Question

The viscosity of a fluid is to be measured by an viscometer constructed of two 5 ft long concentric cylinders. The inner diameter of the outer cylinder is 6 in, and the gap between the two cylinders is 0.035 in. The outer cylinder is rotated at 250 rpm, and the torque is measured to be 1.2lbf *ft.
A) Determine the viscosity of the fluid.

Answer 1

Answer:

$\mu= \frac{1.2 lb ft}{7.85ft^2 * \frac{6.545 ft/s}{0.00292ft} *0.25 ft}=2.728x10^{-4} \frac{lbf s}{ft^2}$

Explanation:

For this case we need to remember first thet the torque T is defined as:

$T = FR$

Where T represent the torque, F the force acting in the inner cylinder and R the radius for the inner cylinder.

For the inner cylinder the force acting can be expressed as:

$F = \mu A \frac{v}{l}$

Where $\mu$ represent the viscosity of the fluid, A the area of the inner cylinder, v represent the tangential velocity and l the thickness of fluid between the two cylinders.

And the tangential velocity for this case can be esxpressed as $v = wR$

The info given is:

$l = 0.035 in *\frac{1ft}{12in}=0.00292 ft$

$R= \frac{D}{2} =\frac{6 in}{2}= 3 in*\frac{1ft}{12 in}=0.25 ft$

$L = 5 ft$ from the info given

N= 250 rpm represent the reveolutions per minute

$T = 1.2 lbf ft$ represent the torque given

We can find the surface area for the cylinder with this formula:

$A= 2\pi R L$

And if we replace we got:

$A= 2\pi 0.25 ft *5 ft= 7.85 ft^2$

Now we can find the tangential velocity like this:

$v=wR= \frac2\pi *250 rpm* \frac{1min}{60s} * 0.25 ft=6.55\frac{ft}{s}$

Now we can set up the following equation for the torque:

$T = FR$

$T = \mu A \frac{v}{l} R$

And we can find the value for the viscosity $\mu$ like this:

$\mu = \frac{T}{A \frac{v}{l} R}$

And if we replace we got:

$\mu= \frac{1.2 lb ft}{7.85ft^2 * \frac{6.545 ft/s}{0.00292ft} *0.25 ft}=2.728x10^{-4} \frac{lbf s}{ft^2}$