Flow Across Banks of Tubes

Evaluate convection coefficients and pressure drop across inline or staggered tube banks using the Zukauskas correlation.

Flow (V)

Aligned vs Staggered Bundles

In crossflow over tube banks, the configuration (inline or staggered) alters the maximum velocity $V_{max}$ inside the bundle channels. Max velocity dictates the Reynolds number ($\text{Re}_{max}$) and the heat transfer rates.

Parameters Setup

📐 Geometry Setup
📐 Flow & Temperatures
💧 Fluid Properties

Results & Curves

4.257e+4
Reynolds Re_max
2.000 m/s
Max Velocity V_max
396.01
Nusselt Nu
12776.5824
h [W/m²K]
1683.40 Pa
Pressure Drop dP

📈 Convection coefficient h vs Velocity

Engine Output

============================================
  FLOW ACROSS BANKS OF TUBES
============================================

--- INPUTS ---
  Configuration           = STAGGERED
  Number of Rows N        =     12
  Tube Diameter D         =     0.0190 m
  Transverse Pitch S_T    =     0.0380 m
  Longitudinal Pitch S_L  =     0.0380 m
  Approach Velocity V     =      1.000 m/s
  T_inf                   =      20.00 C
  T_surface               =      80.00 C

--- GEOMETRY ---
  S_T / D                 =    2.000
  S_L / D                 =    2.000
  Maximum Velocity V_max  =      2.000 m/s

--- RESULTS ---
  Reynolds Re_max         =     4.2569E+04
  C1 (Zukauskas)          =     0.3500
  m  (Zukauskas)          =     0.6000
  C2 (row correction)     =     0.9833
  Nusselt Number Nu       =       396.01
  Convection Coeff h      =   12776.5824 W/m2K
  Pressure Drop dP        =      1683.40 Pa
  dP per row              =       140.28 Pa

--- VELOCITY SWEEP ---
  V[m/s]   Vmax[m/s]  Re_max       Nu        h[W/m2K]  dP[Pa]
  -----------------------------------------------------------------
     0.50       1.00    2.128E+04     261.27    8429.401      470.21
     0.65       1.29    2.749E+04     304.64    9828.494      753.02
     0.79       1.58    3.370E+04     344.22   11105.551     1095.23
     0.94       1.87    3.991E+04     380.97   12291.290     1494.91
     1.08       2.17    4.612E+04     415.49   13405.159     1950.52
     1.23       2.46    5.232E+04     448.20   14460.429     2460.77
     1.38       2.75    5.853E+04     479.39   15466.652     3024.58
     1.52       3.04    6.474E+04     509.28   16430.988     3640.98
     1.67       3.33    7.095E+04     538.04   17358.970     4309.13
     1.81       3.62    7.716E+04     565.82   18254.986     5028.28
     1.96       3.92    8.336E+04     592.71   19122.588     5797.74
     2.10       4.21    8.957E+04     618.81   19964.700     6616.91
     2.25       4.50    9.578E+04     644.19   20783.764     7485.20
     2.40       4.79    1.020E+05     668.93   21581.850     8402.10
     2.54       5.08    1.082E+05     693.07   22360.724     9367.11
     2.69       5.38    1.144E+05     716.67   23121.914    10379.79
     2.83       5.67    1.206E+05     739.75   23866.751    11439.72
     2.98       5.96    1.268E+05     762.37   24596.402    12546.48
     3.12       6.25    1.330E+05     784.54   25311.898    13699.72
     3.27       6.54    1.392E+05     806.31   26014.156    14899.07
     3.42       6.83    1.454E+05     827.69   26703.997    16144.19
     3.56       7.12    1.517E+05     848.71   27382.155    17434.78
     3.71       7.42    1.579E+05     869.39   28049.296    18770.53
     3.85       7.71    1.641E+05     889.75   28706.021    20151.14
     4.00       8.00    1.703E+05     909.80   29352.878    21576.35

--- CORRELATIONS ---
  Zukauskas: Nu = C1*C2*Re_max^m * Pr^0.36 * (Pr/Pr_s)^0.25
  Ref: Incropera Ch.7 Eq.7.58, Kothandaraman Ch.8 Sec.8.6

Calculation Methodology

Mathematical Model & Theory

Heat transfer across tube bundles is modeled using the Zukauskas correlation: $$\text{Nu}_D = C_1 C_2 \text{Re}_{max}^m \text{Pr}^{0.36} \left(\frac{\text{Pr}}{\text{Pr}_s}\right)^{0.25}$$ - $C_1$ and $m$ are constants determined from the Reynolds number range ($\text{Re}_{max} = \rho V_{max} D / \mu$). - $C_2$ is a row correction factor for banks with fewer than 20 rows ($N_L < 20$).

For aligned arrangement, maximum velocity is: $$V_{max} = V \frac{S_T}{S_T - D}$$ For staggered arrangement, if the diagonal flow area is smaller than the transverse area, then: $$V_{max} = V \frac{S_T}{2(S_D - D)}$$ Aerodynamic pressure drop is calculated using friction factors: $$\Delta P = N \chi f \frac{\rho V_{max}^2}{2}$$

Academic References:

  1. Incropera, F. P., & DeWitt, D. P. (2011). Fundamentals of Heat and Mass Transfer. 7th Edition, John Wiley & Sons.
  2. Zukauskas, A. (1972). Heat Transfer from Tubes in Crossflow. Advances in Heat Transfer.

Worked Engineering Example

Problem Statement:
Air at $25^\circ\text{C}$ flows at $5\text{ m/s}$ across an inline tube bank of 10 rows. Tubes of diameter $25\text{ mm}$ are arranged with pitches $S_T = S_L = 50\text{ mm}$. Surface temp is $100^\circ\text{C}$ ($T_f = 62.5^\circ\text{C}$). Find the convection coefficient $h$.

Step-by-step Solution:
1. Evaluate air properties at film temperature:
- $\rho = 1.177\text{ kg/m}^3, \mu = 1.85 \times 10^{-5}\text{ Pa·s}, k = 0.0263\text{ W/m·K}, Pr = 0.71$.
2. Calculate maximum velocity $V_{max}$:
$$V_{max} = V \frac{S_T}{S_T - D} = 5.0 \times \frac{0.05}{0.05 - 0.025} = 10.0\text{ m/s}$$ 3. Calculate maximum Reynolds number:
$$\text{Re}_{max} = \frac{\rho V_{max} D}{\mu} = \frac{1.177 \times 10.0 \times 0.025}{1.85 \times 10^{-5}} = 15,905$$ 4. From Zukauskas constants ($10^3 < \text{Re}_{max} < 2 \times 10^5$, inline):
- $C_1 = 0.27, m = 0.63$. For 10 rows, $C_2 = 0.97$.
$$\text{Nu}_D = 0.27 \times 0.97 \times (15,905)^{0.63} \times (0.71)^{0.36} \times 1.0 = 93.5$$ 5. Calculate convective coefficient $h$:
$$h = \frac{\text{Nu}_D k}{D} = \frac{93.5 \times 0.0263}{0.025} = 98.4\text{ W/m}^2\text{K}$$