Derivation :

The fluid velocity in case of free convection depends upon the following parameters;

1. Temperature difference between solid surface and bulk fluid, $ΔT$

2. Acceleration due to gravity, g

3. Coefficient of volumetric expansion of fluid, β

The change in the volume when temperature changes can be expressed as

$dV = V_1 β (T_2 – T_1) $

where

$dV$ - change in volume (m³ ) $= V_2 – V_1$

β = Coefficient of volumetric expansion of fluid, (m³/m³ °C)

$T_2$ - Final temperature (°C)

$T_1$ - Initial temperature (°C)

Therefore , free convection heat transfer coefficient is a function of following variables

$ \begin{array}{|l|l|l|l|} \hline \text{ Variable } & \text { Symbol } & \text { Dimensions } \\ \hline \text { Fluid density } & \rho & \mathrm{ML}^{-3} \\ \hline \text { Dynamic viscosity } & \mu & \mathrm{ML}^{-1} \mathrm {T}^{-1}\\ \hline \text { Thermal conductivity } & \mathrm{k} & \mathrm{MLT}^{-3} \theta^{-1 }\\ \hline \text { Specific heat } & \mathrm{C}_{\mathrm{p}} & \mathrm{L}^{2} \mathrm{~T}^{-2} \theta^{-1} \\ \hline \text { Characteristic length} & \mathrm{D} & \mathrm{L} \\ \hline \text { Temperature difference } & \Delta \mathrm{T} & \theta \\ \hline \end{array} $

Therefore, convective heat transfer coefficient is expressed as

$h = f(ρ, μ, k, Cp, D, ΔT, β, g) $

However, in free convection, (ΔT β g) will be treated as single parameter as the velocity of fluid particles is a function of these parameters. Therefore, equation (i) can be expressed as

$f(h, ρ, μ, k, Cp, D, (ΔT β g))= 0 $

Convective heat transfer coefficient, h is dependent variable and remaining are independent variables.

Total number of variables, n = 7

Number of fundamental units, m = 4

According to Buckingham’s π-theorem, number of π-terms is given by the difference of total number of variables and number of fundamental units.

Number of π-terms = (n-m) = 7-4 = 3

These non-dimensional π-terms control the forced convection phenomenon and are expressed as,

$f(\pi_1,\pi_2,\pi_3)=0....(i)$

Each $π_1$ -term is expressed as:

$π_1 = μ^a k^b ρ^cD^d h ..(ii)$

Writing down each term in above equation in terms of fundamental dimensions

$M^0L^0T^0 θ^ 0 = (ML^{-1}T^{-1} ) ^a (MLT^{-3}θ^{ -1 }) ^b (ML^{-3 }) ^c (L)^d MT^{-3} θ^ {-1}$

Comparing the powers of M, we get

$0 = a+b+c+1, $

$a+b+c= -1 $

Comparing powers of L, we get

$0 = -a+b+c +d $

Comparing powers of T, we get

$0 = -a- 3b-c -3 $

Comparing powers of θ, we get

$b=-1$

Solving all above equations simultaneously, we get

$a=0,c=0,d=1,b=-1$

Substituting the values of ‘a’, ‘b’, ‘c’ and ‘d’ in equation (ii), we get

$π_1 = μ^0 k^{-1} ρ^0 D^1 h$

$\pi_1=hD/K$

The second $π_2 $–term is expressed as

$π_2 = μ^a k^b ρ^c D^d C_P$

After following same steps we get

$\pi_2=\mu C_p /K=Pr$

The third $π_3$ –term is expressed as

$π_3 = μ^a k^b ρ^c D^d ((ΔT β g)$

After following same steps, we get

$\pi_3=D^3(ΔT β g) / υ^2$

Substituting the values of $π_1, π_2, π_3$ in equation (i), we get

$\begin {aligned} f(\frac{h D }{K} , \frac{μ C_P}{K} , \frac{D^3 (ΔT β g)}{υ^2 }) =0 \end{aligned}$

$\begin {aligned} \frac{h D}{K} = φ(\frac{μ C_P}{K}, \frac{D^3 (ΔT β g) }{ υ^2 }) \end{aligned}$

$Nu = φ(Pr, Gr)$ as

$Gr = D^3(ΔT β g) / υ^2$

The above correlation is generally expressed as,

$Nu = C (Pr)^a(Gr)^b $

The constant C and exponents a and b are determined through experiments.