1. 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

2. 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.

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

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

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

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

7. 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

8. 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 * ft) * 3^½ / 3
Dh = ft

9. Using 477 K and figure 3.4.3 shown below, Viscosity (μ) can be found
Given : 400 F = 5/9 ( ) = 477 K
Using 477 K and figure 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 ) Pa-s
μ = 5.4 * 10-7 lbf s / ft2

10. Re < 2000. Therefore this is Laminar Flow.
Given :
v = 1.67 ft / s ρ = lbm / ft3
Calculated :
μ = 5.4 * 10-7 lbf s/ ft
D = ft
Re = ρνD / μ
Re = ( 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 < Therefore this is Laminar Flow.

11. 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