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

Step-by-step explanation:

The given data is as follows.

T =
`37.8^(o)C`,
`\rho = 1.137 kg/m^(3)`, r =
`1.9 * 10^(-5)` kg/ms

Diameter (D) = 42 mm =
`42 * 10^(-3) m` = 0.042 m

Velocity,
`(\upsilon_(o))` = 23 m/s

Formula for Reynold number is as follows.

`N_(Rl) = (\rho * \upsilon_(o) * D)/(r)`

Putting the given values into the above equation as follows.

`N_(Rl) = (\rho * \upsilon_(o) * D)/(r)`

=
`(1.137 kg/m^(3) * 23 m/s * 0.042 m)/(1.9 * 10^(-5))`

=
`5.781 * 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) * d * C_(D) * (\rho * \upsilon^(2)_(o))/(2)` ......... (1)

`A_(p)` =
`(\pi)/(4) * D^(2)` ....... (2)

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

`F_(D) = A_(p) * d * C_(D) * (\rho * \upsilon^(2)_(o))/(2)`

or,
`F_(D) = (\pi)/(4) * D^(2) * d * C_(D) * (\rho * \upsilon^(2)_(o))/(2)`

=
`0.47 * (((23)^(2))/(2)) * \pi * 1.137 * ((0.042)^(2))/(4)`

= 0.1958 N

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