Viscosity Problem using power law for non-Newtonian fluids

[jetpacks]

Problem Statement:
Blood is a pseudoplastic fluid that has a variable viscosity at 37 °C that depends on the percent composition of hematocrit and plasma. It will usually range between 3 x 10^-3 to 4 x 10^-3 Pa s. A small sample of blood is tested in a viscometer and the following results are obtained:

(Shear stress (Pa), Shear rate (s^-1))
(0.0235, 2.25)
(0.0396, 4.5)
(0.0787, 11.25)
(0.1324, 22.5)
(0.2227, 45)
(0.3746, 90)
(0.7447, 225)
(1.2524, 450)

Determine the viscosity of the blood sample using the power law for non-Newtonian fluids:

τ=ηγ^n

where τ is shear stress, η is viscosity, γ is shear rate, and n is the flow behavior index.
_____

I have no idea where to start, since I have two unknowns in this problem. It also seems like the sample would have a different viscosity at each point, but the problem only asks for one value. I am truly lost. Any help or insight would be greatly appreciated!

 

[Achy47]

Hi there, thank you for bringing this problem to our attention. it is important to be able to analyze and interpret data, especially when it comes to fluid mechanics. Let's break down the problem step by step to help you understand and solve it.

First, let's define the variables in the power law equation. As mentioned, τ is the shear stress, η is the viscosity, γ is the shear rate, and n is the flow behavior index. We also know the values for shear stress and shear rate from the data provided. So, we can rewrite the equation as:

0.0235 = η(2.25)^n
0.0396 = η(4.5)^n
0.0787 = η(11.25)^n
0.1324 = η(22.5)^n
0.2227 = η(45)^n
0.3746 = η(90)^n
0.7447 = η(225)^n
1.2524 = η(450)^n

Now, we have eight equations with two unknowns (η and n). To solve this, we can use a method called linearization. We will take the natural log of both sides of the equation to get:

ln(0.0235) = ln(η) + n*ln(2.25)
ln(0.0396) = ln(η) + n*ln(4.5)
ln(0.0787) = ln(η) + n*ln(11.25)
ln(0.1324) = ln(η) + n*ln(22.5)
ln(0.2227) = ln(η) + n*ln(45)
ln(0.3746) = ln(η) + n*ln(90)
ln(0.7447) = ln(η) + n*ln(225)
ln(1.2524) = ln(η) + n*ln(450)

Now, we can treat this as a linear equation with two unknowns (ln(η) and n). We can use any method to solve this, such as substitution or elimination. Once we have the values for ln(η) and n, we can take the antilog to get the values for η and n.

Once we have