Explain Dash Pot Mechanism

(5 Marks) May-2018, Dec-2018

Subject Fluid Mechanics 2

Topic Laminar Flow

Difficulty Medium

A dashpot is a mechanical device, a damper which resists motion via viscous friction. The resulting force is proportional to the velocity, but acts in the opposite direction, slowing the motion and absorbing energy. It is commonly used in conjunction with a spring (which acts to resist displacement).

Consider a piston moving in a vertical dash pot containing oil as shown in fig.

Let D, L, W, V be diameter, length and weight, velocity of piston.

and $\mu$ is viscosity of oil.

ū = Avg. velocity of oil in clearance.

L = Clearance between the dash pot and piston.

$\triangle$P = Difference of pressure intensities between the two ends of piston.

The flow of oil through clearance is similar to the viscous flow between two parallel plates. The difference of pressure for parallel plates for length 'L' is,

$\triangle$P = $\frac{12ū\mu L}{t^2}$ -------- (1)

Also the difference of pressure at the two ends of piston is given by,

$\triangle$P = $\frac{wt. \ of \ piston}{area \ of \ piston} = \frac{W}{\pi \times D^2} = \frac{4W}{\pi D^2}$ ------- (2)

on Equating (1) and (2),

$\frac{12ū \mu L}{t^2} = \frac{4w}{\pi D^2}$ ------------ (3)

$\therefore ū = \frac{wt^2}{3\pi \mu LD^2}$

$V_p$ is velocity of piston. The rate of flow of oil in dash pot.

$V_p$ = Velocity x area of piston

$V_p = v \times \frac{\pi D_2}{4}$ --------- (4)

Now,

Rate of flow through clearance

= Velocity through clearance x area of clearance

= $ū \times \pi Dt$

$\therefore ū \times \pi D \times t = v_p \times \frac{\pi}{4} D^2$

$\therefore ū = \frac{V_pD}{4t}$ -------- (5)

Equating (3) and (5)

$\therefore \frac{wt^2}{3 \pi \mu L D^2} = \frac{VD}{u t}$

$\therefore \mu = \frac{4t^2w}{3\pi LD^3V_p}$

$\mu = \frac{4wt^3}{3\pi LD^3V_p}$