⚖️ 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)
2.4900e+4
Regime: 🔴 Turbulent
Lewis Number (Le)
68.2287
Pr: 7.007 | Sc: 478.099
Average Heat HTC (h)
1,970.795 W/m²·K
Nusselt Nu: 164.78
Average Mass MTC (hm)
3.7475e-5 m/s
Sherwood Sh: 892.27
Chilton-Colburn Analogy Agreement
100.00% Match
Actual h/hm: 5.26e+7 | Theory: 5.26e+7
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HEAT & MASS TRANSFER ANALOGY SOLVER
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Configuration Type = Internal Flow in Circular Tube
Velocity (U) = 0.5000 m/s
Characteristic Length = 0.0500 m
--- FLUID & DIFFUSION PROPERTIES ---------------------------
Density (rho) = 998.0000 kg/m3
Viscosity (mu) = 1.0020E-03 Pa.s
Thermal Cond (kf) = 0.5980 W/m.K
Specific Heat (Cp) = 4182.00 J/kg.K
Mass Diffusivity D_AB = 2.1000E-09 m2/s
--- DIMENSIONLESS NUMBERS ----------------------------------
Reynolds Number (Re) = 2.4900E+04
Prandtl Number (Pr) = 7.0073
Schmidt Number (Sc) = 478.0991
Lewis Number (Le) = 68.2287
--- ANALOGY RESULTS (AVERAGE VALUES) -----------------------
Skin Friction Coeff (Cf) = 0.006187
Average Nusselt Number = 164.7822
Average Sherwood Number = 892.2720
Average Heat HTC (h) = 1970.7952 W/m2.K
Average Mass MTC (hm) = 3.7475E-05 m/s
--- CHILTON-COLBURN RATIO VERIFICATION ---------------------
Actual Ratio h/hm = 5.2589E+07 J/m3.K
Theoretical rho*Cp*Le^n = 5.2589E+07 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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0.0013 6.2250E+02 2344.26 12693.83 28037.3419 5.3314E-04
0.0025 1.2450E+03 1506.41 8156.98 18016.6369 3.4259E-04
0.0037 1.8675E+03 1174.89 6361.86 14051.6766 2.6720E-04
0.0050 2.4900E+03 990.65 5364.23 11848.1694 2.2530E-04
0.0063 3.1125E+03 871.22 4717.53 10419.7794 1.9814E-04
0.0075 3.7350E+03 786.58 4259.20 9407.4501 1.7889E-04
0.0088 4.3575E+03 722.97 3914.80 8646.7728 1.6442E-04
0.0100 4.9800E+03 673.16 3645.08 8051.0322 1.5309E-04
0.0112 5.6025E+03 632.93 3427.22 7569.8375 1.4394E-04
0.0125 6.2250E+03 599.65 3246.99 7171.7551 1.3637E-04
0.0138 6.8476E+03 571.58 3095.02 6836.0840 1.2999E-04
0.0150 7.4701E+03 547.54 2964.86 6548.5933 1.2452E-04
0.0163 8.0926E+03 526.69 2851.92 6299.1530 1.1978E-04
0.0175 8.7151E+03 508.39 2752.86 6080.3415 1.1562E-04
0.0187 9.3376E+03 492.19 2665.14 5886.5873 1.1194E-04
0.0200 9.9601E+03 477.73 2586.83 5713.6201 1.0865E-04
0.0213 1.0583E+04 464.72 2516.42 5558.1081 1.0569E-04
0.0225 1.1205E+04 452.96 2452.72 5417.4100 1.0301E-04
0.0238 1.1828E+04 442.26 2394.76 5289.4034 1.0058E-04
0.0250 1.2450E+04 432.47 2341.77 5172.3616 9.8354E-05
0.0262 1.3073E+04 423.48 2293.10 5064.8642 9.6310E-05
0.0275 1.3695E+04 415.19 2248.22 4965.7318 9.4425E-05
0.0288 1.4318E+04 407.52 2206.68 4873.9756 9.2681E-05
0.0300 1.4940E+04 400.40 2168.10 4788.7605 9.1060E-05
0.0312 1.5563E+04 393.76 2132.16 4709.3756 8.9551E-05
0.0325 1.6185E+04 387.56 2098.58 4635.2120 8.8140E-05
0.0338 1.6808E+04 381.75 2067.13 4565.7445 8.6819E-05
0.0350 1.7430E+04 376.30 2037.60 4500.5177 8.5579E-05
0.0362 1.8053E+04 371.17 2009.81 4439.1348 8.4412E-05
0.0375 1.8675E+04 366.33 1983.60 4381.2480 8.3311E-05
0.0387 1.9298E+04 361.75 1958.83 4326.5513 8.2271E-05
0.0400 1.9920E+04 357.42 1935.39 4274.7742 8.1286E-05
0.0413 2.0543E+04 353.32 1913.16 4225.6768 8.0353E-05
0.0425 2.1165E+04 349.42 1892.05 4179.0453 7.9466E-05
0.0438 2.1788E+04 345.71 1871.97 4134.6887 7.8623E-05
0.0450 2.2410E+04 342.18 1852.84 4092.4355 7.7819E-05
0.0462 2.3033E+04 338.81 1834.59 4052.1317 7.7053E-05
0.0475 2.3655E+04 335.59 1817.16 4013.6382 7.6321E-05
0.0488 2.4278E+04 332.51 1800.50 3976.8292 7.5621E-05
0.0500 2.4900E+04 329.56 1784.54 3941.5905 7.4951E-05
--- CORRELATIONS USED ---------------------------------------
Laminar Tube (Re <= 2300): Nu = 3.66 (Fully Developed Constant Temperature).
Turbulent Tube (Re > 2300): Dittus-Boelter correlation Nu = 0.023 Re^0.8 Pr^n.
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.