Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 19: Problem 14

(a) Calculate the heat flux through a sheet of steel \(10 \mathrm{~mm}(0.39\) in.) thick if the temperatures at the two faces are 300 and \(100^{\circ} \mathrm{C}\) (572 and \(212^{\circ} \mathrm{F}\) ); assume steady-state heat flow. (b) What is the heat loss per hour if the area of the sheet is \(0.25 \mathrm{~m}^{2}\left(2.7 \mathrm{ft}^{2}\right) ?\) (c) What will be the heat loss per hour if soda-lime glass instead of steel is used? (d) Calculate the heat loss per hour if steel is used and the thickness is increased to \(20 \mathrm{~mm}(0.79\) in.).

Short Answer

Expert verified
Answer: The heat losses per hour are (a) -2500 Wh for the original steel sheet, (b) -50 Wh for the soda-lime glass sheet, and (c) -1250 Wh for the steel sheet with doubled thickness.

Step by step solution

01

Find the thermal conductivity of the materials

For steel, thermal conductivity (k) is approximately \(50 \frac{W}{m \cdot K}\). For soda-lime glass, k is approximately \(1 \frac{W}{m \cdot K}\).
02

Calculate the heat flux for the steel sheet

Use Fourier's Law: \(q = -k \frac{\Delta T}{d}\), where \(q\) is the heat flux, \(k\) is the thermal conductivity, \(\Delta T\) is the temperature difference, and \(d\) is the thickness. For the steel sheet: $$q_{steel} = -50 \frac{W}{m \cdot K} \cdot \frac{(300^{\circ}C - 100^{\circ}C)}{0.01 m} = -50 \frac{W}{m \cdot K} \cdot \frac{200^{\circ}C}{0.01 m} = -10000 W/m^2$$
03

Calculate the heat loss per hour for the steel sheet

To find the heat loss per hour, multiply the heat flux by the area of the sheet: $$Q_{steel} = q_{steel} \cdot A = -10000 \frac{W}{m^2} \cdot 0.25 m^2 = -2500 W$$ Convert the heat loss to watt-hours: $$-2500 W \cdot 1h = -2500 Wh$$
04

Calculate the heat flux for the soda-lime glass sheet

Use Fourier's Law with the thermal conductivity of soda-lime glass: $$q_{glass} = -1 \frac{W}{m \cdot K} \cdot \frac{(300^{\circ}C - 100^{\circ}C)}{0.01 m} = -1 \frac{W}{m \cdot K} \cdot \frac{200^{\circ}C}{0.01 m} = -200 W/m^2$$
05

Calculate the heat loss per hour for the soda-lime glass sheet

Multiply the heat flux by the area of the sheet: $$Q_{glass} = q_{glass} \cdot A = -200 \frac{W}{m^2} \cdot 0.25 m^2 = -50 W$$ Convert the heat loss to watt-hours: $$-50 W \cdot 1 h = -50 Wh$$
06

Calculate the heat flux for the steel sheet with increased thickness

Use Fourier's Law with the new thickness: $$q_{steel2} = -50 \frac{W}{m \cdot K} \cdot \frac{(300^{\circ}C - 100^{\circ}C)}{0.02 m} = -50 \frac{W}{m \cdot K} \cdot \frac{200^{\circ}C}{0.02 m} = -5000 W/m^2$$
07

Calculate the heat loss per hour for the steel sheet with increased thickness

Multiply the heat flux by the area of the sheet: $$Q_{steel2} = q_{steel2} \cdot A = -5000 \frac{W}{m^2} \cdot 0.25 m^2 = -1250 W$$ Convert the heat loss per hour to watt-hours: $$-1250 W \cdot 1 h = -1250 Wh$$ To summarize, the heat losses per hour are: (a) -2500 Wh for the steel sheet, (b) -50 Wh for the soda-lime glass sheet, and (c) -1250 Wh for the steel sheet with increased thickness.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Nonsteady-state heat flow may be described by the following partial differential equation: $$ \frac{\partial T}{\partial t}=D_{T} \frac{\partial^{2} T}{\partial x^{2}} $$ where \(D_{T}\) is the thermal diffusivity; this expression is the thermal equivalent of Fick's second law of diffusion (Equation 5.4b). The thermal diffusivity is defined according to $$ D_{T}=\frac{k}{\rho c_{p}} $$ In this expression, \(k, \rho\), and \(c_{p}\) represent the thermal conductivity, the mass density, and the specific heat at constant pressure, respectively. (a) What are the SI units for \(D_{T}\) ? (b) Determine values of \(D_{T}\) for aluminum, steel, aluminum oxide, soda-lime glass, polystyrene, and nylon 6,6 using the data in Table 19.1. Density values are included in Table B.1, Appendix B. Thermal Stresses

For some ceramic materials, why does the thermal conductivity first decrease and then increase with rising temperature?

A \(0.1 \mathrm{~m}(3.9 \mathrm{in} .)\) rod of a metal elongates \(0.2\) \(\mathrm{mm}\left(0.0079\right.\) in.) on heating from 20 to \(100^{\circ} \mathrm{C}\) \(\left(68\right.\) to \(\left.212^{\circ} \mathrm{F}\right)\). Determine the value of the linear coefficient of thermal expansion for this material.

The two ends of a cylindrical rod of 1025 steel \(75.00 \mathrm{~mm}\) long and \(10.000 \mathrm{~mm}\) in diameter are maintained rigid. If the rod is initially at \(25^{\circ} \mathrm{C}\), to what temperature must it be cooled to have a \(0.008-\mathrm{mm}\) reduction in diameter?

A copper wire is stretched with a stress of \(70 \mathrm{MPa}(10,000 \mathrm{psi})\) at \(20^{\circ} \mathrm{C}\left(68^{\circ} \mathrm{F}\right)\). If the length is held constant, to what temperature must the wire be heated to reduce the stress to \(35 \mathrm{MPa}\) (5000 psi)?
See all solutions

Recommended explanations on Physics Textbooks

Thermodynamics

Read Explanation

Geometrical and Physical Optics

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Rotational Dynamics

Read Explanation

Circular Motion and Gravitation

Read Explanation

Astrophysics

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.

Sign-up for free