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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.
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.
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
ρ = Density of fluid
μ = Coefficient of viscosity
Therefore using Hagen Poiseville's formula
h=32μ¯uLρgD2
But,
¯u=QA=Qπ4D2
Where Q is the rate of liquid flowing through tube
h=32μ×(Qπ4D2)×LρgD2
h=128μ×QLπρgD4
μ=πρgD4128×QL
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 μ for viscous condition is given by,
F=3πμVD……………(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,
ρS = Density of sphere
W = Weight of sphere
ρF = Density of fluid
FB = Buoyant force acting on a sphere
Then,
Constant velocity of sphere,
V=Lt
Weight of sphere, W=volume(sphere) × density of sphere × g
W=π6d3×ρS×d
And buoyant force, FB = weight of fluid displaced
FB=π6d3×ρf×g
For equilibrium,
Drag force = weight of sphere - buoyant force
F=W−FB
Substituting values of F, W and FB, we get,
3πμVd=π6d3×ρS×g−π6d3×ρf×g
3πμVd=π6d3×g[ρS−ρf]
∴μ=πd3×g[ρS−ρf]6×3πVd
∴μ=gd218V[ρS−ρf]
Numericals
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 = 60100=0.6kg/s
ρ=0.9×1000=900kg/m3
Therefore discharge, Q=Mass of oil/secDensity
Q=0.6900=0.000667m3/s
μ=πρgD4128×QL
μ=π×900×9.91×0.5×(0.05)4128×0.000667×2
μ=0.5075NS/m2=0.5075×10 poise
μ=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 7500kg/m3 and liquid is 900kg/m3. Find coefficient of viscosity of the liquid.
Solution:- Given:-
Diameter of sphere, d=2mm=2×10−3m
Distance travelled by sphere =150mm=0.15m
Time taken, t=20seconds
Velocity of sphere, V=0.1520=0.0075m/s
Density of sphere, ρS=7500kg/m3
Density of fluid, ρF=900kg/m3
μ=gD218V[ρS−ρF]
μ=9.81×[2×10−3]218×0.0075[7500−900]
μ=9.81×4×10−6×660018×0.0075
μ=1.917NS/m2=1.917NS/m2×10 poise
μ=19.17 poise