Question

A 0.5m diameter sphere containing pollution monitoring equipment is dragged through the Charles River at a relative velocity of 10m/s. The sphere has a specific gravity of 0.5, is fully submerged, and tethered to the towing device by a 2m cable. What is the angle the towing cable makes with the horizontal? Assume the water is at 10°C.

Answer 1

`\phi = 155.57`

Step-by-step explanation:

from figure

taking summation of force in x direction be zero

`\sum x = 0`

`F_D = Tsin \theta` .....1

`(c_d \rho v^2 A)/(2) =Tsin \theta`

taking summation of force in Y direction be zero

`F_B - W-  Tcos \theta`

`T = (F_B -W)/(cos \theta)` .........2

putting T value in equation 1

`F_D - (F_B -W)/(cos \theta) sin\theta`

`F_D = \rho g V ( 1 -Sg) tan \theta`.........3

`F_D = \rho g [(\pi d^3)/(6)] ( 1 -Sg) tan \theta`

`tan \theta = (6 c_D \rho v&2 A)/( 2 \rho g V \pi D^3 (1- Sg))`

Water at 10 degree C has kinetic viscosity v = 1.3 \times 10^{-6} m^2/s

Reynold number

`Re = ( VD)/(\\u) = (10* 0.5)/(1.3 * 10^(-6)) = 3.84 * 10^6`

so for Re =
`3.84 * 10^6` cd is 0.072

`tan \theta = (3* 0.072 * 10^2 * \pi * 0.5^2)/( 2* 9.81 \pi 0.5^3(1- 1.5))`

`\theta = tan^(-1) [(3* 0.072 * 10^2 * \pi * 0.5^2)/( 2* 9.81 \pi 0.5^3(1- 1.5))]`

`\theta = - 65.57 degree`

`\phi = 90 - (-65.57) = 1557.57` degree