Determining the viscosity of a fluid

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SUMMARY

The discussion focuses on determining the viscosity (μ) of a fluid between two parallel plates, where the top plate moves at speed V and the middle plate at speed V1 = 0.7V. The relationship between shear stress (τ) and viscosity is established using the equation τ = μ(du/dy). By rearranging this equation, the viscosity is expressed as μ = τ(dy/du), leading to μ = τ((V - 0.7V)/(b - 0.3b)). The participants explore the equality of shear stress on both plates and the implications of constant speed on net force.

PREREQUISITES
  • Understanding of fluid mechanics principles, specifically shear stress and viscosity.
  • Familiarity with the equation τ = μ(du/dy) and its application.
  • Knowledge of parallel plate flow dynamics and velocity profiles.
  • Basic algebraic manipulation skills for solving equations.
NEXT STEPS
  • Study the derivation and application of the Navier-Stokes equations in fluid dynamics.
  • Learn about the concept of shear stress in different fluid flow scenarios.
  • Research the effects of varying plate speeds on fluid viscosity measurements.
  • Explore experimental methods for measuring viscosity in laboratory settings.
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Students and professionals in mechanical engineering, chemical engineering, and physics who are studying fluid dynamics and viscosity measurement techniques.

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Homework Statement


Two flat plates are oriented parallel above a fixed lower plate as shown below, The top plate,
located a distance b above the fixed plate, is pulled along with speed V. The other thin plate is
located a distance 0.3b (cb) above the fixed plate. This plate moves with speed V1, where V1 = 0.7V.
The viscosity of the fluid between the middle and the lower plates is μ0. Determine the viscosity (μ) of the fluid between the top and the middle plates.
2hq8hnl.jpg


Homework Equations


\tau=\mu\frac{du}{dy}

The Attempt at a Solution


Using the equation, I rearranged it to make \mu the subject.
This gives \mu=\tau\frac{dy}{du}
I figured that they want to find the viscosity of the top fluid. So, using the information given, with dy being the height and du being the velocity, substituted it into the equation:
\mu=\tau\frac{V-0.7V}{b-0.3b}

Is this the final answer or is more working required? I'm wondering if you can figure out an expression for \tau. Is it correct to say that as V1 is moving at a constant speed, the net force on the middle plate would then be 0. Therefore, the force on the top and bottom plate must be equal to allow for this constant speed. & Thus, the shear stress on the top and bottom plate is also equal. With F=\tauA
I was then thinking that you could equate \taubottom and \tautop, substitute it into the equation and solve for \mu but it would cancel out through being divided. E.g. \mu\frac{0.7b}{0.3V} = \mu\frac{0.3b}{0.7V}

I'm rather confused, any help is appreciated. Thank you.
 
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hints:

\tau(bottom)=\tau(top)

for the bottom fluid:
\tau(bottom)=\mu_o\frac{V_1}{cb}

for the top fluid:
\tau(top)=\mu\frac{V-V_1}{b-cb}
 

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