Question

Air at \(20^{\circ} \mathrm{C}\) with a convection heat transfer coefficient of \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) blows over a pond. The surface temperature of the pond is at \(40^{\circ} \mathrm{C}\). Determine the heat flux between the surface of the pond and the air.

Short answer

Answer: The heat flux between the surface of the pond and the air is 400 W/m².

Step-by-step solution

Write down the convective heat transfer equation

The convective heat transfer equation is: $$ q" = h(T_{s} - T_{\infty}) $$

Convert temperatures to Kelvin

To work with the convective heat transfer equation, we need to convert the given temperatures from Celsius to Kelvin: $$ T_{s} = 40 + 273.15 = 313.15 \mathrm{K} $$ $$ T_{\infty} = 20 + 273.15 = 293.15 \mathrm{K} $$

Insert the values into the equation

Now that we have the temperatures in Kelvin, insert the values into the convective heat transfer equation: $$ q" = 20 \frac{\mathrm{W}}{\mathrm{m}^2 \cdot \mathrm{K}} (313.15 \mathrm{K} - 293.15 \mathrm{K}) $$

Calculate the heat flux

Multiply the convection heat transfer coefficient by the temperature difference: $$ q" = 20 \frac{\mathrm{W}}{\mathrm{m}^2 \cdot \mathrm{K}} \times 20 \mathrm{K} $$ $$ q" = 400 \frac{\mathrm{W}}{\mathrm{m}^2} $$ The heat flux between the surface of the pond and the air is 400 W/m².

Key concepts

Heat Flux

Understanding heat flux is fundamental when studying thermal energy transfer in materials and across surfaces. Heat flux is defined as the rate of heat transfer per unit area, which can be visualized as the amount of heat that passes through a specified area in a certain time frame. In the context of our exercise, it occurs when air moves over the surface of a pond, initiating a transfer of heat.

In a practical example, consider how a warm cup of coffee will cool down as the surrounding air absorbs its heat. The heat flux in this scenario equates to the speed at which the coffee loses its warmth to the air. Mathematically, we express heat flux using the symbol \( q'' \) and measure it in units of watts per square meter (\( W/m^2 \)).

Analyzing our exercise, the air temperature is cooler than the pond’s surface temperature, which causes heat to transfer from the warmer pond to the cooler air. This process is quantifiable as a heat flux value.

Temperature Conversion

Temperature conversion is a simple yet vital step in thermodynamics and heat transfer calculations. Since heat transfer equations often require temperature in an absolute scale, converting Celsius to Kelvin is an essential task. The conversion is straightforward: simply add 273.15 to the Celsius temperature to convert it to Kelvin.

Why Kelvin, though? Kelvin is the SI unit for thermodynamic temperature and is a scale where 0 represents absolute zero, the theoretical temperature at which particles possess no thermal energy. This absolute reference allows for a uniform standard when comparing energy states across different systems.

During the exercise, we converted the pond surface temperature from \( 40^\circ C \) to \( 313.15 K \) and the air temperature from \( 20^\circ C \) to \( 293.15 K \) before substituting them into the heat transfer equation. This correct use of temperature scales ensures accurate calculation of heat flux.

Convection Heat Transfer Coefficient

The convection heat transfer coefficient, symbolized as \( h \), is a quantifier of the convective heat transfer rate between a solid surface and a fluid moving over it. It's measured in watts per square meter per Kelvin (\( W/m^2\cdot K \)). This coefficient essentially illustrates how well the connection between the fluid and the surface facilitates thermal energy exchange.

The value of \( h \) depends on several factors: the nature of the fluid flow (laminar or turbulent), the fluid's thermophysical properties, and the geometry of the surface. For instance, a higher \( h \) value means that the surface and the fluid are exchanging heat more effectively, making it a critical factor in designing heat exchangers, radiators, and similar devices.

In our exercise, the given convection heat transfer coefficient is \( 20 W/m^2\cdot K \). This information, alongside the temperature difference, allows us to compute the heat flux. It’s the \( h \) value that bridges the gap between the mere temperature difference and the actual rate at which heat is transferred – the heat flux we calculated to be \( 400 W/m^2 \).