Question

Air at 37.8 °C and 101.3 kPa absolute pressure flows at a velocity of 23 m/s past a sphere having a diameter of 42 mm. What are the drag coefficient and the force on the sphere?

Answer 1

Explanation:

The given data is as follows.

  T = $37.8^{o}C$,   $\rho = 1.137 kg/m^{3}$,  r = $1.9 \times 10^{-5}$ kg/ms

  Diameter (D) = 42 mm = $42 \times 10^{-3} m$ = 0.042 m

  Velocity, $(\upsilon_{o})$ = 23 m/s

Formula for Reynold number is as follows.

             $N_{Rl} = \frac{\rho \times \upsilon_{o} \times D}{r}$

Putting the given values into the above equation as follows.

            $N_{Rl} = \frac{\rho \times \upsilon_{o} \times D}{r}$

                               = $\frac{1.137 kg/m^{3} \times 23 m/s \times 0.042 m}{1.9 \times 10^{-5}}$

                               = $5.781 \times 10^{4}$  

As it is known that drag coefficient for sphere is $C_{D}$ equals 0.47.

Hence, formula for total drag force is as follows.

             $F_{D} = A_{p} \times d \times C_{D} \times \frac{\rho \times \upsilon^{2}_{o}}{2}$ ......... (1)

                  $A_{p}$ = $\frac{\pi}{4} \times D^{2}$  ....... (2)

Putting equation (2) in equation (1) as follows.

          $F_{D} = A_{p} \times d \times C_{D} \times \frac{\rho \times \upsilon^{2}_{o}}{2}$    

or,          $F_{D} = \frac{\pi}{4} \times D^{2} \times d \times C_{D} \times \frac{\rho \times \upsilon^{2}_{o}}{2}$

                               = $0.47 \times (\frac{(23)^{2}}{2}) \times \pi \times 1.137 \times \frac{(0.042)^{2}}{4}$                

                               = 0.1958 N

Thus, we can conclude that the force on the sphere is 0.1958 N.