/*! 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 An astronaut is at work in the s... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 11

An astronaut is at work in the service bay of a space shuttle and is surrounded by walls that are at \(-100^{\circ} \mathrm{C}\). The outer surface of her space suit has an area of \(3 \mathrm{~m}^{2}\) and is aluminized with an emittance of \(0.05\). Calculate her rate of heat loss when the suit's outer temperature is \(0^{\circ} \mathrm{C}\). Express your answer in watts and \(\mathrm{kcal} / \mathrm{hr}\).

Short Answer

Expert verified
The astronaut's heat loss rate is approximately 466.5 Watts or 401.2 kcal/hr.

Step by step solution

01

Understand the Problem

The astronaut is losing heat from the space suit to the surrounding walls by radiation. To find the rate of heat loss, we need to use the Stefan-Boltzmann law, which calculates the power radiated by a black body based on its temperature and surface area.
02

Calculate the Heat Loss Using Stefan-Boltzmann Law

The Stefan-Boltzmann law is given by the formula: \[P = \varepsilon \cdot \sigma \cdot A \cdot (T_1^4 - T_2^4)\]where:\\(\varepsilon\) = Emittance of the surface = 0.05,\\(\sigma\) = Stefan-Boltzmann constant = \(5.67 \times 10^{-8} \, \text{W/m}^2\,\text{K}^4\),\\(A\) = Area = 3 m虏,\\(T_1\) = Temperature of the suit in Kelvin \((0^{\circ} \text{C} + 273.15 = 273.15\, \text{K})\),\\(T_2\) = Temperature of the walls in Kelvin \((-100^{\circ} \text{C} + 273.15 = 173.15\, \text{K})\).\Let's compute \(P\).
03

Plug in the Values and Calculate Heat Loss in Watts

Substitute the known values into the Stefan-Boltzmann equation:\\[P = 0.05 \times 5.67 \times 10^{-8} \times 3 \times ((273.15)^4 - (173.15)^4)\]Calculate each term step by step. This will give us the power \(P\) in Watts.
04

Simplify the Computation

Firstly calculate \((273.15)^4\) and \((173.15)^4\):\- \((273.15)^4 \approx 556,806,562\)- \((173.15)^4 \approx 8,981,209\)Subtract these: \[ 556,806,562 - 8,981,209 = 547,825,353 \] Now calculate \(P\): \[P = 0.05 \times 5.67 \times 10^{-8} \times 3 \times 547,825,353 \] This gives: \[P \approx 466.495 \text{ Watts (rounding to three decimal places)}\]
05

Convert Watts to kcal/hr

To convert watts to kcal/hr, use the conversion factor: 1 watt = 0.860 kcal/hr. Multiply the power in watts by this factor:\[466.495 \times 0.860 = 401.186 \text{ kcal/hr}\]Thus, the heat loss rate is approximately \(401.186\, \text{kcal/hr}\).

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.

Stefan-Boltzmann Law
The Stefan-Boltzmann law is crucial for understanding radiative heat transfer, especially in space. It helps us calculate the power emitted by an object due to thermal radiation. The formula for this law is given by:\[P = \varepsilon \cdot \sigma \cdot A \cdot (T_1^4 - T_2^4)\]
  • \(P\) is the power radiated, measured in watts (W).
  • \(\varepsilon\) represents the emittance, a measure of how efficiently a surface emits thermal radiation compared to a perfect black body.
  • \(\sigma\) is the Stefan-Boltzmann constant \((5.67 \times 10^{-8} \, \text{W/m}^2\,\text{K}^4)\).
  • \(A\) is the area of the radiating surface.
  • \(T_1\) and \(T_2\) are the absolute temperatures of the object and surroundings in Kelvin.
This law applies to any object in an environment where thermal radiation is the dominant form of heat transfer. In our exercise, we use it to calculate the heat loss from an astronaut's suit to her surroundings.
Emittance
Emittance (or emissivity) is a dimensionless parameter that describes how a real surface emits thermal radiation compared to a perfect black body. A perfect black body has an emittance of 1, meaning it radiates energy perfectly. Most real objects, however, have an emittance lower than 1.
  • For instance, the emittance of the space suit in this exercise is 0.05. Such a low value indicates it's designed to reflect most radiation rather than absorbing it, minimizing thermal radiation loss.
  • The degree of emittance affects the rate of heat transfer. High emittance means more heat is lost, and vice versa.
  • Materials for space gear are typically chosen for their low emittance to efficiently manage thermal conditions.
Understanding emittance is vital in applications like astronaut suits, which must regulate the wearer's temperature while exposed to extreme space conditions.
Temperature Conversion
Temperature conversion is essential when using the Stefan-Boltzmann law, as it requires temperatures to be in Kelvin. Kelvin is an absolute temperature scale, starting from absolute zero, making it ideal for scientific calculations like these.
  • To convert Celsius to Kelvin, simply add 273.15 to the Celsius temperature.
  • In the exercise, the space suit's temperature is given as \(0^{\circ} \text{C}\), converting to \(273.15\, \text{K}\).
  • The surrounding walls are at \(-100^{\circ} \text{C}\), or \(173.15\, \text{K}\).
Converting temperatures correctly is a crucial step that ensures accuracy in heat transfer calculations, especially when space suits are involved.
Space Suit Thermal Management
Space suit thermal management is critical for astronaut safety and comfort. In space, there are extreme temperature variations, so a space suit must efficiently maintain a stable internal temperature.
  • The space suit's thermal management involves reflecting incoming solar radiation while minimizing the wearer's heat loss through radiation. This is achieved through materials like aluminized layers that have low emittance.
  • Efficient thermal regulation ensures astronauts do not overheat or become too cold when they move in or away from the sun's direct exposure.
  • Understanding how space suits manage heat through the principles of radiative heat transfer helps design suits that are safe for all space missions.
These principles ensure astronauts can work safely in space's inhospitable environment, with proper consideration of the surrounding temperatures and suit material properties.

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

Two small blackened spheres of identical size-one of aluminum, the other of an unknown alloy of high conductivity-are suspended by thin wires inside a large cavity in a block of melting ice. It is found that it takes \(4.8\) minutes for the temperature of the aluminum sphere to drop from \(3^{\circ} \mathrm{C}\) to \(1^{\circ} \mathrm{C}\), and \(9.6\) minutes for the alloy sphere to undergo the same change. If the specific gravities of the aluminum and alloy are \(2.7\) and \(5.4\), respectively, and the specific heat of the aluminum is \(900 \mathrm{~J} / \mathrm{kg} \mathrm{K}\), what is the specific heat of the alloy?

A \(1 \mathrm{~cm}\)-diameter sphere is maintained at \(60^{\circ} \mathrm{C}\) in an enclosure with walls at \(35^{\circ} \mathrm{C}\) through which air at \(40^{\circ} \mathrm{C}\) circulates. If the convective heat transfer coefficient is \(11 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\), estimate the rate of heat loss from the sphere when its emittance is (i) \(0.05\). (ii) \(0.85 .\)

A natural convection heat transfer coefficient meter is intended for situations where the air temperature \(T_{e}\) is known but the surrounding surfaces are at an unknown temperature \(T_{w}\). The two sensors that make up the meter each have a surface area of \(1 \mathrm{~cm}^{2}\), one has a surface coating of emittance \(\varepsilon_{1}=0.9\), and the other has an emittance of \(\varepsilon_{2}=0.1\). The rear surface of the sensors is well insulated. When \(T_{e}=300 \mathrm{~K}\) and the test surface is at \(320 \mathrm{~K}\), the power inputs required to maintain the sensor surfaces at \(320 \mathrm{~K}\) are \(\dot{Q}_{1}=21.7 \mathrm{~mW}\) and \(\dot{Q}_{2}=8.28 \mathrm{~mW}\). Determine the heat transfer coefficient at the meter location.

A \(1 \mathrm{~m}\)-high vertical wall is maintained at \(310 \mathrm{~K}\), when the surrounding air is at 1 atm and \(290 \mathrm{~K}\). Plot the local heat transfer coefficient as a function of location up the wall. Take \(v=15.7 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\) for air. Also calculate the convective heat loss per meter width of wall.

A chemical reactor has a \(5 \mathrm{~mm}\)-thick mild steel wall and is lined inside with a 2 \(\mathrm{mm}\)-thick layer of polyvinylchloride. The contents are at \(80^{\circ} \mathrm{C}\), and the ambient air is at \(20^{\circ} \mathrm{C}\). The inside thermal resistance is negligible ( \(h_{c, i}\) very large), and the outside heat transfer coefficient for combined convection and radiation is \(7 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\). (i) Draw the thermal circuit. (ii) Plot a graph of the temperature profile through the wall. (iii) Calculate the rate of heat loss for a surface area of \(10 \mathrm{~m}^{2} .\)
See all solutions

Recommended explanations on Physics Textbooks

Thermodynamics

Read Explanation

Classical Mechanics

Read Explanation

Radiation

Read Explanation

Energy Physics

Read Explanation

Torque and Rotational Motion

Read Explanation

Geometrical and Physical Optics

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.