/*! 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 26 A radioisotope power source for ... [FREE SOLUTION] | 91Ó°ÊÓ

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A radioisotope power source for a space vehicle is in the form of a 10 \(\mathrm{cm}\)-diameter solid sphere, in which heat is generated uniformly. At steady state its surface temperature is \(600 \mathrm{~K}\). It is contained inside a solid spherical shell of inner and outer diameters \(20 \mathrm{~cm}\) and \(40 \mathrm{~cm}\), respectively, and the space between is evacuated. The outer surface of the containment shell sees space, which can be taken to be a blackbody at \(0 \mathrm{~K}\). The thermal conductivities of the power source and containment shell are \(2.0\) and \(3.2 \mathrm{~W} / \mathrm{m} \mathrm{K}\), respectively. All the surfaces are gray with the same emittance \(\varepsilon\). What value of \(\varepsilon\) must be specified for the surface finish in order to limit the heat loss from the assembly to \(100 \mathrm{~W}\) ?

Short Answer

Expert verified
The emittance \( \varepsilon \) required to limit the heat loss to 100 W is determined by balancing conduction and radiation losses with \( \varepsilon \).

Step by step solution

01

Model the Heat Transfer

Let's first understand that the heat transfer occurs via conduction through the spherical shell and radiation from the outer shell to space. Since the space is a blackbody at 0 K, only radiation losses need to be considered for the outer shell.
02

Calculate Conduction Through the Spherical Shell

We use Fourier's law for the conduction heat transfer rate from the inner surface to the outer surface of the shell: \[ Q_{cond} = \frac{4 \pi k (T_{inner} - T_{outer})}{\left(\frac{1}{r_{outer}} - \frac{1}{r_{inner}}\right)} \]where \( k = 3.2 \ \mathrm{W/m \cdot K} \), \( T_{inner} = 600 \ \mathrm{K} \), \( r_{inner} = 0.1 \ \mathrm{m} \), \( r_{outer} = 0.2 \ \mathrm{m} \).
03

Radiation from the Outer Surface

The heat loss by radiation from the outer surface can be calculated using:\[ Q_{rad} = \varepsilon \sigma A (T_{outer}^4 - T_{space}^4) \]where \( \varepsilon \) is the emittance we need to find, \( A = 4 \pi r_{outer}^2 \), \( T_{space} = 0 \ \mathrm{K} \), and \( \sigma = 5.67 \times 10^{-8} \ \mathrm{W/m^2\cdot K^4} \).
04

Set Up the Energy Balance

Apply the condition for steady state: the total heat loss \( Q_{total} \) (sum of conduction \( Q_{cond} \) and radiation \( Q_{rad} \)) should be \(100 \ \mathrm{W}\). Therefore,\[ Q_{cond} + Q_{rad} = 100 \ \mathrm{W} \]
05

Solve for Emittance \( \varepsilon \)

Solve the energy balance equation to find \( \varepsilon \). Substitute expressions from Steps 2 and 3 into the energy balance equation and rearrange to solve for \( \varepsilon \). This involves calculating conduction and rearranging the equation to find the required emittance.

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Key Concepts

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

Radioisotope Power Source
A radioisotope power source is a device that generates electricity using the heat released by the decay of radioactive isotopes. It is particularly useful in applications where solar power is inadequate, such as deep-space missions. The decay process of the isotopes releases thermal energy, which can then be converted into electrical energy.

In this exercise, the radioisotope power source is represented as a solid sphere where heat is generated uniformly throughout its volume. This heat must then be managed effectively to avoid overheating and ensure efficient energy conversion.

Key advantages of radioisotope power sources include:
  • High energy density, allowing for compactness and long-term energy supply.
  • No reliance on solar power, making them ideal for space exploration.
  • Robustness, with little maintenance required over long periods.
Understanding how to evaluate and limit heat loss from these sources is critical to their efficiency and longevity.
Spherical Shell
A spherical shell in scientific problems is often used to model layers of material that surround a core. In our exercise, the power source is enclosed within a spherical shell, which consists of both inner and outer diameters. This shell facilitates the conduction of heat away from the power source before it is radiated into the surrounding space.

The dimensions of the shell—specifically its inner and outer diameters—play a critical role in determining the rate of heat conduction. The evacuated space between the shell and the power source prevents convective heat transfer, allowing for a more straightforward analysis of conductive and radiative losses.

To properly understand the heat transfer dynamics in spherical shells, one should consider:
  • Inner and outer diameters: affecting the surface area exposed for heat conduction.
  • Material properties: influencing how effectively heat can traverse through the shell.
  • Structural role: shielding the power source and managing its thermal output.
Grasping these elements will greatly assist in solving problems involving spherical shell geometries.
Thermal Conductivity
Thermal conductivity is a material's ability to conduct heat. In physics problems, it is a crucial property when analyzing heat transfer through solids. The thermal conductivity coefficient, denoted by \( k \), determines how effectively a material can transfer heat across its temperature gradient.

In the shell described in the exercise, thermal conductivity is utilized to calculate the heat conduction rate from the inner surface to the outer surface. The Fourier's law of heat conduction provides us with the formula that incorporates this coefficient:

\[ Q_{cond} = \frac{4 \pi k (T_{inner} - T_{outer})}{\left(\frac{1}{r_{outer}} - \frac{1}{r_{inner}}\right)} \]

Key factors affecting thermal conductivity include:
  • Material type: different materials have varying abilities to conduct heat, e.g., metals generally have higher thermal conductivities than non-metals.
  • Temperature: thermal conductivity can change with temperature, impacting the rate of heat transfer through the material.
Efficient heat management in engineering often involves selecting materials with suitable thermal conductivities for their intended application.
Emittance
Emittance, often represented by \( \varepsilon \), is a measure of a surface's effectiveness in emitting thermal radiation. It is a crucial parameter in the study of heat transfer, specifically in problems involving radiative heat loss. Emittance is a dimensionless quantity that ranges from 0 (perfect reflector) to 1 (perfect emitter, or blackbody).

In the exercise problem, calculating the required emittance of the shell's surface involves setting up an energy balance that takes into account both conduction and radiation. The formula used to calculate the radiative heat loss is:

\[ Q_{rad} = \varepsilon \sigma A (T_{outer}^4 - T_{space}^4) \]

Understanding emittance is vital for configuring surfaces to either conserve heat or enhance radiative cooling. Here are some critical points about emittance:
  • Surface finish: materials with different textures or coatings can have differing emittances.
  • Temperature effects: a material's emittance might change with its temperature, particularly at higher temperature ranges.
  • System requirements: correctly specifying emittance can help in designing systems that meet specific thermal performance benchmarks.
Mastering this concept can significantly improve accuracy in predicting and controlling heat loss in thermal systems.

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Most popular questions from this chapter

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 \(1 \mathrm{~cm}\)-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). $$ T_{e}=27^{2} \mathrm{C} .50 \mathrm{~cm} \mathrm{RH} $$

Calculate the blackbody emissive power \(E_{b}\) for surfaces at \(300 \mathrm{~K}, 1000 \mathrm{~K}, 2000 \mathrm{~K}\), and \(5000 \mathrm{~K}\). Also estimate the peak wavelength \(\lambda_{\max }\) for each surface using: (i) \(\lambda_{\max } T=2898 \mu \mathrm{m} \mathrm{K}\) (from Eq. \(6.6\) and Fig. \(6.3 a\) ) (ii) \(\lambda_{\max } T=5100 \mu \mathrm{m} \mathrm{K}\) (from Fig. 6.3b)

Two \(20 \mathrm{~cm}-\) diameter parallel disks, \(10 \mathrm{~cm}\) apart, have facing sides at \(1000 \mathrm{~K}\) and \(700 \mathrm{~K}\), and insulated backs. For gray disk surfaces with \(\varepsilon=0.4\), and black surroundings at \(0 \mathrm{~K}\), determine the net radiation exchange between the disks and the heat loss from each disk.

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 \(300 \mathrm{~K}\) ?

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