Question

A solid cylinder is concentric with a straight pipe. The cylinder is 0.5 m long and has an outside diameter of 8 cm. The pipe has an inside diameter of 8.5 cm. The annulus between the cylinder ad the pipe contains stationary oil. The oil has a specific gravity of 0.92 and a kinematic viscosity of 5.57 x 10-4 m2/s. Most nearly, what is the force needed to move the cylinder along the pipe at a constant velocity of 1 m/s?

Answer 1

Answer :  

The force needed to move the cylinder is 25.6 N

Further explanation  

Given that,  

Length of the cylinder, l = 0.5 m  

Outer diameter of the cylinder, d = 8 cm = 0.08 m  

Outer radius of the cylinder, $r=0.04\ m$  

Inside diameter of the pipe, d = 8.5 cm = 0.085 m  

Inside radius of the pipe, $r=0.0425\ m$  

Specific gravity of the oil, $\rho=0.92$  

Density of oil, $d=\rho\times \rho_w$

Kinematic viscosity of the oil, $v=5.57\times 10^{-4}\ m^2/s$  

Velocity of the cylinder, u = 1 m/s  

We need to find the force needed to move the cylinder. Let the force is F.  

Specific gravity is defined as the ratio of the density of the substance to the density of water.  

Kinematic viscosity is the acquired resistance of a fluid when there is no external force is acting except gravity. It is denoted by v.

Absolute viscosity is given by :

$v=\dfrac{\mu}{d}$

Where, $d$ = density of oil

And $d=\rho\times \rho_w$ (density of oil = specific gravity × density of water )

$d=0.92\times 10^3\ kg/m^3$

So,  

$\mu=v\times d$..............(1)

$\mu=5.57\times 10^{-4}\ m^2/s\times 0.92\times 10^3\ kg/m^3$

$\mu=0.512\ Pa-s$

The separation between the cylinder and pipe is given by :

$dy=\dfrac{d_p-d_c}{2}=\dfrac{8.5-8}{2}=0.25\ cm=0.0025\ m$

$d_p\ and\ d_c$ are diameter of pipe and cylinder respectively.  

The mathematical expression for the Newton's law of viscosity can be written as:  

$\tau\propto\dfrac{du}{dy}$  

$\tau=\mu\times \dfrac{du}{dy}$..........(2)  

Where  

$\tau$ = Shear stress, $\tau=\dfrac{F}{A}$............(3)  

$\mu$ = viscosity  

$\dfrac{du}{dy}$ = rate of shear deformation

On rearranging equation (1), (2) and (3) we get :  

$\dfrac{F}{A}=v\times \rho\times \dfrac{du}{dy}$...............(4)  

A is the area of the cylinder, $A=2\pi rl$  

Equation (4) becomes :  

$F=v\times \rho\times \dfrac{du}{dy}\times 2\pi rl$..............(5)

$A=\pi d\times l$

$A=\pi \times 0.08\ m\times 0.5\ m$

$A=0.125\ m^2$

Now, equation (5) becomes :

$F=(v\times \rho)\times \dfrac{du}{dy}\times 2\pi rl$

$F=(0.512\ Pa-s)\times (\dfrac{1}{0.0025\ m})\times \times 0.125\ m^2$

F = 25.6 N

Learn more  

Kinematic viscosity : brainly.com/question/12947932

Keyword :  

Specific gravity, Kinematic viscosity, Area of cylinder, fluid mass density.