Answer:
[tex]P_1-P_2=795.18 Pa[/tex]
Explanation:
Given that diameter of tube= 20 mm
flow rate = 72 lt/hr
The properties of SAE 30 oil at 30°C
[tex]\mu =155.31 mPa[/tex]
[tex]\rho =872.5 \frac{kg}{m^3}[/tex]
We know that volume flow rate Q= AV
[tex]2\times 10^{-5}=\dfrac{\pi}{4}\times{0.02^2}V[/tex]
V= 0.064 m/s
Now lets find Reynolds number to check the type of flow
[tex]Re=\dfrac{\rho VD}{\mu }[/tex]
[tex]Re=\dfrac{872.5\times 0.064\times0.020}{0.15531 }[/tex]
Re=7.18 So we can say that this is laminar flow and we know that pressure drop for laminar flow is given as
[tex]P_1-P_2=\dfrac{32\mu VL}{d^2}[/tex]
Now by putting the values the pressure drop for per unit length
[tex]P_1-P_2=\dfrac{32\times 0.15531\times 0.064 }{0.02^2}[/tex]
[tex]P_1-P_2=795.18 Pa[/tex]
