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 = Equilateral Triangle
Boundary Condition = Uniform Heat Flux (H)
Velocity V = 1.0000 m/s
Inlet Temp T_in = 20.00 C
Wall Temp T_wall = 60.00 C
Pipe Length L = 1.00 m
--- GEOMETRY ---
Hydraulic Diameter D_h = 0.028868 m
Cross Section Area A = 1.0825E-03 m2
Wetted Perimeter P = 0.150000 m
Aspect / Radius Ratio = 1.000
--- FLUID PROPERTIES ---
Density rho = 997.0000 kg/m3
Viscosity mu = 8.9000E-04 Pa.s
Conductivity k = 0.613000 W/mK
Prandtl Pr = 6.1300
--- RESULTS ---
Reynolds Number Re = 3.2338E+04
Flow Regime = Turbulent
Friction Factor f = 0.023214
Nusselt Number Nu = 213.80
Convection Coeff h = 4539.9335 W/m2K
Heat Transfer Q = 27239.60 W
Outlet Temperature T_out= 26.04 C
Entry Length Le = 0.2887 m
Mass Flow Rate = 1.0793E+00 kg/s
--- VELOCITY SWEEP ---
V[m/s] Re Nu h[W/m2K] Regime
--------------------------------------------------------
0.100 3.234E+03 23.59 500.937 Turbulent
0.263 8.489E+03 64.90 1378.154 Turbulent
0.425 1.374E+04 100.91 2142.788 Turbulent
0.588 1.900E+04 134.49 2855.922 Turbulent
0.750 2.425E+04 166.54 3536.533 Turbulent
0.912 2.951E+04 197.49 4193.767 Turbulent
1.075 3.476E+04 227.59 4832.948 Turbulent
1.238 4.002E+04 257.01 5457.529 Turbulent
1.400 4.527E+04 285.85 6069.923 Turbulent
1.563 5.053E+04 314.19 6671.896 Turbulent
1.725 5.578E+04 342.12 7264.797 Turbulent
1.887 6.104E+04 369.66 7849.688 Turbulent
2.050 6.629E+04 396.87 8427.419 Turbulent
2.212 7.155E+04 423.77 8998.691 Turbulent
2.375 7.680E+04 450.39 9564.089 Turbulent
2.538 8.206E+04 476.77 10124.104 Turbulent
2.700 8.731E+04 502.91 10679.161 Turbulent
2.862 9.257E+04 528.83 11229.623 Turbulent
3.025 9.782E+04 554.55 11775.810 Turbulent
3.188 1.031E+05 580.08 12318.002 Turbulent
3.350 1.083E+05 605.44 12856.447 Turbulent
3.512 1.136E+05 630.63 13391.367 Turbulent
3.675 1.188E+05 655.66 13922.960 Turbulent
3.838 1.241E+05 680.55 14451.404 Turbulent
4.000 1.294E+05 705.29 14976.862 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 $h$:
$$h = \frac{\text{Nu} k}{D_h} = \frac{35.8 \times 0.0263}{0.0667} = 14.1\text{ W/m}^2\text{K}$$