Question

Consider a person standing in a room at \(18^{\circ} \mathrm{C}\). Determine the total rate of heat transfer from this person if the exposed surface area and the skin temperature of the person are \(1.7 \mathrm{~m}^{2}\) and \(32^{\circ} \mathrm{C}\), respectively, and the convection heat transfer coefficient is \(5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Take the emissivity of the skin and the clothes to be \(0.9\), and assume the temperature of the inner surfaces of the room to be the same as the air temperature.

Short answer

Answer: The total rate of heat transfer from the person standing in the room is 208 W.

Step-by-step solution

Determine the temperature difference between the person's skin and the room

To find the heat transfer rate, we first need to determine the temperature difference between the person's skin and the room. The room temperature is \(18^{\circ}\mathrm{C}\), and the skin temperature is \(32^{\circ}\mathrm{C}\). Thus, the temperature difference \(\Delta T\) is: \(\Delta T = T_{skin} - T_{room} = 32^{\circ}\mathrm{C} - 18^{\circ}\mathrm{C} = 14\mathrm{K}\)

Calculate the heat transfer rate due to convection

The heat transfer rate due to convection (\(Q_{conv}\)) can be determined using the formula: \(Q_{conv} = h_{conv}A(T_{skin} - T_{room})\) Where \(h_{conv}\) is the convection heat transfer coefficient, \(A\) is the exposed surface area, and \((T_{skin} - T_{room})\) is the temperature difference. Plugging in the given values, we get: \(Q_{conv} = 5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K} \times 1.7 \mathrm{~m}^{2} \times 14\mathrm{K} = 119\mathrm{W}\)

Calculate the heat transfer rate due to radiation

The heat transfer rate due to radiation (\(Q_{rad}\)) can be determined using the formula: \(Q_{rad} = \epsilon A \sigma (T_{skin}^{4} - T_{room}^{4})\) Where \(\epsilon\) is the emissivity (0.9 in this case), \(A\) is the exposed surface area, \(\sigma\) is the Stefan-Boltzmann constant (\(5.67\times10^{-8}\mathrm{W}/\mathrm{m}^{2}\cdot\mathrm{K}^4\)), and \((T_{skin}^{4} - T_{room}^{4})\) is the difference in the temperature raised to the fourth power. First convert temperatures in Kelvin: \(T_{skin} = 32^{\circ}\mathrm{C} + 273.15 = 305.15\mathrm{K}\) \(T_{room} = 18^{\circ}\mathrm{C} + 273.15 = 291.15\mathrm{K}\) Now, plugging in the values, we get: \(Q_{rad} = 0.9 \times 1.7 \mathrm{~m}^{2} \times 5.67\times10^{-8}\mathrm{W}/\mathrm{m}^{2}\cdot\mathrm{K}^4 \times (305.15^{4} - 291.15^{4}) \approx 89\mathrm{W}\)

Calculate the total rate of heat transfer

Now, we can find the total rate of heat transfer by summing up the heat transfer rates due to convection and radiation: \(Q_{total} = Q_{conv} + Q_{rad} = 119\mathrm{W} + 89\mathrm{W} = 208\mathrm{W}\) So, the total rate of heat transfer from the person standing in the room is \(208\mathrm{W}\).

Key concepts

Convection Heat Transfer

When a person stands in a room, their body loses heat to the surrounding air. This process is known as convection heat transfer. It occurs because of the temperature difference between the person's skin and the air.

Convection can be understood as the transfer of heat through a fluid, which in most cases is either a gas or a liquid. The fluid near the warm surface gains energy and, due to its lower density, rises, while the cooler fluid descends to take its place, creating a cycle that continuously transfers heat away from the surface.

The rate of convection heat transfer, denoted as \( Q_{conv} \), is calculated via the formula: \( Q_{conv} = h_{conv} \times A \times (T_{skin} - T_{room}) \), where \( h_{conv} \) is the convection heat transfer coefficient, \( A \) is the exposed surface area, and \( (T_{skin} - T_{room}) \) is the temperature difference.

Understanding convection is crucial for calculating the heat transfer rate, particularly in environments where air flow and temperature can have a significant impact on a person's comfort level.

Radiation Heat Transfer

Apart from convective cooling, our body also loses heat through radiation heat transfer. Radiation does not require any medium to transfer heat—it is the process by which energy is emitted as particles or waves.

Every object emits radiant energy, and this is particularly true for human bodies. Radiant heat transfer depends on the surface temperature of the body and the surface temperature of the surroundings as well as the material's ability to emit radiation, which is characterized by its emissivity \( (\epsilon) \).

The rate of radiation heat transfer, \( Q_{rad} \), can be determined using the formula: \( Q_{rad} = \epsilon \times A \times \sigma \times (T_{skin}^{4} - T_{room}^{4}) \), where \( \sigma \) is the Stefan-Boltzmann constant.

It's important for students to grasp that in radiation heat transfer, temperatures are raised to the fourth power, reflecting the nonlinear relationship between temperature and radiant energy, which means even small temperature differences can result in significant heat transfer.

Temperature Difference

Temperature difference plays a pivotal role in heat transfer calculations for both convection and radiation. It is the driving force that initiates the transfer of heat from a warmer object to a cooler one.

As outlined in the exercise, the temperature difference \( \Delta T \) is the subtraction of the ambient air temperature from the skin temperature. Here, \( \Delta T = T_{skin} - T_{room} \). A larger temperature difference means a higher potential for heat transfer.

This difference not only affects how quickly the person loses heat, but also influences the total heat loss over time. Understanding this concept is vital because it highlights why maintaining a certain temperature in human environments is essential for comfort and energy efficiency.

Stefan-Boltzmann Constant

The Stefan-Boltzmann constant \( (\sigma) \) is fundamental when calculating radiant heat transfer. It is a proportionality constant used in the Stefan-Boltzmann Law, which dictates that the energy radiated from a black body per unit area is directly proportional to the fourth power of its absolute temperature.

In the exercise context, \( \sigma \) is used in the radiation formula: \( Q_{rad} = \epsilon \times A \times \sigma \times (T_{skin}^{4} - T_{room}^{4}) \). With a value of approximately \(5.67 \times 10^{-8} \frac{W}{m^2 \cdot K^4}\), this constant enables the calculation of the radiation heat transfer rate.

Understanding the Stefan-Boltzmann constant is crucial as it relates to the energy emitted by the body. While we might not perceive this energy loss through radiation directly, it has a significant impact on how the body loses heat in various environments.