Question

A \(3-\mathrm{m}^{2}\) black surface at \(140^{\circ} \mathrm{C}\) is losing heat to the surrounding air at \(35^{\circ} \mathrm{C}\) by convection with a heat transfer coefficient of \(16 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and by radiation to the surrounding surfaces at \(15^{\circ} \mathrm{C}\). The total rate of heat loss from the surface is (a) \(5105 \mathrm{~W}\) (b) \(2940 \mathrm{~W}\) (c) \(3779 \mathrm{~W}\) (d) \(8819 \mathrm{~W}\) (e) \(5040 \mathrm{~W}\)

Short answer

Answer: (a) 5105 W

Step-by-step solution

Convert temperatures to Kelvin

To do the calculations, we need to convert the temperatures of the surface, air, and surrounding surfaces from Celsius to Kelvin. K = °C + 273.15 Surface temperature: 140°C + 273.15 = 413.15 K Air temperature: 35°C + 273.15 = 308.15 K Surrounding surfaces temperature: 15°C + 273.15 = 288.15 K

Calculate the heat loss due to convection

Using the formula and given temperature values: Heat loss by convection = heat transfer coefficient × Area × ΔT Heat loss by convection = 16 W/m²·K × 3 m² × (413.15 K - 308.15 K) Heat loss by convection = 16 × 3 × 105 = 5040 W

Calculate the heat loss due to radiation

Apply the Stefan-Boltzmann law with given surface temperature and surrounding surfaces temperature: Heat loss by radiation = εσ × Area × (T1⁴ - T2⁴) Heat loss by radiation = (1)(5.67×10⁻⁸ W/m²K⁴) × 3 m² × (413.15⁴ - 288.15⁴) Heat loss by radiation ≈ 65 W

Calculate the total heat loss

Add the heat loss due to convection and radiation to get the total heat loss: Total heat loss = heat loss by convection + heat loss by radiation Total heat loss = 5040 W + 65 W = 5105 W

Choose the correct option from the given choices

The total rate of heat loss from the surface is: (a) 5105 W Hence, the correct answer is (a) \(5105 \mathrm{~W}\).

Key concepts

Convection

Convection is the process of heat transfer through a fluid (which includes gases and liquids) caused by molecular motion. It occurs when a fluid is heated, causing it to expand and become less dense, thus rising in the surrounding cooler fluid. In the context of our exercise, the hot surface transfers heat to the cooler surrounding air through convection. This is calculated using the equation:

• Heat loss by convection = heat transfer coefficient × Area × ΔT

The heat transfer coefficient is a measure of how effectively heat is transferred between a solid surface and the fluid in contact with it. Convection can be categorized into two types:

• Natural Convection: Caused by buoyancy forces that are due to density differences caused by temperature variations in the fluid.
• Forced Convection: When a fluid is forced to flow over the surface by an external source like a fan or a pump.

In our problem, the surface loses heat to the air with a known heat transfer coefficient, helping us calculate the rate of heat loss efficiently.

Radiation

Radiation is heat transfer that occurs through electromagnetic waves, without the need for a medium. This means heat can be transferred through vacuum or transparent media (like air or glass), making it distinct from conduction and convection. In our exercise, the surface loses heat via radiation to the surrounding surfaces. Radiation heat transfer follows the Stefan-Boltzmann law, highlighting the emission power from a surface depending on its temperature. The formula used is:

• Heat loss by radiation = εσ × Area × (T₁⁴ - T₂⁴)

Where:

• ε is the emissivity of the surface, a measure of how efficiently it radiates energy.
• σ is the Stefan-Boltzmann constant, which is approximately 5.67×10⁻⁸ W/m²K⁴.
• T₁ and T₂ are the absolute temperatures of the surface and the surroundings, respectively.

Unlike convection, radiation does not require a medium and operates at all temperatures. It becomes significant at high temperatures, such as those in our exercise, leading to distinct heat loss computations.

Stefan-Boltzmann Law

The Stefan-Boltzmann Law is foundational in understanding radiation heat transfer. It states that the total energy radiated per unit surface area is directly proportional to the fourth power of the absolute temperature of the surface. The formula we used in our calculation is a direct application of this law. The equations are:

• Energy radiated = σ × Area × T⁴
• For two bodies at different temperatures: Energy exchanged is determined by the temperature difference raised to the fourth power.

The law is instrumental in determining the radiative heat loss in various applications, whether it’s satellites in space or our black surface example. In our case, using this law allows us to calculate the net radiation heat transfer between the surface at a high temperature and its cooler surroundings. It helps highlight the efficiency of heat loss through radiation compared to convection.

Heat Transfer Coefficient

The heat transfer coefficient is a parameter used in calculations of convective heat transfer. It's derived from empirical data and signifies how effectively heat is transferred from a solid surface to a fluid (or vice versa). In this problem, it helps determine the rate at which heat moves from the hot surface into the cooler surrounding air. Defined as:

• Heat transfer coefficient (h) = Heat transfer per unit area per degree of temperature difference

In formulas involving convection, the coefficient h, area (A), and temperature difference (ΔT) together determine the rate of heat transfer:

• Convective Heat Transfer = h × A × ΔT

Values of the heat transfer coefficient can modify based on factors such as the nature of the fluid, flow conditions, and surface characteristics. A higher value indicates more efficient heat transfer, crucial in engineering applications where heat management is vital.