Respuesta :
Answer:
Re = 2.0645
Explanation:
considering pipe with oil flow from 95 degree F to 105 degree F
Average velocity is 2.0 ft./sec
viscosity of oil at 00 degree F is 45 centistoke
[tex]\nu = 45 centi stoke[/tex]
[tex]\nu = 45\times 10^{-4} m^2/s[/tex]
d = 0.6 inch = 0.1524 m
v = 2 ft/sec = 0.609 m/s
Reynold number [tex]= \frac{inertia\ force}{viscous\ force}[/tex]
[tex]= \frac{\rho vd}{\mu}[/tex]
[tex] = \frac{v d}{\nu}[/tex]
[tex]= \frac{ 0.9096 \times 0.1524}{45\times 10^{-4}}[/tex]
Re = 2.0645
Answer:
Re=203.2
Explanation:
Given that
d= 0.6 in
1 in = 0.0254 m
d= 0.01524 m
u= 2 ft/s
1 ft/s = 0.30 m/s
u= 0.60 m/s
kinetic viscosity ν = 45 Centistokes
1 Centistokes = 10⁻⁶ m²/s
ν = 45 x 10⁻⁶ m²/s
We know that Reynolds number given as
[tex]Re=\dfrac{ud}{\nu }[/tex]
Now by putting the values
[tex]Re=\dfrac{ud}{\nu }[/tex]
[tex]Re=\dfrac{0.6\times 0.01524}{45\times 10^{-6}}[/tex]
Re=203.2
This is the Reynolds number inside tube.
