19.09.2019 Views

fluid_mechanics

You also want an ePaper? Increase the reach of your titles

YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.

9.4 Lift 513<br />

Denotes p > p 0<br />

Denotes p < p 0<br />

U, p 0<br />

F I G U R E 9.31<br />

an automobile.<br />

Pressure distribution on the surface of<br />

E XAMPLE 9.14<br />

Lift from Pressure and Shear Stress Distributions<br />

GIVEN When a uniform wind of velocity U blows past the<br />

semicircular building shown in Fig. E9.14a,b, the wall shear<br />

stress and pressure distributions on the outside of the building are<br />

as given previously in Figs. E9.8b and E9.9a, respectively.<br />

FIND If the pressure in the building is atmospheric 1i.e., the<br />

value, p 0 , far from the building2, determine the lift coefficient and<br />

the lift on the roof.<br />

SOLUTION<br />

From Eq. 9.2 we obtain the lift as<br />

As is indicated in Fig. E9.14b, we assume that on the inside of the<br />

building the pressure is uniform, p p 0 , and that there is no<br />

shear stress. Thus, Eq. 1 can be written as<br />

or<br />

where b and D are the length and diameter of the building,<br />

respectively, and dA b1D22du. Equation 2 can be put into<br />

dimensionless form by using the dynamic pressure, rU 2 2, planform<br />

area, A bD, and dimensionless shear stress<br />

to give<br />

l bD 2<br />

l p sin u dA t w cos u dA<br />

p<br />

l 1 p p 0 2 sin u b a D 2 b du<br />

c p<br />

p<br />

1 p p 0 2 sin u du t w cos u du d<br />

0<br />

0<br />

p<br />

t w cos u b a D 2 b du<br />

0<br />

F1u2 t w 1Re2 1 2<br />

1rU 2 22<br />

l 1 2 rU 2 A c 1 2 p 1 p p 0 2<br />

sin u du<br />

1<br />

2 rU 2<br />

1<br />

21Re<br />

p<br />

F1u2 cos u du d<br />

0<br />

From the data in Figs. E9.8b and E9.9a, the values of the two integrals<br />

in Eq. 3 can be obtained by determining the area under the<br />

0<br />

0<br />

(1)<br />

(2)<br />

(3)<br />

curves of 31p p 0 21rU 2 224 sin u versus u and F1u2 cos u<br />

versus u plotted in Figs. E9.14c and E9.14d. The results are<br />

and<br />

Thus, the lift is<br />

or<br />

and<br />

p<br />

0<br />

l 1 2 rU 2 A ca 1 2 b 11.762 1<br />

21Re 13.922d<br />

C L <br />

l<br />

1<br />

2 rU 2 A<br />

(Ans)<br />

(4) (Ans)<br />

COMMENTS Consider a typical situation with D 20 ft,<br />

U 30 fts, b 50 ft, and standard atmospheric conditions<br />

1r 2.38 10 3 slugsft 3 and n 1.57 10 4 ft 2 s2, which<br />

gives a Reynolds number of<br />

Re UD<br />

n 130 ft s2120 ft2<br />

3.82 <br />

1.57 10 4 ft 2 106<br />

s<br />

Hence, the lift coefficient is<br />

1 p p 0 2<br />

sin u du 1.76<br />

1<br />

2 rU 2<br />

p<br />

F1u2 cos u du 3.92<br />

0<br />

l a0.88 1.96<br />

1Re b a1 2 rU 2 Ab<br />

0.88 <br />

1.96<br />

1Re<br />

1.96<br />

C L 0.88 <br />

0.88 0.001 0.881<br />

13.82 10 6 2 1 2

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!