Dashpot mechanism (short notes)

A dash what is a mechanical device, a damper which resist motion via viscous friction (i.e., oil)

The resulting force is proportional to the velocity but acts in the opposite direction, slowing the motion and observing energy.

It is commonly used in conjunction with a spring.

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Dashpot Mechanism

Methods of determination of coefficient of viscosity:-

1) Capillary tube method

In this Hagen Poiseville law is used for calculating viscosity.

In this method, the pressure difference is measured for a given length of capillary tube to determine the viscosity of liquid.

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h = Difference of pressure head for length ‘L’

The pressure at outlet is atmospheric

D=Diameter of capillary tube

L=Length of the tube for which difference of pressure head is known

$\rho$ = Density of fluid

$\mu$ = Coefficient of viscosity

Therefore using Hagen Poiseville's formula

$h=\dfrac{32\mu\overline{u}L}{\rho gD^2}$


$\overline{u}=\dfrac QA=\dfrac Q{\dfrac\pi4D^2}$

Where Q is the rate of liquid flowing through tube

$h=\dfrac{32μ\times\left(\dfrac Q{\dfrac\pi4D^2}\right)\times L}{\rho gD^2}$

$h=\dfrac{128μ\times Q L}{\pi\rho gD^4}$

$\mu=\dfrac{\pi\rho gD^4}{128\times Q L}$

Measurement of D should be done very accurately.

(ii) Falling sphere resistance method

This method is based on stoke’s law, according to which the drag force, ‘F’ on a small sphere moving with a constant velocity ‘V’ through a viscous fluid of viscosity $\mu$ for viscous condition is given by,

$F=3\pi\mu VD……………(1)$

D = Diameter of sphere

V = Velocity of sphere

When the sphere attains a constant velocity ‘V’ the drag force is the difference between the weight of sphere and buoyant force acting on it.

Let L= distance travelled by Sphere in viscous fluid

t = time taken by Sphere to cover distance l,

$\rho_S$ = Density of sphere

W = Weight of sphere

$\rho_F$ = Density of fluid

$F_B$ = Buoyant force acting on a sphere


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Constant velocity of sphere,

$V=\dfrac Lt$

Weight of sphere, W=volume(sphere) $\times$ density of sphere $\times$ g

$W=\dfrac\pi6d^3\times \rho_S\times d$

And buoyant force, $F_B$ = weight of fluid displaced

$F_B=\dfrac\pi6d^3\times\rho_f\times g$

For equilibrium,

Drag force = weight of sphere - buoyant force


Substituting values of F, W and $F_ B$, we get,

$3\pi\mu Vd=\dfrac\pi6d^3\times \rho_S\times g-\dfrac\pi6d^3\times\rho_f\times g$

$3\pi\mu Vd=\dfrac\pi6d^3\times g[\rho_S-\rho_f]$

$\therefore\mu =\dfrac{\pi d^3\times g[\rho_S-\rho_f]}{6\times3\pi Vd}$

$\therefore\mu =\dfrac{gd^2}{18V}[\rho_S-\rho_f]$


Q1) The viscosity of an oil of sp.gr. 0.9 is measured by a capillary tube of diameter 50 mm. The difference of pressure head between 2.2 M apart is 0.5 m of water. The mass of oil collected in a maintaining measuring tank is 60 kg and 100 seconds. Find the viscosity of oil.

Solution:- Given:

Sp.gr. of oil = 0.9

Diameter of capillary tube, D=50mm=5cm=0.05m

Length of tube, L=2m

Difference of pressure head, h=0.5m

Mass of oil, m=60kg

Time, t=100s

Mass of oil/sec = $\dfrac{60}{100}=0.6kg/s$


Therefore discharge, $Q=\dfrac{\text{Mass of oil/sec}}{\text{Density}}$

$Q=\dfrac {0.6}{900}=0.000667m^3/s$

$\mu=\dfrac{\pi\rho gD^4}{128\times Q L}$

$\mu=\dfrac{π\times900\times9.91\times0.5\times(0.05)^4}{128\times0.000667\times 2}$

$\mu=0.5075NS/m^2=0.5075\times10$ poise

$\mu=5.075$ poise

Q2) A sphere of diameter 2 mm Falls 150 mm into 20 seconds in a viscous liquid. The density of the sphere is $7500 kg/m^3$ and liquid is $900 kg/m^3$. Find coefficient of viscosity of the liquid.

Solution:- Given:-

Diameter of sphere, $d=2mm=2\times10^{-3}m$

Distance travelled by sphere =150mm=0.15m

Time taken, t=20seconds

Velocity of sphere, $V=\dfrac{0.15}{20}=0.0075m/s$

Density of sphere, $\rho_S=7500kg/m^3$

Density of fluid, $\rho_F=900kg/m^3$


$\mu=\dfrac{9.81\times [2\times10^{-3}]^2}{18\times0.0075}[7500-900]$

$\mu=\dfrac{9.81\times 4\times10^{-6}\times6600}{18\times0.0075}$

$\mu=1.917NS/m^2=1.917NS/m^2\times10$ poise

$\mu=19.17$ poise

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