Tuesday, January 31, 2012

MrsDrPoe: Examples of Lift and Drag

For this Thesis Tuesday here on the blog, I'd like to share an example of the concepts of lift and drag that we have been talking about for the past two weeks.

Example: The average pressure and shear stress acting on the surface of the 1m square flat plate are as indicated i nthe Figure.  Determine the lift and drag generated if shear stress is and is not neglected.

Given: alpha = 7deg, pTa = -1.2 kN/m*m, pBa = 2.3 kN/m*m tauTa = 5.8e-2 kN/m*m, tauBa = 7.6e-2 kN/m*m
Find: D and L with and without shear
Solution:

A = 1 m*m, Theta = 90deg - alpha = 83deg

D = int(p*cos(theta))dA + int(tauw*sin(theta))dA

Using average values to avoid integration:

DS = (pBa - pTa)*cos(theta)*A + (tauTa + tauBa)*sin(theta)*A = 559.544 N

Dns = (pBa - pTa)*cos(theta)*A = 426.543 N

L = -int(p*sin(theta))dA + int(tauw*cos(theta))dA

Again using average values to avoid integration:

LS = (pBa - pTa)*sin(theta)*A - (tauTa + tauBa)*cos(theta)*A = 3.458e3 N

Lns = (pBa - pTa)*sin(theta)*A = 3.474e3 N

Next week, get ready to study some internal flows!