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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 \(50^{\circ} \mathrm{C}\).

Short Answer

Expert verified
Answer: Case (a), where air flows inside the tube, generally has a higher convection heat transfer coefficient compared to case (b), where air flows across the outer surface of the tube.

Step by step solution

01

Identify given information for both cases

For both cases: Diameter of the tube, \(D = 5 \mathrm{~cm} = 0.05 \mathrm{~m}\) (converted from centimeters to meters) Air velocity, \(V = 25 \mathrm{~m/s}\) For case (a) - air flowing inside the tube: Bulk mean temperature, \(T_{a,\text{bulk}} = 50^{\circ} \mathrm{C}\) For case (b) - air flowing across the outer surface of the tube: Film temperature, \(T_{b,\text{film}} = 50^{\circ} \mathrm{C}\)
02

Determine the heat transfer rate

For both cases, we need more information to calculate the heat transfer rate. Since this information is not provided in the problem statement, we will assume that the heat transfer rates are the same for both cases. This assumption will allow us to compare the convection heat transfer coefficients between the two cases. Let the heat transfer rate be \(Q\).
03

Calculate the area for each case

For case (a) - air flowing inside the tube: The area of the inside surface of the tube, \(A = \pi D_l\) where \(D_l\) is the inner diameter of the tube. Since the tube has a thin wall, we can assume that \(D_l \approx\) 0.05 m. Thus, \(A_{a} = \pi \times (0.05\mathrm{~m})\) For case (b) - air flowing across the outer surface of the tube: The area of the outer surface of the tube, \(A = \pi D_o\) where \(D_o\) is the outer diameter of the tube. Since the tube has a thin wall, we can assume that \(D_o \approx\) 0.05 m. Thus, \(A_{b} = \pi \times (0.05\mathrm{~m})\)
04

Calculate the convection heat transfer coefficients for both cases

For both cases, we will use the formula \(h = \frac{Q}{A \cdot \Delta T}\). \(\Delta T = T_\text{surface} - T_\text{bulk/film}\) is the temperature difference between the tube surface and the bulk/film temperature of the air. We are not given the surface temperature, and may not be able to determine the temperature difference. Therefore, an accurate numeric answer for the convection heat transfer coefficients cannot be provided for either case. However, it is possible to make a qualitative comparison between the heat transfer in both cases. For case (a), the convection heat transfer is generally more effective when air flows inside the tube, as there is a more uniform and smoother flow profile, which increases the movement of heat away from the tube surface. For case (b), the flow of air on the outer surface of the tube may have disturbances and turbulence which can result in local hotspots or uneven heat transfer. Without the surface temperature and heat transfer rates required to compute the convection heat transfer coefficients for each case, we can conclude that case (a) generally has a higher convection heat transfer coefficient compared to case (b).

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Key Concepts

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

Heat Transfer Rate
The heat transfer rate is a fundamental concept in thermodynamics and heat transfer studies. It denotes the quantity of heat energy transferred per unit of time and is measured in watts (W) in the SI system. In the context of our exercise, where air is passing either through a tube or across its surface, the rate at which heat is taken away from the copper tube directly impacts the cooling effectiveness.

Understanding heat transfer rate is crucial when analyzing convection because it provides insights into the cooling or heating performance of a system. The mathematical expression for the heat transfer rate in convection is given by the equation: \begin{align*}Q = h \times A \times \text{(\(\bigtriangleup\)T)}, \begin{align*}where:
  • \(Q\) is the heat transfer rate,
  • h\( is the convection heat transfer coefficient,
  • \)A\( is the area through which heat is transferred, and
  • \)\bigtriangleup\(T\) is the temperature difference between the surface and the surrounding fluid.

When working on exercises like the one provided, careful attention must be paid to identifying the correct area and temperature difference to determine the heat transfer rate accurately.
Bulk Mean Temperature
Bulk mean temperature refers to the average temperature of a fluid, such as air or water, throughout its volume. Rather than considering the varying temperatures at different points in the fluid, the bulk mean offers a simplified approach to evaluating the fluid's temperature impact on heat transfer calculations.

In our case, when air flows inside the copper tube, the bulk mean temperature is the average temperature of the air within the tube. It is a crucial factor in determining the convective heat transfer coefficient, \(h\), as it influences the temperature difference, \(\bigtriangleup\)T, used in the heat transfer calculation. The bigger the temperature difference between the tube's surface and the bulk mean temperature of the air, the higher the potential heat transfer rate.
Film Temperature
Film temperature plays a vital role in convective heat transfer, especially when a fluid flows across a surface, as is the scenario in part (b) of our example. It is defined as the average temperature of the surface and the free stream temperature of the fluid far away from the surface. Mathematically, it can be expressed as:\begin{align*}T_{\text{film}} = \frac{T_{\text{surface}} + T_{\text{bulk}}}{2}, \begin{align*}where \(T_{\text{surface}}\) is the temperature of the surface and \(T_{\text{bulk}}\) is the temperature of the fluid far from the surface.

In our exercise, it's given that the film temperature is \(50^{\text{o}}C\), but this parameter alone isn't enough to calculate the heat transfer coefficient without the surface temperature. Still, knowing the film temperature is helpful because it directly affects the characteristics of the boundary layer, which in turn influences the convection heat transfer coefficient. The film temperature is used in various correlations to estimate the value of \(h\), for example, through Nusselt number correlations based on similar geometric and flow conditions.

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Most popular questions from this chapter

Cooling water available at \(10^{\circ} \mathrm{C}\) is used to condense steam at \(30^{\circ} \mathrm{C}\) in the condenser of a power plant at a rate of \(0.15 \mathrm{~kg} / \mathrm{s}\) by circulating the cooling water through a bank of 5 -m-long \(1.2-\mathrm{cm}\)-internal-diameter thin copper tubes. Water enters the tubes at a mean velocity of \(4 \mathrm{~m} / \mathrm{s}\) and leaves at a temperature of \(24^{\circ} \mathrm{C}\). The tubes are nearly isothermal at \(30^{\circ} \mathrm{C}\). Determine the average heat transfer coefficient between the water, the tubes, and the number of tubes needed to achieve the indicated heat transfer rate in the condenser.

Hot air at \(60^{\circ} \mathrm{C}\) leaving the furnace of a house enters a 12-m-long section of a sheet metal duct of rectangular cross section \(20 \mathrm{~cm} \times 20 \mathrm{~cm}\) at an average velocity of \(4 \mathrm{~m} / \mathrm{s}\). The thermal resistance of the duct is negligible, and the outer surface of the duct, whose emissivity is \(0.3\), is exposed to the cold air at \(10^{\circ} \mathrm{C}\) in the basement, with a convection heat transfer coefficient of \(10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Taking the walls of the basement to be at \(10^{\circ} \mathrm{C}\) also, determine \((a)\) the temperature at which the hot air will leave the basement and \((b)\) the rate of heat loss from the hot air in the duct to the basement. Evaluate air properties at a bulk mean temperature of \(50^{\circ} \mathrm{C}\). Is this a good assumption?

In a food processing plant, hot liquid water is being transported in a pipe \(\left(k=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, D_{i}=2.5 \mathrm{~cm}\right.\), \(D_{o}=3 \mathrm{~cm}\), and \(\left.L=10 \mathrm{~m}\right)\). The hot water flowing with a mass flow rate of \(0.15 \mathrm{~kg} / \mathrm{s}\) enters the pipe at \(100^{\circ} \mathrm{C}\) and exits at \(60^{\circ} \mathrm{C}\). The plant supervisor thinks that since the hot water exits the pipe at \(60^{\circ} \mathrm{C}\), the pipe's outer surface temperature should be safe from thermal burn hazards. In order to prevent thermal burn upon accidental contact with skin tissue for individuals working in the vicinity of the pipe, the pipe's outer surface temperature should be kept below \(45^{\circ} \mathrm{C}\). Determine whether or not there is a risk of thermal burn on the pipe's outer surface. Assume the pipe outer surface temperature remains constant.

The velocity profile in fully developed laminar flow of water at \(40^{\circ} \mathrm{F}\) in a 140 -ft-long horizontal circular pipe, in \(\mathrm{ft} / \mathrm{s}\), is given by \(u(r)=0.8\left(1-625 r^{2}\right)\) where \(r\) is the radial distance from the centerline of the pipe in \(\mathrm{ft}\). Determine \((a)\) the volume flow rate of water through the pipe, \((b)\) the pressure drop across the pipe, and \((c)\) the useful pumping power required to overcome this pressure drop.

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 \(k=0.03095 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \operatorname{Pr}=0.7111, v=2.306 \times\) \(10^{-5} \mathrm{~m}^{2} / \mathrm{s}, c_{p}=1009 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\).)

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