Calculating Shear Stress on Parallel Plates at 15 deg C

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The discussion focuses on calculating shear stress in laminar flow between parallel plates at 15 degrees Celsius. The velocity distribution is given by the equation u/u_max = 1 - (2y/h)^2, with u_max set at 0.10 m/s and h at 0.25 mm. The shear stress on the upper plate is calculated using the formula τ_yx = μ (du/dy), with the dynamic viscosity μ at 1.14 x 10^-3 N*s/m^2. The temperature is converted to Kelvin as 288.15K for accurate calculations. The final calculated shear stress is -1.82 N/m^2, indicating the direction of the stress.
falcon0311
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So I've got this velocity distribution for laminar flow between parallel plates.

\frac{u}{u_{max}} = 1 - (\frac{2y}{h})^2

h is the distance between the plates with the origin placed midway between the plates. I'm assuming this is for water flowing at 15 deg C with u_{max} = 0.10 m/s and h = 0.25 mm.

I'm supposed to calculate the shear stress on the upper plate and give its direction. I'm trying to figure out how to incorporate these into

\tau_{yx} = \mu \frac{du}{dy} at 15 deg C, \mu = 1.14*10^{-3} N*s/m^2 I also know the temperature has to be in Kelvin (288.15K in this case). Anyone willing to give me a push?
 
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I've got it figured out. -1.82 N/m^2 = \tau_{yx}
 
seems correct
 
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