Viscosity of liquid n-pentane based on falling ball velocity measurement​
Housam Binous and Ahmed Bellagi

​​We propose to find the dynamic viscosity of liquid n-pentane based on the falling ball experiment. ​To do so, we consider a glass sphere, which is initially at rest (i.e., at
t=0
we have
v=0
). This ball has a density equal to
2500kg
3
m
and a radius equal to
5mm
. At
t=0
, we let this sphere fall in a container filled with n-pentane.​1/ Considering that this ball reaches
80%
of its terminal fall velocity after
103.83seconds
, determine the viscosity (expressed in
Pa×s
)of liquid n-pentane. Knowing that the viscosity versus temperature follows the empirical law:
η=A×
B
T
e
where
A=0.0191mPa×s
and
B=722.2K
. A and B have been determined using curve fitting of experimental viscosity measurements of liquid n-pentane at different temperatures.​2/ Using the previously determined value of the viscosity, infer the temperature of liquid n-pentane during the falling ball experiment.​
Thegoverningequation,seeRef.1,involvestheweightofthesphere,thedragforceandtheArchimedes'upwardbuoyantforce:​​​​rhos4/3Pir^3v'[t]=-6Piηrv[t]+4/3Pir^3(rhos-rhof)g​​​​Reference:​​[1]J.O.Wilkes,FluidMechanicsforChemicalEngineers,2ndedition,NJ:PrenticeHall,2006.
In[]:=
DSolve[{rhos4/3Pir^3v'[t]==-6Piηrv[t]+4/3Pir^3(rhos-rhof)g,v[0]==0},v,t]
Out[]=
vFunction{t},-
2
-
9tη
2
2
r
rhos

-1+
9tη
2
2
r
rhos

g
2
r
(rhof-rhos)
9η

In[]:=
v[t_]:=
2
-
9tη
2
2
r
rhos

-1+
9tη
2
2
r
rhos

g
2
r
(rhos-rhof)
9η
In[]:=
TerminalFallVelocity=Limit[v[t],{t->+Infinity}][[1]]
Out[]=
2g
2
r
(-rhof+rhos)
9η
In[]:=
Ratio=v[t]/TerminalFallVelocity//FullSimplify
Out[]=
1-
-
9tη
2
2
r
rhos

rhos=2500;r=0.005;t=103.831;(*givenarethesphere'sdensityandradiusaswellasthetimewhenratio=0.8*)
Viscosity of liquid n-pentane is found equal to 0.000215 Pa*s. This value is in agreement with what is found using NISTWebbook.​​
In[]:=
η=η/.FindRoot[Ratio==0.8,{η,0.004}]
Out[]=
0.000215285
Temperature of liquid n-pentane in Kelvin
In[]:=
A=0.0191;B=722.2;
In[]:=
FindRoot[η==0.001AExp[B/T],{T,280}]
Out[]=
{T298.149}