/*! 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 48 A thin electrical heater is wrap... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 48

A thin electrical heater is wrapped around the outer surface of a long cylindrical tube whose inner surface is maintained at a temperature of \(5^{\circ} \mathrm{C}\). The tube wall has inner and outer radii of 25 and \(75 \mathrm{~mm}\), respectively, and a thermal conductivity of \(10 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The thermal contact resistance between the heater and the outer surface of the tube (per unit length of the tube) is \(R_{t, c}^{\prime}=\) \(0.01 \mathrm{~m} \cdot \mathrm{K} / \mathrm{W}\). The outer surface of the heater is exposed to a fluid with \(T_{\infty}=-10^{\circ} \mathrm{C}\) and a convection coefficient of \(h=100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the heater power per unit length of tube required to maintain the heater at \(T_{o}=25^{\circ} \mathrm{C} .\)

Short Answer

Expert verified
The heater power per unit length required to maintain the heater at \(T_o = 25^\circ\mathrm{C}\) is approximately \(892.86\,\mathrm{W}/\mathrm{m}\).

Step by step solution

01

Identify the given parameters

In this problem, we already have the following given: Inner radius of the tube, \(r_i = 25\,\mathrm{mm} = 0.025\,\mathrm{m}\) Outer radius of the tube, \(r_o = 75 \,\mathrm{mm}= 0.075\,\mathrm{m}\) Thermal conductivity of the tube, \(k = 10\, \mathrm{W}/\mathrm{m}\cdot\mathrm{K}\) Inner surface temperature, \(T_i = 5^\circ\mathrm{C}\) Thermal contact resistance per unit length, \(R_{t,c}' = 0.01 \, \mathrm{m}\cdot\mathrm{K}/\mathrm{W}\) Fluid temperature outside the heater, \(T_\infty = -10^\circ\mathrm{C}\) Convection coefficient, \(h = 100 \, \mathrm{W}/\mathrm{m}^{2}\cdot\mathrm{K}\) Target temperature of the heater, \(T_o = 25^\circ\mathrm{C}\) Now, let's find the heater power per unit length of the tube.
02

Calculate the conduction resistance through the tube

The first step is to compute the conduction resistance through the tube wall. The general equation for conduction resistance in a cylindrical system is: \[R_\text{cond} = \frac{\ln(r_o/r_i)}{2\pi k L}\] Where \(L\) is the length of the tube. In this case, we need to find the resistance per unit length, so we can divide by \(L\): \[R_\text{cond}' = \frac{R_\text{cond}}{L} = \frac{\ln(r_o/r_i)}{2\pi k}\] Now, we can plug in the given values: \[R_\text{cond}' = \frac{\ln(0.075/0.025)}{2\pi(10)} = 0.008\,\mathrm{m}\cdot\mathrm{K}/\mathrm{W}\]
03

Calculate the convection resistance between the heater and the fluid

Next, we need to determine the convection resistance between the heater and the fluid. The general equation for convection resistance in a cylindrical system is: \[R_\text{conv} = \frac{1}{h(2\pi r_o L)}\] Again, we need to find the resistance per unit length, so we can divide by \(L\): \[R_\text{conv}' = \frac{R_\text{conv}}{L} = \frac{1}{h(2\pi r_o)}\] Now, we can plug in the given values: \[R_\text{conv}' = \frac{1}{100(2\pi(0.075))} = 0.0212\,\mathrm{m}\cdot\mathrm{K}/\mathrm{W}\]
04

Calculate the total thermal resistance

To find the total thermal resistance, we add the conduction resistance, convection resistance, and contact resistance per unit length: \[R_\text{total}' = R_\text{cond}' + R_{t,c}' + R_\text{conv}'\] \[R_\text{total}' = 0.008 + 0.01 + 0.0212 = 0.0392\,\mathrm{m}\cdot\mathrm{K}/\mathrm{W}\]
05

Calculate the heater power per unit length required

Now that we have the total thermal resistance, we can use it to calculate the heater power per unit length required to maintain the heater at \(T_o = 25^\circ\mathrm{C}\). We use the formula: \[\dot{Q}' = \frac{T_o - T_\infty}{R_\text{total}'}\] Now, we can plug in the given values: \[\dot{Q}' = \frac{25 - (-10)}{0.0392} = 892.86\,\mathrm{W}/\mathrm{m}\] Therefore, the heater power per unit length required to maintain the heater at \(T_o = 25^\circ\mathrm{C}\) is approximately \(892.86\,\mathrm{W}/\mathrm{m}\).

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 Resistance
Thermal resistance is a crucial concept in understanding heat transfer between various components. It's akin to electrical resistance but applies to the flow of heat rather than electricity. The thermal resistance of a material indicates how effectively it resists the flow of heat. The less resistance, the more heat can pass through it. In our exercise, we calculate thermal resistance to understand how heat moves through the cylindrical tube.
A cylindrical tube's thermal resistance depends on several factors including the material's thermal conductivity, its geometry, and thickness. For cylindrical systems, the conduction thermal resistance can be expressed as:
  • \[ R_{cond} = \frac{\ln(r_o/r_i)}{2\pi k L} \]
Where:
  • \(r_o\) and \(r_i\) are the outer and inner radii respectively.
  • \(k\) is the thermal conductivity.
  • \(L\) represents the length.

This formula helps us evaluate how well the tube will conduct heat from the hotter inside to the cooler outside.
Convection
Convection is a heat transfer mechanism where heat moves through a fluid such as air or liquid. This process occurs when a fluid moves across the surface of a solid that's either heated or cooled. In our exercise, the cylinder鈥檚 outer surface conducts heat to the surrounding fluid through convection.
We measure how effectively convection transfers this heat using the convection coefficient, \(h\). This coefficient varies based on the fluid properties and flow conditions. The convection resistance in cylindrical systems is given by:
  • \[ R_{conv} = \frac{1}{h(2\pi r_o L)} \]
Where:
  • \(h\) is the convection coefficient.
  • \(r_o\) is the outer radius of the cylinder.
  • \(L\) is the length.

The convection process demonstrates how heat energy continues to move from the heater's outer surface into the fluid, significantly affecting the overall energy balance.
Thermal Conductivity
Thermal conductivity, symbolized by \(k\), is a material property that describes its ability to conduct heat. A high thermal conductivity means the material allows heat to pass through it quickly, whereas a low value indicates resistance to heat flow. In our problem, the tube's material has a thermal conductivity of 10 W/m路K.
This property plays a critical role in determining the conduction thermal resistance, which depends directly on \(k\). The better a material conducts heat, the lower its conduction resistance.
  • Materials with high thermal conductivity include metals like copper and aluminum.
  • Materials like rubber or insulation tend to have low thermal conductivity.

Understanding thermal conductivity helps in selecting the best materials for insulating or conducting heat in various applications. In engineering problems, it allows for precise calculations of how heat will distribute across different sections of a system.

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

Consider uniform thermal energy generation inside a one-dimensional plane wall of thickness \(L\) with one surface held at \(T_{s, 1}\) and the other surface insulated. (a) Find an expression for the conduction heat flux to the cold surface and the temperature of the hot surface \(T_{s, 2}\), expressing your results in terms of \(k, \dot{q}, L\), and \(T_{s, 1}\). (b) Compare the heat flux found in part (a) with the heat flux associated with a plane wall without energy generation whose surface temperatures are \(T_{s, 1}\) and \(T_{s, 2}\).

A nuclear fuel element of thickness \(2 L\) is covered with a steel cladding of thickness \(b\). Heat generated within the nuclear fuel at a rate \(\dot{q}\) is removed by a fluid at \(T_{\infty}\), which adjoins one surface and is characterized by a convection coefficient \(h\). The other surface is well insulated, and the fuel and steel have thermal conductivities of \(k_{f}\) and \(k_{s}\), respectively. (a) Obtain an equation for the temperature distribution \(T(x)\) in the nuclear fuel. Express your results in terms of \(\dot{q}, k_{f}, L, b, k_{s}, h\), and \(T_{\infty}\). (b) Sketch the temperature distribution \(T(x)\) for the entire system.

A probe of overall length \(L=200 \mathrm{~mm}\) and diameter \(D=\) \(12.5 \mathrm{~mm}\) is inserted through a duct wall such that a portion of its length, referred to as the immersion length \(L_{i}\), is in contact with the water stream whose temperature, \(T_{\infty, i}\) is to be determined. The convection coefficients over the immersion and ambient-exposed lengths are \(h_{i}=1100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(h_{o}=10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. The probe has a thermal conductivity of \(177 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and is in poor thermal contact with the duct wall. (a) Derive an expression for evaluating the measurement error, \(\Delta T_{\mathrm{err}}=T_{\text {tip }}-T_{\infty, i}\), which is the difference between the tip temperature, \(T_{\text {lip }}\), and the water temperature, \(T_{\infty, i \cdot}\) Hint: Define a coordinate system with the origin at the duct wall and treat the probe as two fins extending inward and outward from the duct, but having the same base temperature. Use Case A results from Table 3.4. (b) With the water and ambient air temperatures at 80 and \(20^{\circ} \mathrm{C}\), respectively, calculate the measurement error, \(\Delta T_{\mathrm{er}}\), as a function of immersion length for the conditions \(L_{i} / L=0.225,0.425\), and \(0.625\). (c) Compute and plot the effects of probe thermal conductivity and water velocity \(\left(h_{i}\right)\) on the measurement error.

Consider cylindrical and spherical shells with inner and outer surfaces at \(r_{1}\) and \(r_{2}\) maintained at uniform temperatures \(T_{s, 1}\) and \(T_{s, 2}\), respectively. If there is uniform heat generation within the shells, obtain expressions for the steady-state, one-dimensional radial distributions of the temperature, heat flux, and heat rate. Contrast your results with those summarized in Appendix C.

Radioactive wastes \(\left(k_{\mathrm{rw}}=20 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\) are stored in a spherical, stainless steel \(\left(k_{\mathrm{ss}}=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\) container of inner and outer radii equal to \(r_{i}=0.5 \mathrm{~m}\) and \(r_{o}=0.6 \mathrm{~m}\). Heat is generated volumetrically within the wastes at a uniform rate of \(\dot{q}=10^{5} \mathrm{~W} / \mathrm{m}^{3}\), and the outer surface of the container is exposed to a water flow for which \(h=\) \(1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(T_{\infty}=25^{\circ} \mathrm{C}\). (a) Evaluate the steady-state outer surface temperature, \(T_{s, o}\) (b) Evaluate the steady-state inner surface temperature, \(T_{s, i^{*}}\) (c) Obtain an expression for the temperature distribution, \(T(r)\), in the radioactive wastes. Express your result in terms of \(r_{i}, T_{s, i}, k_{\mathrm{rw}}\), and \(\dot{q}\). Evaluate the temperature at \(r=0\). (d) A proposed extension of the foregoing design involves storing waste materials having the same thermal conductivity but twice the heat generation \(\left(\dot{q}=2 \times 10^{5} \mathrm{~W} / \mathrm{m}^{3}\right)\) in a stainless steel container of equivalent inner radius \(\left(r_{i}=0.5 \mathrm{~m}\right)\). Safety considerations dictate that the maximum system temperature not exceed \(475^{\circ} \mathrm{C}\) and that the container wall thickness be no less than \(t=0.04 \mathrm{~m}\) and preferably at or close to the original design \((t=0.1 \mathrm{~m})\). Assess the effect of varying the outside convection coefficient to a maximum achievable value of \(h=5000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (by increasing the water velocity) and the container wall thickness. Is the proposed extension feasible? If so, recommend suitable operating and design conditions for \(h\) and \(t\), respectively.
See all solutions

Recommended explanations on Physics Textbooks

Conservation of Energy and Momentum

Read Explanation

Waves Physics

Read Explanation

Space Physics

Read Explanation

Atoms and Radioactivity

Read Explanation

Linear Momentum

Read Explanation

Famous Physicists

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.