Surface Pressure Distribution and Lift

The surface pressure is calculated from the Bernoulli equation as

$\displaystyle p_s = p_\infty + {1 \over 2} \rho U_\infty ^2 - 2 U_\infty ^2
 \left( \sin \theta
 -  \sin \beta \right )^2$ (4.133)
or
   
$\displaystyle C_p = 1 - {\left( {v_\theta \over U_\infty}
 \right)^2} = 1 - 4 \left(\sin \theta - \sin \beta \right)^2$ (4.134)

Figure 4.35A: Cp distribution for a lifting cylinder, $ \beta$=-150.

Figure 4.35B: Cp distribution for a lifting cylinder plottd around the cylinder, .$ \beta$=-150


 

The Cp distribution is plotted in Fig.4.35A and is also shown plotted along the cylinder surface in Fig.4.35B . Asymmetry about x-axis is evident indicating the generation of lift. Drag however is zero. Magnitude of the lift force is calculated by integration as in Eqn. 4.118 and 4.119 .

We have from 4.133,

$\displaystyle p_s = p_\infty + {1 \over 2} \rho U_\infty ^2 \left ( 1 - 4 \left(
 \sin \theta
 -  \sin \beta \right )^2 \right)$ (4.135)
$\displaystyle \noalign{i.e.,}$    
$\displaystyle p_s = p_\infty + {1 \over 2} \rho U_\infty ^2 \left (1 - 4 \sin^2 \theta
 -4 \sin^2 \beta + 8 \sin \theta \sin \beta \right)$ (4.136)

Lift is now given by

$\displaystyle L = -\int_0 ^ {2 \pi} a \sin \theta \left[ p_\infty + {1 \over 2}...
...1 - 4 \sin^2 \theta
 -4 \sin^2 \beta + 8 \sin \theta \sin \beta \right) \right]$ (4.137)

giving,

$\displaystyle L = -\int_0 ^ {2 \pi} a \left[ p_\infty + {1 \over 2} \rho
 U_\in...
...theta
 -4 \sin^2 \beta \sin \theta + 8 \sin^2 \theta \sin \beta \right) \right]$ (4.138)

This reduces to

$\displaystyle L $ $\displaystyle = -\int_0 ^ {2 \pi} a p_\infty \sin \theta - {1 \over 2} \rho
 U_\infty ^2 a \int_0 ^ {2 \pi} 4 \sin^3 \theta$ (4.139)
  $\displaystyle + {1 \over 2} \rho U_\infty ^2 a \int_0 ^ {2 \pi}\sin^2 \beta
 \s...
...eta +{1 \over 2} \rho U_\infty ^2 a \int_0 ^{2 \pi}
 8 \sin^2 \theta \sin \beta$ (4.140)

As we had in Eqn. 4.121 and 4.124 , the first three integrals in 4.139 and 4.140 are each zero. It follows that Lift is now given by,

$\displaystyle L $ $\displaystyle ={1 \over 2} \rho U_\infty ^2 a \int_0 ^{2 \pi}
 8 \sin^2 \theta \sin \beta$ (4.141)
  $\displaystyle =4 \rho U_\infty ^2 \sin \beta \left[ \frac{\theta}{2} -
 \frac{\sin^2 \theta}{4} \right ]_0 ^{2 \pi}$ (4.142)
  $\displaystyle = 4 \pi \rho U_\infty ^2 a \sin \beta $ (4.143)

Upon substituting for $ \sin \beta $ we have


Thus lift developed by a rotating circular cylinder is equal to the product of density, freestream speed and circulation.

(c) Aerospace, Mechanical & Mechatronic Engg. 2005
University of Sydney