/*! 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 52 A thin electrical heater is inse... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 52

A thin electrical heater is inserted between a long circular rod and a concentric tube with inner and outer radii of 20 and \(40 \mathrm{~mm}\). The rod (A) has a thermal conductivity of \(k_{\mathrm{A}}=0.15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), while the tube (B) has a thermal conductivity of \(k_{\mathrm{B}}=1.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and its outer surface is subjected to convection with a fluid of temperature \(T_{\infty}=-15^{\circ} \mathrm{C}\) and heat transfer coefficient \(50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The thermal contact resistance between the cylinder surfaces and the heater is negligible. (a) Determine the electrical power per unit length of the cylinders \((\mathrm{W} / \mathrm{m})\) that is required to maintain the outer surface of cylinder \(\mathrm{B}\) at \(5^{\circ} \mathrm{C}\). (b) What is the temperature at the center of cylinder A?

Short Answer

Expert verified
To maintain the outer surface of cylinder B at 5掳C and determine the electrical power per unit length of the cylinders, first, find the resistances to heat flow in both cylinders A and B, and the convection resistance at the outer surface of cylinder B. Then, calculate the total resistance, R_total, and heat transfer rate, q, using the given temperatures. The electrical power per unit length, P, can be found by dividing q by the length, L. To find the temperature at the center of cylinder A, T_center, use the formula T_center = T_s + q鈰匯_A, and plug in the values obtained from the previous steps.

Step by step solution

01

(Step 1: Find the resistance to heat flow in cylinders A and B)

We can use the formula for the thermal resistance of a cylindrical shell: \(R = \frac{\ln{(r_{2}/r_{1})}}{2 \pi k L}\) where \(R\) is the thermal resistance, \(r_{1}\) and \(r_{2}\) are the inner and outer radii, \(k\) is the thermal conductivity, and \(L\) is the length of the cylinder. For cylinder A: \(R_{A} = \frac{\ln{(0.02/0.04)}}{2 \pi (0.15) L}\) For cylinder B: \(R_{B} = \frac{\ln{(0.04/0.02)}}{2 \pi (1.5) L}\)
02

(Step 2: Find the convection resistance at the outer surface of cylinder B)

The convection resistance is given by the formula: \(R_{conv} = \frac{1}{h \cdot A}\) where \(h\) is the heat transfer coefficient and \(A\) is the surface area. For the outer surface of cylinder B: \(R_{conv} = \frac{1}{50 \cdot 2 \pi (0.04) L}\)
03

(Step 3: Calculate the heat transfer rate)

The total resistance \(R_{total}\) to heat flow can be found by adding the resistances in series: \(R_{total} = R_{A} + R_{B} + R_{conv}\) The heat transfer rate (\(q\)) can be calculated using the formula: \(q = \frac{T_{\infty} - T_{s}}{R_{total}}\) where \(T_{\infty} = -15^{\circ}C\) and \(T_{s} = 5^{\circ}C\). Calculating the heat transfer rate: \(q = \frac{-15 - 5}{R_{A} + R_{B} + R_{conv}}\) Since we need to determine the electrical power (\(P\)) per unit length of the cylinders, we'll divide the heat transfer rate (\(q\)) by the length of the cylinders (\(L\)): \(P = \frac{q}{L}\)
04

(Step 4: Find the temperature at the center of cylinder A)

To find the temperature at the center of cylinder A (\(T_{center}\)), we can use the following formula: \(T_{center} = T_{s} + q \cdot R_{A}\) Substitute the value of \(q\) from Step 3: \(T_{center} = 5^{\circ}C + q \cdot R_{A}\) Now, we have all the necessary information to find the electrical power per unit length and temperature at the center of cylinder A. Plug in the numbers to find the values for electrical power and central temperature.

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 91影视!

Key Concepts

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

Thermal Conductivity
Thermal conductivity is a fundamental property of materials that describes their ability to conduct heat. It is represented by the symbol
  • \(k\), which has units of \(\mathrm{W} / \mathrm{m} \cdot \mathrm{K}\).
The higher the thermal conductivity, the more efficiently a material can transfer heat. Conversely, materials with low thermal conductivities are good insulators. In the given exercise, the problem involves two materials:
  • the rod A with \(k_{A} = 0.15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\)
  • the tube B with \(k_{B} = 1.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).
This indicates that tube B is much more effective at conducting heat than rod A. By understanding thermal conductivity in this context, we can better comprehend how each component of the system impacts the overall heat transfer process.
Convection Resistance
Convection resistance occurs when heat is transferred between a solid surface and a fluid. This is often calculated to understand how heat moves from the outer surface of a material to the surrounding environment. The formula used to determine convection resistance, \(R_{conv}\), is:
  • \(R_{conv} = \frac{1}{h \cdot A}\)
where \(h\) stands for the heat transfer coefficient and \(A\) is the surface area involved. In our scenario, the cylinder B is in contact with a fluid, and has
  • a heat transfer coefficient of \(50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).
This coefficient represents how well heat moves from the cylinder surface to the fluid. A high value means efficient heat transfer, lowering the convection resistance.
Thermal Resistance
Thermal resistance is a measure used to quantify how a material resists heat flow. It is crucial in analyzing systems such as this one, where you have thermal contact between different materials. The thermal resistance for a cylindrical shell is given by:
  • \(R = \frac{\ln(r_{2}/r_{1})}{2 \pi k L}\)
where
  • \(r_{1}\) and \(r_{2}\) are the inner and outer radii,
  • \(k\) is the thermal conductivity, and
  • \(L\) is the length.
Calculating these resistances helps to understand the overall heat transfer in the system. By summing the resistances from different sections, like in cylinders A and B, and considering convection, we obtain the total resistance for the entire heat path. The lower the overall thermal resistance, the more effectively heat can transfer through the system.
Electrical Power Calculation
Calculating electrical power is essential when dealing with heat systems that involve electrical heaters. In this exercise, we calculated the heat transfer rate first, then determined the necessary electrical power. The heat transfer rate \(q\) is computed using:
  • \(q = \frac{T_{\infty} - T_{s}}{R_{total}}\)
Here, \(T_{\infty}\) is the ambient temperature and \(T_{s}\) is the desired surface temperature. Once \(q\) is determined, the electrical power per unit length \(P\) is calculated as
  • \(P = \frac{q}{L}\)
This value tells us the power required to maintain the desired temperature difference between the surfaces. Understanding this calculation allows us to ensure the system will provide the necessary amount of heat to counteract any losses and maintain efficient operation.

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

An air heater consists of a steel tube \((k=20 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\), with inner and outer radii of \(r_{1}=13 \mathrm{~mm}\) and \(r_{2}=16\) \(\mathrm{mm}\), respectively, and eight integrally machined longitudinal fins, each of thickness \(t=3 \mathrm{~mm}\). The fins extend to a concentric tube, which is of radius \(r_{3}=\) \(40 \mathrm{~mm}\) and insulated on its outer surface. Water at a temperature \(T_{\infty, i}=90^{\circ} \mathrm{C}\) flows through the inner tube, while air at \(T_{\infty, o}=25^{\circ} \mathrm{C}\) flows through the annular region formed by the larger concentric tube. (a) Sketch the equivalent thermal circuit of the heater and relate each thermal resistance to appropriate system parameters. (b) If \(h_{i}=5000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(h_{o}=200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), what is the heat rate per unit length? (c) Assess the effect of increasing the number of fins \(N\) and/or the fin thickness \(t\) on the heat rate, subject to the constraint that \(N t<50 \mathrm{~mm}\).

A truncated solid cone is of circular cross section, and its diameter is related to the axial coordinate by an expression of the form \(D=a x^{3 / 2}\), where \(a=1.0 \mathrm{~m}^{-1 / 2}\). The sides are well insulated, while the top surface of the cone at \(x_{1}\) is maintained at \(T_{1}\) and the bottom surface at \(x_{2}\) is maintained at \(T_{2}\). (a) Obtain an expression for the temperature distribution \(T(x)\). (b) What is the rate of heat transfer across the cone if it is constructed of pure aluminum with \(x_{1}=0.075 \mathrm{~m}\), \(T_{1}=100^{\circ} \mathrm{C}, x_{2}=0.225 \mathrm{~m}\), and \(T_{2}=20^{\circ} \mathrm{C}\) ?

The fin array of Problem \(3.142\) is commonly found in compact heat exchangers, whose function is to provide a large surface area per unit volume in transferring heat from one fluid to another. Consider conditions for which the second fluid maintains equivalent temperatures at the parallel plates, \(T_{o}=T_{L}\), thereby establishing symmetry about the midplane of the fin array. The heat exchanger is \(1 \mathrm{~m}\) long in the direction of the flow of air (first fluid) and \(1 \mathrm{~m}\) wide in a direction normal to both the airflow and the fin surfaces. The length of the fin passages between adjoining parallel plates is \(L=8 \mathrm{~mm}\), whereas the fin thermal conductivity and convection coefficient are \(k=200 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) (aluminum) and \(h=150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. (a) If the fin thickness and pitch are \(t=1 \mathrm{~mm}\) and \(S=4 \mathrm{~mm}\), respectively, what is the value of the thermal resistance \(R_{t, o}\) for a one-half section of the fin array? (b) Subject to the constraints that the fin thickness and pitch may not be less than \(0.5\) and \(3 \mathrm{~mm}\), respectively, assess the effect of changes in \(t\) and \(S\).

One method that is used to grow nanowires (nanotubes with solid cores) is to initially deposit a small droplet of a liquid catalyst onto a flat surface. The surface and catalyst are heated and simultaneously exposed to a higher- temperature, low-pressure gas that contains a mixture of chemical species from which the nanowire is to be formed. The catalytic liquid slowly absorbs the species from the gas through its top surface and converts these to a solid material that is deposited onto the underlying liquid-solid interface, resulting in construction of the nanowire. The liquid catalyst remains suspended at the tip of the nanowire. Consider the growth of a 15 -nm-diameter silicon carbide nanowire onto a silicon carbide surface. The surface is maintained at a temperature of \(T_{s}=2400 \mathrm{~K}\), and the particular liquid catalyst that is used must be maintained in the range \(2400 \mathrm{~K} \leq T_{c} \leq 3000 \mathrm{~K}\) to perform its function. Determine the maximum length of a nanowire that may be grown for conditions characterized by \(h=10^{5} \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(T_{\infty}=8000 \mathrm{~K}\). Assume properties of the nanowire are the same as for bulk silicon carbide.

Finned passages are frequently formed between parallel plates to enhance convection heat transfer in compact heat exchanger cores. An important application is in electronic equipment cooling, where one or more air-cooled stacks are placed between heat-dissipating electrical components. Consider a single stack of rectangular fins of length \(L\) and thickness \(t\), with convection conditions corresponding to \(h\) and \(T_{\infty}\). (a) Obtain expressions for the fin heat transfer rates, \(q_{f, o}\) and \(q_{f, L}\), in terms of the base temperatures, \(T_{o}\) and \(T_{L}\). (b) In a specific application, a stack that is \(200 \mathrm{~mm}\) wide and \(100 \mathrm{~mm}\) deep contains 50 fins, each of length \(L=12 \mathrm{~mm}\). The entire stack is made from aluminum, which is everywhere \(1.0 \mathrm{~mm}\) thick. If temperature limitations associated with electrical components joined to opposite plates dictate maximum allowable plate temperatures of \(T_{o}=400 \mathrm{~K}\) and \(T_{L}=350 \mathrm{~K}\), what are the corresponding maximum power dissipations if \(h=150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(T_{\infty}=300 \mathrm{~K} ?\)
See all solutions

Recommended explanations on Physics Textbooks

Kinematics Physics

Read Explanation

Electricity

Read Explanation

Fields in Physics

Read Explanation

Dynamics

Read Explanation

Nuclear Physics

Read Explanation

Particle Model of Matter

Read Explanation
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.