Cone-and-Plate Viscometer: Dynamic Viscosity Calculation
Authors: Housam Binous and Ahmed Bellagi
Cone-and-Plate Viscometer: Dynamic Viscosity Calculation
Authors: Housam Binous and Ahmed Bellagi
Authors: Housam Binous and Ahmed Bellagi
Here, we determine the dynamic viscosity 𝜇 of a Newtonian fluid using a cone-and-plate viscometer. The setup consists of a circular plate and a very shallow cone placed above it, with the cone rotating at a known angular velocity ω. The fluid occupies the small gap between the cone and the plate, creating a nearly uniform shear rate.For a Newtonian fluid, the shear stress 𝜏 is proportional to the shear rate : 𝜏 = 𝜇 with =ω/α where α is the cone angle in radians.The total torque 𝑇 required to rotate the cone is obtained by integrating the local torque over the cone surface:By measuring the torque 𝑇 experimentally and knowing the geometric parameters 𝑅 and 𝛼, as well as the angular velocity ω, we can solve for the fluid viscosity:Note that a good temperature control is essential for the exact experimental determination of the dynamic viscosity.Nomenclature:R (cone radius)α (cone angle, radians)ω (angular velocity, rad/s)μ (dynamic viscosity, Pa·s)Reference:J. O. Wilkes, Fluid Mechanics for Chemical Engineers, 2nd edition, NJ: Prentice Hall, 2006.
𝛾
𝛾
𝛾
T=
2πωμ
3
R
3α
μ=
3TMeasuredα
2πω
3
R
In[]:=
γdot=w/α(*Shearrateisconstantforcone-and-plate*)
Out[]=
w
α
In[]:=
dA=2πrdr(*annulararea*)
Out[]=
2drπr
In[]:=
dF=μγdotdA(*differentialforce*)
Out[]=
2drπrwμ
α
In[]:=
(*differentialtorque*)dT=rdF
Out[]=
2drπwμ
2
r
α
In[]:=
Ttotal=Integrate[dT/dr,{r,0,R}](*totaltorqueT*)
Out[]=
2πwμ
3
R
3α
In[]:=
(*viscosityfrommeasuredtorqueorTMeasured*)μSol=μ/.Solve[Ttotal==TMeasured,μ][[1]]
Out[]=
3TMeasuredα
2πw
3
R
In[]:=
R=0.05;(*Coneradiusinm*)αdeg=2;(*Coneangleindegrees*)Ω=10;(*Angularvelocityinrad/s*)Tmeasured=0.05;(*MeasuredtorqueinN·m*)
In[]:=
(*convertangletoradians*)α=αdeg*Pi/180
Out[]=
π
90
In[]:=
(*dynamicviscosityinPa·s*)μ=(3αTmeasured)/(2PiΩR^3)
Out[]=
0.666667
In[]:=
(*dynamicviscosityincPoiseknowingthat1Pa·s=1000cPoise*)μ=1000(3αTmeasured)/(2PiΩR^3)
Out[]=
666.667