Below is the diagram for the question. Here is the EDITED wording of the questio

Question

Below is the diagram for the question. Here is the EDITED wording of the question. Do NOT use what's written in the picture below, just use the diagram and answer the questions. Here is what my professor meant to ask:

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"Notice in the figure of the rotational viscometer, the velocity profile at any point is exactly like the two plates that were used to discuss Newton

Viscosity is often measured in Petroleum Engineering using a Rotational Viscometer. Notice in the figure of the rotational viscometer, the velocity profile at any point is exactly like the two plates that were used to discuss Newton's Law of Viscosity. The outer cylinder is stationary while the inner cylinder is rotated with a set speed. The torque necessary to spin the inner cylinder with that set speed is used to determine the viscosity of the fluid. In order to maintain a linear velocity profile between the two cylinders, the gap must stay pretty small. When you have this situation, tau = mu du/dy = mu du/dr If Ohm = Rotational speed, T = Torque = Force x moment arm, the velocity of the inner cylinder = R Ohm, and the velocity of the outer cylinder = 0, then Derive an expression for torque in terms of R, Ohm, h, viscosity, and the length of the cylinder Using the formula you derived, what is the torque required to rotate the Viscometer at 20 ft/s while measuring SAE 10W oil @ 100 degree F?

Explanation / Answer

Shear stress at wall of inner cylinder = mu*du/dr

Now assuming that u varies linearly with r, and noting that u(r) = 0 and u(R) = R*Omega we get

du / dr = R*Omega / h

Shear stress = mu*(R*Omega / h)

Torque = Shear stress * surface area of inner cylinder

= mu*(R*Omega / h) * pi*r*L

b)

For SAE 10 W oil at 100 deg F, we have viscosity mu = 0.03 Ns/m^2

Also, R*Omega = 20 ft/s = 6.1 m/s

and r = 9" = 0.229 m

Torque = 0.03*(6.1 / h)*3.14*0.229*L

T = 0.1316*L/h Nm

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