Question

An oil of viscosity 8 poise flows between two parallel fixed plates which are kept at 60 mm distance. The pressure drop over the length of 150 cm is observed to be 4.5 kPa. The width of the plates is 0.2 m. The discharge of flow (in lps) is_______.[Assume laminar flow.]

• A. 13.5

Answer 1

The correct option is A 13.5

$\mu =8Poise=0.8Ns/m^2$

t = 60 mm = 0.06 m

$-\frac{\Delta P}{L}=-\frac{dP}{dx}=\frac{4.5\times {10}^3}{1.5}$

$=3\times {10}^{3}N/m^2/m$

For a system of two parallel fixed plates and flow is taking place through it then velocity at any point is given as

$\mu =\frac{1}{2\mu }(-\frac{dP}{dx})(ty-y^2)$

Since, discharge is asked in question, we use average velocity $\overline{u}$

$\overline{u}=\frac{1}{12\mu }(-\frac{dP}{dx})t^2$

$=\frac{1}{12\times 0.8}(3\times {10}^3)(0.06{)}^2$

= 1.125 m/s

$\therefore Discharge,Q=\overline{u}A$

$=1.125\times (0.2\times 0.06)$

$=0.0135m^3/s$

= 13.5 litres/sec