Chemistry, 22.08.2019 03:10, citiboy45

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?

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answered: Geo777

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.

answered: Guest

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answered: Guest

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