,y_center,xx,yy) u_vortex_lower, v_vortex_lower = get_velocity_vortex(-gamma,x_center,lower_image_y,xx,yy) u_doublet_lower, v_doublet_lower = get_velocity_doublet(k,x_center,lower_image_y,xx,yy) u = u_freestream + u_doublet + u_vortex + u_vortex_lower + u_doublet_lower v = v_freestream + v_doublet + v_vortex + v_vortex_lower + v_doublet_lower plt.figure(2, figsize= (4,4)) plt.streamplot(xx, yy, u, v,density=0.8, linewidth=1, arrowsize=1.5, arrowstyle=’->’, broken_streamlines=False) plt.plot(ball_shape_x, ball_shape_y) plt.plot([-3*h,3*h],[-h,-h], ’k’) plt.show() 8

Alternatively, if you wanted to solve the problem analytically, these are the first few steps you would take. We will set up the problem in the same way as in the code, building up a velocity field from the 5 different elementary flows. The ball will be centered at (0,0). Here is the terms for the freestream flow. u freestream = U inf v freestream = 0 Here are the terms for the ball. u doublet,ball = − k 2 π x 2 − y 2 x 2 + y 2 v doublet,ball = − k 2 π 2 xy x 2 + y 2 u vortex,ball = Γ 2 π y x 2 + y 2 v vortex,ball = − Γ 2 π x x 2 + y 2 Here are the terms for the image of the ball, at (0 , − 2 h ). We flip the direction of the vortex to mirror 9

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

the ball. u doublet,image = − k 2 π x 2 − ( y + 2 h ) 2 x 2 + ( y + 2 h ) 2 v doublet,image = − k 2 π 2 x ( y + 2 h ) x 2 + ( y + 2 h ) 2 u vortex,image = − Γ 2 π y + 2 h x 2 + ( y + 2 h ) 2 v vortex,image = Γ 2 π x x 2 + ( y + 2 h ) 2 Adding these flows together: u = u freestream + u doublet,ball + u vortex,ball + u doublet,image + u vortex,image v = v freestream + v doublet,ball + v vortex,ball + v doublet,image + v vortex,image Next we would plug in all the variables we know, and leave h as an unknown. Then we would calculate the magnitude of the velocity, and use that to find the pressure distribution around the surface of the ball. That expression could then be integrated to and set equal to our desired lift. At that point we would be able to solve for the exact minimum height required. 10