Answer:
f_D = =3.24 N/m
Explanation:
data given
properties of air[tex]\nu\ of air =19.31*10^{-6} m2/s[/tex]
[tex]\rho = 1.048 kg/m3[/tex]
k = 0.0288 W/m.K
WE KNOW THAT
Reynold's number is given as
[tex]Re =\frac{VD}{\nu}[/tex]
[tex]= \frac{ 15*0.025}{19.31*10^{-6}}[/tex]
= 1.941 *10^4
drag coffecient is given as
[tex]C_D = \frac{f_D}{A_f\frac{\rho v^2}{2}}[/tex]
solving for f_D
[tex]f_D = C_D A_f*\frac{\rho v^2}{2}[/tex]
[tex]=C_D D*\frac{\rho v^2}{2}[/tex]
Drag coffecient for smooth circular cylinder is 1.1
therefore Drag force is
[tex]f_D = 1.1*0.025 *\frac{1.048*15^2}{2}[/tex]
f_D = =3.24 N/m