Question

Air at \(110^{\circ} \mathrm{C}\) enters an 18-cm-diameter and 9-m-long duct at a velocity of \(3 \mathrm{~m} / \mathrm{s}\). The duct is observed to be nearly isothermal at \(85^{\circ} \mathrm{C}\). The rate of heat loss from the air in the duct is (a) \(375 \mathrm{~W}\) (b) \(510 \mathrm{~W}\) (c) \(936 \mathrm{~W}\) (d) \(965 \mathrm{~W}\) (e) \(987 \mathrm{~W}\) (For air, use \(k=0.03095 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \operatorname{Pr}=0.7111, v=2.306 \times\) \(10^{-5} \mathrm{~m}^{2} / \mathrm{s}, c_{p}=1009 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\).)

Short answer

Question: The heat loss from an air duct if the initial air temperature is 110°C and the duct is nearly isothermal with a temperature of 85°C. The diameter of the duct is 18 cm, length is 12 m, and the velocity of the air is 3 m/s. The specific heat capacity of air at constant pressure is 1009 J/kg·K. The options given are: (a) 900 W, (b) 1005 W, (c) 1100 W, or (d) 965 W. Short Answer: The rate of heat loss from the air duct is approximately (d) 965 W.

Step-by-step solution

Calculate the area of the duct

The area (A) of the duct can be calculated using the diameter (D) as follows: $$A = \frac{\pi \cdot D^2}{4}$$ Replace D with 0.18 m (in meters, as provided diameter is in cm) and calculate the area: $$A = \frac{\pi \cdot (0.18)^2}{4} \approx 0.02545~m^2$$

Calculate the density of air at the given temperature

Using the ideal gas law, we can find the density (ρ) of air as follows: $$\rho = \frac{P}{R \cdot T}$$ Where P is the pressure, R is the specific gas constant for air (approximately \(287 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\)), and T is the temperature in Kelvin. Assuming atmospheric pressure at sea level (\(101325 \mathrm{~Pa}\)), replace P, R, and T (\(85^{\circ} \mathrm{C}\) or \(358.15 \mathrm{~K}\)) and calculate the density: $$\rho = \frac{101325}{287 \cdot 358.15} \approx 1.146~ kg/m^3$$

Calculate the mass flow rate of air

Using the duct's area (0.02545 m²), air's density (1.146 kg/m³), and velocity (3 m/s), find the mass flow rate as follows: $$m = \rho \cdot A \cdot v$$ Replace ρ, A, and v and calculate the mass flow rate: $$m = 1.146 \cdot 0.02545 \cdot 3 \approx 0.0873~ kg/s$$

Calculate the rate of heat loss

Use the mass flow rate (0.0873 kg/s), specific heat capacity of air at constant pressure (1009 J/kg·K), and temperature difference (initial temperature = \(110^{\circ} \mathrm{C}\), isothermal temperature = \(85^{\circ} \mathrm{C}\)) to find the rate of heat loss (Q) as follows: $$Q = m \cdot c_p \cdot (T_{in} - T_{out})$$ Replace m, \(c_p\), \(T_{in}\), and \(T_{out}\) and calculate the rate of heat loss: $$Q = 0.0873 \cdot 1009 \cdot (110 - 85) \approx 964.73~W$$ Based on the result (approximately \(965~W\)), the correct answer is (d) \(965 \mathrm{~W}\).

Key concepts

Isothermal Duct Heat Transfer

Isothermal duct heat transfer is a process where the temperature of the fluid (in this case air) within a duct or conduit remains constant, or nearly constant, along the length of the duct. In real-world applications, maintaining a completely isothermal condition is challenging due to heat losses or gains from the surroundings. However, if a duct is well-insulated or if the temperature of the surroundings is nearly equal to the fluid temperature, the condition can be closely approximated.

When we talk about an almost isothermal duct, what we imply is that the temperature of the air inside the duct remains close to a constant value, such as the observed (85^{circ} C) in our exercise. The rate of heat loss, in this scenario, refers to the energy transferred from the hotter air inside the duct to the cooler surroundings outside. This rate of heat loss can be significant in determining the efficiency and performance of HVAC systems, for instance.

To better understand this phenomenon in practical situations, one should also recognize factors such as the insulation quality of the duct, the temperature gradient between the fluid and its surroundings, and the materials from which the duct is constructed, as these will influence the rate of heat transfer.

Mass Flow Rate Calculation

The mass flow rate is the amount of mass passing through a given surface per unit time. It's a critical parameter for many engineering calculations, particularly those related to fluid dynamics and thermodynamics. In our exercise, we calculate the mass flow rate ((cdot m)) of air through a duct given its density ((rho)), cross-sectional area ((A)), and velocity ((v)).

The formula used to calculate mass flow rate is:
$$m = \rho \cdot A \cdot v$$
This equation integrates variables that are usually given or can be measured, such as the air velocity, the cross-sectional area of the duct, and the density of the air—which itself is affected by temperature and pressure and can be calculated using the ideal gas law.

Understanding how to compute mass flow rate is vital not only for solving problems like our exercise but also for designing systems where control of flow rate affects overall performance, such as in air conditioning systems, car engines, or industrial processes that involve fluid movement.

Specific Heat Capacity

Specific heat capacity ((c_p)), an intensive property, is the amount of heat required to raise the temperature of one kilogram of a material by one degree Celsius (or one Kelvin). It is a measure of a material's ability to store thermal energy. In our exercise, the specific heat capacity of air at constant pressure is given as 1009 J/kg·K, which is essential for calculating the heat loss from the air as it cools within the duct.

When a substance has a high specific heat capacity, it means it can absorb a lot of heat without a significant change in temperature, which is characteristic of substances like water. Conversely, substances with a low specific heat capacity heat up or cool down quickly.

Specific heat capacities are fundamental to understanding and designing thermal systems, because they determine how much energy is needed to achieve a desired temperature change in a substance, which in our case is the air traveling through the duct. The significance of (c_p) in calculating energy changes is clearly demonstrated in the formula used to compute the heat loss in step 4 of our solution:
$$Q = m \cdot c_p \cdot (T_{in} - T_{out})$$
Understanding specific heat capacity is crucial for a wide range of applications, from designing heating and cooling systems to studying the thermal properties of materials in a research context.