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Air ( 1 atm) entered into a 5-mm-diameter circular tube at an average velocity of \(5 \mathrm{~m} / \mathrm{s}\). The tube wall is maintained at a constant surface temperature. Determine the convection heat transfer coefficient for \((a)\) a 10 -cm-long tube and \((b)\) a \(50-\mathrm{cm}-\) long tube. Evaluate the air properties at \(50^{\circ} \mathrm{C}\).

Short Answer

Expert verified
Answer: The convection heat transfer coefficient for both a 10 cm long tube and a 50 cm long tube at an air temperature of 50掳C and an average velocity of 5 m/s is approximately 21.96 W/(m虏路K).

Step by step solution

01

Calculate Reynolds number

First, we need to calculate the Reynolds number (Re) for the laminar flow of air inside the circular tube. The formula for Reynolds number is: $$Re = \frac{蟻VD}{渭}$$ where: \(蟻\) = density of the air (\(\mathrm{kg/m^3}\)), \(V\) = average velocity of the air (\(\mathrm{m/s}\)), \(D\) = diameter of the tube (\(\mathrm{m}\)), \(渭\) = dynamic viscosity of the air (\(\mathrm{kg/(m路s)}\)). To calculate the air properties at \(50^{\circ} \mathrm{C}\), we can refer to standard air property tables or use online property calculators. At \(50^{\circ} \mathrm{C}\), we have: \(蟻 = 1.164 \mathrm{~kg/m^3}\) \(渭 = 2.04 脳 10^{-5} \mathrm{~kg/(m路s)}\) Therefore, the Reynolds number is: $$Re = \frac{(1.164)(5.0)(0.005)}{2.04 脳 10^{-5}} \approx 1426$$
02

Determine Nusselt number

Since the flow is laminar, we can use the following formula to determine the Nusselt number (Nu) for a circular tube with a constant surface temperature: $$Nu = 3.66$$
03

Calculate the convection heat transfer coefficient

To find the convection heat transfer coefficient (h), we can use the formula: $$h = \frac{Nu 脳 k}{D}$$ where: \(k\) = thermal conductivity of the air (\(\mathrm{W/(m路K)}\)). At \(50^{\circ} \mathrm{C}\), the thermal conductivity of air is: \(k = 0.030 \mathrm{~W/(m路K)}\) For both part \((a)\) and \((b)\), the diameter (D) of the tube is the same (\(0.005 \mathrm{~m}\)), and the Nusselt number (Nu) remains constant since it does not depend on the tube length. Hence, the convection heat transfer coefficient (h) will be the same for both cases. $$h = \frac{3.66 脳 0.030}{0.005} \approx 21.96 \mathrm{~W/(m^2路K)}$$ In conclusion, the convection heat transfer coefficient for both \((a)\) a \(10 \mathrm{~cm}\) long tube and \((b)\) a \(50 \mathrm{~cm}\) long tube is approximately \(21.96 \mathrm{~W/(m^2路K)}\).

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Key Concepts

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

Reynolds Number
When air or any fluid moves through a pipe, the nature of its flow鈥攚hether it's smooth, called laminar flow, or rough and mixed up, known as turbulent flow, is critically important in understanding how heat is transferred. This is where the Reynolds number (Re) comes into play.

Reynolds number is a dimensionless quantity used in fluid mechanics to predict the flow regime of a fluid in a conduit. It is defined by the formula:
\[\begin{equation}Re = \frac{\rho VD}{\mu}\end{equation}\]where \(\rho\) is the density of the fluid, V is the velocity, D is a characteristic dimension such as the diameter of the tube, and \(\mu\) is the dynamic viscosity of the fluid.

For the given exercise, the calculated Reynolds number suggests a laminar flow, as it is below the threshold value which is generally around 2000. In laminar flows, the motion of the particles of the fluid is very orderly with all particles moving in straight lines parallel to the walls of the pipe. This orderly flow affects how heat is dispersed and thus influences the heat transfer rate.
Nusselt Number
The Nusselt number (Nu) is a dimensionless parameter closely related to convection heat transfer. It characterizes the enhancement of heat transfer through a fluid layer as a result of convection relative to conduction across the same fluid layer.

In simple terms, the Nusselt number can be understood as a ratio that compares the actual heat transfer to the heat transfer it would be if it occurred purely through conduction. The formula for the Nusselt number in a laminar flow inside a tube, which is independent of the length of the tube, is given as:
\[\begin{equation}Nu = 3.66\end{equation}\]
This value indicates a balance between convection and conduction processes in the heat transfer mechanism. For the given problem, since the flow is laminar and the surface temperature of the tube is constant, the use of the above Nusselt number is appropriate.
Laminar Flow
The term laminar flow describes a flow regime characterized by high order and low turbulence. In a laminar flow, fluid particles move in smooth paths in layers, with little or no mixing between the layers. Imagine a deck of cards smoothly sliding over one another鈥攖hat's how the fluid layers move in a laminar flow.

Laminar flow is typically found when fluids move slowly or through smaller non-complex passages. It can significantly influence the heat transfer process. The presence of laminar flow tends to facilitate a predictable pattern of heat distribution, which can be advantageous in precise thermal management scenarios. Comparatively, turbulent flow, which is characterized by chaotic changes in pressure and flow velocity, leads to enhanced mixing and potentially higher rates of heat transfer, but with reduced predictability and control.
Thermal Conductivity
An important player in the heat transfer process is thermal conductivity (k), a physical property that indicates a material's ability to conduct heat. Essentially, it represents how easily heat passes through a material. The thermal conductivity of a substance depends on its molecular structure and temperature.

In our given exercise, the thermal conductivity of air at 50 degrees Celsius is a known constant, allowing us to calculate the convection heat transfer coefficient using the Nusselt number. Materials with higher thermal conductivity, such as metals, transfer heat quickly, while materials with low thermal conductivity, such as gases or insulators, transfer heat more slowly.

Understanding thermal conductivity is crucial because it directly impacts the convection heat transfer coefficient, which represents how efficiently a surface (like the tube wall in our exercise) can transfer heat to a fluid (the air in this case). Higher thermal conductivity typically leads to potentially higher heat transfer coefficients, facilitating more effective heat dissipation.

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Most popular questions from this chapter

Water at \(15^{\circ} \mathrm{C}\) is flowing through a 5 -cm-diameter smooth tube with a length of \(200 \mathrm{~m}\). Determine the Darcy friction factor and pressure loss associated with the tube for (a) mass flow rate of \(0.02 \mathrm{~kg} / \mathrm{s}\) and \((b)\) mass flow rate of \(0.5 \mathrm{~kg} / \mathrm{s}\).

Hot water at \(90^{\circ} \mathrm{C}\) enters a \(15-\mathrm{m}\) section of a cast iron pipe \((k=52 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) whose inner and outer diameters are 4 and \(4.6 \mathrm{~cm}\), respectively, at an average velocity of \(1.2 \mathrm{~m} / \mathrm{s}\). The outer surface of the pipe, whose emissivity is \(0.7\), is exposed to the cold air at \(10^{\circ} \mathrm{C}\) in a basement, with a convection heat transfer coefficient of \(12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Taking the walls of the basement to be at \(10^{\circ} \mathrm{C}\) also, determine \((a)\) the rate of heat loss from the water and \((b)\) the temperature at which the water leaves the basement.

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) \(900 \mathrm{~W}\)

Laid water is flowing between two very thin parallel 1 -m-wide and 10 -m-long plates with a spacing of \(12.5 \mathrm{~mm}\). The water enters the parallel plates at \(20^{\circ} \mathrm{C}\) with a mass flow rate of \(0.58 \mathrm{~kg} / \mathrm{s}\). The outer surface of the parallel plates is subjected to hydrogen gas (an ideal gas at \(1 \mathrm{~atm}\) ) flow width-wise in parallel over the upper and lower surfaces of the two plates. The free-stream hydrogen gas has a velocity of \(5 \mathrm{~m} / \mathrm{s}\) at a temperature of \(155^{\circ} \mathrm{C}\). Determine the outlet mean temperature of the water, the surface temperature of the parallel plates, and the total rate of heat transfer. Evaluate the properties for water at \(30^{\circ} \mathrm{C}\) and the properties of \(\mathrm{H}_{2}\) gas at \(100^{\circ} \mathrm{C}\). Is this a good assumption?

Water at \(15^{\circ} \mathrm{C}\) is flowing through a \(200-\mathrm{m}\)-long standard 1-in Schedule 40 cast iron pipe with a mass flow rate of \(0.5 \mathrm{~kg} / \mathrm{s}\). If accuracy is an important issue, use the appropriate equation to determine \((a)\) the pressure loss and \((b)\) the pumping power required to overcome the pressure loss. Assume flow is fully developed. Is this a good assumption?

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