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

Answer:

$\phi = 155.57$

Explanation:

from figure

taking summation of force in x direction be zero

$\sum x = 0$

$F_D = Tsin \theta$  .....1

$\frac{c_d \rho v^2 A}{2} =Tsin \theta$

taking summation of force in Y direction be zero

$F_B - W- Tcos \theta$

$T = \frac{F_B -W}{cos \theta}$ .........2

putting T value in equation 1

$F_D - \frac{F_B -W}{cos \theta} sin\theta$

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

$F_D = \rho g [\frac{\pi d^3}{6}] ( 1 -Sg) tan \theta$

$tan \theta = \frac{6 c_D \rho v&2 A}{ 2 \rho g V \pi D^3 (1- Sg)}$

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

Reynold number

$Re = \frac{ VD}{\nu} = \frac{10\times 0.5}{1.3 \times 10^{-6}} = 3.84 \times 10^6$

so for Re =$3.84 \times 10^6$  cd is 0.072

$tan \theta = \frac{3\times 0.072 \times 10^2 \times \pi \times 0.5^2}{ 2\times 9.81 \pi 0.5^3(1- 1.5)}$

$\theta = tan^{-1} [\frac{3\times 0.072 \times 10^2 \times \pi \times 0.5^2}{ 2\times 9.81 \pi 0.5^3(1- 1.5)}]$

$\theta = - 65.57 degree$

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