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
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NON-CIRCULAR DUCT & ANNULUS CONVECTION
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--- INPUTS ---
Geometry = Annulus
Boundary Condition = Uniform Wall Temp (T)
Velocity V = 5.0000 m/s
Inlet Temp T_in = 25.00 C
Wall Temp T_wall = 100.00 C
Pipe Length L = 1.50 m
--- GEOMETRY ---
Hydraulic Diameter D_h = 0.020000 m
Cross Section Area A = 1.2566E-03 m2
Wetted Perimeter P = 0.251327 m
Aspect / Radius Ratio = 0.600
--- 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 = 6.3622E+03
Flow Regime = Turbulent
Friction Factor f = 0.035885
Nusselt Number Nu = 20.67
Convection Coeff h = 27.1751 W/m2K
Heat Transfer Q = 768.36 W
Outlet Temperature T_out= 128.18 C
Entry Length Le = 0.2000 m
Mass Flow Rate = 7.3953E-03 kg/s
--- VELOCITY SWEEP ---
V[m/s] Re Nu h[W/m2K] Regime
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0.100 1.272E+02 5.74 7.548 Laminar
0.929 1.182E+03 5.74 7.548 Laminar
1.758 2.237E+03 5.74 7.548 Laminar
2.587 3.292E+03 11.13 14.634 Turbulent
3.417 4.347E+03 14.69 19.322 Turbulent
4.246 5.403E+03 17.92 23.569 Turbulent
5.075 6.458E+03 20.93 27.523 Turbulent
5.904 7.513E+03 23.77 31.264 Turbulent
6.733 8.568E+03 26.49 34.839 Turbulent
7.562 9.623E+03 29.11 38.283 Turbulent
8.392 1.068E+04 31.65 41.617 Turbulent
9.221 1.173E+04 34.11 44.858 Turbulent
10.050 1.279E+04 36.52 48.019 Turbulent
10.879 1.384E+04 38.87 51.109 Turbulent
11.708 1.490E+04 41.17 54.137 Turbulent
12.537 1.595E+04 43.43 57.109 Turbulent
13.367 1.701E+04 45.65 60.031 Turbulent
14.196 1.806E+04 47.84 62.907 Turbulent
15.025 1.912E+04 49.99 65.741 Turbulent
15.854 2.017E+04 52.12 68.537 Turbulent
16.683 2.123E+04 54.22 71.296 Turbulent
17.512 2.228E+04 56.29 74.023 Turbulent
18.342 2.334E+04 58.34 76.718 Turbulent
19.171 2.439E+04 60.37 79.384 Turbulent
20.000 2.545E+04 62.38 82.023 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 $h$:
$$h = \frac{\text{Nu} k}{D_h} = \frac{35.8 \times 0.0263}{0.0667} = 14.1\text{ W/m}^2\text{K}$$