Flow through tubes is the subject of many fluid dynamics problems Flow through tubes is the subject of many fluid dynamics problems. If you know certain parameters, other parameters can easily be calculated. Known Parameters *Geometry of the Pipe *Average Velocity Parameters Solved For *Reynolds’ Number *Volumetric Flowrate *Mass Flowrate

PROBLEM: Air flows through a pipe in the shape of an equilateral triangle, with sides of 0.5 inch. The air flows with an average velocity of 20 in/s at 400 oF. Find its volumetric flowrate, mass flowrate, and determine whether this flow is laminar.

Volumetric Flowrate The general formula to calculate the volumetric flowrate is : Q = vave * A v: velocity A: cross-sectional area

Given: vave = 20 in/s = 1.67 ft/s Triangular cross-section: a= 0.25 in = 0.021 ft 2a h a

Solution h 2a a Find cross-section area: A = ½ (2a * h) h = ( 4a2 – a2) = a3 Given : a = 0.021 ft A = ½ (2a * a3) = a2 3 = (0.021ft)2 * 30.5 A= 7.6 * 10-4 ft2 h 2a a

Calculate Q Q = vave * A Q = (1.67 ft/s) *( 7.6 * 10-4 ft2)

Mass Flow Rate Given : 400o F Density of air at 400o F = .0462 lbm/ft3 Mass flow rate = (Density of air) * Q = (.0462 lbm/ft3) * (1.3 * 10-3 ft3/s) = 5.9 * 10-5 lbm/s

Determining whether the flow is Laminar or Turbulent Flow Re = ρνD / μ D must be hydraulic diameter Dh Dh = 4(Cross-sectional area) / Wetted perimeter Cross-sectional area = ½ (2a * (a * 3^½)) = a^2 * 3^½ = ½ (base * height) 2a a h H = a * 3^½ Wetted perimeter = 2a * 3 = 6a Dh = 4 ( a^2 * 3^½) / 6a = (2a * 3^½) / 3 Given : a = .021 ft Dh = (2 * 0.021 ft) * 3^½ / 3 Dh = 0.024 ft

Using 477 K and figure 3.4.3 shown below, Viscosity (μ) can be found Given : 400 F = 5/9 ( 400-212) + 373 = 477 K Using 477 K and figure 3.4.3 shown below, Viscosity (μ) can be found Since the entire problem is in English unit system, the viscosity unit must be changed to English unit as well 2.6 * 10-5 Pa-s μ = 2.6 * 10-5 Pa-s ( 1 lbf s / ft2 ) 47.9 Pa-s μ = 5.4 * 10-7 lbf s / ft2

Re < 2000. Therefore this is Laminar Flow. Given : v = 1.67 ft / s ρ = 0.0462 lbm / ft3 Calculated : μ = 5.4 * 10-7 lbf s/ ft D = 0.024 ft Re = ρνD / μ Re = (0.0462 lbm / ft3) * (1.67 ft / s) * (0.024 ft) (5.4 * 10-7 lbf s/ ft) * (32.2 lbm ft / s2 / 1 lbf) Re = 106.5 **Laminar Flow ( Re < 2000 ) Transition Period ( 2000 < Re < 3500) Turbulent Flow ( Re > 3500)** Re < 2000. Therefore this is Laminar Flow.

This problem illustrated how to solve for the following information from given parameters and reference data: Volumetric Flowrate Mass Flowrate Reynolds’ Number Laminar or Turbulent Flow