Respuesta :
Answer:
[tex] \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}[/tex]
Explanation:
For this case we need to remember first thet the torque T is defined as:
[tex] T = FR[/tex]
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:
[tex] F = \mu A \frac{v}{l}[/tex]
Where [tex] \mu[/tex] 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 [tex] v = wR[/tex]
The info given is:
[tex] l = 0.035 in *\frac{1ft}{12in}=0.00292 ft[/tex]
[tex] R= \frac{D}{2} =\frac{6 in}{2}= 3 in*\frac{1ft}{12 in}=0.25 ft[/tex]
[tex] L = 5 ft[/tex] from the info given
N= 250 rpm represent the reveolutions per minute
[tex] T = 1.2 lbf ft[/tex] represent the torque given
We can find the surface area for the cylinder with this formula:
[tex] A= 2\pi R L[/tex]
And if we replace we got:
[tex] A= 2\pi 0.25 ft *5 ft= 7.85 ft^2[/tex]
Now we can find the tangential velocity like this:
[tex] v=wR= \frac2\pi *250 rpm* \frac{1min}{60s} * 0.25 ft=6.55\frac{ft}{s}[/tex]
Now we can set up the following equation for the torque:
[tex] T = FR[/tex]
[tex] T = \mu A \frac{v}{l} R[/tex]
And we can find the value for the viscosity [tex]\mu[/tex] like this:
[tex] \mu = \frac{T}{A \frac{v}{l} R}[/tex]
And if we replace we got:
[tex] \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}[/tex]
