Question

Consider an \(L \times L\) horizontal plate that is placed in quiescent air with the hot surface facing up. If the film temperature is \(20^{\circ} \mathrm{C}\) and the average Nusselt number in natural convection is of the form \(\mathrm{Nu}=C \mathrm{Ra}_{L}^{n}\), show that the average heat transfer coefficient can be expressed as $$ \begin{array}{ll} h=1.95(\Delta T / L)^{1 / 4} & 10^{4}<\mathrm{Ra}_{L}<10^{7} \\ h=1.79 \Delta T^{1 / 3} & 10^{7}<\mathrm{Ra}_{L}<10^{11} \end{array} $$

Short answer

Question: Show that the average heat transfer coefficient (h) can be expressed in two different forms for two different ranges of the Rayleigh number (Ra) for the given average Nusselt number expression, \(Nu = C \cdot Ra_{L}^{n}\), and film temperature of 20°C. Answer: The two different forms of the average heat transfer coefficient (h) for different ranges of the Rayleigh number (Ra) are: $$ \begin{array}{ll} h=1.95(\Delta T / L)^{1 / 4} & 10^{4}<\mathrm{Ra}_{L}<10^{7} \\\ h=1.79 \Delta T^{1 / 3} & 10^{7}<\mathrm{Ra}_{L}<10^{11} \end{array} $$

Step-by-step solution

Relate Nusselt number with Heat Transfer Coefficient

The average Nusselt number is defined as the ratio of the convective heat transfer to the conductive heat transfer which can be expressed as: $$ Nu=\frac{h L}{k} $$ Where h is the heat transfer coefficient, L is the characteristic length, and k is the thermal conductivity of the fluid.

Express Rayleigh number in terms of temperature difference

The Rayleigh number (Ra) is a dimensionless number that characterizes fluid flow in a gravitational field due to temperature differences. It is defined as: $$ Ra=\frac{g \beta (\Delta T) L^{3}}{\nu \alpha} $$ Where g is the gravitational acceleration, \(\beta\) is the coefficient of thermal expansion, \(\Delta T\) is the temperature difference, \(\nu\) is the kinematic viscosity, and \(\alpha\) is the thermal diffusivity.

Re-write the Nusselt number expression

We can rewrite the expression of \(Nu = C \cdot Ra_{L}^{n}\) by substituting the relationships from Step 1 and Step 2 into the expression: $$ \frac{h L}{k} = C\left(\frac{g \beta (\Delta T) L^{3}}{\nu \alpha}\right)^{n} $$

Express heat transfer coefficient for two different ranges of Ra

We need to find the average heat transfer coefficient (h) for two different ranges of Ra: (1) For \(10^{4} < Ra_{L} < 10^{7}\): We are given that \(C=1.95\), and \(n=1/4\). Substituting these values into the expression obtained in Step 3 and isolating h, we get: $$ h = 1.95\left(\frac{g \beta (\Delta T) L^{3}}{\nu \alpha k}\right)^{1/4} = 1.95(\Delta T / L)^{1 / 4} $$ (2) For \(10^{7} < Ra_{L} < 10^{11}\): We are given that \(C=1.79\), and \(n=1/3\). Substituting these values into the expression obtained in Step 3 and isolating h, we get: $$ h = 1.79\left(\frac{g \beta (\Delta T) L^{3}}{\nu \alpha k}\right)^{1/3} = 1.79 \Delta T^{1 / 3} $$ Thus, we have the two expressions for the average heat transfer coefficient (h) as required: $$ \begin{array}{ll} h=1.95(\Delta T / L)^{1 / 4} & 10^{4}<\mathrm{Ra}_{L}<10^{7} \\\ h=1.79 \Delta T^{1 / 3} & 10^{7}<\mathrm{Ra}_{L}<10^{11} \end{array} $$