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 = 2
Gap L = 0.0500 m
Height H = 1.0000 m
Width W = 1.0000 m
T_hot = 80.00 C
T_cold = 20.00 C
Delta T = 60.00 C
--- FLUID ---
rho = 997.0000 kg/m3
mu = 8.9000E-04 Pa.s
k = 0.613000 W/mK
Pr = 6.1300
beta = 2.1000E-04 1/K
--- RESULTS ---
Rayleigh Number Ra = 1.1763E+08
Nusselt Number Nu = 38.6628
Effective h = 474.0063 W/m2K
k_effective = 0.6130 W/mK
Heat Transfer Q = 28440.3762 W (or W/m)
Aspect H/L = 1.00
--- DELTA-T SWEEP ---
dT[C] Ra Nu Q[W]
---------------------------------------------------
1.00 1.960E+06 9.876 121.0785
8.46 1.658E+07 20.122 2086.6362
15.92 3.120E+07 24.842 4847.7085
23.38 4.583E+07 28.238 8092.2476
30.83 6.045E+07 30.968 11706.5154
38.29 7.507E+07 33.287 15626.8951
45.75 8.969E+07 35.322 19811.7026
53.21 1.043E+08 37.145 24231.1045
60.67 1.189E+08 38.805 28862.4934
68.12 1.336E+08 40.335 33688.0358
75.58 1.482E+08 41.756 38693.2365
83.04 1.628E+08 43.087 43866.0382
90.50 1.774E+08 44.340 49196.2246
97.96 1.920E+08 45.526 54675.0077
105.42 2.067E+08 46.653 60294.7329
112.88 2.213E+08 47.728 66048.6615
120.33 2.359E+08 48.757 71930.8070
127.79 2.505E+08 49.744 77935.8091
135.25 2.652E+08 50.694 84058.8352
142.71 2.798E+08 51.609 90295.5016
150.17 2.944E+08 52.493 96641.8104
157.62 3.090E+08 53.348 103094.0978
165.08 3.236E+08 54.176 109648.9917
172.54 3.383E+08 54.980 116303.3760
180.00 3.529E+08 55.761 123054.3610
--- 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 $h_{eff}$:
$$h_{eff} = \frac{\text{Nu} k}{L} = \frac{1.99 \times 0.0271}{0.02} = 2.70\text{ W/m}^2\text{K}$$