Problem Statement
Known
Three large plates are separated by thin layers of two Newtonian fluids, as shown in the figure. The thicknesses of the two fluid layers are \(a\) and \(b\), respectively.
The top plate is moving to the right at a constant velocity \(V_t = 0.5~\mathrm{m/s}\).
Find
Determine the velocity of the bottom plate moving to the left, \(V_b\), to keep the middle plate at rest.
Analysis
FBD for the middle plate in the \(x\) direction
To keep the middle plate at rest, the force balance in the \(x\) direction is
\[ F_{\text{shear},1} = F_{\text{shear},2} \]
\[
F_{\text{shear},1} = \tau_1 A = \left(\mu_1 \frac{du_1}{dy}\right)A
= \mu_1\left(\frac{V_t-0}{a}\right)A = \mu_1\frac{V_t}{a}A
\]
\[
F_{\text{shear},2} = \tau_2 A = \left(\mu_2 \frac{du_2}{dy}\right)A
= \mu_2\left(\frac{V_b-0}{b}\right)A = \mu_2\frac{V_b}{b}A
\]
Thus,
\[ \mu_1\frac{V_t}{a}A = \mu_2\frac{V_b}{b}A \] \[ V_b = \frac{\mu_1}{\mu_2}\frac{b}{a}\,V_t \]Inputs & Outputs
Fluid 1 is water, with \(\mu_1 = 0.00100~\mathrm{Pa\cdot s}\).
Pa · s
Bottom plate speed, \(V_b\) (m/s)
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Plot
Solid line: \(V_b(b/a)\). Dashed line: \(V_t = 0.5~\mathrm{m/s}\). Drag the red point left/right to change \(b/a\).