/*! 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 18 You've experienced convection co... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 18

You've experienced convection cooling if you've ever extended your hand out the window of a moving vehicle or into a flowing water stream. With the surface of your hand at a temperature of \(30^{\circ} \mathrm{C}\), determine the convection heat flux for (a) a vehicle speed of \(35 \mathrm{~km} / \mathrm{h}\) in air at \(-5^{\circ} \mathrm{C}\) with a convection coefficient of 40 \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and (b) a velocity of \(0.2 \mathrm{~m} / \mathrm{s}\) in a water stream at \(10^{\circ} \mathrm{C}\) with a convection coefficient of \(900 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Which condition would feel colder? Contrast these results with a heat loss of approximately \(30 \mathrm{~W} / \mathrm{m}^{2}\) under normal room conditions.

Short Answer

Expert verified
The convection heat flux for the vehicle in air scenario (Case a) is \(1400 \,\mathrm{W}/\mathrm{m}^2\), and for the hand in the water stream scenario (Case b) is \(18000 \,\mathrm{W}/\mathrm{m}^2\). Since \(q_b > q_a\), the hand in the water stream would feel colder than the hand out of the vehicle in the air. Both scenarios feel significantly colder than normal room conditions, where the heat loss is approximately \(30 \,\mathrm{W}/\mathrm{m}^2\).

Step by step solution

01

Case (a): Vehicle in Air

First, let's find the convection heat flux for the vehicle in air scenario: Given values: - \(T_s = 30^\circ \mathrm{C}\) - \(T_f = -5^\circ \mathrm{C}\) - \(h = 40 \,\mathrm{W} / \mathrm{m}^2 \cdot \mathrm{K}\) We can now calculate the convection heat flux: \(q_a = h \cdot (T_s - T_f)\) \(q_a = 40 \,\mathrm{W} / \mathrm{m}^2 \cdot \mathrm{K} \cdot (30^\circ \mathrm{C} - (-5^\circ \mathrm{C}))\) \(q_a = 40 \,\mathrm{W} / \mathrm{m}^2 \cdot \mathrm{K} \cdot (35 \mathrm{K})\) \(q_a = 1400 \,\mathrm{W} / \mathrm{m}^2\)
02

Case (b): Hand in Water Stream

Now let's find the convection heat flux for the hand in the water stream scenario: Given values: - \(T_s = 30^\circ \mathrm{C}\) - \(T_f = 10^\circ \mathrm{C}\) - \(h = 900 \,\mathrm{W} / \mathrm{m}^2 \cdot \mathrm{K}\) We can now calculate the convection heat flux: \(q_b = h \cdot (T_s - T_f)\) \(q_b = 900 \,\mathrm{W} / \mathrm{m}^2 \cdot \mathrm{K} \cdot (30^\circ \mathrm{C} - 10^\circ \mathrm{C})\) \(q_b = 900 \,\mathrm{W} / \mathrm{m}^2 \cdot \mathrm{K} \cdot (20 \mathrm{K})\) \(q_b = 18000 \,\mathrm{W} / \mathrm{m}^2\)
03

Comparing Results

We can now compare both scenarios and determine which condition would feel colder: - Case (a): \(q_a = 1400 \,\mathrm{W} / \mathrm{m}^2\) - Case (b): \(q_b = 18000 \,\mathrm{W} / \mathrm{m}^2\) Since \(q_b > q_a\), the condition with the hand in the water stream would feel colder than the condition with the hand out of the vehicle in the air. Now, let's contrast this with a heat loss of approximately \(30 \,\mathrm{W} / \mathrm{m}^2\) under normal room conditions: Under normal room conditions, the heat loss is much less than in either of the other two scenarios, making both scenarios feel significantly colder than normal room conditions.

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.

Heat Flux Calculation
When talking about heat transfer, it's important to understand what heat flux is. Heat flux is the rate of heat energy transferred per unit area, typically measured in watts per square meter (\( ext{W/m}^2 \)). In convection scenarios, like when your hand is sticking out of a moving vehicle or dipped in a flowing stream, calculating the heat flux gives you an idea of how cold it might feel.

The calculation involves a simple formula:
  • \( q = h \cdot (T_s - T_f) \)
where:
  • \( q \) is the heat flux,
  • \( h \) is the convection coefficient,
  • \( T_s \) is the surface temperature,
  • \( T_f \) is the fluid's temperature.
By substituting values for these variables, you can find the heat flux in different conditions, such as in air or water. Understanding this helps determine why some environments might feel colder than others.
Convection Coefficient
The convection coefficient, often symbolized as \( h \), is crucial in determining how effectively heat is transferred in convection. It tells us how easily heat flows across a boundary between a solid surface and a surrounding fluid.

  • A higher \( h \) value means more efficient heat transfer, making the surrounding fluid feel cooler to the touch.
  • A lower \( h \) value indicates reduced heat transfer, often perceived as warmer.
In our exercise, air had a lower convection coefficient (40 \( ext{W/m}^2 \/ \cdot \/ ext{K} \)) compared to water (900 \( ext{W/m}^2 \/ \cdot \/ ext{K} \)), making the hand in water dissipate heat more quickly, thus feeling colder. Recognizing how this coefficient varies with medium and conditions is essential for accurately predicting thermal comfort in different environments.
Temperature Difference in Convection
The temperature difference between a surface and the fluid around it is vital in convection heat transfer. It's essentially what drives the cooling or heating effect observable in everyday situations.

  • The larger the temperature difference, the greater the potential for heat transfer.
  • This difference is represented in our formula as \( T_s - T_f \), where \( T_s \) is the temperature of the surface (like your hand) and \( T_f \) is the temperature of the fluid (air or water).
For instance, in our example, the vehicle's air had a temperature difference of 35掳C, while the water stream had a 20掳C difference. Although the air had a larger temperature difference, the heat flux was higher in water due to its significantly higher convection coefficient. This indicates how both temperature difference and convection coefficient combine to affect how you perceive thermal conditions.

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 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?

A vertical slab of Wood's metal is joined to a substrate on one surface and is melted as it is uniformly irradiated by a laser source on the opposite surface. The metal is initially at its fusion temperature of \(T_{f}=72^{\circ} \mathrm{C}\), and the melt runs off by gravity as soon as it is formed. The absorptivity of the metal to the laser radiation is \(\alpha_{1}=0.4\), and its latent heat of fusion is \(h_{s f}=33 \mathrm{~kJ} / \mathrm{kg}\). (a) Neglecting heat transfer from the irradiated surface by convection or radiation exchange with the surroundings, determine the instantaneous rate of melting in \(\mathrm{kg} / \mathrm{s} \cdot \mathrm{m}^{2}\) if the laser irradiation is \(5 \mathrm{~kW} / \mathrm{m}^{2}\). How much material is removed if irradiation is maintained for a period of \(2 \mathrm{~s}\) ? (b) Allowing for convection to ambient air, with \(T_{\infty}=20^{\circ} \mathrm{C}\) and \(h=15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and radiation exchange with large surroundings \((\varepsilon=0.4\), \(T_{\text {sur }}=20^{\circ} \mathrm{C}\) ), determine the instantaneous rate of melting during irradiation.

A concrete wall, which has a surface area of \(20 \mathrm{~m}^{2}\) and is \(0.30 \mathrm{~m}\) thick, separates conditioned room air from ambient air. The temperature of the inner surface of the wall is maintained at \(25^{\circ} \mathrm{C}\), and the thermal conductivity of the concrete is \(1 \mathrm{~W} / \mathrm{m}=\mathrm{K}\). (a) Determine the heat loss through the wall for outer surface temperatures ranging from \(-15^{\circ} \mathrm{C}\) to \(38^{\circ} \mathrm{C}\), which correspond to winter and summer extremes, respectively. Display your results graphically. (b) On your graph, also plot the heat loss as a function of the outer surface temperature for wall materials having thermal conductivities of \(0.75\) and \(1.25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Explain the family of curves you have obtained.

Consider a surface-mount type transistor on a circuit board whose temperature is maintained at \(35^{\circ} \mathrm{C}\). Air at \(20^{\circ} \mathrm{C}\) flows over the upper surface of dimensions \(4 \mathrm{~mm} \times\) \(8 \mathrm{~mm}\) with a convection coefficient of \(50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Three wire leads, each of cross section \(1 \mathrm{~mm} \times 0.25 \mathrm{~mm}\) and length \(4 \mathrm{~mm}\), conduct heat from the case to the circuit board. The gap between the case and the board is \(0.2 \mathrm{~mm}\). (a) Assuming the case is isothermal and neglecting radiation, estimate the case temperature when \(150 \mathrm{~mW}\) is dissipated by the transistor and (i) stagnant air or (ii) a conductive paste fills the gap. The thermal conductivities of the wire leads, air, and conductive paste are \(25,0.0263\), and \(0.12 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), respectively. (b) Using the conductive paste to fill the gap, we wish to determine the extent to which increased heat dissipation may be accommodated, subject to the constraint that the case temperature not exceed \(40^{\circ} \mathrm{C}\). Options include increasing the air speed to achieve a larger convection coefficient \(h\) and/or changing the lead wire material to one of larger thermal conductivity. Independently considering leads fabricated from materials with thermal conductivities of 200 and \(400 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), compute and plot the maximum allowable heat dissipation for variations in \(h\) over the range \(50 \leq h \leq 250 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

A transmission case measures \(W=0.30 \mathrm{~m}\) on a side and receives a power input of \(P_{i}=150 \mathrm{hp}\) from the engine. 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 \(T_{\infty}=50^{\circ} \mathrm{C}\) ?
See all solutions

Recommended explanations on Physics Textbooks

Mechanics and Materials

Read Explanation

Medical Physics

Read Explanation

Physics of Motion

Read Explanation

Electric Charge Field and Potential

Read Explanation

Astrophysics

Read Explanation

Modern 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.