Air is forced through a smooth horizontal pipe that is 15 m

Air is forced through a smooth horizontal pipe that is 15 m long and has an inner diameter of 0.086 m. Assume that the density and viscosity of air are 1.16 kg/m3 and 1.85 x 10–5 N·s/m2, respectively. (a) If the average airflow velocity is 0.23 m/s, is the flow laminar or turbulent? What is the pressure drop (P) across the pipe length? (b) Repeat part (1) for an average velocity of 2.3 m/s, indicating the type of flow and pressure drop P.

Solution

For flow in pipes, laminar and turbulent flows are differentiated by the Reynold\'s number as

For Re < 2000, flow is laminar

For. Re > 4000, flow is turbulent

(a) Given data :

Length of the pipe, L = 15 m

Diameter of the pipe, D = 0.086 m

Viscosity of air = 1.85 * 10^-5 NS/m^2

Density of air = 1.16 kg/m^3

Average flow velocity, V = 0.23 m/s

Reynold\'s number is given by

Reynold\'s number, Re = Density * Average flow velocity * Diameter of pipe / Viscosity

===> Re = 1.16 * 0.23 * 0.086 / (1.85 * 10^-5)

===> Re = 1240.259

As Reynold\'s number Re < 2000, the flow is laminar

Pressure difference in the pipe is given by

  Pi - Po = 32 * Viscosity * Average flow velocity * Length of pipe / (Diameter)^2

===> Pi - Po = 32 * 1.85 * 10^-5 * 0.23 * 15 / 0.086^2

===> Pi - Po = 0.276 N/m^2 or Pascal

(b)

For an average flow velocity of 2.3 m/s, we have

Reynold\'s number Re = 1.16 * 2.3 * 0.086 / (1.85 * 10^-5)

===> Re = 12402.595

As Reynold\'s number Re > 4000, the flow is turbulent

Pi - Po = 32 * Viscosity * Average flow velocity * Length of pipe / (Diameter)^2

===> Pi - Po = 32 * 1.85 * 10^-5 * 2.3 * 15 / 0.086^2

===> Pi - Po = 2.76 N/m^2 or Pascals