Messiah Sutton

Answered

2022-11-09

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 $h=0.50\text{}mm$ . Calculate the shear stress on the upper plate and give its direction.

$\frac{u}{{u}_{max}}=1-{\left(\frac{2y}{h}\right)}^{2}$

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 $h=0.50\text{}mm$ . Calculate the shear stress on the upper plate and give its direction.

Answer & Explanation

dobradisamgn

Expert

2022-11-10Added 17 answers

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,

${u}_{max}=0.30\text{}m/s$

The value of the distance between the plates is,

$h=0.50\text{}mm$

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

$y=\frac{h}{2}=0.25\text{}mm$

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 $1.14\times {10}^{-3}\text{}kg\times {m}^{-1}\times {s}^{-1}$ at given temperature.

Substitute the expression for u to find the expression for shear stress.

Therefore,

$\tau =\mu \frac{d}{dy}\left({u}_{max}(1-\frac{4{y}^{2}}{{h}^{2}})\right)\phantom{\rule{0ex}{0ex}}=-\frac{8\mu {u}_{max}y}{{h}^{2}}$

Substitute the known values,

$\tau =-\frac{8\times 1.14\times {10}^{-3}kg\text{}{m}^{-1}{s}^{-1}0.30m/s\times 0.25\text{}mm\left(\frac{{10}^{-3}m}{mm}\right)}{{(0.50\text{}mm\times \left(\frac{{10}^{-3}m}{mm}\right))}^{2}}\phantom{\rule{0ex}{0ex}}=-2.74kg\times {m}^{-1}\times {s}^{-2}\left(\frac{N}{kg\times m\times {s}^{-2}}\right)\phantom{\rule{0ex}{0ex}}=-2.74N\times {m}^{-2}\phantom{\rule{0ex}{0ex}}=-2.74Pa$

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.

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,

${u}_{max}=0.30\text{}m/s$

The value of the distance between the plates is,

$h=0.50\text{}mm$

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

$y=\frac{h}{2}=0.25\text{}mm$

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 $1.14\times {10}^{-3}\text{}kg\times {m}^{-1}\times {s}^{-1}$ at given temperature.

Substitute the expression for u to find the expression for shear stress.

Therefore,

$\tau =\mu \frac{d}{dy}\left({u}_{max}(1-\frac{4{y}^{2}}{{h}^{2}})\right)\phantom{\rule{0ex}{0ex}}=-\frac{8\mu {u}_{max}y}{{h}^{2}}$

Substitute the known values,

$\tau =-\frac{8\times 1.14\times {10}^{-3}kg\text{}{m}^{-1}{s}^{-1}0.30m/s\times 0.25\text{}mm\left(\frac{{10}^{-3}m}{mm}\right)}{{(0.50\text{}mm\times \left(\frac{{10}^{-3}m}{mm}\right))}^{2}}\phantom{\rule{0ex}{0ex}}=-2.74kg\times {m}^{-1}\times {s}^{-2}\left(\frac{N}{kg\times m\times {s}^{-2}}\right)\phantom{\rule{0ex}{0ex}}=-2.74N\times {m}^{-2}\phantom{\rule{0ex}{0ex}}=-2.74Pa$

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.

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