/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 68 In a power plant, pipes transpor... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In a power plant, pipes transporting superheated vapor are very common. Superheated vapor is flowing at a rate of \(0.3 \mathrm{~kg} / \mathrm{s}\) inside a pipe with \(5 \mathrm{~cm}\) in diameter and \(10 \mathrm{~m}\) in length. The pipe is located in a power plant at \(20^{\circ} \mathrm{C}\), and has a uniform pipe surface temperature of \(100^{\circ} \mathrm{C}\). If the temperature drop between the inlet and exit of the pipe is \(30^{\circ} \mathrm{C}\), and the specific heat of the vapor is \(2190 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), determine the heat transfer coefficient as a result of convection between the pipe surface and the surrounding.

Short Answer

Expert verified
Answer: The heat transfer coefficient is 157.3 W/m²·K.

Step by step solution

01

Calculate the heat loss from the pipe due to the temperature drop

Using the specific heat of the vapor, the mass flow rate, and the temperature drop, we can calculate the heat loss from the pipe as follows: \(q = mc\Delta T\) where \(q\) is the heat loss, \(m\) is the mass flow rate (\(0.3\,\mathrm{kg/s}\)), \(c\) is the specific heat of the vapor (\(2190\,\mathrm{J/kg\cdot K}\)), and \(\Delta T\) is the temperature drop (\(30^\circ \mathrm{C}\)). Plugging the given values into the equation, we get: \(q = (0.3\,\mathrm{kg/s}) (2190\,\mathrm{J/kg\cdot K}) (30\,\mathrm{K})\) Calculating the result, we obtain: \(q = 19710\,\mathrm{W}\) (Watts)
02

Calculate the area of the pipe's surface

Next, we need to find the area of the pipe's surface to determine the heat transfer due to convection. The surface area of a cylinder can be calculated as: \(A = 2\pi rl\) where \(A\) is the area, \(r\) is the radius of the pipe, and \(l\) is the length of the pipe. Given the diameter of the pipe is \(5\,\mathrm{cm}\), we can calculate the radius as: \(r = \frac{d}{2} = \frac{5\,\mathrm{cm}}{2} = 2.5\,\mathrm{cm} = 0.025\,\mathrm{m}\) (converted to meters) Now, we can calculate the surface area of the pipe: \(A = 2\pi (0.025\,\mathrm{m})(10\,\mathrm{m})\) Calculating the result, we obtain: \(A = 1.57\,\mathrm{m^2}\)
03

Determine the heat transfer coefficient

Finally, we can use the formula for the heat transfer due to convection to find the heat transfer coefficient. The formula is: \(q = hA\Delta T_s\) where \(h\) is the heat transfer coefficient, \(A\) is the area of the pipe's surface, and \(\Delta T_s\) is the temperature difference between the pipe surface and the surroundings. Since we have already calculated the heat loss (\(q\)) and the surface area (\(A\)), we can rearrange the formula to solve for the heat transfer coefficient (\(h\)): \(h = \frac{q}{A\Delta T_s}\) The temperature difference between the pipe surface and the surroundings is: \(\Delta T_s = (100^\circ \mathrm{C}) - (20^\circ \mathrm{C}) = 80\,\mathrm{K}\) Now, we can plug the known values into the formula and solve for \(h\): \(h = \frac{19710\,\mathrm{W}}{(1.57\,\mathrm{m^2})(80\,\mathrm{K})}\) Calculating the result, we obtain: \(h = 157.3\,\mathrm{W/m^2\cdot K}\) Thus, the heat transfer coefficient as a result of convection between the pipe surface and the surrounding is \(157.3\,\mathrm{W/m^2\cdot K}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Convection
Convection is one of the primary modes of heat transfer and it occurs when heat is carried away by the movement of fluids or gases. In our exercise involving superheated vapor inside a pipe at a power plant, convection is the process by which heat is transferred from the outer surface of the pipe to the surrounding air.

In technical terms, convection is classified into two types: natural or free convection and forced convection. The former occurs due to buoyancy forces that result from density differences caused by temperature variations in the fluid. The latter, which is more applicable to this exercise, involves the movement of fluid by external means such as a pump or fan. Here, the mass flow rate of the vapor is driven through the pipe, which enhances the heat transfer coefficient, impacting the efficiency of heat exchange.

Role of Convection in Heat Transfer Coefficient

The heat transfer coefficient, a measure of how well heat is transferred by convection, can vary greatly depending on the velocity of the fluid, its properties, and the temperature difference between the fluid and the surroundings. It serves as an indicator of the convection heat transfer capability of the system, and is a crucial value in calculating the energy balance in power plant operations.
Thermal Conductivity
Thermal conductivity represents a material's ability to conduct heat and it is a vital property in the study of thermodynamics and heat transfer. In the context of our exercise, the pipe's material will have its own thermal conductivity value, although it isn't directly provided or required to solve for the heat transfer coefficient due to convection. However, it indirectly affects the overall heat transfer process.

Materials with high thermal conductivity, such as metals, are efficient at transferring heat, whereas materials with low thermal conductivity, such as plastics or ceramics, are less efficient and often serve as insulators. The effectiveness of the pipe in transferring heat from the superheated vapor to its walls before losing it to the surroundings by convection depends on its thermal conductivity.

Understanding Heat Flow

Inside the pipe, the heat initially conducts through the pipe material itself before being convected away. The greater the thermal conductivity of the pipe material, the quicker and more effectively this process occurs. While this property is not the focus of the exercise, knowledge of thermal conductivity is essential when designing and analyzing real-world heat transfer systems in power plants.
Mass Flow Rate
Mass flow rate is a critical concept in thermodynamics and fluid mechanics, indicating the amount of mass passing through a cross-section of a pipe or conduit per unit time. It is typically expressed in units such as kilograms per second (kg/s) and plays a central role in the calculation of heat loss in our exercise.

Understanding the mass flow rate is crucial to determining how much heat is being transported by the vapor. This has direct implications on the heating or cooling processes within any system, including the power plant scenario outlined in the exercise.

Impact on Heat Transfer

A higher mass flow rate generally implies that more heat energy is being carried by the fluid, thus potentially increasing the heat transfer to the surroundings provided the pipe's surface is large enough for effective heat dissipation. Consequently, accurately calculating the mass flow rate is essential for evaluating the cooling needs of the pipe and ensuring optimal operational conditions for the power plant.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A transistor with a height of \(0.4 \mathrm{~cm}\) and a diameter of \(0.6 \mathrm{~cm}\) is mounted on a circuit board. The transistor is cooled by air flowing over it with an average heat transfer coefficient of \(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the air temperature is \(55^{\circ} \mathrm{C}\) and the transistor case temperature is not to exceed \(70^{\circ} \mathrm{C}\), determine the amount of power this transistor can dissipate safely. Disregard any heat transfer from the transistor base.

Eggs with a mass of \(0.15 \mathrm{~kg}\) per egg and a specific heat of \(3.32 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\) are cooled from \(32^{\circ} \mathrm{C}\) to \(10^{\circ} \mathrm{C}\) at a rate of 200 eggs per minute. The rate of heat removal from the eggs is (a) \(7.3 \mathrm{~kW}\) (b) \(53 \mathrm{~kW}\) (c) \(17 \mathrm{~kW}\) (d) \(438 \mathrm{~kW}\) (e) \(37 \mathrm{~kW}\)

A 2-kW electric resistance heater in a room is turned on and kept on for 50 minutes. The amount of energy transferred to the room by the heater is (a) \(2 \mathrm{~kJ}\) (b) \(100 \mathrm{~kJ}\) (c) \(6000 \mathrm{~kJ}\) (d) \(7200 \mathrm{~kJ}\) (e) \(12,000 \mathrm{~kJ}\)

One way of measuring the thermal conductivity of a material is to sandwich an electric thermofoil heater between two identical rectangular samples of the material and to heavily insulate the four outer edges, as shown in the figure. Thermocouples attached to the inner and outer surfaces of the samples record the temperatures. During an experiment, two \(0.5-\mathrm{cm}\) thick samples \(10 \mathrm{~cm} \times\) \(10 \mathrm{~cm}\) in size are used. When steady operation is reached, the heater is observed to draw \(25 \mathrm{~W}\) of electric power, and the temperature of each sample is observed to drop from \(82^{\circ} \mathrm{C}\) at the inner surface to \(74^{\circ} \mathrm{C}\) at the outer surface. Determine the thermal conductivity of the material at the average temperature.

A person standing in a room loses heat to the air in the room by convection and to the surrounding surfaces by radiation. Both the air in the room and the surrounding surfaces are at \(20^{\circ} \mathrm{C}\). The exposed surface of the person is \(1.5 \mathrm{~m}^{2}\) and has an average temperature of \(32^{\circ} \mathrm{C}\), and an emissivity of \(0.90\). If the rates of heat transfer from the person by convection and by radiation are equal, the combined heat transfer coefficient is (a) \(0.008 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (b) \(3.0 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (c) \(5.5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (d) \(8.3 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (e) \(10.9 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\)

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.