Answer:
R_x = 49.78 lb/ft
Explanation:
Given:
- The velocity u_1 upstream = 10 ft/s
- Pressure field p_1 = p_2 = 10 psig
- The velocity profile downstream is given by:
u_2 = 10 ft/s ...... |y| >= 2
u_2 = 10 - 3/8*(2 - |y| )^2 ....... |y| < 2
- The profiles are given in attachment.
Find:
Determine the drag force per unit length.
Solution:
- Develop a control volume around the object see attachment which only consists of water. Then apply principle of linear momentum along the x-direction. It can be expressed as:
[tex]-p_w*u_1^2*A_1 + 2 \int\limits^2_0 {u_2^2*p_w} \, dy = -F_x\\p_w*u_1^2*h - 2*p_w \int\limits^2_0 {u_2^2} \, dy = R_x[/tex]
Where, p_w: Density of water = 1.94 slugs/ft^3
h : Vertical height of control volume
R_x: The reaction force exerted by object on control volume(Drag)
- To determine h, The conservation of mass principle is applied at sections 1 and 2:
[tex]p_w*u_1*h = 2\int\limits^2_0 ({p_w*u_2}) \, dy \\h = \frac{2}{u_1} \int\limits^2_0 ({10-\frac{3*(2-y)^2}{8} } )\, dy \\\\h = \frac{2}{u_1} \limits^2_0 ({10y+\frac{(2-y)^3}{8} } )\,\\\\h = \frac{2}{10}*({20 + 0 - 0-1 ) = 3.8 ft[/tex]
- Now for Drag force per unit length we have:
[tex]p_w*u_1^2*h - 2*p_w \int\limits^2_0 {(10-\frac{3*(2-y)^2}{8} )^2} \, dy = R_x\\\\p_w*u_1^2*h - 2*p_w \int\limits^2_0 ({100+\frac{9*(2-y)^4}{64}-\frac{15*(2-y)^2}{2}}) \, dy = R_x\\\\p_w*u_1^2*h - 2*p_w * ({100y-\frac{9*(2-y)^5}{320}+\frac{5*(2-y)^3}{2}}) \limits^2_0 = R_x\\\\1.94*10^2*3.8 - 2*1.9*({200+\frac{9}{10}-20}) = R_x\\\\R_x = 737.2 - 687.42\\\\R_x = 49.78 lb/ft[/tex]
- The drag force per unit length on the object is given by R_x = 49.78 lb/ft. It is also the reaction developed due to change in momentum of fluid.
