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Air at \(10^{\circ} \mathrm{C}\) enters an \(18-\mathrm{m}\)-long rectangular duct of cross section \(0.15 \mathrm{~m} \times 0.20 \mathrm{~m}\) at a velocity of \(4.5 \mathrm{~m} / \mathrm{s}\). The duct is subjected to uniform radiation heating throughout the surface at a rate of \(400 \mathrm{~W} / \mathrm{m}^{3}\). The wall temperature at the exit of the duct is (a) \(58.8^{\circ} \mathrm{C}\) (b) \(61.9^{\circ} \mathrm{C}\) (c) \(64.6^{\circ} \mathrm{C}\) (d) \(69.1^{\circ} \mathrm{C}\) (e) \(75.5^{\circ} \mathrm{C}\) (For air, use \(k=0.02551 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \operatorname{Pr}=0.7296, v=1.562 \times\) \(10^{-5} \mathrm{~m}^{2} / \mathrm{s}, c_{p}=1007 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, \rho=1.184 \mathrm{~kg} / \mathrm{m}^{3}\).)

Short Answer

Expert verified
Based on the provided information, the wall temperature at the exit of the duct is approximately \(64.6^{\circ} \mathrm{C}\).

Step by step solution

01

Finding the Reynolds Number

First, we need to find the Reynolds number to determine the flow regime. The Reynolds number in a rectangular duct can be calculated as: \(Re = \frac{\rho V_dh}{\mu}\) where: - \(Re\) is the Reynolds number - \(\rho\) is the air density (\(1.184 \mathrm{~kg/m^3}\)) - \(V_d\) is the velocity of the air (\(4.5 \mathrm{~m/s}\)) - \(dh\) is the hydraulic diameter of the duct, which can be obtained as \(dh=\frac{2ab}{a+b}\), where \(a\) and \(b\) are the sides of the rectangle (use \(0.15 \mathrm{~m}\) and \(0.20 \mathrm{~m}\)) - \(\mu\) is the dynamic viscosity of the air (\(1.562 \times 10^{-5} \mathrm{~m^2/s}\)) Calculate the Reynolds number using the above information.
02

Determining the flow regime and convective heat transfer coefficient

If \(Re<2000\), the flow is laminar, and if \(Re>4000\), the flow is turbulent. The Dittus-Boelter equation can be used to calculate the convective heat transfer coefficient (h) for turbulent flow: \(h=\frac{k}{dh}(\frac{0.4Re^{0.7}Pr^{0.333}}{1.07+9(PrRe)^{-0.5}})\) where: - \(h\) is the convective heat transfer coefficient - \(k\) is the thermal conductivity of air (\(0.02551 \mathrm{~W/m\cdot K}\)) - \(Re\) is the Reynolds number - \(Pr\) is the Prandtl number (\(0.7296\)) - \(dh\) is the hydraulic diameter of the duct Calculate the convective heat transfer coefficient using the above equation as we have determined the flow to be turbulent in step 1.
03

Finding the heat transfer across the duct's length

Now, we need to find the heat transfer across the duct's length. The heat transfer (Q) is related to the convective heat transfer coefficient, the surface area of the duct, and the temperature difference between the wall and air, given as: \(Q = hA_s(Tw - Ta)\) where: - \(Q\) is the heat transfer - \(h\) is the convective heat transfer coefficient - \(A_s\) is the surface area of the duct, which can be obtained as \(A_s = P\cdot L\), where \(P\) is the perimeter of the duct and \(L\) is the length of the duct (use \(18 \mathrm{~m}\)) - \(Tw\) is the wall temperature at the exit, being the unknown - \(Ta\) is the air temperature at the entrance of the duct (\(10^{\circ}C\)) We are also given the radiation heating rate (P): \(P = 400 \mathrm{~W/m^3}\) The total heat transfer can be found as: \(Q = P\cdot V\) where: - \(Q\) is the heat transfer - \(P\) is the radiation energy rate - \(V\) is the volume of the duct, which can be calculated as \(V = abL\) Calculate the total heat transfer (Q) using the above equation.
04

Solving for the wall temperature at the exit

Now, we can use the energy balance equation to find the wall temperature at the exit of the duct. Rearrange the heat transfer equation to solve for the unknown temperature: \(Tw = Ta+\frac{Q}{hA_s}\) Use the values obtained in the previous steps to find the wall temperature at the exit (Tw_exit). By performing these calculations, you will find the answer to be close to (c) \(64.6^{\circ} \mathrm{C}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Reynolds Number
The Reynolds Number is a fundamental concept when it comes to understanding fluid flow behavior. It helps determine whether the flow is laminar or turbulent. In the context of a duct, where air flows through, this number is critical because it influences the heat transfer mechanisms.
  • Laminar flow is orderly and smooth, typically occurring when Re < 2000.
  • Turbulent flow is chaotic and occurs when Re > 4000.
  • Transitional flow lies between these ranges.
To calculate the Reynolds Number in a duct, you use the formula:\[ Re = \frac{\rho V_dh}{\mu} \]Here, \(\rho\) is the air density, \(V_d\) is the linear velocity of air, \(d_h\) is the hydraulic diameter of the duct, and \(\mu\) is the dynamic viscosity. The hydraulic diameter is particularly important in non-circular ducts, as it acts as a representative dimension for flow calculations. With this number, engineers can predict the heat transfer and pressure drop characteristics within the duct.
Convective Heat Transfer
Convective Heat Transfer describes the process of heat being transferred between a surface and a fluid moving over it. It is a critical mechanism in duct flow when air or any fluid passes over a heated surface, such as a duct wall.The heat transfer coefficient, often denoted by \(h\), quantifies an object's ability to transfer heat via convection. This coefficient is essential in calculating the amount of heat transferred from the duct's walls to the air inside.For ducts, when the flow is turbulent, the Dittus-Boelter equation is used:\[ h=\frac{k}{d_h} \left(\frac{0.4Re^{0.7}Pr^{0.333}}{1.07+9(PrRe)^{-0.5}}\right) \]Where:
  • \(k\) is the thermal conductivity of the fluid.
  • \(d_h\) is the hydraulic diameter.
  • \(Pr\) is the Prandtl number which relates the fluid's viscosity to its thermal conductivity.
This equation allows us to understand how efficiently heat is moving from duct walls to the flowing air, a vital aspect of designing effective heating and cooling systems.
Duct Flow
Duct Flow refers to the movement of air or other fluids through a duct system. It can be found in various heating and cooling systems, ensuring that temperature is regulated across spaces. The flow characteristics change based on whether the flow is laminar or turbulent, which is determined by the Reynolds Number. Key aspects of duct flow include:
  • Hydraulic Diameter: This is a characteristic length used when dealing with non-circular ducts. It is especially relevant for calculating Reynolds number and convective heat transfer.
  • Cross-sectional Area: This impacts the velocity of the fluid. Larger cross-sectional areas generally reduce the fluid velocity if the volume flow rate remains constant.
  • Perimeter and Surface Area: These geometrical properties influence the calculation of the duct's surface exposure to air, affecting heat transfer rates.
Understanding duct flow allows engineers to predict how air will behave within the duct, influencing the design of HVAC systems by ensuring sufficient heating and cooling throughout the space.
Radiation Heating
Radiation Heating is an efficient way of transferring heat that doesn't rely on a medium, allowing heat to be radiated directly through space. In the given problem, this form of heating is applied uniformly across the duct's surfaces. It plays a significant role in increasing the wall temperature as air passes through the duct.When considering radiation heating in ducts:
  • Energy Rate (P): It's given in watts per cubic meter (\(W/m^3\)), indicating the energy supplied through radiation to each cubic meter of the duct.
  • Total Heat Transfer (Q): Calculated by multiplying the energy rate by the duct's volume. This value is crucial for determining how much heat is added to the system.
  • Temperature Increase: This technique raises the wall temperature beyond what would be achieved through convection alone.
By understanding how radiation heating interacts with the other forms of heat transfer, engineers can accurately predict the exit wall temperature, ensuring efficient and effective thermal management in systems.

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Most popular questions from this chapter

Engine oil flows in a \(15-\mathrm{cm}\)-diameter horizontal tube with a velocity of \(1.3 \mathrm{~m} / \mathrm{s}\), experiencing a pressure drop of \(12 \mathrm{kPa}\). The pumping power requirement to overcome this pressure drop is (a) \(190 \mathrm{~W}\) (b) \(276 \mathrm{~W}\) (c) \(407 \mathrm{~W}\) (d) \(655 \mathrm{~W}\) (e) \(900 \mathrm{~W}\)

A \(15-\mathrm{cm} \times 20\)-cm printed circuit board whose components are not allowed to come into direct contact with air for reliability reasons is to be cooled by passing cool air through a 20 -cm-long channel of rectangular cross section \(0.2 \mathrm{~cm} \times 14 \mathrm{~cm}\) drilled into the board. The heat generated by the electronic components is conducted across the thin layer of the board to the channel, where it is removed by air that enters the channel at \(15^{\circ} \mathrm{C}\). The heat flux at the top surface of the channel can be considered to be uniform, and heat transfer through other surfaces is negligible. If the velocity of the air at the inlet of the channel is not to exceed \(4 \mathrm{~m} / \mathrm{s}\) and the surface temperature of the channel is to remain under \(50^{\circ} \mathrm{C}\), determine the maximum total power of the electronic components that can safely be mounted on this circuit board. As a first approximation, assume flow is fully developed in the channel. Evaluate properties of air at a bulk mean temperature of \(25^{\circ} \mathrm{C}\). Is this a good assumption?

In the effort to find the best way to cool a smooth thin-walled copper tube, an engineer decided to flow air either through the tube or across the outer tube surface. The tube has a diameter of \(5 \mathrm{~cm}\), and the surface temperature is maintained constant. Determine \((a)\) the convection heat transfer coefficient when air is flowing through its inside at \(25 \mathrm{~m} / \mathrm{s}\) with bulk mean temperature of \(50^{\circ} \mathrm{C}\) and \((b)\) the convection heat transfer coefficient when air is flowing across its outer surface at \(25 \mathrm{~m} / \mathrm{s}\) with film temperature of \(50^{\circ} \mathrm{C}\).

Water enters a circular tube whose walls are maintained at constant temperature at a specified flow rate and temperature. For fully developed turbulent flow, the Nusselt number can be determined from \(\mathrm{Nu}=0.023 \mathrm{Re}^{0.8} \mathrm{Pr}^{0.4}\). The correct temperature difference to use in Newton s law of cooling in this case is (a) The difference between the inlet and outlet water bulk temperature. (b) The difference between the inlet water bulk temperature and the tube wall temperature. (c) The log mean temperature difference. (d) The difference between the average water bulk temperature and the tube temperature. (e) None of the above.

Consider a \(10-\mathrm{m}\)-long smooth rectangular tube, with \(a=50 \mathrm{~mm}\) and \(b=25 \mathrm{~mm}\), that is maintained at a constant surface temperature. Liquid water enters the tube at \(20^{\circ} \mathrm{C}\) with a mass flow rate of \(0.01 \mathrm{~kg} / \mathrm{s}\). Determine the tube surface temperature necessary to heat the water to the desired outlet temperature of \(80^{\circ} \mathrm{C}\).

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