Question

Fluids ofviscosities μ1 = .15 N*s/m^2, μ2 = 0.5 Ns/m^2, μ3 = 0.2 Ns/m^2 are contained between two parallel plates (each plate is 1 m^2 in area). Thethicknesses are h1 = 0.5mm, h2 = 0.25mm, h3 0.2mm. Find the steady speed V of the upper plate and the velocities at the two interfaces due to a force F = 100N.Plot the velocity distribution.

Answer 1

Given:

\(F_{1}=F_{2}=F_{3}=100 \mathrm{~N}\)

\(\mathrm{A}_{1}=A_{2}=A_{3}=1 \mathrm{~m}^{2}\)

Shear stress distribution plate 1 is \(\tau_{1}=\mu \frac{d u}{d y}\)

\(\tau_{1}=\mu_{1} \frac{\left(V_{1}-0\right)}{h_{1}}\)

\(\tau_{1}=0.15 \frac{\left(V_{1}-0\right)}{(0.5 / 1000)}\)

\(\tau_{1}=300 \mathrm{~V}_{1}\)

Shear stress distribution plate 2 is \(\tau_{2}=\mu \frac{d u}{d y}\)

\(\tau_{2}=\mu_{2} \frac{\left(V_{2}-V_{1}\right)}{h_{2}}\)

\(\tau_{2}=0.5 \frac{\left(V_{2}-V_{1}\right)}{(0.25 / 1000)}\)

\(\tau_{2}=2000\left(V_{2}-V_{1}\right)\)

Shear stress distribution plate 3 is \(\tau_{3}=\mu \frac{d u}{d y}\)

\(\tau_{3}=\mu_{3} \frac{\left(V_{3}-V_{2}\right)}{h_{3}}\)

\(\tau_{3}=0.2 \frac{\left(V_{3}-V_{2}\right)}{(0.2 / 1000)}\)

\(\tau_{3}=1000\left(V_{3}-V_{2}\right) \ldots \ldots \ldots \ldots(3)\)

Divided the equation 1 by 2 we gett \(\frac{\tau_{1}}{\tau_{2}}=\frac{300 V_{1}}{2000\left(V_{2}-V_{1}\right)}\)

\(\frac{\left(\frac{F_{1}}{A_{1}}\right)}{\left(\frac{F_{2}}{A_{2}}\right)}=\frac{300 V_{1}}{2000\left(V_{2}-V_{1}\right)}\)

\(\frac{\left(\frac{100}{1}\right)}{\left(\frac{100}{1}\right)}=\frac{300 V_{1}}{2000\left(V_{2}-V_{1}\right)}\)

\(2000 \mathrm{~V}_{2}-2000 \mathrm{~V}_{1}=300 \mathrm{~V}_{1}\)

\(2300 V_{1}=2000 V_{2} \ldots \ldots \ldots \ldots . .(4)\)

Divided 1 by 3 we get \(\frac{\tau_{1}}{\tau_{3}}=\frac{300 V_{1}}{1000\left(V_{3}-V_{2}\right)}\)

\(0.3 V_{3}-0.3 V_{2}=300 V_{1} \ldots \ldots \ldots \ldots . .(5)\)

Divided equation 2 by 3 we get \(\frac{\tau_{2}}{\tau_{3}}=\frac{2000\left(V_{2}-V_{1}\right)}{1000\left(V_{3}-V_{2}\right)}\)

\(V_{3}-V_{2}=2 V_{2}-2 V_{1}\)

\(3 V_{2}-2 V_{1}=V_{3} \ldots \ldots \ldots . .(6)\)

\(\tau_{1}=\mu_{1} \frac{\left(V_{1}-0\right)}{h_{1}}\)

\(F \times A_{1}=\mu_{1} \frac{\left(V_{1}-0\right)}{h_{1}}\)

\(100 \times 1=(0.15) \frac{\left(V_{1}-0\right)}{(0.5 / 1000)}\)

\(V_{1}=0.33 \mathrm{~m} / \mathrm{s}\)

from the equation 1 we get \(\mathrm{V}_{2}=\frac{0.869}{0.33}=0.379 \mathrm{~m} / \mathrm{s}\)

\(\mathrm{V}_{3}=0.479 \mathrm{~m} / \mathrm{s}\)