/*! 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 36 As permanent space stations incr... [FREE SOLUTION] | 91Ó°ÊÓ

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As permanent space stations increase in size, there is an attendant increase in the amount of electrical power they dissipate. To keep station compartment temperatures from exceeding prescribed limits, it is necessary to transfer the dissipated heat to space. A novel heat rejection scheme that has been proposed for this purpose is termed a Liquid Droplet Radiator (LDR). The heat is first transferred to a high vacuum oil, which is then injected into outer space as a stream of small droplets. The stream is allowed to traverse a distance \(L\), over which it cools by radiating energy to outer space at absolute zero temperature. The droplets are then collected and routed back to the space station. Consider conditions for which droplets of emissivity \(\varepsilon=0.95\) and diameter \(D=0.5 \mathrm{~mm}\) are injected at a temperature of \(T_{i}=500 \mathrm{~K}\) and a velocity of \(V=0.1 \mathrm{~m} / \mathrm{s}\). Properties of the oil are \(\rho=885 \mathrm{~kg} / \mathrm{m}^{3}, c=1900 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and \(k=0.145 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Assuming each drop to radiate to deep space at \(T_{\text {sur }}=0 \mathrm{~K}\), determine the distance \(L\) required for the droplets to impact the collector at a final temperature of \(T_{f}=300 \mathrm{~K}\). What is the amount of thermal energy rejected by each droplet?

Short Answer

Expert verified
To find the required distance \(L\) for the Liquid Droplet Radiator, we first calculated the surface area of the droplet as \(A = 4 \pi r^2\), where \(r = \frac{D}{2}\), and then found the radiation heat transfer using the Stefan-Boltzmann law, given by \(q_{rad} = \varepsilon A \sigma (T_i^4 - T_{sur}^4)\). Next, we used an energy balance to determine the cooling time: \(q_{rad}\cdot t = m c (T_i - T_f)\). Solving for \(t\), we calculated the mass of a single droplet using the volume and density of the oil, with \(m = \rho V\). Finally, we determined the required distance \(L\) as \(L = V \cdot t\). The amount of thermal energy rejected by each droplet was calculated as \(Q = m c (T_i - T_f)\).

Step by step solution

01

Calculate the area of the droplet

First, find the surface area of the droplet, which is modeled as a sphere: \[A = 4 \pi r^2\] where \(r\) is the radius of the droplet, which can be calculated as: \[r = \frac{D}{2}\]
02

Calculate the radiation heat transfer

Next, we need to find the heat transfer by radiation from each droplet, using the Stefan-Boltzmann law: \[q_{rad} = \varepsilon A \sigma (T_i^4 - T_{sur}^4)\] where \(\varepsilon\) is the emissivity, \(A\) is the area calculated in Step 1, \(\sigma\) is the Stefan-Boltzmann constant \(\left(5.67 \times 10^{-8} \frac{W}{m^2 K^4}\right)\), and \(T_i\) and \(T_{sur}\) are droplet and surrounding temperatures, respectively.
03

Calculate the cooling time

Now, we need to find the time required to cool the droplet from the initial temperature \(T_i\) to the final temperature \(T_f\). To do this, we will use an energy balance equation. The amount of heat removed by radiation is equal to the amount of heat lost by the droplet: \[q_{rad}\cdot t = m c (T_i - T_f)\] where \(t\) is the cooling time, \(m\) is the mass of a single droplet, \(c\) is the specific heat capacity of the oil and \(T_i\) and \(T_f\) are initial and final temperatures, respectively. To find the mass of a single droplet, use the volume formula for a sphere: \[V = \frac{4}{3} \pi r^3\] Then, use the density of the oil \(\rho\) to find the mass of a single droplet: \[m = \rho V\] Solve for \(t\) in the energy balance equation.
04

Calculate the required distance

Finally, calculate the distance \(L\) required for radiative cooling, using the droplet velocity &\(V\)& and the cooling time &\(t\)& as follows: \[L = V \cdot t\] Now we have calculated the distance \(L\) for the droplets to impact the collector at a final temperature of \(300 \mathrm{~K}\).
05

Determine the amount of thermal energy rejected by each droplet

To calculate the amount of thermal energy rejected by each droplet, use the energy balance equation from Step 3: \[Q = m c (T_i - T_f)\] Simplify the equation and solve for the amount of thermal energy rejected by each droplet, \(Q\).

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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 serves as a cornerstone for understanding the thermal radiation emitted by objects in space. According to this law, the power radiated per unit area of a black body is proportional to the fourth power of its absolute temperature. It's given by the equation:
\[ P = \frac{dQ}{dt} = \'epsilon sigma T^4 \'
\], where \( P \) is the power radiated per unit area, \( \epsilon \) is the emissivity of the material, \( \sigma \) represents the Stefan-Boltzmann constant, and \( T \) is the absolute temperature of the body.\
\
\In the context of space applications, where vacuum conditions and the absence of an atmosphere mean that convective and conductive heat transfer are effectively non-existent, the Stefan-Boltzmann law allows us to calculate how much heat an object like a liquid droplet radiator will radiate into space. It helps in designing cooling systems that manage the heat generated by a spacecraft effectively, preventing overheating of onboard systems and equipment.

Emissivity and Its Role in Radiative Cooling

The emissivity \( \epsilon \) is a measure of how effectively a material emits thermal radiation compared to a perfect black body. A perfect black body, which absorbs all incident radiation, would have an emissivity of 1. In reality, all materials have emissivities less than 1, meaning they radiate less energy than a perfect black body at the same temperature. The liquid droplet radiator's high emissivity of 0.95 used in the exercise indicates it’s quite effective at radiating energy, a crucial characteristic for a system designed to dissipate heat in the challenging environmental conditions of outer space.
Energy Balance Equation
Another fundamental concept in thermodynamics and heat transfer is the energy balance equation, which is crucial to calculating the thermal energy transfer in systems like the Liquid Droplet Radiator (LDR) mentioned in the exercise. For the droplets in LDR systems, the energy balance can be represented as:

\[ Energy_{\text{in}} - Energy_{\text{out}} = \Delta Energy_{\text{system}} \]
In simpler terms, the change in energy of the system is the difference between what energy comes in and what goes out. For the droplets cooling in space, all of the heat they lose (Energy_out) is due to radiation. There's no 'in' energy since space is at absolute zero, or \(0\mathrm{~K}\), and cannot supply heat to the droplets. The equation reflects the fact that energy cannot be created or destroyed (according to the first law of thermodynamics), only transformed or transferred.

In the context of the textbook exercise, the energy balance equation takes the form:
\[ q_{rad} \times t = m \times c \times (T_{i} - T_{f}) \]
Here, \( q_{rad} \) is the heat radiated per second (found using the Stefan-Boltzmann law), \( t \) is the time it takes for the droplet to cool, \( m \) is the mass of the droplet, \( c \) is the specific heat capacity of the oil, and \( T_{i} \) and \( T_{f} \) are the initial and final temperatures. This equation allows us to solve for the unknown in the system, such as the time taken for the droplet to cool — a critical factor when engineering the cooling system for space applications.
Liquid Droplet Radiator
The Liquid Droplet Radiator (LDR) is a novel concept designed to manage waste heat in space through the principles of radiative cooling. It comprises streams of liquid droplets that absorb heat from a spacecraft and radiate it away into space. The droplets play a key role mainly because liquids can absorb a significant amount of heat before they reach high temperatures, and as small droplets, they provide a large surface area relative to their volume, optimizing the radiation process.

In practice, as outlined in our exercise, the LDR system injects oil droplets into space where they cool down by radiating their heat away. These droplets, which have a high surface-to-mass ratio, are then recollected and recycled back into the system, creating a closed-loop that’s efficient at dispatching heat without losing fluid. This process can be seen as an elegant application of the Stefan-Boltzmann law; by controlling the droplets' diameter, emissivity, and travel distance, the system can be designed to keep space station compartments within prescribed temperature limits.

Efficiency and Optimization of LDR Systems

Efficient LDR systems are characterized by their ability to reject thermal energy rapidly into the harsh environment of space. Optimization involves adjusting parameters like droplet size, emissivity, travel distance, and even fluid properties such as specific heat capacity and thermal conductivity to ensure that the heat generated by onboard equipment is managed correctly so that the station remains within safe operating temperatures. The exercise calculation informs this balance, determining the distance required for the droplets to achieve the desired temperature before recollection. The ability of LDRs to seamlessly integrate with spacecraft systems makes them a promising solution for future space missions with high thermal loads.

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Most popular questions from this chapter

A sphere \(30 \mathrm{~mm}\) in diameter initially at \(800 \mathrm{~K}\) is quenched in a large bath having a constant temperature of \(320 \mathrm{~K}\) with a convection heat transfer coefficient of \(75 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The thermophysical properties of the sphere material are: \(\rho=400 \mathrm{~kg} / \mathrm{m}^{3}, c=1600 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and \(k=1.7 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). (a) Show, in a qualitative manner on \(T \dashv\) coordinates, the temperatures at the center and at the surface of the sphere as a function of time. (b) Calculate the time required for the surface of the sphere to reach \(415 \mathrm{~K}\). (c) Determine the heat flux \(\left(\mathrm{W} / \mathrm{m}^{2}\right)\) at the outer surface of the sphere at the time determined in part (b). (d) Determine the energy (J) that has been lost by the sphere during the process of cooling to the surface temperature of \(415 \mathrm{~K}\). (e) At the time determined by part (b), the sphere is quickly removed from the bath and covered with perfect insulation, such that there is no heat loss from the surface of the sphere. What will be the temperature of the sphere after a long period of time has elapsed? (f) Compute and plot the center and surface temperature histories over the period \(0 \leq t \leq 150 \mathrm{~s}\). What effect does an increase in the convection coefficient to \(h=200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) have on the foregoing temperature histories? For \(h=75\) and \(200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), compute and plot the surface heat flux as a function of time for \(0 \leq t \leq 150 \mathrm{~s}\).

As part of a heat treatment process, cylindrical, 304 stainless steel rods of \(100-\mathrm{mm}\) diameter are cooled from an initial temperature of \(500^{\circ} \mathrm{C}\) by suspending them in an oil bath at \(30^{\circ} \mathrm{C}\). If a convection coefficient of \(500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) is maintained by circulation of the oil, how long does it take for the centerline of a rod to reach a temperature of \(50^{\circ} \mathrm{C}\), at which point it is withdrawn from the bath? If 10 rods of length \(L=1 \mathrm{~m}\) are processed per hour, what is the nominal rate at which energy must be extracted from the bath (the cooling load)?

Special coatings are often formed by depositing thin layers of a molten material on a solid substrate. Solidification begins at the substrate surface and proceeds until the thickness \(S\) of the solid layer becomes equal to the thickness \(\delta\) of the deposit. (a) Consider conditions for which molten material at its fusion temperature \(T_{f}\) is deposited on a large substrate that is at an initial uniform temperature \(T_{i}\). With \(S=0\) at \(t=0\), develop an expression for estimating the time \(t_{d}\) required to completely solidify the deposit if it remains at \(T_{f}\) throughout the solidification process. Express your result in terms of the substrate thermal conductivity and thermal diffusivity \(\left(k_{s}, \alpha_{s}\right)\), the density and latent heat of fusion of the deposit \(\left(\rho, h_{s f}\right)\), the deposit thickness \(\delta\), and the relevant temperatures \(\left(T_{f}, T_{i}\right)\). (b) The plasma spray deposition process of Problem \(5.25\) is used to apply a thin \((\delta=2 \mathrm{~mm})\) alumina coating on a thick tungsten substrate. The substrate has a uniform initial temperature of \(T_{i}=300 \mathrm{~K}\), and its thermal conductivity and thermal diffusivity may be approximated as \(k_{s}=120 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\alpha_{s}=4.0 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\), respectively. The density and latent heat of fusion of the alumina are \(\rho=3970 \mathrm{~kg} / \mathrm{m}^{3}\) and \(h_{s f}=3577 \mathrm{~kJ} / \mathrm{kg}\), respectively, and the alumina solidifies at its fusion temperature \(\left(T_{f}=2318 \mathrm{~K}\right)\). Assuming that the molten layer is instantaneously deposited on the substrate, estimate the time required for the deposit to solidify.

A tile-iron consists of a massive plate maintained at \(150^{\circ} \mathrm{C}\) by an embedded electrical heater. The iron is placed in contact with a tile to soften the adhesive, allowing the tile to be easily lifted from the subflooring. The adhesive will soften sufficiently if heated above \(50^{\circ} \mathrm{C}\) for at least \(2 \mathrm{~min}\), but its temperature should not exceed \(120^{\circ} \mathrm{C}\) to avoid deterioration of the adhesive. Assume the tile and subfloor to have an initial temperature of \(25^{\circ} \mathrm{C}\) and to have equivalent thermophysical properties of \(k=0.15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\rho c_{p}=1.5 \times 10^{6}\) \(\mathrm{J} / \mathrm{m}^{3} \cdot \mathrm{K}\) Tile, 4-mm thickness Subflooring (a) How long will it take a worker using the tile-iron to lift a tile? Will the adhesive temperature exceed \(120^{\circ} \mathrm{C} ?\) (b) If the tile-iron has a square surface area \(254 \mathrm{~mm}\) to the side, how much energy has been removed from it during the time it has taken to lift the tile?

The heat transfer coefficient for air flowing over a sphere is to be determined by observing the temperature-time history of a sphere fabricated from pure copper. The sphere, which is \(12.7 \mathrm{~mm}\) in diameter, is at \(66^{\circ} \mathrm{C}\) before it is inserted into an airstream having a temperature of \(27^{\circ} \mathrm{C}\). A thermocouple on the outer surface of the sphere indicates \(55^{\circ} \mathrm{C} 69 \mathrm{~s}\) after the sphere is inserted into the airstream. Assume and then justify that the sphere behaves as a spacewise isothermal object and calculate the heat transfer coefficient.

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