Answer:
[tex]\phi = 155.57[/tex]
Explanation:
from figure
taking summation of force in x direction be zero
[tex]\sum x = 0 [/tex]
[tex]F_D = Tsin \theta[/tex] .....1
[tex]\frac{c_d \rho v^2 A}{2} =Tsin \theta[/tex]
taking summation of force in Y direction be zero
[tex]F_B - W- Tcos \theta[/tex]
[tex]T = \frac{F_B -W}{cos \theta}[/tex] .........2
putting T value in equation 1
[tex]F_D - \frac{F_B -W}{cos \theta} sin\theta[/tex]
[tex]F_D = \rho g V ( 1 -Sg) tan \theta [/tex].........3
[tex]F_D = \rho g [\frac{\pi d^3}{6}] ( 1 -Sg) tan \theta [/tex]
[tex]tan \theta = \frac{6 c_D \rho v&2 A}{ 2 \rho g V \pi D^3 (1- Sg)}[/tex]
Water at 10 degree C has kinetic viscosity v = 1.3 \times 10^{-6} m^2/s
Reynold number
[tex]Re = \frac{ VD}{\nu} = \frac{10\times 0.5}{1.3 \times 10^{-6}} = 3.84 \times 10^6[/tex]
so for Re =[tex] 3.84 \times 10^6 [/tex] cd is 0.072
[tex]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)}[/tex]
[tex]\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)}][/tex]
[tex]\theta = - 65.57 degree[/tex]
[tex]\phi = 90 - (-65.57) = 1557.57[/tex] degree
