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 C_p 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

$\displaystyle L = \rho U_\infty K = -\rho U_\infty \Gamma$

(4.144)

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