Two parallel plates kept 10cm apart have laminar flow of oil between them with a maximum velocity of 2m/s. Calculate.

i) The discharge per meter width

ii) The shear stress at the plate.

iii) The difference in pressure between two points 20 m apart.

iv) The velocity gradient at the plates.

v) The velocity at 20 mm from plate. Assume viscosity of oil as 20 poise.

Given :

$u_{\text {max }}=2 \mathrm{~m}/\mathrm{s}$

$u_{\text {avg }}=\frac{2}{3}u_{\text {max }}=1.334 \mathrm{~m} / \mathrm{s}$

$D=100mm$

$\mu =20 Poise = 2Pa.sec$

(1 Poise = 0.1 Pa.sec = 0.1 Ns/m²)

$\space $

(a) Discharge per unit width

$\begin {aligned} Q &= Area \times Velocity_{avg}\\ &=D \times 1 \times u_{avg}\\ &= 0.1 \times 1 \times 1.334 \\ &=0.1334 \mathrm{~m} / \mathrm{s} / \mathrm{m} width \end{aligned}$

(b) Shear stress at the plate:

$ \begin {aligned} u_{avg} &=( \frac{D^{2}} {12 \mu}) (\frac{-\partial p}{ \partial x})\\ (\frac {-\partial \mathrm{p}}{ \partial \mathrm{x}}) &=\frac{12 \mu \mathrm{u}_{\mathrm{avg}}} {\mathrm{D}^{2}}\\ &=\frac{12 \times 2\times 1.334} {\mathrm{0.1}^{2}}\\ &=3202 \mathrm{~N} / \mathrm{m}^{2} / \mathrm{m}\\ \space \\ \tau_{0} &=(\frac{-\partial p }{ \partial x}) (\frac{D}{ 2})\\ \left(\tau_{0}\right) &=160 \mathrm{~N} / \mathrm{m}^{2} \end{aligned} $

(c) Pressure difference between two points 20(L) m apart:

$ \begin {aligned}\Delta \mathrm{p} &=\left[\frac{12 \mu u_{avg}}{\mathrm{D}^{2}}\right] \mathrm{L} \\ &=\frac{12 \times 2\times 1.334}{\mathrm{0.1}^{2}} \times 20\\ &=64032 \mathrm{~N} / \mathrm{m}^{2} \end{aligned}$

(d) Velocity gradient at the plates:

$ \begin {aligned} \tau_{0} &=\mu (\frac{du}{dy})_{y=0}\\ (\frac{du}{dy})_{y=0} &=\frac{\tau_{0}}{\mu}=\frac{160 }{2}=80 / \mathrm{s} \end{aligned}$

(e) Velocity at 20 mm from plates:

$ \begin {aligned} u(at \space Y=20 mm) &=\frac{1}{2 \mu}(\frac {-\partial \mathrm{p}}{ \partial \mathrm{x}})(DY-{Y}^{2}) \\ &=\frac{1}{2 \times 2}(3202)(0.1 \times 0.02-{0.02}^{2}) \\ &=1.28 \mathrm{~m} / \mathrm{s} \end{aligned}$