Question

Flow rate of a fluid (density = 1000 kg/m3) in a small diameter tube is 800 mm3/s. The length and the diameter of the tube are 2 m and 0.5 mm, respectively. The pressure drop in 2 m length is equal to 2.0 mpa. The viscosity of the fluid is

Answer 1

Explanation:

Given Flow rate of a fluid (density = 1000 kg/m 3) in a small diameter tube is 800 m m 3/s. The length and the diameter of the tube are 2 m and 0.5 mm, respectively. The pressure drop in 2 m length is equal to 2.0 m pa. The viscosity of the fluid is

• Given density ρ = 1000 kg/m^3, Q = 800 m^3/s = 800 x 10^-9 m^3/s, L = 2 m, d = 0.5 mm = 5 x 10^-4 m
• So ΔP / L = 2 x 10^6 / 2 N/m^2
• So we need to find the viscosity.
• So pressure drop per unit length will be Δp / L = 8 μ u / R^2
•                 So 2 x 10^6 / 2 = 8 μ Q/A / R^2
•                  = 8μQ / π R^4
•       So μ = 2 π (2.5)^4 x 10^-6 x 10^6 / 2 x 8 x 800 x 10^-9

                                              So μ = 0.0192 N – S / m^2

Answer 2

Answer:

0.00192N-s/m²

Explanation:

The pressure drop in laminar flow through pipe is given by:

∆P = 32μVL / D²

and we know that

Velocity (V) = Discharge (Q)/Area(A)

V= 4.07m/s

putting velocity in the equation

we get

μ=0.00192N-s/m²