/*! 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 62 A sliding door on a furnace does... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A sliding door on a furnace does not close entirely but leaves a gap \(1.6 \mathrm{~cm}\) wide and \(1 \mathrm{~m}\) high. The door is \(20 \mathrm{~cm}\) thick and the emittance of the refractory ceramic is \(0.8\). What is the heat loss through the gap when the furnace is at \(1800 \mathrm{~K}\) and the surroundings are at \(300 \mathrm{~K}\) ?

Short Answer

Expert verified
The heat loss through the gap is approximately 7580 W.

Step by step solution

01

Understand the Problem

We need to calculate the heat loss through a gap left by a sliding door of a furnace. The dimensions of the gap are given, as well as the temperatures of the furnace and the surroundings. The emittance of the material is also provided. We will use the Stefan-Boltzmann law to calculate the radiative heat loss.
02

Identify Relevant Formula

The relevant formula for radiative heat transfer is:\[ q = \varepsilon \sigma A (T_1^4 - T_2^4) \]where \( q \) is the heat transfer per time unit (W), \( \varepsilon \) is the emittance of the material, \( \sigma \) is the Stefan-Boltzmann constant (\( 5.67 \times 10^{-8} \, \text{W/m}^2\text{K}^4 \)), \( A \) is the area of the gap, and \( T_1 \) and \( T_2 \) are the temperatures of the furnace and the surroundings, respectively.
03

Calculate the Area of the Gap

The area \( A \) of the gap is calculated using the formula for the area of a rectangle. The gap width is \( 1.6 \, \text{cm} \) (or \( 0.016 \, \text{m} \)) and the height is \( 1 \, \text{m} \):\[ A = 0.016 \, \text{m} \times 1 \, \text{m} = 0.016 \, \text{m}^2 \]
04

Insert Known Values into Formula

Now we can substitute the known values into the retrieved formula:- \( \varepsilon = 0.8 \) (emittance)- \( \sigma = 5.67 \times 10^{-8} \, \text{W/m}^2\text{K}^4 \)- \( A = 0.016 \, \text{m}^2 \)- \( T_1 = 1800 \, \text{K} \)- \( T_2 = 300 \, \text{K} \)\[ q = 0.8 \times 5.67 \times 10^{-8} \, \text{W/m}^2\text{K}^4 \times 0.016 \, \text{m}^2 \times (1800^4 - 300^4) \]
05

Solve the Equation

Calculate each component:1. Temperatures in Kelvin to the fourth power: \[ 1800^4 = 1.04976 \times 10^{12} \, \text{K}^4 \] \[ 300^4 = 8.1 \times 10^{7} \, \text{K}^4 \]2. Substitute these into the equation: \[ q = 0.8 \times 5.67 \times 10^{-8} \, \times 0.016 \, \text{m}^2 \times (1.04976 \times 10^{12} - 8.1 \times 10^7) \]3. Calculate the difference in powers and further computations: \[ q = 0.8 \times 5.67 \times 10^{-8} \times 0.016 \times 1.048952 \times 10^{12} \]4. The final result: \[ q \approx 7.58 \times 10^{3} \, \text{W} \]
06

Conclusion

The heat loss through the gap is approximately \( 7580 \, \text{W} \).

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 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Stefan-Boltzmann Law
The Stefan-Boltzmann Law is a fundamental principle in thermal physics. It describes how the total energy radiated per unit surface area of a black body is proportional to the fourth power of its temperature. This means that even small changes in temperature can lead to large changes in the amount of heat radiated.
The law is expressed mathematically as:
  • \[ q = \varepsilon \sigma A (T_1^4 - T_2^4) \]
Here, \( q \) is the heat transfer per unit time, \( \varepsilon \) is the material's emittance, \( A \) is the area through which heat is transferred, and \( T_1 \) and \( T_2 \) are the absolute temperatures of the emitting and absorbing surfaces, respectively.
In our application with the furnace, this law helps calculate how much heat escapes through the door gap by considering both temperatures and the area of the gap. This approach is useful in various thermal management situations, such as in engineering or environmental science.
Emittance
Emittance, also known as emissivity, describes how effectively a material radiates energy compared to a perfect black body. It is a dimensionless value that ranges from 0 to 1.
  • A perfect black body has an emittance of 1, meaning it radiates energy as efficiently as possible.
  • A material with an emittance of 0 does not emit any radiation.
In practical applications, emittance values for materials are determined experimentally and are used to predict heat transfer rates through radiation.
For instance, the refractory ceramic of the furnace door has an emittance of 0.8, which indicates it radiates 80% as efficiently as a perfect black body. This high emittance value contributes significantly to the heat loss calculated through the gap.
Thermal Conductivity
While the main focus of our problem is on radiative heat transfer, it's worthwhile to briefly touch on the concept of thermal conductivity. Thermal conductivity is a material property that indicates how well a material conducts heat through direct contact.
  • Materials with high thermal conductivity, like metals, transfer heat quickly.
  • Materials with low thermal conductivity, like insulating materials, transfer heat slowly.
In our scenario, although thermal conductivity might not be directly used for the calculation of radiative heat loss, it's essential to consider when designing materials that can affect heat retention or loss in practical applications. For the furnace, the thick door is likely designed to minimize conductive losses, complementing efforts to manage radiative losses as calculated with the Stefan-Boltzmann Law.
Temperature Difference
Temperature difference is a crucial factor in determining the rate of heat transfer between two bodies. It is the difference in temperature between the hot body (the furnace) and the cooler surroundings.
  • In our case, the furnace operates at \( 1800 \, \text{K} \) and the surroundings are at \( 300 \, \text{K} \), resulting in a significant temperature difference.
  • A larger temperature difference generally increases the rate of heat transfer.
This concept is pivotal in many heat exchange systems, as it drives the energy flow from the hotter to the cooler areas. In the furnace example, the large temperature difference amplifies the heat loss through the gap, underscoring the importance of controlling temperature to effectively manage energy efficiency and minimize unwanted heat loss.

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 \(1 \mathrm{~cm}\)-thick redwood \((k=0.1 \mathrm{~W} / \mathrm{m} \mathrm{K})\) patio cover is surfaced with a layer of black tar paper ( \(\varepsilon=0.90 . \alpha_{s}=0.95\) ), and the underside is stained with redwood sealer-stain \((\varepsilon=0.88)\). Typical summer noon conditions include a solar irradiation of \(950 \mathrm{~W} / \mathrm{m}^{2}\), a clear sky, and ambient air at \(27^{\circ} \mathrm{C}\) and \(50 \% \mathrm{RH}\). Wind blowing through the patio gives an estimated convective heat transfer coefficient on both sides of the cover of \(8 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\). An important factor determining the comfort of people sitting on the patio is the radiosity of the underside of the cover, since it is this radiosity that determines the radiant heating experienced by the people. Estimate the radiosity for the following situations: (i) For the cover as designed. (ii) If the underside is painted with aluminum paint ( \(\varepsilon=0.22\) ). (iii) If, instead, the top of the cover is painted with a white paint \(\left(\alpha_{s}=0.30, \varepsilon=0.85\right)\) and kept clean. (iv) The combination of (ii) and (iii). $$ T_{e}=27^{2} \mathrm{C} .50 \mathrm{~cm} \mathrm{RH} $$

A long furnace with a \(2 \mathrm{~m}-\) square cross section has three refractory walls, and one wall cooled to \(400 \mathrm{~K}\). Combustion gases at \(1200 \mathrm{~K}\) flow through the furnace. The cooled wall is gray, with an emittance of \(0.7\), and the gases can be modeled as isothermal and gray, with a total absorption coefficient \(\kappa=0.15 \mathrm{~m}^{-1}\). Determine the radiant heat transfer to the cooled wall.

A thermistor is used to measure the temperature of a hot-air stream in a forced-air heating system of a hospital. It is located in a \(1 \mathrm{~m}\)-square duct through which air flows at \(1.6 \mathrm{~m} / \mathrm{s}\). The thermistor can be modeled as a \(3 \mathrm{~mm}\)-diameter sphere of emittance \(0.8\) and is located inside a radiation shield in the form of a 1 \(\mathrm{cm}-\) diameter, \(5 \mathrm{~cm}\)-long tube of emittance \(0.7\) (the axis of the tube is in the flow direction). If the thermistor records a temperature of \(46.8^{\circ} \mathrm{C}\) when the duct walls are at \(43.5^{\circ} \mathrm{C}\), determine the true temperature of the air. Convective heat transfer coefficients on the thermistor and shield can be taken as \(93 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\) and \(22 \mathrm{~W} / \mathrm{m}^{2}\) \(\mathrm{K}\), respectively. (Hint: First estimate the shield temperature.)

A circular cross-section heating duct of \(50 \mathrm{~cm}\) diameter runs horizontally through the basement garage of a large apartment complex. The ambient air and walls of the garage are at \(17^{\circ} \mathrm{C} .\) At a location where the surface temperature of the duct is \(40^{\circ} \mathrm{C}\), determine the radiation contribution to the total heat loss per meter length if (i) the duct has an emittance of \(0.7 .\) (ii) the duct is painted with aluminum paint with \(\varepsilon=0.2\).

A long furnace used for an enameling process has a \(3 \mathrm{~m} \times 3 \mathrm{~m}\) cross section with the roof maintained at \(1900 \mathrm{~K}\), and the side walls are well insulated. What is the radiant heat transfer to the floor when it is at \(600 \mathrm{~K}\) ? All surfaces may be taken to be gray and diffuse, with emittances of \(0.5\). (i) Assume the side walls are isothermal. (ii) Divide the side walls into two surfaces and allow these surfaces to be isothermal at different temperatures.

See all solutions

Recommended explanations on Physics Textbooks

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.