Answer:
μ = 0.038 Pa.s
Explanation:
Given that
V= 1.5 m/s
r= 2 mm
L = 18 cm
If we assume that flow inside the tube is laminar ,then the pressure drop ΔP given as
[tex]\Delta P=\dfrac{32\mu LV^2}{d^2}[/tex]
[tex]\Delta P=\dfrac{32\mu LV^2}{d^2}[/tex]
[tex]\mu = \dfrac{\Delta P D^2}{LV^2}[/tex]
[tex]\mu = \dfrac{967\times 0.004^2}{0.18\times 1.5^2}\ Pa.s[/tex]
μ = 0.038 Pa.s
The viscosity of the fluid = 0.038 Pa.s
