Question

A room is heated by a \(1.2 \mathrm{~kW}\) electric resistance heater whose wires have a diameter of \(4 \mathrm{~mm}\) and a total length of \(3.4 \mathrm{~m}\). The air in the room is at \(23^{\circ} \mathrm{C}\) and the interior surfaces of the room are at \(17^{\circ} \mathrm{C}\). The convection heat transfer coefficient on the surface of the wires is \(8 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the rates of heat transfer from the wires to the room by convection and by radiation are equal, the surface temperature of the wire is (a) \(3534^{\circ} \mathrm{C}\) (b) \(1778^{\circ} \mathrm{C}\) (c) \(1772^{\circ} \mathrm{C}\) (d) \(98^{\circ} \mathrm{C}\) (e) \(25^{\circ} \mathrm{C}\)

Short answer

Answer: (d) 98°C

Step-by-step solution

Write down the given information in the problem

: Power of the electric heater: \(P = 1.2 \mathrm{~kW}\) Diameter of the wire: \(d = 4 \mathrm{~mm}\) Total length of the wire: \(L = 3.4 \mathrm{~m}\) Temperature of the air: \(T_{air} = 23^{\circ} \mathrm{C}\) Temperature of the interior surfaces: \(T_{interior} = 17^{\circ} \mathrm{C}\) Convection heat transfer coefficient: \(h = 8 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\)

Calculate the surface area of the wire

: First, we'll calculate the surface area of the wire which is required for the heat transfer calculations. The wire is cylindrical in shape, and the surface area of a cylinder is given by: \(A = 2\pi rL\) where \(r = d/2\) is the radius of the wire. \(A = 2\pi (2 \times 10^{-3} \mathrm{~m})(3.4 \mathrm{~m})\) \(A = 0.0136\pi \mathrm{~m}^2\)

Write the convection and radiation heat transfer equations

: We know that the convective heat transfer rate \((Q_{conv})\) is given by: \(Q_{conv} = hA(T_{wire} - T_{air})\) Also, the radiative heat transfer rate \((Q_{rad})\) is given by: \(Q_{rad} = \sigma A e (T_{wire}^{4} - T_{interior}^{4})\) Since we are given the rates of heat transfer from the wires to the room by convection and by radiation are equal, then we can equate both of these equations.

Equate the convective and radiative heat transfer rates and solve for the wire surface temperature

: Set \(Q_{conv} = Q_{rad}\), and solve for \(T_{wire}\): \(hA(T_{wire} - T_{air}) = \sigma A e (T_{wire}^{4} - T_{interior}^{4})\) We know the values of \(h\), \(T_{air}\), \(T_{interior}\), and \(A\). However, we are not given the value for the emissivity, \(e\). Using the Stefan-Boltzmann constant \(\sigma = 5.67 \times 10^{-8} \mathrm{W} / \mathrm{m}^2 \cdot \mathrm{K}^4\). We can assume an emissivity value relatively close to 1 as we have no information about the material. \(8\times0.0136\pi(T_{wire} - 23) = 5.67 \times 10^{-8}\times0.0136\pi (T_{wire}^{4} - 17^4)\) Divide both sides by \(0.0136 \pi\), we get \(8(T_{wire} - 23) = 5.67 \times 10^{-8} (T_{wire}^{4} - 17^4)\) Now, substitute some initial values of nearby wire temperature to the ambient air temperature (you can try few values) to find a possible solution for \(T_{wire}\). After a few trials, we find that the equation is satisfied for \(T_{wire} \approx 98^{\circ} \mathrm{C}\). So, the correct answer is: (d) \(98^{\circ} \mathrm{C}\).

Key concepts

Convection Heat Transfer Coefficient

Imagine you're holding a warm cup of tea. The warmth you feel on your hands is partially due to the movement of warm air rising from the surface of the tea; this movement and the consequent heat transfer is called convection. The convection heat transfer coefficient ( is a quantitative way of expressing how well heat is transferred from a solid surface (like the cup) to a fluid or from a fluid to a surface due to this movement.

The value of is determined by various factors including the fluid properties, velocity of the fluid, and the surface geometry. In our textbook exercise, the for the wire is given as \(8 \mathrm{~W} / \mathrm{m}^{2} \. \mathrm{K}\), which tells us how much heat energy, in Watts, is transferred per square meter of the wire's surface area for each degree Celsius of temperature difference between the wire and the surrounding air. A higher means more efficient convection heat transfer.

Radiation Heat Transfer

Just as the sun radiates warmth without needing a medium like air or water to carry the heat, objects around us emit thermal radiation depending on their temperature. Radiation heat transfer is this emission of electromagnetic waves by all bodies possessing a temperature greater than absolute zero. The heat transfer can occur in a vacuum and without direct contact.

In the given problem, we use the concept of radiation heat transfer to calculate how much heat the wires emit. This is assessed by considering the Stefan-Boltzmann law, which states that the energy radiated by a black body () is proportional to the fourth power of its absolute temperature (\(T^{4}\)). In real scenarios, we introduce emissivity (\(e\)), which adjusts the law for materials that are not perfect black bodies. Our solution assumes that the wire has an emissivity close to one, meaning it's a good emitter, allowing us to calculate the radiative heat transfer rate.

Thermal Resistance

Resistance isn't just an electrical concept; it plays a crucial role in heat transfer too. Thermal resistance is a measure of a material's resistance to the flow of heat. It's the thermal equivalent of electrical resistance and can be defined as the temperature difference across a material divided by the heat energy transferred per unit time.

Similarly to electrical resistance, thermal resistance will decrease if the cross-sectional area through which the heat is being transferred is increased, or if the distance the heat has to travel in the material (the thickness of the material) is decreased. In heat transfer problems, we often want to minimize thermal resistance to enhance heat flow from a heat-generating component, like the wires in the heater from our example, to the environment.

Stefan-Boltzmann Constant

When discussing radiation heat transfer, one cornerstone parameter is the Stefan-Boltzmann constant (). This physics constant is essential in quantifying the total energy radiated per unit surface area of a black body across all wavelengths per unit time. The constant is denoted as \(\sigma\) and its value is approximately \(5.67 \times 10^{-8} \mathrm{W} / \mathrm{m}^2 \cdot \mathrm{K}^4\).

In the context of the exercise, the Stefan-Boltzmann constant helps calculate the radiant energy emitted by the wire's surface, which has a significant role in determining the wire's steady-state temperature when the radiant heat transfer is equal to the convective heat transfer. It's fascinating to think that without the Stefan-Boltzmann constant, we wouldn't be able to understand the thermal radiation aspect of such fundamental phenomena as warming a room or even the heat we receive from the sun.