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 = 3.0000 m/s
Inlet Temp T_in = 25.00 C
Wall Temp T_wall = 80.00 C
Pipe Length L = 2.00 m
--- GEOMETRY ---
Hydraulic Diameter D_h = 0.066667 m
Cross Section Area A = 5.0000E-03 m2
Wetted Perimeter P = 0.300000 m
Aspect / Radius Ratio = 2.000
--- FLUID PROPERTIES ---
Density rho = 1.1770 kg/m3
Viscosity mu = 1.8500E-05 Pa.s
Conductivity k = 0.026300 W/mK
Prandtl Pr = 0.7100
--- RESULTS ---
Reynolds Number Re = 1.2724E+04
Flow Regime = Turbulent
Friction Factor f = 0.029457
Nusselt Number Nu = 36.37
Convection Coeff h = 14.3490 W/m2K
Heat Transfer Q = 473.52 W
Outlet Temperature T_out= 51.63 C
Entry Length Le = 0.6667 m
Mass Flow Rate = 1.7655E-02 kg/s
--- VELOCITY SWEEP ---
V[m/s] Re Nu h[W/m2K] Regime
--------------------------------------------------------
0.100 4.241E+02 3.39 1.337 Laminar
0.596 2.527E+03 8.20 3.235 Turbulent
1.092 4.630E+03 15.59 6.149 Turbulent
1.588 6.733E+03 21.69 8.556 Turbulent
2.083 8.836E+03 27.17 10.718 Turbulent
2.579 1.094E+04 32.26 12.729 Turbulent
3.075 1.304E+04 37.09 14.631 Turbulent
3.571 1.515E+04 41.70 16.452 Turbulent
4.067 1.725E+04 46.15 18.207 Turbulent
4.562 1.935E+04 50.47 19.909 Turbulent
5.058 2.145E+04 54.66 21.565 Turbulent
5.554 2.356E+04 58.76 23.182 Turbulent
6.050 2.566E+04 62.78 24.765 Turbulent
6.546 2.776E+04 66.71 26.318 Turbulent
7.042 2.987E+04 70.58 27.843 Turbulent
7.537 3.197E+04 74.38 29.344 Turbulent
8.033 3.407E+04 78.13 30.823 Turbulent
8.529 3.618E+04 81.83 32.281 Turbulent
9.025 3.828E+04 85.48 33.721 Turbulent
9.521 4.038E+04 89.08 35.143 Turbulent
10.017 4.249E+04 92.65 36.549 Turbulent
10.512 4.459E+04 96.17 37.939 Turbulent
11.008 4.669E+04 99.66 39.316 Turbulent
11.504 4.879E+04 103.12 40.679 Turbulent
12.000 5.090E+04 106.54 42.030 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 $h$:
$$h = \frac{\text{Nu} k}{D_h} = \frac{35.8 \times 0.0263}{0.0667} = 14.1\text{ W/m}^2\text{K}$$