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Chapter 8: Problem 133

Water \(\left(\mu=9.0 \times 10^{-4} \mathrm{~kg} / \mathrm{m} \cdot \mathrm{s}, \rho=1000 \mathrm{~kg} / \mathrm{m}^{3}\right)\) enters a 2-cm- diameter and 3-m-long tube whose walls are maintained at \(100^{\circ} \mathrm{C}\). The water enters this tube with a bulk temperature of \(25^{\circ} \mathrm{C}\) and a volume flow rate of \(3 \mathrm{~m}^{3} / \mathrm{h}\). The Reynolds number for this internal flow is (a) 59,000 (b) 105,000 (d) 236,000 (e) 342,000 (c) 178,000

Short Answer

Expert verified
Based on the given information, the Reynolds number for internal flow of water in the tube is 236,000. This value is essential in determining the type of flow (laminar, transitional, or turbulent) and can be calculated using the formula Re = (蟻VD) / 渭.

Step by step solution

01

Calculating the flow velocity (V)

We will first convert the volume flow rate of water from cubic meters per hour(m^3/h) to cubic meters per second(m^3/s). $$ Q = 3 \frac{\mathrm{m}^3}{\mathrm{h}} \cdot \frac{1\mathrm{h}}{3600\mathrm{s}} = 8.33\times 10^{-4}\frac{\mathrm{m}^3}{\mathrm{s}} $$ The flow velocity can be determined using the equation: $$ V = \frac{Q}{A} $$ Where V is the flow velocity, Q is the volume flow rate, and A is the cross-sectional area of the tube. The area A of the tube can be calculated using the formula for the area of a circle: $$ A = \pi\left(\frac{D}{2}\right)^2 $$ Where D is the diameter of the tube (2 cm).
02

Calculating the cross-sectional area (A)

Let's first convert the diameter from centimeters to meters, and then calculate the area of the tube. $$ D = 2 \mathrm{cm} \cdot \frac{1 \mathrm{m}}{100 \mathrm{cm}} = 0.02 \mathrm{m} $$ $$ A = \pi\left(\frac{0.02 \mathrm{m}}{2}\right)^2 = 3.14 \times 10^{-4} \mathrm{m}^2 $$
03

Calculating the flow velocity (V)

Now we can calculate the flow velocity using the volume flow rate (Q) and the cross-sectional area (A). $$ V = \frac{8.33\times 10^{-4}\frac{\mathrm{m}^3}{\mathrm{s}}}{3.14 \times 10^{-4} \mathrm{m}^2} = 2.65 \frac{\mathrm{m}}{\mathrm{s}} $$
04

Calculating the Reynolds number (Re)

Finally, we can calculate the Reynolds number(Re) using the formula: $$ \begin{aligned} \operatorname{Re} &=\frac{\rho \mathrm{V} \cdot{\mathrm{D}}}{\mu} \\ &=\frac{1000 \frac{\mathrm{kg}}{\mathrm{m}^{3}} \times 2.65 \frac{\mathrm{m}}{\mathrm{s}} \times 0.02 \mathrm{m}}{9.0 \times 10^{-4} \frac{\mathrm{kg}}{\mathrm{m} \cdot \mathrm{s}}} \\ &=236000 \end{aligned} $$ The correct answer is (d) 236,000.

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Key Concepts

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

Fluid Mechanics
Fluid mechanics is a branch of physics that studies the behavior of fluids (liquids, gases, and plasmas) and the forces on them. One of its central concepts is the Reynolds number, which is a dimensionless quantity used to predict flow patterns in different fluid flow situations. The Reynolds number helps to determine whether the flow will be laminar or turbulent.

In the case of the exercise, water flowing through a tube constitutes an internal flow problem in fluid mechanics. By calculating the Reynolds number, we can infer the characteristics of the flow within the tube, which is crucial for tasks such as selecting the proper equipment for fluid transport systems, predicting heat transfer rates, and designing piping systems.
Internal Flow
Internal flow refers to the study of fluid motion within a confined boundary, such as pipes, tubes, or any channel. Unlike external flow where fluid streams over a body, internal flow is confined by the physical boundaries of the conduit. This feature critically impacts the distribution of velocity, pressure, and other properties of the fluid within it.

In our solved problem, we analyzed the water movement inside a tube, which is a classic internal flow scenario. Knowing the flow velocity and tube dimensions, we've utilized these to comprehend the behavior of the water as affected by the tube's walls and the heat transfer occurring due to the water's contact with these walls.
Engineering Education
Engineering education combines theoretical knowledge with practical skills to equip individuals with the competencies required to solve complex real-world problems. It's crucial for engineering students to learn how to apply principles of fluid mechanics, like calculation of the Reynolds number, to ascertain the performance of systems involving fluid flow.

Exercises such as the one provided help bridge the gap between theory and application. The step-by-step solution demonstrates how to perform essential calculations that will be common in many engineering tasks, ensuring students understand the underlying principles and can replicate similar computations independently.
Heat Transfer
Heat transfer is an area of thermal engineering that involves the generation, use, conversion, and exchange of thermal energy (heat) between physical systems. Heat transfer is classified into three basic types: conduction, convection, and radiation. In the context of internal flow, convection is the main mode of heat transfer, happening between the fluid (water in this case) and the tube wall.

In the solution, the water enters a tube that is maintained at a higher temperature, and this temperature difference drives the heat to transfer from the tube walls to the cooler water. Understanding this interaction and being able to quantify it using principles of fluid mechanics and heat transfer is crucial for designing efficient thermal systems, such as heating, ventilation, and air conditioning (HVAC) systems, and for analyzing energy conversion processes.

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Most popular questions from this chapter

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.

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) \(90.7 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\)

The components of an electronic system dissipating \(180 \mathrm{~W}\) are located in a \(1-\mathrm{m}\)-long horizontal duct whose cross section is \(16 \mathrm{~cm} \times 16 \mathrm{~cm}\). The components in the duct are cooled by forced air, which enters at \(27^{\circ} \mathrm{C}\) at a rate of \(0.65 \mathrm{~m}^{3} / \mathrm{min}\). Assuming 85 percent of the heat generated inside is transferred to air flowing through the duct and the remaining 15 percent is lost through the outer surfaces of the duct, determine \((a)\) the exit temperature of air and \((b)\) the highest component surface temperature in the duct. As a first approximation assume flow is fully developed in the channel. Evaluate properties of air at a bulk mean temperature of \(35^{\circ} \mathrm{C}\). Is this a good assumption?

Water is to be heated from \(10^{\circ} \mathrm{C}\) to \(80^{\circ} \mathrm{C}\) as it flows through a 2 -cm-internal-diameter, 13 -m-long tube. The tube is equipped with an electric resistance heater, which provides uniform heating throughout the surface of the tube. The outer surface of the heater is well insulated, so that in steady operation all the heat generated in the heater is transferred to the water in the tube. If the system is to provide hot water at a rate of \(5 \mathrm{~L} / \mathrm{min}\), determine the power rating of the resistance heater. Also, estimate the inner surface temperature of the pipe at the exit.

A 10 -m-long and 10 -mm-inner-diameter pipe made of commercial steel is used to heat a liquid in an industrial process. The liquid enters the pipe with \(T_{i}=25^{\circ} \mathrm{C}, V=0.8 \mathrm{~m} / \mathrm{s}\). A uniform heat flux is maintained by an electric resistance heater wrapped around the outer surface of the pipe, so that the fluid exits at \(75^{\circ} \mathrm{C}\). Assuming fully developed flow and taking the average fluid properties to be \(\rho=1000 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=\) \(4000 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, \mu=2 \times 10^{-3} \mathrm{~kg} / \mathrm{m} \cdot \mathrm{s}, k=0.48 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(\operatorname{Pr}=10\), determine: (a) The required surface heat flux \(\dot{q}_{s}\), produced by the heater (b) The surface temperature at the exit, \(T_{s}\) (c) The pressure loss through the pipe and the minimum power required to overcome the resistance to flow.
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