The viscosity of a liquid is determined by rotating cylinder method, in which case the inner cylinder of diameter 20 cm is stationary. The outer cylinder of diameter 20.5 cm, contains the liquid upto a height of 30 cm. The clearance at the bottom of the two cylinders is 0.5 cm. The outer cylinder is rotated at 400 r.P.M. The torque registered on the torsion meter attached to the inner cylinder is 5.886 nm. Find the viscosity of fluid.
Answers
The viscosity of fluid is 1.9286 poise
Explanation:
Given data:
Diameter of inner cylinder = 20 cm
Radius of inner cylinder R1= 10 cm = 0.1 m
Diameter of outer cylinder = 20.5 cm
Radius of inner cylinder R2 = 10.25 cm = .1025 m
Height of liquid = 30 cm
Clearance at the bottom of the two cylinders = 0.5 cm = 0.005 m
Height of inner cylinder immersed in liquid = 30 - h = 30 - 0.5 = 29.5 cm
Height = 29.5 cm = .295 m
Speed of outer cylinder N = 400 r.m.p
ω = 2πN / 60 = 2 x 3.14 x 400 / 60 = 41.88
Torque measured = 5.886 Nm
μ = 2(R2 - R1) x h x T / πR1^2ω [4HhR2 + R1^2(R2 - R1)]
μ = 2(.1025 - 0.1 ) x .005 x 5.886/ π(0.1)^2 x 41.88 [ 4 x .295 x .005 x .1025 + (0.1)^2 (.1025 - 0.1 )
= 2 x 0.0025 x 0.005 x 5.886 / π x 0.01 x 41.88 [ 0.0006047 - 0.000025]
μ = 0.19286 Ns/m^2 = v x 10 = 1.9286 poise
Thus the viscosity of fluid is 1.9286 poise
Also learn more
What is viscosity..?
https://brainly.in/question/6562636