Free Convection in Enclosures
Calculate heat transfer inside cavities, rectangular slots, double-pane windows, and concentric cylinders or spheres.
Cavity Natural Convection
In enclosures, fluid movement is confined. For low Rayleigh numbers ($\text{Ra} \le 10^3$), heat transfer is purely conductive ($\text{Nu} \approx 1.0$). At higher Rayleigh numbers, convective circulation cells form, which significantly increases heat exchange.
Parameters Setup
Results & Curves
📈 Nusselt number vs Delta T (dT)
Engine Output
============================================
FREE CONVECTION IN ENCLOSURES
============================================
--- INPUTS ---
Enclosure Type = 5
Gap L = 0.0000 m
Height H = 0.0000 m
Width W = 0.0000 m
T_hot = 70.00 C
T_cold = 25.00 C
Delta T = 45.00 C
--- FLUID ---
rho = 870.0000 kg/m3
mu = 5.0000E-02 Pa.s
k = 0.140000 W/mK
Pr = 500.0000
beta = 7.0000E-04 1/K
--- RESULTS ---
Rayleigh Number Ra = 5.3443E+05
Nusselt Number Nu = 19.9994
Effective h = 139.9961 W/m2K
k_effective = 2.7999 W/mK
Heat Transfer Q = 63.3327 W (or W/m)
--- DELTA-T SWEEP ---
dT[C] Ra Nu Q[W]
---------------------------------------------------
1.00 1.188E+04 7.722 0.5434
6.58 7.819E+04 12.369 5.7302
12.17 1.445E+05 14.421 12.3474
17.75 2.108E+05 15.849 19.7975
23.33 2.771E+05 16.971 27.8666
28.92 3.434E+05 17.906 36.4374
34.50 4.097E+05 18.714 45.4346
40.08 4.760E+05 19.429 54.8046
45.67 5.423E+05 20.073 64.5077
51.25 6.087E+05 20.660 74.5126
56.83 6.750E+05 21.201 84.7943
62.42 7.413E+05 21.704 95.3319
68.00 8.076E+05 22.174 106.1081
73.58 8.739E+05 22.616 117.1081
79.17 9.402E+05 23.033 128.3189
84.75 1.007E+06 23.429 139.7292
90.33 1.073E+06 23.805 151.3292
95.92 1.139E+06 24.165 163.1099
101.50 1.205E+06 24.509 175.0633
107.08 1.272E+06 24.840 187.1824
112.67 1.338E+06 25.157 199.4605
118.25 1.404E+06 25.463 211.8917
123.83 1.471E+06 25.759 224.4706
129.42 1.537E+06 26.044 237.1922
135.00 1.603E+06 26.321 250.0517
--- CORRELATIONS ---
Vertical rect: Catton (1978) correlations
Horizontal hot bottom: Globe-Dropkin Nu=0.069*Ra^(1/3)*Pr^0.074
Concentric cyl: k_eff/k=0.386(Pr/(0.861+Pr))^0.25*Ra^0.25
Concentric sph: k_eff/k=0.74(Pr/(0.861+Pr))^0.25*Ra^0.25
Ref: Incropera Ch.9 Sec.9.8, Kothandaraman Ch.10
Calculation Methodology
Mathematical Model & Theory
Natural convection in enclosures is governed by Rayleigh number $\text{Ra}_L = g \beta \Delta T L^3 / (\nu \alpha)$. Different correlations apply:
- Vertical Rectangular Cavities: Catton's correlations based on aspect ratio $H/L$:
$$\text{Nu} = C(Pr) \text{Ra}^{0.28-0.29} (H/L)^{-0.25-0.30}$$
- Horizontal Cavities (Hot Bottom): Globe-Dropkin correlation:
$$\text{Nu} = 0.069 \text{Ra}^{1/3} \text{Pr}^{0.074}$$
- Concentric Annulus: Raithby & Hollands formulation:
$$\frac{k_{eff}}{k} = 0.386 \left(\frac{\text{Pr}}{0.861 + \text{Pr}}\right)^{1/4} \text{Ra}_L^{1/4}$$
Academic References:
- Incropera, F. P., & DeWitt, D. P. (2011). Fundamentals of Heat and Mass Transfer. 7th Edition, John Wiley & Sons.
- Catton, I. (1978). Natural Convection in Enclosures. Proceedings of the 6th International Heat Transfer Conference.
Worked Engineering Example
A vertical double-pane window has a gap $L = 20\text{ mm}$ and height $H = 0.5\text{ m}$. The inner glass is at $60^\circ\text{C}$ and the outer glass is at $20^\circ\text{C}$. Find the heat transfer coefficient of the air slot ($T_f = 40^\circ\text{C}$).
Step-by-step Solution:
1. Evaluate properties at $40^\circ\text{C}$:
- Air: $\nu = 1.7 \times 10^{-5}\text{ m}^2\text{/s}$, $\alpha = 2.4 \times 10^{-5}\text{ m}^2\text{/s}$, $Pr = 0.707$, $\beta = 1/313.15 = 0.00319\text{ K}^{-1}$, $k = 0.0271\text{ W/m·K}$.
2. Calculate Rayleigh number $\text{Ra}_L$:
$$\text{Ra}_L = \frac{9.81 \times 0.00319 \times (60-20) \times 0.02^3}{1.7 \times 10^{-5} \times 2.4 \times 10^{-5}} = 24,534$$ 3. Evaluate aspect ratio and choose Catton correlation ($H/L = 0.5/0.02 = 25$):
Since $H/L > 10$, we use: $$\text{Nu} = 0.42 \text{Ra}_L^{0.25} \text{Pr}^{0.012} (H/L)^{-0.3}$$ $$\text{Nu} = 0.42 \times (24,534)^{0.25} \times (0.707)^{0.012} \times (25)^{-0.3} = 0.42 \times 12.52 \times 0.996 \times 0.38 = 1.99$$ 4. Convection coefficient $h_{eff}$:
$$h_{eff} = \frac{\text{Nu} k}{L} = \frac{1.99 \times 0.0271}{0.02} = 2.70\text{ W/m}^2\text{K}$$