/*! 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 1 Consider a thin electrical heate... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 1

Consider a thin electrical heater attached to a plate and backed by insulation. Initially, the heater and plate are at the temperature of the ambient air, \(T_{\infty}\). Suddenly, the power to the heater is activated, yielding a constant heat flux \(q_{o}^{\prime \prime}\left(\mathrm{W} / \mathrm{m}^{2}\right)\) at the inner surface of the plate. (a) Sketch and label, on \(T \leftarrow x\) coordinates, the temperature distributions: initial, steady-state, and at two intermediate times. (b) Sketch the heat flux at the outer surface \(q_{x}^{\prime \prime}(L, t)\) as a function of time.

Short Answer

Expert verified
(a) The temperature distributions on T鈫恱 coordinates are as follows: 1. Initial state: Horizontal line at \(T_{\infty}\). 2. Steady-state: Linear temperature distribution with highest temperature at the inner surface and lowest at the outer surface. 3. Intermediate time 1: Exponential temperature distribution with most significant difference between initial and steady-state distributions near the heater's surface. 4. Intermediate time 2: Exponential temperature distribution closer to the steady-state distribution, with the largest differences near the heater's surface. (b) The heat flux at the outer surface \(q_{x}^{\prime \prime}(L, t)\) as a function of time will show an exponential increase until reaching a steady constant value, indicating a constant heat conduction rate through the plate at the steady-state.

Step by step solution

01

a) Sketching the temperature distributions on T鈫恱 coordinates

We need to sketch the temperature distributions at four different stages: initial, steady-state, and two intermediate times. 1. Initial state: The heater and plate are at the ambient temperature, \(T_{\infty}\), which means that the whole plate will be at the same temperature, so the graph will be a horizontal line at \(T_{\infty}\). 2. Steady-state: After some time, the system reaches a steady state where the temperature distribution within the plate becomes linear, with the highest temperature at the inner surface (where the heater is located) and lowest at the outer surface (backed by insulation. 3. Intermediate time 1: At the first intermediate time, when the heater is activated, we will see an exponential temperature distribution with the most significant difference between the initial and steady-state distributions near the heater's surface. 4. Intermediate time 2: At the second intermediate time, the temperature distribution is still exponential but closer to the steady-state distribution, meaning the plate has started to reach equilibrium, but still, the largest differences will be seen near the heater's surface.
02

b) Sketching the heat flux at the outer surface \(q_{x}^{\prime \prime}(L, t)\) as a function of time.

Initially, the heat flux at the outer surface is zero because the heater is not yet activated. After the heater is activated, heat will be conducted through the plate. As time progresses, we will see an exponential increase in the heat flux at the outer surface of the plate, indicating that more heat is being conducted through the plate until it reaches a maximum value. Once the maximum value of the heat flux at the outer surface is reached, the system reaches a steady-state, and the heat flux remains constant. The heat flux will be constant as a function of time, indicating that the heat is being conducted through the plate at a steady rate. So, the graph of the heat flux at the outer surface \(q_{x}^{\prime \prime}(L, t)\) as a function of time will show an exponential increase until reaching a steady constant value.

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.

Temperature Distribution
Temperature distribution describes how temperature varies across different points on a material or object. Consider a thin plate with a heater attached to one side. Initially, the entire plate is at the ambient temperature, denoted as \( T_{\infty} \), so the temperature distribution is a flat line on a temperature vs. position graph.

Once the heater is turned on, the inner surface starts to warm up. At this stage, the temperature distribution becomes exponential. It has a higher temperature at the heater's surface and gradually decreases as you move away from the heater. As time passes, this distribution approaches a steady-state.
  • Initial State: Uniform temperature \( T_{\infty} \).
  • Intermediate Times: Exponential curve starting to form.
  • Steady-State: Linear temperature drop from inside to outside.
This progression is crucial for understanding how materials respond to thermal inputs over time. The intermediate stages show how heat is conducted through the plate until it achieves a homogeneous temperature gradient in the steady-state.
Steady-State Heat Transfer
Steady-state heat transfer occurs when the temperatures in a system no longer change over time. Once the plate reaches this state, its temperature distribution becomes linear. This linearity results from a consistent flow of heat across the plate, maintaining a balance between heat entering at the heater and heat leaving at the outer insulated surface.

In the context of our plate:
  • The surface closest to the heater will have the highest temperature.
  • There will be a gradual, linear temperature drop across the thickness of the plate.
  • The outer surface will maintain the lowest temperature, determined by the insulation and ambient conditions.
Reaching steady-state is essential in thermal system design, as it allows for the prediction of system behavior and performance. It ensures that temperatures remain consistent, avoiding fluctuations that could lead to inefficiencies or damage.
Heat Flux
The term 'heat flux' refers to the rate of heat flow through a surface per unit area, measured typically in \( \mathrm{W/m}^2 \). In our heater-plate scenario, the heat flux has dynamic behaviors over time.

Initially, before the heater is activated, no heat is conducted through the plate, giving a zero heat flux at the outer surface.
  • Initial: Zero heat flux.
  • Post Activation: As the heater turns on, heat starts conducting through the plate, and so the heat flux increases.
  • Exponential Increase: Heat flux rises steeply over time until the system reaches equilibrium.
  • Steady-state: When equilibrium is achieved, the heat flux becomes constant, reflecting a persistent heat transfer across the plate from the heater to the outer surface.
Understanding the behavior of heat flux is vital in informing how quickly a system can respond to changes and how much energy is required to maintain desired conditions. It aids in the effective design and management of thermal systems.

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

In a manufacturing process, long rods of different diameters are at a uniform temperature of \(400^{\circ} \mathrm{C}\) in a curing oven, from which they are removed and cooled by forced convection in air at \(25^{\circ} \mathrm{C}\). One of the line operators has observed that it takes \(280 \mathrm{~s}\) for a \(40-\mathrm{mm}\) diameter rod to cool to a safe-to-handle temperature of \(60^{\circ} \mathrm{C}\). For an equivalent convection coefficient, how long will it take for an 80 -mm-diameter rod to cool to the same temperature? The thermophysical properties of the rod are \(\rho=2500 \mathrm{~kg} / \mathrm{m}^{3}, c=900 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and \(k=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Comment on your result. Did you anticipate this outcome?

To determine which parts of a spider's brain are triggered into neural activity in response to various optical stimuli, researchers at the University of Massachusetts Amherst desire to examine the brain as it is shown images that might evoke emotions such as fear or hunger. Consider a spider at \(T_{i}=20^{\circ} \mathrm{C}\) that is shown a frightful scene and is then immediately immersed in liquid nitrogen at \(T_{\infty}=77 \mathrm{~K}\). The brain is subsequently dissected in its frozen state and analyzed to determine which parts of the brain reacted to the stimulus. Using your knowledge of heat transfer, determine how much time elapses before the spider's brain begins to freeze. Assume the brain is a sphere of diameter \(D_{b}=1 \mathrm{~mm}\), centrally located in the spider's cephalothorax, which may be approximated as a spherical shell of diameter \(D_{c}=3 \mathrm{~mm}\). The brain and cephalothorax properties correspond to those of liquid water. Neglect the effects of the latent heat of fusion and assume the heat transfer coefficient is \(h=100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

A one-dimensional slab of thickness \(2 L\) is initially at a uniform temperature \(T_{i}\). Suddenly, electric current is passed through the slab causing uniform volumetric heating \(\dot{q}\left(\mathrm{~W} / \mathrm{m}^{3}\right)\). At the same time, both outer surfaces \((x=\pm L)\) are subjected to a convection process at \(T_{\infty}\) with a heat transfer coefficient \(h\). Write the finite-difference equation expressing conservation of energy for node 0 located on the outer surface at \(x=-L\). Rearrange your equation and identify any important dimensionless coefficients.

As permanent space stations increase in size, there is an attendant increase in the amount of electrical power they dissipate. To keep station compartment temperatures from exceeding prescribed limits, it is necessary to transfer the dissipated heat to space. A novel heat rejection scheme that has been proposed for this purpose is termed a Liquid Droplet Radiator (LDR). The heat is first transferred to a high vacuum oil, which is then injected into outer space as a stream of small droplets. The stream is allowed to traverse a distance \(L\), over which it cools by radiating energy to outer space at absolute zero temperature. The droplets are then collected and routed back to the space station. Consider conditions for which droplets of emissivity \(\varepsilon=0.95\) and diameter \(D=0.5 \mathrm{~mm}\) are injected at a temperature of \(T_{i}=500 \mathrm{~K}\) and a velocity of \(V=0.1 \mathrm{~m} / \mathrm{s}\). Properties of the oil are \(\rho=885 \mathrm{~kg} / \mathrm{m}^{3}, c=1900 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and \(k=0.145 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Assuming each drop to radiate to deep space at \(T_{\text {sur }}=0 \mathrm{~K}\), determine the distance \(L\) required for the droplets to impact the collector at a final temperature of \(T_{f}=300 \mathrm{~K}\). What is the amount of thermal energy rejected by each droplet?

A soda lime glass sphere of diameter \(D_{1}=25 \mathrm{~mm}\) is encased in a bakelite spherical shell of thickness \(L=\) \(10 \mathrm{~mm}\). The composite sphere is initially at a uniform temperature, \(T_{i}=40^{\circ} \mathrm{C}\), and is exposed to a fluid at \(T_{\infty}=10^{\circ} \mathrm{C}\) with \(h=30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the center temperature of the glass at \(t=200 \mathrm{~s}\). Neglect the thermal contact resistance at the interface between the two materials.
See all solutions

Recommended explanations on Physics Textbooks

Modern Physics

Read Explanation

Electric Charge Field and Potential

Read Explanation

Physical Quantities And Units

Read Explanation

Energy Physics

Read Explanation

Atoms and Radioactivity

Read Explanation

Further Mechanics and Thermal Physics

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.