A horizontal tube 20 cm long and 1.0 cm in diameter is connected at one end to a tank with a fluid whose constant level height is 2 m. Calculate the viscosity if 500 cm^3 flows through the tube in 5 min.
Work:
d = pipe diameter = 1.0 cm = 0.01 m
r = radius of pipe = 0.005 m
h = height of fluid = 2 m
Q = flow rate = 500 cm^3/5 min = (5 x 10^-4 m^3/300 s) = 1.67 x 10^-6 m^3/s
A = pi*r^2 = pi*(0.5 cm)^2 = pi*(0.005 m^2) = 7.85 x 10^-5 m^2
mu = absolute (dynamic) viscosity
mu = (F*h)/(A*U)
U = Q/A = (1.67 x 10^-6 m^3/s)/(7.85 x 10^-5 m^2) = 0.02 m/s
mu = [(F*2 m)/(7.85 x 10^-5 m^2)*(0.02 m/s)]
mu = F/(1.57 x 10^-6) Ns/m^2
Solution:
mu = F/(1.57 x 10^-6) Ns/m^2