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.
Given:\(F_=F_=F_=100 \mathrm{~N}\)\(\mathrm_=A_=A_=1 \mathrm{~m}^\)Shear stress distribution plate 1 is \(\tau_=\mu \frac\)\(\tau_=\mu_ \frac{\left(V_-0\right)}}\)\(\tau_=0. 15 \frac{\left(V_-0\right)}{(0. 5 / 1000)}\)\(\tau_=300 \mathrm{~V}_\)Shear stress distribution plate 2 is \(\tau_=\mu \frac\)\(\tau_=\mu_ \frac{\left(V_-V_\right)}}\)\(\tau_=0. 5 \frac{\left(V_-V_\right)}{(0. 25 / 1000)}\)\(\tau_=2000\left(V_-V_\right)\)Shear stress distribution plate 3 is \(\tau_=\mu \frac\)\(\tau_=\mu_ \frac{\left(V_-V_\right)}}\)\(\tau_=0. 2 \frac{\left(V_-V_\right)}{(0. 2 / 1000)}\)\(\tau_=1000\left(V_-V_\right) \ldots \ldots \ldots \ldots(3)\)Divided the equation 1 by 2 we gett \(\frac{\tau_}{\tau_}=\frac}-V_\right)}\)\(\frac{\left(\frac}}\right)}{\left(\frac}}\right)}=\frac}-V_\right)}\)\(\frac{\left(\frac\right)}{\left(\frac\right)}=\frac}-V_\right)}\)\(2000 \mathrm{~V}_-2000 \mathrm{~V}_=300 \mathrm{~V}_\)\(2300 V_=2000 V_ \ldots \ldots \ldots \ldots . . (4)\)Divided 1 by 3 we get \(\frac{\tau_}{\tau_}=\frac}-V_\right)}\)\(0. 3 V_-0. 3 V_=300 V_ \ldots \ldots \ldots \ldots . . (5)\)Divided equation 2 by 3 we get \(\frac{\tau_}{\tau_}=\frac-V_\right)}-V_\right)}\)\(V_-V_=2 V_-2 V_\)\(3 V_-2 V_=V_ \ldots \ldots \ldots . . (6)\)\(\tau_=\mu_ \frac{\left(V_-0\right)}}\)\(F \times A_=\mu_ \frac{\left(V_-0\right)}}\)\(100 \times 1=(0. 15) \frac{\left(V_-0\right)}{(0. 5 / 1000)}\)\(V_=0. 33 \mathrm{~m} / \mathrm\)from the equation 1 we get \(\mathrm_=\frac=0. 379 \mathrm{~m} / \mathrm\)\(\mathrm_=0. 479 \mathrm{~m} / \mathrm\)
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Engineering 2022-05-15 19:04:59