/*! 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 11 The heat flux that is applied to... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 11

The heat flux that is applied to one face of a plane wall is \(q^{\prime \prime}=20 \mathrm{~W} / \mathrm{m}^{2}\). The opposite face is exposed to air at temperature \(30^{\circ} \mathrm{C}\), with a convection heat transfer coefficient of \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The surface temperature of the wall exposed to air is measured and found to be \(50^{\circ} \mathrm{C}\). Do steady-state conditions exist? If not, is the temperature of the wall increasing or decreasing with time?

Short Answer

Expert verified
Steady-state conditions do not exist in this case since the heat transfer from the wall to the air (\(q^{\prime\prime}_{out} = 400 \, \mathrm{W/m^2}\)) is greater than the heat flux entering the wall (\(q^{\prime\prime}_{in} = 20 \, \mathrm{W/m^2}\)). Since the wall is losing heat faster than it is gaining it, the temperature of the wall is decreasing with time.

Step by step solution

01

Calculate the heat transfer from the wall to the air

Using the convection heat transfer formula, we can calculate the heat transfer rate between the wall and the air: \[q' = hA(T_s - T_\infty)\], where: - \(q'\) is the heat transfer per unit area, - \(h\) is the convection heat transfer coefficient, - \(A\) is the area of the surface, - \(T_s\) is the surface temperature, - \(T_\infty\) is the ambient air temperature. In this case, we are interested in calculating the heat transfer per unit area (\(q^{\prime\prime}\)). So we can write the formula: \[q^{\prime\prime} = h(T_s - T_\infty)\]
02

Plug in the given values

We are given: - \(h = 20 \, \mathrm{W/m^2 \cdot K}\) - \(T_s = 50^{\circ}\mathrm{C}\) - \(T_\infty = 30^{\circ}\mathrm{C}\) - \(q^{\prime\prime}_{in} = 20 \, \mathrm{W/m^2}\) (the given heat flux entering the wall) Plugging these values in the formula, we get: \[q^{\prime\prime}_{out} = 20(50 - 30)\]
03

Calculate the heat transfer per unit area

We calculate the heat transfer per unit area (the heat transfer rate between the wall and the air) using the formula derived in step 1: \[q^{\prime\prime}_{out} = 20(50 - 30) = 20 \times 20 = 400 \, \mathrm{W/m^2}\]
04

Compare the heat transfer into and out of the wall

We now compare the heat transfer into and out of the wall: - \(q^{\prime\prime}_{in} = 20 \, \mathrm{W/m^2}\) - \(q^{\prime\prime}_{out} = 400 \, \mathrm{W/m^2}\) Since \(q^{\prime\prime}_{out} > q^{\prime\prime}_{in}\), the heat transfer away from the wall to the air is greater than the heat flux entering the wall; thus, steady-state conditions do not exist.
05

Determine if the temperature of the wall is increasing or decreasing with time

Since the heat transfer from the wall to the air is greater than the heat flux entering the wall, the wall is losing heat faster than it is gaining it. Therefore, the temperature of the wall is decreasing with time.

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.

Steady-State Conditions
In the context of heat transfer, steady-state conditions occur when the temperature profile within a material does not change over time. This means that the amount of heat entering the system is equal to the amount of heat leaving it, resulting in no net change in energy. Think about it like a perfectly tuned seesaw; both sides are balanced, and there is no movement in either direction.

For the exercise given, we need to verify whether the heat entering the wall equals the heat leaving it. The heat flux into the wall is specified as 20 W/m虏, and via calculation, it鈥檚 found that the heat flux out of the wall is 400 W/m虏. Clearly, the heat leaving the wall is much greater than the heat entering it.

This imbalance implies that the conditions are not steady-state. Instead, the system is in a transient state, meaning the temperature within the wall will be changing over time as it seeks equilibrium.
Heat Flux
Heat flux is a measure of the rate of heat energy transfer per unit area. It essentially describes how much heat flows through a surface in a specific amount of time. The units for heat flux are typically watts per square meter (W/m虏), which indicates power per area.

In our exercise, there are two key heat flux values: the input heat flux and the output heat flux.
  • The **input heat flux** is given as 20 W/m虏, representing the heat applied to the wall.
  • The **output heat flux** calculated as 400 W/m虏 illustrates how much heat is being transferred from the wall to the surrounding air.
    This calculation uses the formula: \(q'' = h(T_s - T_\infty)\) where \(h\) is the convection heat transfer coefficient, \(T_s\) is the wall temperature, and \(T_\infty\) is the air temperature.
The significant difference between these two values highlights that more heat is leaving the wall than entering it. This differential triggers changes in the wall's temperature over time.
Temperature Gradient
A temperature gradient describes the rate at which temperature changes over a certain distance. In heat transfer problems, understanding the temperature gradient is crucial as it drives the flow of heat from warmer regions to cooler ones.

Temperature gradients are steep when the difference in temperature over distance is large, leading to faster heat transfer. Conversely, a gentle gradient indicates slower heat transfer.
  • In this problem, the temperature on the surface of the wall is 50掳C, while the air temperature is only 30掳C. This 20掳C difference across the wall-air boundary creates a significant temperature gradient, propelling a strong heat flow from the wall to the cooler air.
This gradient is what ultimately results in the calculated heat flux of 400 W/m虏 leaving the wall. Understanding how the temperature changes across surfaces helps predict how rapidly energy balance will be achieved or why non-steady-state conditions might occur, as it does in this scenario.

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

During its manufacture, plate glass at \(600^{\circ} \mathrm{C}\) is cooled by passing air over its surface such that the convection heat transfer coefficient is \(h=5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). To prevent cracking, it is known that the temperature gradient must not exceed \(15^{\circ} \mathrm{C} / \mathrm{mm}\) at any point in the glass during the cooling process. If the thermal conductivity of the glass is \(1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and its surface emissivity is \(0.8\), what is the lowest temperature of the air that can initially be used for the cooling? Assume that the temperature of the air equals that of the surroundings.

A square silicon chip \((k=150 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is of width \(w=5 \mathrm{~mm}\) on a side and of thickness \(t=1 \mathrm{~mm}\). The chip is mounted in a substrate such that its side and back surfaces are insulated, while the front surface is exposed to a coolant. If \(4 \mathrm{~W}\) are being dissipated in circuits mounted to the back surface of the chip, what is the steady-state temperature difference between back and front surfaces?

The roof of a car in a parking lot absorbs a solar radiant flux of \(800 \mathrm{~W} / \mathrm{m}^{2}\), and the underside is perfectly insulated. The convection coefficient between the roof and the ambient air is \(12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Neglecting radiation exchange with the surroundings, calculate the temperature of the roof under steadystate conditions if the ambient air temperature is \(20^{\circ} \mathrm{C}\). (b) For the same ambient air temperature, calculate the temperature of the roof if its surface emissivity is \(0.8\). (c) The convection coefficient depends on airflow conditions over the roof, increasing with increasing air speed. Compute and plot the roof temperature as a function of \(h\) for \(2 \leq h \leq 200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

The temperature controller for a clothes dryer consists of a bimetallic switch mounted on an electrical heater attached to a wall-mounted insulation pad. The switch is set to open at \(70^{\circ} \mathrm{C}\), the maximum dryer air temperature. To operate the dryer at a lower air temperature, sufficient power is supplied to the heater such that the switch reaches \(70^{\circ} \mathrm{C}\left(T_{\text {set }}\right)\) when the air temperature \(T\) is less than \(T_{\text {set. }}\). If the convection heat transfer coefficient between the air and the exposed switch surface of \(30 \mathrm{~mm}^{2}\) is \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), how much heater power \(P_{e}\) is required when the desired dryer air temperature is \(T_{\infty}=50^{\circ} \mathrm{C}\) ?

Electronic power devices are mounted to a heat sink having an exposed surface area of \(0.045 \mathrm{~m}^{2}\) and an emissivity of \(0.80\). When the devices dissipate a total power of \(20 \mathrm{~W}\) and the air and surroundings are at \(27^{\circ} \mathrm{C}\), the average sink temperature is \(42^{\circ} \mathrm{C}\). What average temperature will the heat sink reach when the devices dissipate \(30 \mathrm{~W}\) for the same environmental condition?
See all solutions

Recommended explanations on Physics Textbooks

Kinematics Physics

Read Explanation

Engineering Physics

Read Explanation

Classical Mechanics

Read Explanation

Radiation

Read Explanation

Medical Physics

Read Explanation

Force

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.