An oil of dynamic viscosity 1.5 poise and relative density 0.9 flows through a 3 cm diameter vertical pipe. Two pressure guages are fixed 20 m apart. The guage A fixed at the top records 200kPa and the guage B fixed at the bottom records 500 kPa. Find the direction of flow and rate of flow.

Data:-

$u=1.5 poise=0.15 N-s/m^2$

$S_{oil}=0.9,L=20m$

$D=3cm=3\times 10^{-2}m$

$P_A=200kPa=200\times 10^3N/m^2$

$P_B=500kPa=500\times 10^3N/m^2$

To find:-

(a) Direction of flow=?

(b)Rate of flow=?

Solution:-

(a)Direction of flow:

Total energy at point A

$T.E_A=\frac{P_A}{S_{oil}.9}+Z_A$

Here $Z_A=20m$

$S_{oil}=S_{oil}\times S_{water}$

$=0.9\times 1000=900\frac{kg}{m^3}$

$\therefore T.E_A=\frac{200\times 10^3}{900\times 9.81}+20$

$=42.652m$................(1)

Total energy at point B

$T.E_B=\frac{P_B}{S_{oil}.9}+Z_B$

Here $Z_B=0......(Datum)$

$\therefore T.E_B=\frac{500\times 10^3}{900\times 9.81}+0$

$=56.63 m$..............(2)

From (1) and (2)

$T.E_B\gt T.E_A$

$\therefore \text{water flows from 'B' to 'A'}$

(b) Rate of flow

$h_F=\frac{32.u.v.L}{S_{oil}.9.D^2}$............loss of head

Also $h_F=T.E_B-T.E_A$............piczometric head

$=56.63-42.652$

$=13.979m$

$\therefore$ Above equation becomes

$13.979=\frac{32\times 0.15\times v\times 20}{900\times 9.81\times 0.03^2}$

$v=1.157 m/s$

Now, using continuity equation

$Q=A.v$

$=\frac{\pi}{4}\times 0.03^2\times 1.157$

$=8.1783\times 10^{-4} m^3/sec$