Crude oil, with density = 850 kg/m63 and viscosity 2 10^-2 Pa s, flows steadily

Question

Crude oil, with density = 850 kg/m63 and viscosity 2 10^-2 Pa s, flows steadily down a surface inclined theta = 30 degree below the horizontal in a film of thickness h = 2 mm. The velocity profile is given by u_1(x_2) g/mu(hx_2 - x^2_2/2) sin theta (Coordinate x_1 is along the surface and x_2 is normal to the surface). Plot the velocity profile. Determine the magnitude of the shear stress that acts on the surface. We can find the shear stress, tau through: tau = mu u/y = mu rho g(h - y)sin theta/mu =rho g sin theta(h - y) To find tau at the surface, we plug in y = 0 which gives:

Explanation / Answer

shear Stress = viscosity * dU1 / dx2

                    = rho * g * (h - x2) * sin(theta)

                    = 850 * 9.8 * 2 * 10^-3 * sin(30)      =   8.33 N/m^2

Matlab Code to plot

rho = 850;
g = 9.8;
h = 2*10^-3;
theta = pi/6;
mu = 0.02 ;
i=1;
for x2=0:0.0002:h
    f(i) = rho*g*(h*x2 - x2^2/2)*sin(theta) / mu;
    i = i+1;
end
plot(f)

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.