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

5.220e+2
Reynolds Re_D
114.24
Nu (Churchill-Bernstein)
100.12
Nu (Hilpert)
159.9425
h [W/m²K]
6029.69
Heat rate Q/L [W/m]
4.5189
Drag Force F/L [N/m]

📈 Nusselt number vs Velocity

Engine Terminal Output

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

--- INPUTS ---
  Cylinder Diameter D     =     0.100000 m
  Free Stream Velocity V  =       0.3000 m/s
  Free Stream Temp T_inf  =      30.00 C
  Surface Temp T_s        =     150.00 C
  Film Temperature T_f    =      90.00 C

--- 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_D    =     5.2200E+02
  Nu Churchill-Bernstein  =       114.24
  Nu Hilpert              =       100.12
  Convection Coeff h      =     159.9425 W/m2K
  Heat Transfer Q/L       =      6029.69 W/m
  Drag Coefficient Cd     =     1.1542
  Drag Force F/L          =       4.5189 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    1.740E+02       65.58      91.816       3461.39
     0.148    2.580E+02       79.97     111.955       4220.59
     0.197    3.420E+02       92.20     129.081       4866.22
     0.245    4.260E+02      103.05     144.265       5438.66
     0.293    5.100E+02      112.90     158.063       5958.85
     0.341    5.940E+02      122.01     170.809       6439.34
     0.390    6.780E+02      130.51     182.719       6888.36
     0.438    7.620E+02      138.53     193.947       7311.61
     0.486    8.460E+02      146.14     204.601       7713.29
     0.534    9.300E+02      153.41     214.768       8096.55
     0.583    1.014E+03      160.36     224.510       8463.84
     0.631    1.098E+03      167.06     233.882       8817.14
     0.679    1.182E+03      173.52     242.924       9158.01
     0.728    1.266E+03      179.77     251.671       9487.78
     0.776    1.350E+03      185.82     260.153       9807.53
     0.824    1.434E+03      191.71     268.394      10118.20
     0.872    1.518E+03      197.44     276.414      10420.58
     0.921    1.602E+03      203.02     284.233      10715.35
     0.969    1.686E+03      208.48     291.866      11003.10
     1.017    1.770E+03      213.81     299.327      11284.36
     1.066    1.854E+03      219.02     306.628      11559.59
     1.114    1.938E+03      224.13     313.779      11829.20
     1.162    2.022E+03      229.14     320.791      12093.55
     1.210    2.106E+03      234.05     327.673      12352.97
     1.259    2.190E+03      238.88     334.431      12607.75
     1.307    2.274E+03      243.62     341.073      12858.16
     1.355    2.358E+03      248.29     347.606      13104.44
     1.403    2.442E+03      252.88     354.035      13346.81
     1.452    2.526E+03      257.40     360.366      13585.48
     1.500    2.610E+03      261.86     366.603      13820.62

--- 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 $h$:
$$h = \frac{\text{Nu}_D k}{D} = \frac{227.1 \times 0.613}{0.025} = 5,572\text{ W/m}^2\text{K}$$