Answer:
The dynamic viscosity and kinematic viscosity are [tex]1.3374\times 10^{-6}[/tex] lb-s/in2 and [tex]1.4012\times 10^{-3}[/tex] in2/s.
Explanation:
Step1
Given:
Inner diameter is 2.00 in.
Gap between cups is 0.2 in.
Length of the cylinder is 2.5 in.
Rotation of cylinder is 10 rev/min.
Torque is 0.00011 in-lbf.
Density of the fluid is 850 kg/m3 or 0.00095444 slog/in³.
Step2
Calculation:
Tangential force is calculated as follows:
T= Fr
[tex]0.00011 = F\times(\frac{2}{2})[/tex]
F = 0.00011 lb.
Step3
Tangential velocity is calculated as follows:
[tex]V=\omega r[/tex]
[tex]V=(\frac{2\pi N}{60})r[/tex]
[tex]V=(\frac{2\pi \times10}{60})\times1[/tex]
V=1.0472 in/s.
Step4
Apply Newton’s law of viscosity for dynamic viscosity as follows:
[tex]F=\mu A\frac{V}{y}[/tex]
[tex]F=\mu (\pi dl)\frac{V}{y}[/tex]
[tex]0.00011=\mu (\pi\times2\times2.5)\frac{1.0472}{0.2}[/tex]
[tex]\mu =1.3374\times 10^{-6}[/tex]lb-s/in².
Step5
Kinematic viscosity is calculated as follows:
[tex]\upsilon=\frac{\mu}{\rho}[/tex]
[tex]\upsilon=\frac{1.3374\times 10^{-6}}{0.00095444}[/tex]
[tex]\upsilon=1.4012\times 10^{-3}[/tex] in2/s.
Thus, the dynamic viscosity and kinematic viscosity are [tex]1.3374\times 10^{-6}[/tex] lb-s/in2 and [tex]1.4012\times 10^{-3}[/tex] in2/s.