Non-Circular Duct & Annulus Convection
Calculate forced convection coefficients and outlet temperatures for non-circular pipe geometries.
Hydraulic Diameter Concept
For non-circular ducts, fluid flow parameters are evaluated by using the hydraulic diameter ($D_h$) defined as: $$D_h = \frac{4 A_c}{P}$$ where $A_c$ is the cross-sectional area and $P$ is the wetted perimeter.
Parameters Setup
Results & Curves
📈 Nusselt number vs Velocity
Engine Output
============================================
NON-CIRCULAR DUCT & ANNULUS CONVECTION
============================================
--- INPUTS ---
Geometry = Rectangular
Boundary Condition = Uniform Wall Temp (T)
Velocity V = 0.5000 m/s
Inlet Temp T_in = 40.00 C
Wall Temp T_wall = 120.00 C
Pipe Length L = 3.00 m
--- GEOMETRY ---
Hydraulic Diameter D_h = 0.133333 m
Cross Section Area A = 2.0000E-02 m2
Wetted Perimeter P = 0.600000 m
Aspect / Radius Ratio = 2.000
--- FLUID PROPERTIES ---
Density rho = 870.0000 kg/m3
Viscosity mu = 5.0000E-02 Pa.s
Conductivity k = 0.140000 W/mK
Prandtl Pr = 500.0000
--- RESULTS ---
Reynolds Number Re = 1.1600E+03
Flow Regime = Laminar
Friction Factor f = 0.055172
Nusselt Number Nu = 3.39
Convection Coeff h = 3.5595 W/m2K
Heat Transfer Q = 512.57 W
Outlet Temperature T_out= 40.03 C
Entry Length Le = 7.7333 m
Mass Flow Rate = 8.7000E+00 kg/s
--- VELOCITY SWEEP ---
V[m/s] Re Nu h[W/m2K] Regime
--------------------------------------------------------
0.100 2.320E+02 3.39 3.560 Laminar
0.179 4.157E+02 3.39 3.560 Laminar
0.258 5.993E+02 3.39 3.560 Laminar
0.337 7.830E+02 3.39 3.560 Laminar
0.417 9.667E+02 3.39 3.560 Laminar
0.496 1.150E+03 3.39 3.560 Laminar
0.575 1.334E+03 3.39 3.560 Laminar
0.654 1.518E+03 3.39 3.560 Laminar
0.733 1.701E+03 3.39 3.560 Laminar
0.812 1.885E+03 3.39 3.560 Laminar
0.892 2.069E+03 3.39 3.560 Laminar
0.971 2.252E+03 3.39 3.560 Laminar
1.050 2.436E+03 70.18 73.692 Turbulent
1.129 2.620E+03 78.15 82.058 Turbulent
1.208 2.803E+03 85.99 90.290 Turbulent
1.288 2.987E+03 93.72 98.403 Turbulent
1.367 3.171E+03 101.34 106.409 Turbulent
1.446 3.354E+03 108.88 114.319 Turbulent
1.525 3.538E+03 116.33 122.141 Turbulent
1.604 3.722E+03 123.70 129.883 Turbulent
1.683 3.905E+03 131.00 137.552 Turbulent
1.762 4.089E+03 138.24 145.152 Turbulent
1.842 4.273E+03 145.42 152.688 Turbulent
1.921 4.456E+03 152.54 160.166 Turbulent
2.000 4.640E+03 159.61 167.588 Turbulent
--- CORRELATIONS ---
Laminar: Nu from geometry tables (Incropera Table 8.1)
Turbulent: Gnielinski Nu=(f/8)(Re-1000)Pr/[1+12.7sqrt(f/8)(Pr^2/3-1)]
f = (0.790 ln Re - 1.64)^-2
Ref: Incropera Ch.8 Sec.8.6, Kothandaraman Ch.9 Sec.9.6
Calculation Methodology
Mathematical Model & Theory
Forced convection in non-circular ducts uses the hydraulic diameter:
$$D_h = \frac{4 A_c}{P}$$
- Laminar Regime ($\text{Re} \le 2300$): Nusselt numbers are constants interpolated from numerical geometry tables (Incropera Table 8.1 & 8.3) based on aspect ratio (rectangular) or radius ratio (annular).
- Turbulent Regime ($\text{Re} > 2300$): Gnielinski correlation is employed:
$$\text{Nu} = \frac{(f/8)(\text{Re} - 1000)\text{Pr}}{1 + 12.7\sqrt{f/8}\left(\text{Pr}^{2/3} - 1\right)}$$
where friction factor $f$ is:
$$f = (0.790 \ln \text{Re} - 1.64)^{-2}$$
Academic References:
- Incropera, F. P., & DeWitt, D. P. (2011). Fundamentals of Heat and Mass Transfer. 7th Edition, John Wiley & Sons.
- Gnielinski, V. (1976). New Equations for Heat and Mass Transfer in Turbulent Pipe and Channel Flow. Int. Chem. Eng.
Worked Engineering Example
Air at $25^\circ\text{C}$ flows at $3.0\text{ m/s}$ through a rectangular duct of $100\text{ mm} \times 50\text{ mm}$ wetted walls at $80^\circ\text{C}$. Find the convection coefficient ($T_f = 52.5^\circ\text{C}$).
Step-by-step Solution:
1. Evaluate air properties at film temp:
- $\rho = 1.177\text{ kg/m}^3, \mu = 1.85 \times 10^{-5}\text{ Pa·s}, k = 0.0263\text{ W/m·K}, Pr = 0.71$.
2. Calculate wetted perimeter, wetted area, and hydraulic diameter:
- $A_c = 0.1 \times 0.05 = 0.005\text{ m}^2$
- $P = 2 \times (0.1 + 0.05) = 0.3\text{ m}$
- $D_h = 4 \times 0.005 / 0.3 = 0.0667\text{ m}$
3. Calculate Reynolds number:
$$\text{Re} = \frac{\rho V D_h}{\mu} = \frac{1.177 \times 3.0 \times 0.0667}{1.85 \times 10^{-5}} = 12,735 \quad \text{(Turbulent)}$$ 4. Compute friction factor and Nusselt number (Gnielinski):
- $f = (0.790 \ln(12,735) - 1.64)^{-2} = 0.029$
$$\text{Nu} = \frac{(0.029/8) \times (12,735 - 1000) \times 0.71}{1 + 12.7\sqrt{0.029/8}(0.71^{2/3} - 1)} = 35.8$$ 5. Calculate $h$:
$$h = \frac{\text{Nu} k}{D_h} = \frac{35.8 \times 0.0263}{0.0667} = 14.1\text{ W/m}^2\text{K}$$