⚖️ Heat & Mass Transfer Analogy Calculator
Verify boundary layer analogies and predict mass transfer coefficients from thermal coefficients. Computes Sherwood ($Sh$), Schmidt ($Sc$), Lewis ($Le$) numbers, and Chilton-Colburn $j$-factors.
🌊 Boundary Layers & Relative Thickness Visualizer
📝 Configuration
Dimensionless Numbers:
Quick Presets:
- Wet plate drying in air flow (20°C)
- Oxygen transport in water pipe flow
Prandtl Number: $$Pr = \frac{\nu}{\alpha}$$
Schmidt Number: $$Sc = \frac{\nu}{D_{AB}}$$
Lewis Number: $$Le = \frac{\alpha}{D_{AB}} = \frac{Sc}{Pr}$$
Chilton-Colburn Analogy:
$$\frac{h}{h_m} = \rho C_p Le^{2/3}$$
Quick Presets:
- Wet plate drying in air flow (20°C)
- Oxygen transport in water pipe flow
Prandtl Number: $$Pr = \frac{\nu}{\alpha}$$
Schmidt Number: $$Sc = \frac{\nu}{D_{AB}}$$
Lewis Number: $$Le = \frac{\alpha}{D_{AB}} = \frac{Sc}{Pr}$$
Chilton-Colburn Analogy:
$$\frac{h}{h_m} = \rho C_p Le^{2/3}$$
📊 Results & Analogy Verification
📊 Calculation Output
📈 Convective Heat HTC h(x)
📈 Mass Transfer Coefficient hm(x)
Reynolds Number (Re)
1.6493e+5
Regime: 🟢 Laminar
Lewis Number (Le)
0.7975
Pr: 0.731 | Sc: 0.583
Average Heat HTC (h)
12.214 W/m²·K
Nusselt Nu: 242.92
Average Mass MTC (hm)
1.1714e-2 m/s
Sherwood Sh: 225.27
Chilton-Colburn Analogy Agreement
100.00% Match
Actual h/hm: 1.04e+3 | Theory: 1.04e+3
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HEAT & MASS TRANSFER ANALOGY SOLVER
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Configuration Type = External Flow over Flat Plate
Velocity (U) = 5.0000 m/s
Characteristic Length = 0.5000 m
--- FLUID & DIFFUSION PROPERTIES ---------------------------
Density (rho) = 1.2040 kg/m3
Viscosity (mu) = 1.8250E-05 Pa.s
Thermal Cond (kf) = 0.0251 W/m.K
Specific Heat (Cp) = 1007.00 J/kg.K
Mass Diffusivity D_AB = 2.6000E-05 m2/s
--- DIMENSIONLESS NUMBERS ----------------------------------
Reynolds Number (Re) = 1.6493E+05
Prandtl Number (Pr) = 0.7310
Schmidt Number (Sc) = 0.5830
Lewis Number (Le) = 0.7975
--- ANALOGY RESULTS (AVERAGE VALUES) -----------------------
Skin Friction Coeff (Cf) = 0.003270
Average Nusselt Number = 242.9194
Average Sherwood Number = 225.2722
Average Heat HTC (h) = 12.2140 W/m2.K
Average Mass MTC (hm) = 1.1714E-02 m/s
--- CHILTON-COLBURN RATIO VERIFICATION ---------------------
Actual Ratio h/hm = 1.0427E+03 J/m3.K
Theoretical rho*Cp*Le^n = 1.0427E+03 J/m3.K
Analogy Deviation = 0.00 %
--- LOCAL PROFILE ALONG SURFACE ----------------------------
x [m] Re_x Nu_x Sh_x h_x [W/m2.K] hm_x [m/s]
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0.0125 4.1233E+03 19.20 17.81 38.6240 3.7043E-02
0.0250 8.2466E+03 27.16 25.19 27.3113 2.6194E-02
0.0375 1.2370E+04 33.26 30.85 22.2996 2.1387E-02
0.0500 1.6493E+04 38.41 35.62 19.3120 1.8522E-02
0.0625 2.0616E+04 42.94 39.82 17.2732 1.6566E-02
0.0750 2.4740E+04 47.04 43.62 15.7682 1.5123E-02
0.0875 2.8863E+04 50.81 47.12 14.5985 1.4001E-02
0.1000 3.2986E+04 54.32 50.37 13.6557 1.3097E-02
0.1125 3.7110E+04 57.61 53.43 12.8747 1.2348E-02
0.1250 4.1233E+04 60.73 56.32 12.2140 1.1714E-02
0.1375 4.5356E+04 63.69 59.07 11.6456 1.1169E-02
0.1500 4.9479E+04 66.53 61.69 11.1498 1.0694E-02
0.1625 5.3603E+04 69.24 64.21 10.7124 1.0274E-02
0.1750 5.7726E+04 71.86 66.64 10.3227 9.9003E-03
0.1875 6.1849E+04 74.38 68.98 9.9727 9.5646E-03
0.2000 6.5973E+04 76.82 71.24 9.6560 9.2609E-03
0.2125 7.0096E+04 79.18 73.43 9.3677 8.9843E-03
0.2250 7.4219E+04 81.48 75.56 9.1038 8.7312E-03
0.2375 7.8342E+04 83.71 77.63 8.8610 8.4983E-03
0.2500 8.2466E+04 85.88 79.65 8.6366 8.2832E-03
0.2625 8.6589E+04 88.01 81.61 8.4285 8.0835E-03
0.2750 9.0712E+04 90.08 83.53 8.2347 7.8977E-03
0.2875 9.4836E+04 92.10 85.41 8.0537 7.7241E-03
0.3000 9.8959E+04 94.08 87.25 7.8841 7.5615E-03
0.3125 1.0308E+05 96.02 89.05 7.7248 7.4087E-03
0.3250 1.0721E+05 97.92 90.81 7.5748 7.2648E-03
0.3375 1.1133E+05 99.79 92.54 7.4332 7.1290E-03
0.3500 1.1545E+05 101.62 94.24 7.2993 7.0005E-03
0.3625 1.1958E+05 103.42 95.91 7.1723 6.8788E-03
0.3750 1.2370E+05 105.19 97.55 7.0517 6.7632E-03
0.3875 1.2782E+05 106.93 99.16 6.9371 6.6532E-03
0.4000 1.3195E+05 108.64 100.74 6.8278 6.5484E-03
0.4125 1.3607E+05 110.32 102.31 6.7236 6.4484E-03
0.4250 1.4019E+05 111.98 103.85 6.6240 6.3529E-03
0.4375 1.4432E+05 113.62 105.36 6.5287 6.2615E-03
0.4500 1.4844E+05 115.23 106.86 6.4373 6.1739E-03
0.4625 1.5256E+05 116.82 108.33 6.3498 6.0899E-03
0.4750 1.5668E+05 118.38 109.78 6.2656 6.0092E-03
0.4875 1.6081E+05 119.93 111.22 6.1848 5.9317E-03
0.5000 1.6493E+05 121.46 112.64 6.1070 5.8571E-03
--- CORRELATIONS USED ---------------------------------------
Flat Plate Laminar (Re <= 5e5): Nu_x = 0.332 Re_x^0.5 Pr^1/3, Sh_x = 0.332 Re_x^0.5 Sc^1/3.
Flat Plate Turbulent (Re > 5e5): Nu_x = 0.0296 Re_x^0.8 Pr^1/3, Sh_x = 0.0296 Re_x^0.8 Sc^1/3.
Boundary layers assumed analogous under Chilton-Colburn definition.
📘 Calculation Methodology
Mathematical Model & Theory
Under boundary layer assumptions, the Nusselt ($Nu$) and Sherwood ($Sh$) numbers represent the dimensionless temperature and concentration gradients at the wall. The Chilton-Colburn analogy correlates these parameters directly using the Lewis number:
$$Sh = Nu \cdot \left(\frac{Sc}{Pr}\right)^{1/3} = Nu \cdot Le^{1/3}$$
$$\frac{h}{h_m} = \rho C_p Le^{2/3}$$
Worked Engineering Example
Wet Plate Drying:
Air at 20°C ($\rho = 1.2\text{ kg/m}^3$, $C_p = 1007\text{ J/kg·K}$, $Pr = 0.73$, $\nu = 1.5 \times 10^{-5}\text{ m}^2/\text{s}$) flows over a flat plate ($L = 0.5\text{ m}$) at $U = 5\text{ m/s}$. Water binary diffusivity $D_{AB} = 2.6 \times 10^{-5}\text{ m}^2/\text{s}$. Calculate $h$ and $h_m$.
Solution:
1. Calculate Reynolds number:
$$Re = \frac{U L}{\nu} = \frac{5 \times 0.5}{1.5 \times 10^{-5}} = 1.67 \times 10^5 \quad (\text{Laminar})$$ 2. Calculate Nusselt number ($Nu_L$):
$$Nu_L = 0.664 Re^{0.5} Pr^{1/3} = 0.664 \times (1.67 \times 10^5)^{0.5} \times 0.73^{1/3} = 244.1$$ 3. Calculate Schmidt & Lewis numbers:
$$Sc = \frac{1.5 \times 10^{-5}}{2.6 \times 10^{-5}} = 0.577, \quad Le = \frac{0.577}{0.73} = 0.79$$ 4. Compute Sherwood number ($Sh_L$):
$$Sh_L = 0.664 Re^{0.5} Sc^{1/3} = 225.8$$ 5. Compute average transfer coefficients:
$$h = \frac{Nu_L k_f}{L} = 12.3\text{ W/m}^2\text{K}$$ $$h_m = \frac{Sh_L D_{AB}}{L} = 0.0117\text{ m/s}$$
Air at 20°C ($\rho = 1.2\text{ kg/m}^3$, $C_p = 1007\text{ J/kg·K}$, $Pr = 0.73$, $\nu = 1.5 \times 10^{-5}\text{ m}^2/\text{s}$) flows over a flat plate ($L = 0.5\text{ m}$) at $U = 5\text{ m/s}$. Water binary diffusivity $D_{AB} = 2.6 \times 10^{-5}\text{ m}^2/\text{s}$. Calculate $h$ and $h_m$.
Solution:
1. Calculate Reynolds number:
$$Re = \frac{U L}{\nu} = \frac{5 \times 0.5}{1.5 \times 10^{-5}} = 1.67 \times 10^5 \quad (\text{Laminar})$$ 2. Calculate Nusselt number ($Nu_L$):
$$Nu_L = 0.664 Re^{0.5} Pr^{1/3} = 0.664 \times (1.67 \times 10^5)^{0.5} \times 0.73^{1/3} = 244.1$$ 3. Calculate Schmidt & Lewis numbers:
$$Sc = \frac{1.5 \times 10^{-5}}{2.6 \times 10^{-5}} = 0.577, \quad Le = \frac{0.577}{0.73} = 0.79$$ 4. Compute Sherwood number ($Sh_L$):
$$Sh_L = 0.664 Re^{0.5} Sc^{1/3} = 225.8$$ 5. Compute average transfer coefficients:
$$h = \frac{Nu_L k_f}{L} = 12.3\text{ W/m}^2\text{K}$$ $$h_m = \frac{Sh_L D_{AB}}{L} = 0.0117\text{ m/s}$$
Standard Assumptions & References
Assumptions: Constant fluid and species properties, negligible viscous dissipation, negligible radiative heat transfer, and low mass transfer rate (so bulk diffusion velocity does not disturb velocity boundary layer profile).
References:
- Chilton, T. H., & Colburn, A. P. Mass transfer coefficients; correlation of data. Industrial & Engineering Chemistry, 1934.
- Incropera, F. P., et al. Fundamentals of Heat and Mass Transfer. Wiley.