The viscosity of an oil of sp.gr. 0.9 is measured by a capillary tube of diameter 50 mm. The difference of pressure head between 2.2 M apart is 0.5 m of water. The mass of oil collected in a maintaining measuring tank is 60 kg and 100 seconds. Find the viscosity of oil.

 The viscosity of an oil of sp.gr. 0.9 is measured by a capillary tube of diameter 50 mm. The difference of pressure head between 2.2 M apart is 0.5 m of water. The mass of oil collected in a maintaining measuring tank is 60 kg and 100 seconds. Find the viscosity of oil.

Solution:- 

Given:

Sp.gr. of oil = 0.9

Diameter of capillary tube, D=50mm=5cm=0.05m

Length of tube, L=2m

Difference of pressure head, h=0.5m

Mass of oil, m=60kg

Time, t=100s

Mass of oil/sec = 60100=0.6kg/s${\displaystyle \frac{60}{100}}=0.6kg/s$

ρ=0.9×1000=900kg/m3$\rho =0.9\times 1000=900kg/m^3$

Therefore discharge, Q=Mass of oil/secDensity$Q={\displaystyle \frac{\text{Mass of oil/sec}}{\text{Density}}}$

${\displaystyle \frac{\text{<span style="color: \#351c75; font-family: georgia; font-size: medium;"><i><br></i></span>}}{}}$

Q=0.6900=0.000667m3/s$Q={\displaystyle \frac{0.6}{900}}=0.000667m^3/s$

μ=πρgD4128×QL$\mu ={\displaystyle \frac{\pi \rho g{D}^4}{128\times QL}}$

μ=π×900×9.91×0.5×(0.05)4128×0.000667×2$\mu ={\displaystyle \frac{\pi \times 900\times 9.91\times 0.5\times (0.05{)}^4}{128\times 0.000667\times 2}}$

μ=0.5075NS/m2=0.5075×10$\mu =0.5075NS/m^2=0.5075\times 10$ poise

μ=5.075$\mu =5.075$ poise