Flow Across Banks of Tubes
Evaluate convection coefficients and pressure drop across inline or staggered tube banks using the Zukauskas correlation.
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
Results & Curves
📈 Convection coefficient h vs Velocity
Engine Output
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FLOW ACROSS BANKS OF TUBES
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--- INPUTS ---
Configuration = INLINE
Number of Rows N = 8
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) = 0.2700
m (Zukauskas) = 0.6300
C2 (row correction) = 0.9567
Nusselt Number Nu = 409.20
Convection Coeff h = 13202.0983 W/m2K
Pressure Drop dP = 1065.17 Pa
dP per row = 133.15 Pa
--- VELOCITY SWEEP ---
V[m/s] Vmax[m/s] Re_max Nu h[W/m2K] dP[Pa]
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0.50 1.00 2.128E+04 264.42 8530.885 297.52
0.65 1.29 2.749E+04 310.68 10023.487 476.47
0.79 1.58 3.370E+04 353.20 11395.271 693.00
0.94 1.87 3.991E+04 392.90 12676.077 945.90
1.08 2.17 4.612E+04 430.36 13884.911 1234.19
1.23 2.46 5.232E+04 466.01 15034.803 1557.05
1.38 2.75 5.853E+04 500.11 16135.174 1913.79
1.52 3.04 6.474E+04 532.90 17193.108 2303.82
1.67 3.33 7.095E+04 564.55 18214.098 2726.59
1.81 3.62 7.716E+04 595.18 19202.515 3181.63
1.96 3.92 8.336E+04 624.92 20161.903 3668.51
2.10 4.21 8.957E+04 653.85 21095.190 4186.83
2.25 4.50 9.578E+04 682.04 22004.826 4736.24
2.40 4.79 1.020E+05 709.57 22892.890 5316.41
2.54 5.08 1.082E+05 736.48 23761.162 5927.02
2.69 5.38 1.144E+05 762.83 24611.184 6567.79
2.83 5.67 1.206E+05 788.65 25444.299 7238.46
2.98 5.96 1.268E+05 813.98 26261.691 7938.76
3.12 6.25 1.330E+05 838.86 27064.404 8668.47
3.27 6.54 1.392E+05 863.32 27853.371 9427.35
3.42 6.83 1.454E+05 887.37 28629.424 10215.21
3.56 7.12 1.517E+05 911.05 29393.313 11031.82
3.71 7.42 1.579E+05 934.37 30145.715 11877.01
3.85 7.71 1.641E+05 957.35 30887.245 12750.59
4.00 8.00 1.703E+05 980.02 31618.463 13652.39
--- 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:
- Incropera, F. P., & DeWitt, D. P. (2011). Fundamentals of Heat and Mass Transfer. 7th Edition, John Wiley & Sons.
- Zukauskas, A. (1972). Heat Transfer from Tubes in Crossflow. Advances in Heat Transfer.
Worked Engineering Example
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 $h$:
$$h = \frac{\text{Nu}_D k}{D} = \frac{93.5 \times 0.0263}{0.025} = 98.4\text{ W/m}^2\text{K}$$