Respuesta :
Explanation:
The given data is as follows.
T = [tex]37.8^{o}C[/tex], [tex]\rho = 1.137 kg/m^{3}[/tex], r = [tex]1.9 \times 10^{-5}[/tex] kg/ms
Diameter (D) = 42 mm = [tex]42 \times 10^{-3} m[/tex] = 0.042 m
Velocity, [tex](\upsilon_{o})[/tex] = 23 m/s
Formula for Reynold number is as follows.
[tex]N_{Rl} = \frac{\rho \times \upsilon_{o} \times D}{r}[/tex]
Putting the given values into the above equation as follows.
[tex]N_{Rl} = \frac{\rho \times \upsilon_{o} \times D}{r}[/tex]
= [tex]\frac{1.137 kg/m^{3} \times 23 m/s \times 0.042 m}{1.9 \times 10^{-5}}[/tex]
= [tex]5.781 \times 10^{4}[/tex]
As it is known that drag coefficient for sphere is [tex]C_{D}[/tex] equals 0.47.
Hence, formula for total drag force is as follows.
[tex]F_{D} = A_{p} \times d \times C_{D} \times \frac{\rho \times \upsilon^{2}_{o}}{2}[/tex] ......... (1)
[tex]A_{p}[/tex] = [tex]\frac{\pi}{4} \times D^{2}[/tex] ....... (2)
Putting equation (2) in equation (1) as follows.
[tex]F_{D} = A_{p} \times d \times C_{D} \times \frac{\rho \times \upsilon^{2}_{o}}{2}[/tex]
or, [tex]F_{D} = \frac{\pi}{4} \times D^{2} \times d \times C_{D} \times \frac{\rho \times \upsilon^{2}_{o}}{2}[/tex]
= [tex]0.47 \times (\frac{(23)^{2}}{2}) \times \pi \times 1.137 \times \frac{(0.042)^{2}}{4}[/tex]
= 0.1958 N
Thus, we can conclude that the force on the sphere is 0.1958 N.
