/*! 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

The stability criterion for the explicit method requires that the coefficient of the \(T_{m}^{p}\) term of the one-dimensional, finite-difference equation be zero or positive. Consider the situation for which the temperatures at the two neighboring nodes \(\left(T_{\mathrm{m}-1}^{p}, T_{\mathrm{m}+1}^{p}\right)\) are \(100^{\circ} \mathrm{C}\) while the center node \(\left(T_{m}^{p}\right)\) is at \(50^{\circ} \mathrm{C}\). Show that for values of \(F o>\frac{1}{2}\) the finite-difference equation will predict a value of \(T_{m}^{p+1}\) that violates the second law of thermodynamics.

A plate of thickness \(2 L=25 \mathrm{~mm}\) at a temperature of \(600^{\circ} \mathrm{C}\) is removed from a hot pressing operation and must be cooled rapidly to achieve the required physical properties. The process engineer plans to use air jets to control the rate of cooling, but she is uncertain whether it is necessary to cool both sides (case 1 ) or only one side (case 2) of the plate. The concern is not just for the time-to-cool, but also for the maximum temperature difference within the plate. If this temperature difference is too large, the plate can experience significant warping. The air supply is at \(25^{\circ} \mathrm{C}\), and the convection coefficient on the surface is \(400 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The thermophysical properties of the plate are \(\rho=3000 \mathrm{~kg} / \mathrm{m}^{3}, c=750 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and \(k=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). (a) Using the IHT software, calculate and plot on one graph the temperature histories for cases 1 and 2 for a \(500-s\) cooling period. Compare the times required for the maximum temperature in the plate to reach \(100^{\circ} \mathrm{C}\). Assume no heat loss from the unexposed surface of case 2 . (b) For both cases, calculate and plot on one graph the variation with time of the maximum temperature difference in the plate. Comment on the relative magnitudes of the temperature gradients within the plate as a function of time.

An electrical cable, experiencing uniform volumetri generation \(\dot{q}\), is half buried in an insulating materia while the upper surface is exposed to a convection process \(\left(T_{\infty}, h\right)\). (a) Derive the explicit, finite-difference equations for an interior node \((m, n)\), the center node \((m=0)\), and the outer surface nodes \((M, n)\) for the convection and insulated boundaries. (b) Obtain the stability criterion for each of the finitedifference equations. Identify the most restrictive criterion.

In heat treating to harden steel ball bearings \(\left(c=500 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, \rho=7800 \mathrm{~kg} / \mathrm{m}^{3}, k=50 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\), it is desirable to increase the surface temperature for a short time without significantly warming the interior of the ball. This type of heating can be accomplished by sudden immersion of the ball in a molten salt bath with \(T_{\infty}=1300 \mathrm{~K}\) and \(h=5000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assume that any location within the ball whose temperature exceeds \(1000 \mathrm{~K}\) will be hardened. Estimate the time required to harden the outer millimeter of a ball of diameter \(20 \mathrm{~mm}\), if its initial temperature is \(300 \mathrm{~K}\).

Thermal stress testing is a common procedure used to assess the reliability of an electronic package. Typically, thermal stresses are induced in soldered or wired connections to reveal mechanisms that could cause failure and must therefore be corrected before the product is released. As an example of the procedure, consider an array of silicon chips \(\left(\rho_{\text {ch }}=2300 \mathrm{~kg} / \mathrm{m}^{3}\right.\), \(\left.c_{\text {ch }}=710 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) joined to an alumina substrate \(\left(\rho_{\mathrm{sb}}=4000 \mathrm{~kg} / \mathrm{m}^{3}, c_{\mathrm{sb}}=770 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) by solder balls \(\left(\rho_{\mathrm{sd}}=11,000 \mathrm{~kg} / \mathrm{m}^{3}, c_{\mathrm{sd}}=130 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\). Each chip of width \(L_{\mathrm{ch}}\) and thickness \(t_{\mathrm{ch}}\) is joined to a unit substrate section of width \(L_{\mathrm{sb}}\) and thickness \(t_{\mathrm{sb}}\) by solder balls of diameter \(D\). A thermal stress test begins by subjecting the multichip module, which is initially at room temperature, to a hot fluid stream and subsequently cooling the module by exposing it to a cold fluid stream. The process is repeated for a prescribed number of cycles to assess the integrity of the soldered connections. (a) As a first approximation, assume that there is negligible heat transfer between the components (chip/solder/substrate) of the module and that the thermal response of each component may be determined from a lumped capacitance analysis involving the same convection coefficient \(h\). Assuming no reduction in surface area due to contact between a solder ball and the chip or substrate, obtain expressions for the thermal time constant of each component. Heat transfer is to all surfaces of a chip, but to only the top surface of the substrate. Evaluate the three time constants for \(L_{\mathrm{ch}}=15 \mathrm{~mm}\), \(t_{\mathrm{ch}}=2 \mathrm{~mm}, L_{\mathrm{sb}}=25 \mathrm{~mm}, t_{\mathrm{sb}}=10 \mathrm{~mm}, D=2 \mathrm{~mm}\), and a value of \(h=50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), which is characteristic of an airstream. Compute and plot the temperature histories of the three components for the heating portion of a cycle, with \(T_{i}=20^{\circ} \mathrm{C}\) and \(T_{\infty}=80^{\circ} \mathrm{C}\). At what time does each component experience \(99 \%\) of its maximum possible temperature rise, that is, \(\left(T-T_{i}\right) /\left(T_{\infty}-T_{i}\right)=0.99\) ? If the maximum stress on a solder ball corresponds to the maximum difference between its temperature and that of the chip or substrate, when will this maximum occur? (b) To reduce the time required to complete a stress test, a dielectric liquid could be used in lieu of air to provide a larger convection coefficient of \(h=200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). What is the corresponding savings in time for each component to achieve \(99 \%\) of its maximum possible temperature rise?

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