# The velocity distribution for laminar flow between parallel plates is given by: u/(u_(max)) = 1- ((2y)/h)2 where h is the distance separating the plates and the origin is placed midway between the plates.

The velocity distribution for laminar flow between parallel plates is given by:
$\frac{u}{{u}_{max}}=1-{\left(\frac{2y}{h}\right)}^{2}$
where h is the distance separating the plates and the origin is placed midway between the plates. Consider a flow of water at ${15}^{\circ }C$ , with ${u}_{max}=0.30\frac{m}{sec}$ and . Calculate the shear stress on the upper plate and give its direction.
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Step 1
Given data:
The velocity distribution for laminar flow is,
$\frac{u}{{u}_{max}}=1-{\left(\frac{2y}{h}\right)}^{2}$
The temperature at which the flow is to be considered is,
$T={15}^{\circ }C$
The value of maximum velocity is,

The value of the distance between the plates is,

Origin is midway between the plates, so the distance of upper plate from the axis is,

Step 2
The shear stress in terms of velocity for a Newtonian fluid like water is given by formula,
$\tau =\mu \frac{du}{dy}$
Here, $\tau$ represents the shear stress and $\mu$ is the dynamic viscosity of the flow of water which is known to be at given temperature.
Substitute the expression for u to find the expression for shear stress.
Therefore,
$\tau =\mu \frac{d}{dy}\left({u}_{max}\left(1-\frac{4{y}^{2}}{{h}^{2}}\right)\right)\phantom{\rule{0ex}{0ex}}=-\frac{8\mu {u}_{max}y}{{h}^{2}}$
Substitute the known values,

The negative sign shows that the shear stress on the upper plate is opposite to the direction of the flow of water.
The shear stress on the upper plate is, $\tau =-2.74Pa$ acting opposite to the direction of laminar flow of water.