Question

Heat is generated in a long \(0.3-\mathrm{cm}\)-diameter cylindrical electric heater at a rate of \(180 \mathrm{~W} / \mathrm{cm}^{3}\). The heat flux at the surface of the heater in steady operation is (a) \(12.7 \mathrm{~W} / \mathrm{cm}^{2}\) (b) \(13.5 \mathrm{~W} / \mathrm{cm}^{2}\) (c) \(64.7 \mathrm{~W} / \mathrm{cm}^{2}\) (d) \(180 \mathrm{~W} / \mathrm{cm}^{2}\) (e) \(191 \mathrm{~W} / \mathrm{cm}^{2}\)

Short answer

The heat flux at the surface of the cylindrical electric heater in steady operation is 13.5 W/cm².

Step-by-step solution

Calculate the radius of the cylindrical electric heater

Given the diameter of the cylindrical heater is \(0.3\mathrm{~cm}\), so the radius (r) will be half of it: \(r = \frac{0.3}{2} = 0.15\mathrm{~cm}\)

Calculate the heat generated per unit length of the heater

The heat is generated at a rate of \(180\mathrm{~W}/\mathrm{cm}^3\). To find the heat generated per unit length, we can multiply by cross-sectional area of the heater: Heat generated per unit length (Q_l) \(= 180 \times \pi r^2 = 180 \times \pi \times (0.15)^2 \mathrm{~W}/\mathrm{cm}\)

Calculate the surface area of the heater per unit length

The surface area of the heater per unit length can be found by calculating the lateral surface area of a cylinder with unit height: Surface area per unit length (A_l) \(= 2 \times \pi \times r = 2 \times \pi \times 0.15\mathrm{~cm}^2/\mathrm{cm}\)

Calculate the heat flux at the surface of the heater

The heat flux (HF) at the surface of the heater can be found by dividing the heat generated per unit length by the surface area per unit length: \(HF = \frac{Q_l}{A_l}=\frac{180 \times \pi \times (0.15)^2}{2 \times \pi \times 0.15}\) Now, we can cancel out \(\pi\) and \(0.15\) from the numerator and the denominator. Hence, we have: \(HF = \frac{180 \times (0.15)}{2}\) Solve for HF: \(HF = 13.5\mathrm{~W}/\mathrm{cm}^2\) So, the heat flux at the surface of the heater in steady operation is: (b) \(13.5\mathrm{~W}/\mathrm{cm}^2\).

Key concepts

Thermal Analysis

Thermal analysis is the process used to determine the heat transfer characteristics of materials and systems. This is essential for understanding how heat moves through objects, such as the cylindrical electric heater in our example. By using the principles of thermal analysis, we calculate the heat generation rate and how it's distributed across the material's surface. This involves understanding how heat is conducted through materials and what the thermal properties, such as thermal conductivity and specific heat capacity, mean practically for different substances and scenarios. In our case, for the cylindrical heater, the main goal of thermal analysis was to ascertain the heat flux at the heater's surface during steady operation, a critical factor in ensuring its safe and efficient performance.

By analyzing the heater's dimensional data and the rate of heat generation, the appropriate mathematical model is used to calculate the heat flux. This is vital in real-world applications where this information influences designs, safety measures, and the performance of heating systems.

Cylindrical Heater

A cylindrical heater, like the one described in our exercise, is a type of heat generating device with a circular cross-section. The design is robust, provides uniform heating, and is common in many industrial and domestic settings. Understanding the specifics of cylindrical heaters help with predicting their behavior under various operational conditions.

For our calculations, the dimension of the heater (its diameter and resulting radius) were important figures as they directly influenced the determination of cross-sectional and surface areas, both of which are key factors in computing the heat generated per unit length and the surface heat flux. Notably, in the context of our problem-solving steps, we emphasized the heater's surface area as it relates to its efficacy in dissipating heat energy, an insight pivotal to thermal management in practical applications.

Heat Transfer

Heat transfer, in the framework of our cylindrical heater problem, refers to the way heat energy moves from the heater to its environment. There are three fundamental modes of heat transfer: conduction, convection, and radiation. In a solid material like our cylindrical heater, conduction is the primary mode, where heat diffuses through the material itself.

In calculating the heat flux, which is the rate of heat transfer per unit area, we effectively measure how well the heater transfers heat to its surroundings through its surface. This quantity is pivotal as it impacts the heater's temperature and operational safety. High heat flux might mean that the heater could reach high surface temperatures, necessitating detailed analysis to prevent overheating or damage to adjacent materials. Our solution took into account these aspects to find a dependable value for the heat flux for steady-state operation.

Steady Operation

Steady operation, in the context of our analysis, means that the heat generation and transfer are constant over time. For our cylindrical electric heater, this implies a stable operating condition where thermal inputs and outputs are balanced. When a system remains in steady operation, the temperature distribution within it does not change with time. This is critical for long-term reliability and performance as it helps prevent thermal fatigue and material failure.

The assumption of steady operation simplifies our calculations by ensuring that the heat flux across the heater's surface remains uniform, making our analysis less complex but still meaningful. This underlines the importance of achieving and maintaining steady operation in practical thermal systems for both safety and functionality. The calculated heat flux can then be used to guide the design of thermal management systems, ensuring that equipment operates within safe temperature limits.