Respuesta :
Answer:
0.25916 lb-s/ft^2
Explanation:
Given:-
- The specific gravity of oil, SGo = 0.8
- The specific gravity of sphere, SGo = 8
- Terminal velocity of sphere, v = 2.5 fpm
- The diameter of sphere, D = 0.25 in
Find:-
What is the viscosity of the oil?
Solution:-
- Consider a sphere completely submerged into oil and travelling with terminal velocity ( v ).
- Develop a free body diagram for the sphere. There are forces acting on the sphere.
- The downward acting force is due to the weight of the sphere ( W ):
[tex]W = m_s*g[/tex]
Where,
The mass ( m_s ) of the sphere is given as:
[tex]m_s = S.G_s*p_w*V_s[/tex]
Where,
ρ_w : Density of water = 1.940 slugs/ft3
V_s: The volume of object ( sphere )
- The volume of sphere is expressed as a function of radius:
[tex]V_s = \frac{\pi *D^3}{6}[/tex]
Hence,
[tex]W = S.G_s*p_w*\frac{\pi*D^3 }{6}* g\\\\W = 8*1.940*\frac{\pi*(0.25/12)^3 }{6}*32\\\\W = 0.00235 lb[/tex]
- One of the upward acting force is the buoyant force ( Fb ) that is proportional to the volume of fluid displaced by the immersed object.
- The buoyant force ( Fb ) is given by:
[tex]F_b = S.G_o*p_w*V_s*g[/tex]
- Therefore the buoyant force ( Fb ) becomes:
[tex]F_b = 0.8*1.94*\frac{\pi*(0.25/12)^3 }{6} *32\\\\F_b = (4.73451*10^-^6)*(49.664)\\\\F_b = 0.00023 lb[/tex]
- The other upward acting force is the frictional drag ( F_d ) i.e the resistive frictional force acting on the contact points of the sphere and the fluid oil.
- From stokes formulations the drag force acting on a spherical object which is completely immersed in a fluid is given as:
[tex]F_d = 3*\pi*D*u*v[/tex]
Where,
μ: The viscosity of fluid
v : The velocity of object
Therefore,
[tex]F_d = 3*\pi*\frac{0.25}{12} *u*0.041666\\\\F_d = 0.00818*u\\[/tex]
- Apply Newton's second law of motion for the sphere travelling in the fluid:
[tex]F_n_e_t = m_s*a[/tex]
Where,
a: Acceleration of object = 0 ( Terminal velocity condition )
[tex]F_n_e_t = 0[/tex]
- Plug in the three forces acting on the metal sphere:
[tex]F_d + F_b - W = 0\\\\F_d = W - F_b\\\\0.00818*u = 0.00235 - 0.00023\\\\u = \frac{0.00212}{0.00818} = 0.25916 \frac{lb-s}{ft^2}[/tex]
