Viscosity of a liquid

Question

The viscosity of a liquid can also be determined using a so-called capillary viscometer. For this

the volume flow ∆V/∆t is measured through a tube with a radius of r = 0.5 mm and a length of s = 24 cm, between the ends of which there is a pressure difference of ∆p, which is kept constant during the measurement.

This ensures that the filling level is kept constant in the left flask by regularly adding liquid. In the right flask, the flowing liquid is simply allowed to overflow and in this way the filling level stays constant.

a) What is the mathematical relationship between the difference in filling levels ∆h and the pressure difference ∆p between the two pipe ends?

b) In an experiment, a mass of m = 95 g has to be refilled in the time ∆t = 5.0 min in order to keep the difference in filling levels constant at ∆h = 0.25 m. The density of the liquid is ρF liquid = 1.7g/cm3.

Determine the viscosity η of the liquid!

Answer 1

Hello Maya;

First pressure difference;

∆P = ρ * g * ∆h

∆P = 1700 kg/m³ x 9.8 m/s² x 0.25m = 4165 Pa

Then let's calculate the volume (∆V) of liquid that is added to maintain the constant filling level:

∆V = m / ρ_liquid

∆V = 95/1.7 = 55.88 cm³ ~= 5.6 x 10^-2 m³

Next, let's use the volume flow rate formula from the capillary viscometer setup:

∆V/∆t = (π * r⁴ * ∆P) / (8 * η * s)

Solve for the viscosity (η):

η = (π * r^4 * ∆P * ∆t) / (8 * ∆V * s)

η = (3.14 x 0.5⁴ x 10^-12 x 300) / ( 8 x 5.6 x 10^-2 x 24 x 10^-2)

η = 5.88 x 10^-11 / 0.10752 = Please double check the result, I may have done mistake,