Question

Air \(\left(c_{p}=1007 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) enters a 17 -cm-diameter and 4-m-long tube at \(65^{\circ} \mathrm{C}\) at a rate of \(0.08 \mathrm{~kg} / \mathrm{s}\) and leaves at \(15^{\circ} \mathrm{C}\). The tube is observed to be nearly isothermal at \(5^{\circ} \mathrm{C}\). The average convection heat transfer coefficient is (a) \(24.5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (b) \(46.2 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (c) \(53.9 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (d) \(67.6 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (e) \(90.7 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\)

Short answer

Based on the calculated average convection heat transfer coefficient of 18.8 W/m²·K, the closest answer choice is: (a) 24.5 W/m²·K

Step-by-step solution

Calculate the heat transfer rate

(Q) Using the formula: Q = m_dot * cp * (T_inlet - T_outlet), where m_dot is the mass flow rate, cp is the specific heat capacity of air, T_inlet and T_outlet are the inlet and outlet temperatures of the air. Q = 0.08 kg/s * 1007 J/kg·K * (65°C - 15°C) Q = 0.08 * 1007 * (50) Q = 4028 W

Calculate the surface area of the tube (A)

Using the formula: A = 2 * pi * r * L, where r is the radius and L is the length of the tube. L = 4 m D = 0.17 m R = D / 2 = 0.085 m A = 2 * pi * 0.085 m * 4 m A ≈ 2.1376 m²

Compute the temperature difference

Calculate the temperature difference between the air outlet temperature and the tube wall temperature. ΔT = T_outlet - T_wall = 15°C - 5°C ΔT = 10 K

Calculate the average convection heat transfer coefficient

(h) Using the formula: Q = h * A * ΔT, we can solve for h by dividing Q by the product of A and ΔT. h = Q / (A * ΔT) h = 4028 W / (2.1376 m² * 10 K) h ≈ 18.8 W/m²·K Comparing this value with the given answer choices, we find the closest one to be: (a) 24.5 W/m²·K

Key concepts

Heat Transfer Rate

Understanding the heat transfer rate is crucial when dealing with thermodynamics and heating or cooling processes. The heat transfer rate, denoted as Q, represents the amount of thermal energy transferred per unit of time. It is calculated using the simple but fundamental equation:

\[ Q = m_\dot \times c_p \times (T_{\text{inlet}} - T_{\text{outlet}}) \]
where \( m_\dot \) is the mass flow rate of the fluid, \( c_p \) is the specific heat capacity of the fluid at constant pressure, and \( T_{\text{inlet}} \) and \( T_{\text{outlet}} \) are the inlet and outlet temperatures of the fluid, respectively. This equation allows us to calculate the energy required to heat or cool a fluid as it moves through a system—a key concept in design and analysis of heating, ventilation, and air conditioning (HVAC) systems, among others.

In practical terms, the greater the mass flow rate or the larger the temperature change, the higher the heat transfer rate. For example, if a heating system is too slow, increasing the flow rate of the hot fluid or the temperature difference can improve its performance.

Surface Area Calculation

The surface area of any object is a measure of the total area that the surface of the object occupies. Calculating the surface area is essential when you need to determine how much material is required to cover a given shape or the amount of heat that can be exchanged between a surface and its environment.

For our discussion, we focus on a cylindrical tube. The surface area \( A \) of a cylinder can be calculated with the formula:
\[ A = 2 \pi r L \]
where \( r \) is the radius of the tube's cross-section, and \( L \) is its length. This equation is particularly important when dealing with problems of heat transfer, as the rate at which heat is exchanged depends on the surface area in contact with the fluid inside the tube or with the surrounding atmosphere.

From an educational standpoint, ensuring students understand how to properly determine and apply the surface area of objects is vital to correctly solving physics and engineering problems, such as understanding how different shapes and sizes affect heat transfer characteristics in a system.

Temperature Difference

Temperature difference, often represented as \( \Delta T \), is a key driver in heat transfer processes. It refers to the difference in temperature between two points—in our case, the air outlet temperature and the tube wall temperature. The temperature difference compels heat to flow from the hotter to the cooler region.

Mathematically, it's expressed as: \[ \Delta T = T_{\text{outlet}} - T_{\text{wall}} \]
where \( T_{\text{outlet}} \) is the temperature of the fluid leaving the system, and \( T_{\text{wall}} \) is the temperature of the tube's wall. In thermal systems, the larger the temperature difference, the greater the potential for heat transfer. In real-world applications, engineers must optimize this temperature difference to ensure efficient energy use without compromising the system’s integrity or performance.

Specific Heat Capacity

Specific heat capacity, denoted as \( c_p \), is a vital concept in thermodynamics that measures the amount of heat one unit of mass of a substance can absorb before its temperature rises by one degree Celsius (or one Kelvin). Essentially, it is an intrinsic property of substances that tells us how 'thermally resistant' they are to changing temperature.

Adding heat to an object with high specific heat capacity will result in a slower temperature increase compared to an object with a low specific heat capacity, assuming equal mass and heat input. For instance, water has a high \( c_p \), which is why it's used as a coolant in many applications: it can absorb a lot of heat with minimal temperature change. This property is crucial when designing systems where temperature regulation is essential. An accurate understanding of the specific heat capacity allows for more effective thermal management in systems ranging from automotive engines to climate control systems.

To enrich students' learning experience, it's beneficial to conduct practical experiments where they can measure and compare specific heat capacities of different materials and observe how this property influences thermal energy transfer.