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Chapter 6: Problem 51

Two \(5 \mathrm{~mm}\)-thick rectangular aluminum alloy plates are attached to the exterior [ - surface of a spacecraft to form a channel \(12 \mathrm{~cm}\) long, \(2 \mathrm{~cm}\) wide, and \(6 \mathrm{~cm}\) deep. If the plates and wall are at a uniform temperature of \(310 \mathrm{~K}\), calculate the radiation heat transfer from the inside of the channel to space (taken as black at \(0 \mathrm{~K}\) ). Take \(\varepsilon=0.23\) for the aluminum.

Short Answer

Expert verified
The radiation heat transfer rate is approximately 0.205 W.

Step by step solution

01

Understand the Problem

We need to find the rate of radiation heat transfer from the inside surface of a channel formed by aluminum plates at a temperature of 310 K into space, which is considered as a black body at 0 K. The area of the channel is 12 cm long, 2 cm wide, and 6 cm deep.
02

Calculate the Surface Area of the Channel

The channel has three surfaces that will radiate heat: the base, and two sides. The area of the base is \( A_1 = 12 imes 2 = 24 \; \text{cm}^2 \). Each side has an area of \( A_2 = 12 imes 6 = 72 \; \text{cm}^2 \). Thus, the total radiating area is \( A = A_1 + 2A_2 = 24 + 144 = 168 \; \text{cm}^2 \). Convert this into square meters: \( A = 0.0168 \; \text{m}^2 \).
03

Use Stefan-Boltzmann Law to Calculate Radiative Heat Transfer

The Stefan-Boltzmann law for radiative heat transfer is given by: \[ Q = \varepsilon \sigma A (T_s^4 - T_{\text{space}}^4) \]where \( Q \) is the heat transfer, \( \varepsilon = 0.23 \) is the emissivity, \( \sigma = 5.67 \times 10^{-8} \; \text{W/m}^2\text{K}^4 \) is the Stefan-Boltzmann constant, \( A \) is the area, \( T_s = 310 \; \text{K} \) is the surface temperature, and \( T_{\text{space}} = 0 \; \text{K} \).
04

Calculate the Heat Transfer Rate

Substitute the known values into the Stefan-Boltzmann formula:\[ Q = 0.23 \times 5.67 \times 10^{-8} \times 0.0168 \times (310^4 - 0^4) \]\[ Q = 0.23 \times 5.67 \times 10^{-8} \times 0.0168 \times 9.299\times 10^9 \]\[ Q \approx 0.205 \; \text{W} \]

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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 cornerstone concept in understanding radiation heat transfer. It states that the power radiated by a black body in thermal equilibrium per unit area is proportional to the fourth power of the temperature of that body. This is mathematically expressed as: \[ Q = \varepsilon \sigma A (T_s^4 - T_{\text{space}}^4) \] Here's the breakdown:
  • \( Q \) is the heat transfer rate you wish to calculate.
  • \( \varepsilon \) represents the emissivity of the surface, denoting its efficiency in emitting thermal radiation compared to a perfect black body.
  • \( \sigma \) is the Stefan-Boltzmann constant, a fixed value \(5.67 \times 10^{-8} \, \ ext{W/m}^2\text{K}^4 \).
  • \( T_s \) and \( T_{\text{space}} \) stand for the surface and surrounding space temperatures in Kelvin, respectively.
In our exercise, this principle allows us to determine the radiation heat transfer from the spacecraft's aluminum plates into the chilling vacuum of space. By utilizing this law, we're able to mathematically capture the thermal energy exchange inherent in nearly every thermal system.
Emissivity
Emissivity is a crucial factor in the calculation of radiation heat transfer. It measures a material's ability to emit thermal radiation compared to an ideal black body. Values range from 0 to 1, where 1 indicates a perfect black body that perfectly emits and absorbs all thermal radiation. In this exercise, the emissivity of the aluminum plates is given as \( \varepsilon = 0.23 \). This implies that aluminum is not a great emitter compared to a black body, making it an important parameter in computing the heat lost through radiation. Key points about emissivity:
  • It influences how much radiation a surface will emit at a given temperature.
  • Emissivity varies based on material type, surface texture, and condition.
  • Low emissivity materials (like our aluminum plates) will emit less heat and retain more energy, which can be beneficial in certain applications.
Aluminum Plates
In our scenario, aluminum plates are used to form a part of the spacecraft's exterior channel. Aluminum is often chosen in thermal applications because of its excellent thermal conductivity and relatively low emissivity. Why use aluminum plates?
  • Lightweight: Aluminum is light, which is crucial for spacecraft, where minimizing weight is paramount.
  • Heat dissipation: Aluminum alloys effectively distribute heat across their surface, aiding in the even radiation of thermal energy.
  • Corrosion resistance: Aluminum resists oxidation, making it suitable for the harsh environment of space.
In the exercise, aluminum plates form a channel that serves as a key area for radiative heat transfer from the spacecraft to space. Their emissivity and low heat emission quality ensure that energy loss is minimized, which is critical when dealing with the limited energy resources available on spacecraft.
Spacecraft Heat Transfer
Heat transfer in spacecraft revolves around managing thermal energy to ensure the functionality and safety of onboard systems and instruments. In the vacuum of space, heat transfer occurs primarily through radiation since conduction and convection, common on Earth, are nearly nonexistent in space. Here's why efficient heat management is vital:
  • Temperature regulation: Ensuring equipment operates within safe temperature limits is essential for performance and longevity.
  • Energy management: Minimizing unwanted heat loss helps conserve the energy precious for running the spacecraft systems.
  • Survival of materials: Different parts of the spacecraft need to withstand extreme temperatures found in space.
The problem involving aluminum plates captures the principles of heat transfer via radiation to space. By calculating how much heat is emitted from the channel formed by these plates, we gain insights into managing this crucial aspect of spacecraft design and operation.

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Most popular questions from this chapter

A square room of sides \(8 \mathrm{~m}\) has a \(3 \mathrm{~m}\)-diameter radiant heating panel centrally located on the ceiling, \(3 \mathrm{~m}\) above the floor. The panel can be taken to be black and is maintained at \(60^{\circ} \mathrm{C}\). Calculate the irradiation on the floor due to the heater, at the center of the room, and in one corner.

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 kiln has an average inside surface temperature of \(2330 \mathrm{~K}\) and has a small \(15 \times\) \(15 \mathrm{~cm}\) square opening in its \(20 \mathrm{~cm}\)-thick walls. If the sides of the opening can be assumed to be adiabatic, determine the rate of radiant energy loss through the opening.

A long, square evacuated duct has \(1 \mathrm{~m}\) sides. Two adjacent sides are perfectly insulated and have an emittance of \(0.4\). The other two sides have an emittance of \(0.5\). One of these sides is cooled and has a thermostat temperature control; the other contains an embedded electrical heater. If the thermostat maintains the cooled surface at \(1000 \mathrm{~K}\), at what rate must power be supplied to the electrical heater to maintain the two insulated sides at \(2000 \mathrm{~K}\) ? Also, what is the temperature of the heated side? (i) Assume that the two insulated sides can be represented by a single isothermal surface. (ii) Can the problem be solved if the two insulated sides are represented by separate isothermal surfaces? (iii) Comment on the validity of the approach used in part (i).

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 \(100 \mathrm{~W}\) ?
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