Question

Consider a thin layer of liquid water on a concrete surface. The surrounding air is dry with a convection heat transfer coefficient of \(50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The liquid water has an emissivity of \(0.95\), and the air and surrounding temperature is \(30^{\circ} \mathrm{C}\). If the layer of liquid water has a uniform temperature of \(20^{\circ} \mathrm{C}\), determine the conduction heat flux through the concrete.

Short answer

Answer: The conduction heat flux through the concrete layer is approximately -618.93 W/m².

Step-by-step solution

Calculate convective heat transfer

To find the convective heat transfer, we will use the formula: \(q_{conv} = h \times A \times (T_{water} - T_{air})\), where \(h\) is the convection heat transfer coefficient, \(A\) is the surface area, and \(T_{water}\) and \(T_{air}\) are the temperatures of the water and air, respectively. Since we are interested in heat flux, we can ignore the surface area and find the convective heat transfer per unit area: \(q_{conv} = h \times (T_{water} - T_{air})\). Given values: \(h = 50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), \(T_{water} = 20^{\circ} \mathrm{C}\), and \(T_{air} = 30^{\circ} \mathrm{C}\). Plugging in the values, we get: \(q_{conv} = 50 \times (20 - 30) = -500 \mathrm{~W/m^2}\)

Calculate radiative heat transfer

To find the radiative heat transfer, we will use the formula: \(q_{rad} = \epsilon \sigma A (T_{water}^4 - T_{air}^4)\), where \(\epsilon\) is the emissivity of the liquid water, \(\sigma\) is the Stefan-Boltzmann constant (\(5.67 \times 10^{-8} \mathrm{~W/m^2 \cdot K^4}\)), and \(T_{water}\) and \(T_{air}\) are the temperatures of the water and air in Kelvin. Since we are interested in heat flux, we can ignore the surface area and find the radiative heat transfer per unit area: \(q_{rad} = \epsilon \sigma (T_{water}^4 - T_{air}^4)\). Given values: \(\epsilon = 0.95\), \(T_{water} = 293 \mathrm{K}\), and \(T_{air} = 303 \mathrm{K}\). Plugging in the values, we get: \(q_{rad} = 0.95 \times 5.67 \times 10^{-8} \times (293^4 - 303^4) \approx -118.93 \mathrm{~W/m^2}\)

Determine the total heat transfer

Now, we will add the convective and radiative heat transfers to find the total heat transfer per unit area. \(q_{total} = q_{conv} + q_{rad} = -500 + (-118.93) = -618.93 \mathrm{~W/m^2}\)

Calculate conduction heat flux

Since there is no heat generation within the layer, the conduction heat flux through the concrete is equal to the total heat transfer per unit area. \(q_{cond} = q_{total} = -618.93\mathrm{~W/m^2}\)

Key concepts

Convective Heat Transfer

Convective heat transfer is a mechanism of heat movement involving the motion of fluid molecules. It occurs when a temperature difference exists between a solid surface and a moving fluid, such as air or water. This kind of heat transfer contributes to the warming or cooling of the surface.

In our exercise, the convective heat transfer coefficient, denoted by 'h', plays a crucial role. It represents the amount of heat transferred per unit area and per degree temperature difference between the surface and the fluid. A higher 'h' value indicates more efficient heat transfer, whereas a lower value suggests less efficient transfer. The negative sign obtained in the solution indicates that heat flows from the water to the air, due to the air being warmer than the liquid water surface.

Radiative Heat Transfer

Radiative heat transfer is energy transfer in the form of electromagnetic waves, typically infrared radiation when we talk about thermal radiation emitted by objects at typical temperatures found on Earth. Unlike convection, radiative heat transfer does not require a medium; it can even occur in the vacuum of space.

Every object that has a temperature above absolute zero emits thermal radiation, and the intensity of that radiation is a function of its temperature. In this context, when the surface of the water cools by emitting radiation, the rate of heat loss is determined through the Stefan-Boltzmann law, using its emissivity and temperature to calculate the radiative heat flux.

Stefan-Boltzmann Constant

The Stefan-Boltzmann constant is a physical constant denoted by \( \sigma \) and is fundamental when calculating radiative heat transfer. Its value is \( 5.67 \times 10^{-8} \mathrm{~W/m^2 \cdot K^4} \), indicating the power per unit area (aka intensity) emitted by a black body in thermal equilibrium through radiation.

Even though real materials are not perfect black bodies, the Stefan-Boltzmann law can still be applied to them by introducing the concept of emissivity. This constant ensures that the thermal radiation exchanged between the water and its surroundings is accurately quantified. Since the value is standardized, it provides a reliable basis to calculate radiative heat flux for different temperature scenarios.

Emissivity

Emissivity, denoted by \( \epsilon \), is a measure of an object's ability to emit thermal radiation compared to that of a perfect black body. It is dimensionless and ranges between 0 and 1, with a black body having an emissivity of 1.

The emissivity value is crucial because it modifies the Stefan-Boltzmann law to account for the radiative heat transfer properties of real materials. In our exercise, the water has an emissivity of 0.95, suggesting it is a very good emitter of thermal radiation, nearly as effective as a perfect black body. The high emissivity results in significant radiative heat loss, as evident from the calculation of radiative heat flux.