Answer:
Flow rate is [tex]1.82\times 10^{-8} m^{3}/s[/tex]
Explanation:
Given information
Density of oil, [tex]\rho_{oil}= 850 Kg/m^{3}[/tex]
kinematic viscosity, [tex]v= 0.00062 m^{2} /s[/tex]
Diameter of pipe, D= 5 mm= 0.005 m
Length of pipe, L=40 m
Height of liquid, h= 3 m
Volume flow rate for horizontal pipe will be given by
[tex]\bar v=\frac {\triangle P\pi D^{4}}{128\mu L}[/tex] where [tex]\mu[/tex] is dynamic viscosity and [tex]\triangle P[/tex] is pressure drop
At the bottom of the tank, pressure is given by
[tex]P_{bottom}=\rho_{oil} gh=850 Kg/m^{3}\times 9.81 m/s^{2}\times 3 m= 25015.5 N/m^{2}[/tex]
Since at the top pressure is zero, therefore change in pressure a difference between the pressure at the bottom and the top. It implies that change in pressure is still [tex]25015.5 N/m^{2}[/tex]
Dynamic viscosity, [tex]\mu=\rho_{oil}v= 850 Kg/m^{3}\times 0.00062 m^{2}/s=0.527 Kg/m.s[/tex]
Now the volume flow rate will be
[tex]\bar v=\frac {25015.5 N/m^{2}\times \pi \times 0.005^{4}}{128\times 0.527 Kg/m.s \times 40}=1.82037\times 10^{-8} m^{3}/s\approx 1.82\times 10^{-8} m^{3}/s[/tex]
Proof of flow being laminar
The velocity of flow is given by
[tex]V_{flow}=\frac {\bar v}{A}=\frac {1.82\times 10^{-8} m^{3}/s}{0.25\times \pi\times 0.005^{2}}=0.000927104 m/s[/tex]
Reynolds number, [tex]Re=\frac {\rho_{oil} v_{flow} D}{\mu}=\frac {850 Kg/m^{3}\times 0.000927104 m/s\times 0.005}{0.527 kg/m.s}=0.007476648[/tex]
Since the Reynolds number is less than 2300, the flow is laminar and the assumption is correct.
