Question

A person standing in a room loses heat to the air in the room by convection and to the surrounding surfaces by radiation. Both the air in the room and the surrounding surfaces are at \(20^{\circ} \mathrm{C}\). The exposed surface of the person is \(1.5 \mathrm{~m}^{2}\) and has an average temperature of \(32^{\circ} \mathrm{C}\), and an emissivity of \(0.90\). If the rates of heat transfer from the person by convection and by radiation are equal, the combined heat transfer coefficient is (a) \(0.008 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (b) \(3.0 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (c) \(5.5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (d) \(8.3 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (e) \(10.9 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\)

Short answer

a) 4.5 W/m²·K b) 5.0 W/m²·K c) 5.5 W/m²·K d) 6.0 W/m²·K Answer: c) 5.5 W/m²·K

Step-by-step solution

\(T_{person} = 32^{\circ} \mathrm{C} + 273.15 = 305.15 \mathrm{K}\)

\(T_{surroundings} = 20^{\circ} \mathrm{C} + 273.15 = 293.15 \mathrm{K}\) #Step 2: Set convection heat loss equal to radiation heat loss# We set the equations for convection and radiation heat loss equal to each other:

\(h \cdot A \cdot \Delta T = \epsilon \cdot \sigma \cdot A \cdot \left( T^4_{person} - T^4_{surroundings} \right)\) #Step 3: Simplify and isolate \(h\)# We simplify the equation and isolate the heat transfer coefficient \(h\):

\(h = \frac{\epsilon \cdot \sigma \cdot A \cdot \left( T^4_{person} - T^4_{surroundings} \right)}{A \cdot \Delta T} \) Since \(A\) is a common factor in both numerator and denominator, we can simplify the equation further:

\(h = \frac{\epsilon \cdot \sigma \cdot \left( T^4_{person} - T^4_{surroundings} \right)}{\Delta T} \) #Step 4: Input the given data and solve for \(h\)# Now, input the given values and solve for the heat transfer coefficient \(h\):

\(h = \frac{0.9 \cdot 5.67 \times 10^{-8} \mathrm{Wm}^{-2}\mathrm{K}^{-4} \cdot \left( (305.15)^4 - (293.15)^4 \right)}{ (305.15 - 293.15)}\) Solve for \(h\):

\(h = 5.53 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) Comparing the calculated value of \(h\) with the given answer choices, we can see that the closest option is:

(c) \(5.5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\)

Key concepts

Convection

Understanding how convection works is essential in grasping the concept of heat transfer in the given exercise. Convection is the process by which heat is transferred through fluids (liquids or gases) by the movement of the fluid itself. It happens when warmer areas of a liquid or gas rise to cooler areas in the fluid. The heat is then transferred because of this motion.

For example, when a person is in a room, their body transfers heat to the cooler surrounding air through convection. The air gets warmer and rises, making way for cooler air to come into contact with the skin, continuing the cycle. This is why, in the exercise, the person loses heat to the room through convection.

Key points to remember about convection:

• It requires a medium, like air or water, to transfer heat.
• The heat moves because of the fluid motion, rising up when heated.
• This cycle helps distribute heat evenly throughout the fluid.

Radiation

Radiation is another way heat is transferred, as explained in the exercise. Unlike convection, radiation does not require a medium to transfer heat. Instead, heat is transferred in the form of electromagnetic waves. These waves can transfer heat through a vacuum or transparent medium like air.

The human body loses heat to its surroundings through radiation. In the exercise, heat is lost to the surfaces around, like the walls and furniture, without the need for air movement. This happens through thermal radiation, where the heat emitted by the warm skin transfers to cooler surfaces.

Essential aspects of radiation include:

• Heat transfer through electromagnetic waves.
• Does not require physical contact or a medium.
• Can occur across distances and even in a vacuum.

Emissivity

Emissivity is a measure of how efficiently a surface emits thermal radiation. In the context of the exercise, it's crucial because it affects how much heat the person can radiate. The person's skin has a given emissivity value of 0.90, meaning it emits 90% of the heat energy compared to a perfect black body, which has an emissivity of 1. This value significantly impacts the radiation heat loss calculation, as seen in the solution steps. Important points about emissivity:

• Emissivity ranges from 0 to 1, with 1 being a perfect emitter.
• Higher emissivity means more efficient heat radiation.
• It influences how much heat is radiated at a given temperature.

Heat Transfer Coefficient

The heat transfer coefficient is a crucial parameter in determining the rate of heat transfer between a surface and a fluid (such as air, in the exercise). It measures how easily heat is transferred from the person's skin to the surrounding environment.In the exercise, the heat transfer coefficient combines the heat losses by convection and radiation into one value. The heat transfer coefficient (\( h \)) depends on:

• The nature of fluid flow around the surface (e.g., still or moving air).
• Material properties, such as surface emissivity.
• The temperature difference between the surface and the fluid.

Understanding the heat transfer coefficient helps us understand how efficiently heat is lost from the body. The final calculated value in the exercise is 5.5 W/m²·K, indicating the specific conditions under which both convection and radiation losses balance.