Determination of Viscosity: Experiment 2 - Materials Science

This assignment presents a comprehensive solution to an experiment focused on measuring the viscosity of a fluid, specifically Potassium Chloride, using an Ostwald viscometer. The solution begins by deriving the Poiseuille equation, a fundamental formula in fluid dynamics, under the assumptions of steady, laminar flow. The derivation involves force analysis on a differential element of the fluid within a pipe, considering pressure and shear forces. The solution then applies Newton's law of viscosity to relate shear stress to the velocity gradient, leading to the development of a parabolic velocity profile. The volumetric flow rate is calculated and integrated across the pipe's cross-section. The solution concludes by using the Poiseuille equation to prove the relationship between the viscosities, flow times, and densities of two liquids, highlighting the practical application of the derived equation in determining viscosity. The assignment references key literature sources for further context.

1) Derive the Poiseuille Equation:
The free flow of liquid from upper mark to lower mark in the bulb, see figure (Vesztergom,
2014), can be consider as flow of liquid in a pipe.
Consider flow a fluid through a section of pipe as shown in figure below (Bulu, 2010).
Assumptions: Flow is fully developed, steady and laminar. For practical purposes pressure
can be considered uniform over any cross section of the pipe.
We will perform force analysis on the differential element.
Pressure at the left face is p. Pressure at the right face, which is at a distance dx, is:
p+ ∂ p
∂ x dx
Since the radius of differential element is r, pressure force on each face is:
Left: p ( πr 2 )

Right: ( p+ ∂ p
∂ x dx ) ( πr2 )
The shear force acts over the surface area of the differential element and it’s given by:
Shear force =τ· ( 2 πr )
where, τ is shear stress.
Applying force balance:
Net pressure force on the element = Net shear force on the element
⇒ ( p+ ∂ p
∂ x dx ) ( πr2 )− p ( πr2 ) =−τ ( 2 πr )
⇒ τ=−∂ p
∂ x
r
2
Using Newton’s law of viscosity:
τ =−η ∂u
∂ r
where, u is velocity, η is viscosity and ∂u
∂ r is rate of shear strain.
Therefore,
τ =−η ∂u
∂ r =−∂ p
∂ x
r
2
or,
∂ u= 1
2 η
dp
dx rdr
p is not a function of r (uniform across c/s assumption), therefore integrating the above
equation results in:
u= 1
4 η
dp
dx r2 +C
At the outer radius (r =r0), the fluid is stationary (u=0). Therefore,
0= 1
4 η
dp
dx r0
2 +C

⇒C=−1
4 η
dp
dx r0
2
⇒ u= 1
4 η (−dp
dx ) ( r0
2−r 2
)
This equation represents a parabolic velocity profile, as shown in figure below (Bulu, 2010).
Volumetric flow rate through thru an annular disk located at radius r and with thickness dr is
given by:
dq=udA
dq= 1
4 η ( −dp
dx ) ( r0
2−r2
) (2 π r )
Volumetric flow rate through the entire cross section can be obtained by integration:
Q=∫
0
r 0
1
4 η (−dp
dx ) (r 0
2−r2
) · 2 π dr
¿ π
8 η ( −dp
dx ) r0
4
The pressure drop (difference) per unit length in the bulb of Oswald’s setup is the difference
in the hydrostatic pressure and the atmospheric pressure:
−dp
dx = ( p0 +gdh ) −p0
l = gdh
l = f
l
Volumetric flow rate of liquid volume V in time t is: Q= V
t

therefore,
V
t = π
8 η
f
l r 0
4
⇒η= t·f· π r4
8Vl
here, the outer radius is labelled as r instead of r0
2) To prove: η1
η2
= f 1 · t1
f 2 · t2
From the Poiseuille equation we have:
η=t·f· π r 4
8Vl
π r 4
8 Vl = constant, since these terms represent geometrical properties of the setup.
implies,
η ∝ f·t
⇒η1 ∝ f 1 · t1 , η2 ∝ f 2 · t2
therefore,
η1
η2
= f 1 · t1
f 2 · t2
References:
1) Bulu A, 2010, ‘Fluid Mechanics – Lecture Notes’, Istanbul Technical University,
Istanbul.
https://web.itu.edu.tr/~bulu/fluid_mechanics_files/lecture_notes_07.pdf

2) Vesztergom S, 2014, ‘Determination of viscosity with Ostwald viscometer’, Pázmány
Péter promenade, Budapest.
http://phys.chem.elte.hu/turi/SysPhysChem/Materials/Viscosity_Ostwald.pdf