Question

In one test, between two concentric cylinders (20 cm long, 1 and 2 cm radius), with very thin walls, glycerin is placed; keeping the outer cylinder fixed, the other is rotated by applying a moment of 26 kdina.cm to it. Under such conditions, the velocity in the glycerin (V, in m / s) changes with the distance to the axis (r, in m), according to the equation V = 0.4 / r - 1000 * r

At what temperature was the glycerin during essay?

Answer 1

moment of force= 26 k dyne .cm= 0.0026 N m

force applied= moment / radius = 0.0026/0.02=0.13 N

stress apllied = force/ area= 0.13/(2 x 3.14* r₀* L)= 5.17 N/m²

newtons law of viscosity,

shear stress= viscosity x velocity gradient

velocity gradient at r will be 0.4/r²-1000

velocity gradient at r= 0.01 will be 3000

viscosity = 5.17/3000=0.00172 Pa-s

now find the temperature at which viscosity of glycerin( from nomo graph) is equal to calculated above.

it will come 145 C