Flow Over Cylinder in Cross Flow

Evaluate convective heat transfer and aerodynamic drag for cylinders under external cross-flow conditions.

Flow (V) Ts Diameter D Stagnation Wake / Sep

Boundary Layer Separation

As fluid passes over a circular cylinder, the boundary layer grows under a favorable pressure gradient on the front side, then faces an adverse pressure gradient on the back, causing flow separation. The separation angle shifts from $\approx 80^\circ$ (laminar separation) to $\approx 140^\circ$ (turbulent separation) at higher Reynolds numbers ($\text{Re}_D \gtrsim 2 \times 10^5$).

Parameters Setup

📐 Geometry & Flow
🌡️ Thermal Conditions
💧 Fluid Properties

Results & Curves

1.590e+4
Reynolds Re_D
69.55
Nu (Churchill-Bernstein)
68.00
Nu (Hilpert)
36.5829
h [W/m²K]
430.98
Heat rate Q/L [W/m]
0.3292
Drag Force F/L [N/m]

📈 Nusselt number vs Velocity

Engine Terminal Output

============================================
  FLOW OVER CYLINDER IN CROSS FLOW
============================================

--- INPUTS ---
  Cylinder Diameter D     =     0.050000 m
  Free Stream Velocity V  =       5.0000 m/s
  Free Stream Temp T_inf  =      25.00 C
  Surface Temp T_s        =     100.00 C
  Film Temperature T_f    =      62.50 C

--- 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_D    =     1.5905E+04
  Nu Churchill-Bernstein  =        69.55
  Nu Hilpert              =        68.00
  Convection Coeff h      =      36.5829 W/m2K
  Heat Transfer Q/L       =       430.98 W/m
  Drag Coefficient Cd     =     0.4476
  Drag Force F/L          =       0.3292 N/m
  BL Separation           ~ 80 deg (laminar)

--- VELOCITY SWEEP ---
  V[m/s]     Re_D         Nu_CB      h[W/m2K]   Q/L[W/m]
  -----------------------------------------------------------
     0.100    3.181E+02        9.06       4.767         56.15
     0.959    3.049E+03       28.38      14.928        175.86
     1.817    5.781E+03       39.81      20.938        246.67
     2.676    8.512E+03       49.09      25.820        304.18
     3.534    1.124E+04       57.23      30.102        354.63
     4.393    1.397E+04       64.64      33.999        400.54
     5.252    1.671E+04       71.52      37.621        443.21
     6.110    1.944E+04       78.01      41.035        483.44
     6.969    2.217E+04       84.19      44.286        521.73
     7.828    2.490E+04       90.12      47.402        558.44
     8.686    2.763E+04       95.83      50.407        593.84
     9.545    3.036E+04      101.36      53.316        628.12
    10.403    3.309E+04      106.74      56.144        661.43
    11.262    3.583E+04      111.98      58.899        693.89
    12.121    3.856E+04      117.09      61.592        725.61
    12.979    4.129E+04      122.11      64.227        756.66
    13.838    4.402E+04      127.02      66.812        787.11
    14.697    4.675E+04      131.85      69.351        817.02
    15.555    4.948E+04      136.59      71.848        846.44
    16.414    5.221E+04      141.27      74.307        875.41
    17.272    5.494E+04      145.88      76.730        903.96
    18.131    5.768E+04      150.42      79.121        932.13
    18.990    6.041E+04      154.91      81.482        959.94
    19.848    6.314E+04      159.34      83.815        987.42
    20.707    6.587E+04      163.73      86.122       1014.60
    21.566    6.860E+04      168.07      88.404       1041.48
    22.424    7.133E+04      172.36      90.663       1068.10
    23.283    7.406E+04      176.62      92.901       1094.46
    24.141    7.680E+04      180.83      95.118       1120.59
    25.000    7.953E+04      185.01      97.316       1146.48

--- CORRELATIONS ---
  Churchill-Bernstein (Re*Pr>0.2):
  Nu = 0.3 + 0.62*Re^0.5*Pr^(1/3)/[1+(0.4/Pr)^(2/3)]^0.25
       * [1+(Re/282000)^(5/8)]^(4/5)
  Ref: Incropera Ch.7 Eq.7.54

Calculation Methodology

Mathematical Model & Theory

Convection over a circular cylinder is determined by boundary layer development. The average Nusselt number is computed using two primary models:

1. Churchill-Bernstein Correlation (All $\text{Re}_D \text{Pr} > 0.2$): $$\text{Nu}_D = 0.3 + \frac{0.62 \text{Re}_D^{1/2} \text{Pr}^{1/3}}{\left[1 + (0.4/\text{Pr})^{2/3}\right]^{1/4}} \left[1 + \left(\frac{\text{Re}_D}{282,000}\right)^{5/8}\right]^{4/5}$$ 2. Hilpert Correlation: $$\text{Nu}_D = C \text{Re}_D^m \text{Pr}^{1/3}$$ where $C$ and $m$ are constants evaluated based on the Reynolds number range.

Aerodynamic drag force per unit length is computed as: $$F_D/L = \frac{1}{2} C_d \rho V^2 D$$

Academic References:

  1. Incropera, F. P., & DeWitt, D. P. (2011). Fundamentals of Heat and Mass Transfer. 7th Edition, John Wiley & Sons.
  2. Churchill, S. W., & Bernstein, M. (1977). A Correlating Equation for Forced Convection from Gases and Liquids to a Circular Cylinder in Crossflow. J. Heat Transfer.

Worked Engineering Example

Problem Statement:
Water at $20^\circ\text{C}$ flows at $1.0\text{ m/s}$ perpendicular to a circular tube of diameter $25\text{ mm}$ with surface temperature maintained at $80^\circ\text{C}$. Calculate the convective heat transfer coefficient.

Step-by-step Solution:
1. Evaluate properties at film temperature $T_f = (20+80)/2 = 50^\circ\text{C}$:
- Water: $\rho = 997\text{ kg/m}^3, \mu = 8.9 \times 10^{-4}\text{ Pa·s}, k = 0.613\text{ W/m·K}, Pr = 6.13$.
2. Calculate Reynolds number:
$$\text{Re}_D = \frac{\rho V D}{\mu} = \frac{997 \times 1.0 \times 0.025}{8.9 \times 10^{-4}} = 28,005$$ 3. Evaluate Nusselt number via Churchill-Bernstein correlation:
$$\text{Nu}_D = 0.3 + \frac{0.62 \times (28,005)^{1/2} \times (6.13)^{1/3}}{\left[1 + (0.4/6.13)^{2/3}\right]^{1/4}} \left[1 + \left(\frac{28,005}{282,000}\right)^{5/8}\right]^{4/5}$$ $$\text{Nu}_D = 0.3 + \frac{0.62 \times 167.3 \times 1.83}{1.037} \times 1.17 = 227.1$$ 4. Calculate convection coefficient $h$:
$$h = \frac{\text{Nu}_D k}{D} = \frac{227.1 \times 0.613}{0.025} = 5,572\text{ W/m}^2\text{K}$$