An oil is being pumped inside a 45.0-cm-diameter pipe at a Reynolds number of 1300. The oil density is 855 kg/m3 and the viscosity is 2.1 × 10−1 Pa · s. (i) What is the velocity in the pipe? (ii) It is desired to maintain the same Reynolds number of 1300 and the same velocity as in part (i) using a second fluid with a density of 955 kg/m3 and a viscosity of 1.5 × 10−3 Pa · s. What pipe diameter should be used?

(i) The velocity in the pipe is 0.71 m/s

(ii) A pipe diameter of 0.0029 m (0.29 cm) should be used

Explanation:

(i) Velocity (V) = Reynolds number × viscosity/density × diameter of pipe

Reynolds number = 1300

Viscosity = 0.21 Pa.s

Density = 855 kg/m^3

Diameter of pipe = 45 cm = 45/100 = 0.45 m

V = 1300×0.21/855×0.45 = 0.71m/s

(ii) Pipe diameter (D) = Reynolds number × viscosity/density×velocity

Reynolds number = 1300

Viscosity = 0.0015 Pa.s

Density = 955 kg/m^3

Velocity = 0.71 m/s

D = 1300×0.0015/955×0.71 = 0.0029 m (0.0029×100 = 0.29 cm)

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