/*! 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 101 An instrumentation transmitter p... [FREE SOLUTION] | 91Ó°ÊÓ

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An instrumentation transmitter pod is a box containing electronic circuitry and a power supply for sending sensor signals to a base receiver for recording. Such a pod is placed on a conveyor system, which passes through a large vacuum brazing furnace as shown in the sketch. The exposed surfaces of the pod have a special diffuse, opaque coating with spectral emissivity as shown. To stabilize the temperature of the pod and prevent overheating of the electronics, the inner surface of the pod is surrounded by a layer of a phase- change material (PCM) having a fusion temperature of \(87^{\circ} \mathrm{C}\) and a heat of fusion of \(25 \mathrm{~kJ} / \mathrm{kg}\). The pod has an exposed surface area of \(0.040 \mathrm{~m}^{2}\) and the mass of the PCM is \(1.6 \mathrm{~kg}\). Furthermore, it is known that the power dissipated by the electronics is \(50 \mathrm{~W}\). Consider the situation when the pod enters the furnace at a uniform temperature of \(87^{\circ} \mathrm{C}\) and all the \(\mathrm{PCM}\) is in the solid state. How long will it take before all the PCM changes to the liquid state?

Short Answer

Expert verified
It will take \(800 \mathrm{~s}\) for all the phase-change material to change from the solid state to the liquid state.

Step by step solution

01

Identify the energy input to the PCM

The energy input to the PCM comes from the electronic power dissipation inside the pod. We are given this value as \(50 \mathrm{~W}\).
02

Calculate the energy needed for the phase change of PCM

The energy needed to change the phase of the PCM can be calculated using the formula: Energy needed = mass (PCM) \(\times\) heat of fusion Where: mass (PCM) = \(1.6 \mathrm{~kg}\) heat of fusion = \(25 \mathrm{~kJ} / \mathrm{kg}\) Now we need to convert the heat of fusion from \(\mathrm{kJ}\) to \(\mathrm{J}\): heat of fusion = \(25 \mathrm{~kJ/kg} \times 1000 \mathrm{~J/ kJ} = 25000 \mathrm{~J/kg}\) Therefore, the energy needed is: Energy needed = \(1.6 \mathrm{~kg} \times 25000 \mathrm{~J/kg} = 40000 \mathrm{~J}\)
03

Apply the energy conservation concept

Now we will use the energy conservation concept to find the time necessary for the phase change: Energy input = Energy needed Since energy is power multiplied by time, we can write: Power (electronics) \(\times\) time = Energy needed
04

Calculate the time for the phase change

Now, we will calculate the time necessary for the phase change: \(\text{time} = \frac{\text{Energy needed}}{\text{Power (electronics)}}\) Where: Energy needed = \(40000~\text{J}\) Power (electronics) = \(50 \mathrm{~W}\) So, the time required is: \(\text{time} = \frac{40000 \mathrm{~J}}{50 \mathrm{~W}} = 800 \mathrm{~s}\) Thus, it will take 800 seconds for all the PCM to change from the solid state to the liquid state.

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Key Concepts

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

Energy Conservation
Energy conservation is a fundamental principle that governs the behavior of energy in any system. The essence of energy conservation is that energy cannot be created or destroyed; it can only be transferred from one form to another. In the context of phase change materials (PCMs), this principle is vital for understanding how energy is managed and utilized.

PCMs store and release energy through phase changes, such as from solid to liquid or liquid to gas. The energy absorbed during this process is known as the "heat of fusion". When considering a system like the instrumentation transmitter pod, we need to estimate how long it will take for the PCM to absorb enough energy to change state fully.

The calculation involves quantifying the energy input (in this case, from the electronics power dissipation) and matching it to the energy required for the phase change. This approach allows us to predict the time required for a complete phase change, demonstrating energy conservation's role in thermal management strategies.
Heat of Fusion
The heat of fusion is a material property that indicates the amount of energy required to change a unit mass of a solid to a liquid at a constant temperature. It's a key concept for understanding how phase change materials help in energy regulation, especially in thermal management systems.

To find the heat required for a phase change, we multiply the mass of the PCM by its heat of fusion. This calculation indicates the total energy needed for the entire phase change process. For example, if the PCM has a heat of fusion of 25 kJ/kg and a mass of 1.6 kg, the energy needed becomes:
  • Converted to joules: Heat of fusion = 25 kJ/kg x 1000 J/kJ = 25000 J/kg
  • Total energy needed = 1.6 kg x 25000 J/kg = 40000 J

By determining these values, we can effectively use PCMs in applications that require precise thermal control, ensuring devices like transmitter pods operate efficiently within their specified temperature ranges.
Thermal Management
Thermal management is the process of controlling and regulating the temperature of a device to ensure its functionality and longevity. In electronic applications, like the instrumentation transmitter pod, avoiding overheating is crucial.

Phase change materials play an essential role in thermal management by absorbing excess heat and stabilizing temperatures. The fusion process, where PCMs absorb heat without a significant temperature increase, allows devices to maintain safe operating conditions. This is particularly useful in environments with fluctuating temperatures or high thermal loads.

The heat absorbed by the PCM helps delay temperature rises that could damage sensitive electronic components. Additionally, when the environment cools, the PCM can revert to its solid state, releasing stored energy and providing consistent thermal management.

By incorporating PCMs into thermal management systems, we leverage their latent heat properties to protect electronics, improve energy efficiency, and enhance the overall reliability of devices.

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Most popular questions from this chapter

The oxidized-aluminum wing of an aircraft has a chord length of \(L_{c}=4 \mathrm{~m}\) and a spectral, hemispherical emissivity characterized by the following distribution. (a) Consider conditions for which the plane is on the ground where the air temperature is \(27^{\circ} \mathrm{C}\), the solar irradiation is \(800 \mathrm{~W} / \mathrm{m}^{2}\), and the effective sky temperature is \(270 \mathrm{~K}\). If the air is quiescent, what is the temperature of the top surface of the wing? The wing may be approximated as a horizontal, flat plate. (b) When the aircraft is flying at an elevation of approximately \(9000 \mathrm{~m}\) and a speed of \(200 \mathrm{~m} / \mathrm{s}\), the air temperature, solar irradiation, and effective sky temperature are \(-40^{\circ} \mathrm{C}, 1100 \mathrm{~W} / \mathrm{m}^{2}\), and \(235 \mathrm{~K}\), respectively. What is the temperature of the wing's top surface? The properties of the air may be approximated as \(\rho=0.470 \mathrm{~kg} / \mathrm{m}^{3}, \mu=1.50 \times\) \(10^{-5} \mathrm{~N} \cdot \mathrm{s} / \mathrm{m}^{2}, k=0.021 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(P r=0.72\).

Solar radiation incident on the earth's surface may be divided into the direct and diffuse components described in Problem 12.9. Consider conditions for a day in which the intensity of the direct solar radiation is \(I_{\text {dir }}=210 \times 10^{7} \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{sr}\) in the solid angle subtended by the sun with respect to the earth, \(\Delta \omega_{s}=6.74 \times 10^{-5} \mathrm{sr}\). The intensity of the diffuse radiation is \(I_{\text {dif }}=70 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{sr}\). (a) What is the total solar irradiation at the earth's surface when the direct radiation is incident at \(\theta=30^{\circ}\) ? (b) Verify the prescribed value for \(\Delta \omega_{s}\), recognizing that the diameter of the sun is \(1.39 \times 10^{9} \mathrm{~m}\) and the distance between the sun and the earth is \(1.496 \times 10^{11} \mathrm{~m}\) (1 astronomical unit).

Two small surfaces, \(A\) and \(B\), are placed inside an isothermal enclosure at a uniform temperature. The enclosure provides an irradiation of \(6300 \mathrm{~W} / \mathrm{m}^{2}\) to each of the surfaces, and surfaces A and B absorb incident radiation at rates of 5600 and \(630 \mathrm{~W} / \mathrm{m}^{2}\), respectively. Consider conditions after a long time has elapsed. (a) What are the net heat fluxes for each surface? What are their temperatures? (b) Determine the absorptivity of each surface. (c) What are the emissive powers of each surface? (d) Determine the emissivity of each surface.

A thin sheet of glass is used on the roof of a greenhouse and is irradiated as shown. $$ G_{5} $$ The irradiation comprises the total solar flux \(G_{S}\), the flux \(G_{\mathrm{am}}\) due to atmospheric emission (sky radiation), and the flux \(G_{i}\) due to emission from interior surfaces. The fluxes \(G_{\text {atm }}\) and \(G_{i}\) are concentrated in the far IR region \((\lambda \gtrless 8 \mu \mathrm{m})\). The glass may also exchange energy by convection with the outside and inside atmospheres. The glass may be assumed to be totally transparent for \(\lambda<1 \mu \mathrm{m}\left(\tau_{\lambda}=1.0\right.\) for \(\left.\lambda<1 \mu \mathrm{m}\right)\) and opaque, with \(\alpha_{\lambda}=1.0\) for \(\lambda \geq 1 \mu \mathrm{m}\). (a) Assuming steady-state conditions, with all radiative fluxes uniformly distributed over the surfaces and the glass characterized by a uniform temperature \(T_{g}\), write an appropriate energy balance for a unit area of the glass. (b) For \(T_{z}=27^{\circ} \mathrm{C}, \quad h_{i}=10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}, \quad G_{S}=1100\) \(\mathrm{W} / \mathrm{m}^{2}, T_{\infty \rho}=24^{\circ} \mathrm{C}, \quad h_{o}=55 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}, \quad G_{\operatorname{tam}}=\) \(250 \mathrm{~W} / \mathrm{m}^{2}\), and \(G_{i}=440 \mathrm{~W} / \mathrm{m}^{2}\), calculate the temperature of the greenhouse ambient air, \(T_{x j}\).

An enclosure has an inside area of \(100 \mathrm{~m}^{2}\), and its inside surface is black and is maintained at a constant temperature. A small opening in the enclosure has an area of \(0.02 \mathrm{~m}^{2}\). The radiant power emitted from this opening is \(70 \mathrm{~W}\). What is the temperature of the interior enclosure wall? If the interior surface is maintained at this temperature, but is now polished, what will be the value of the radiant power emitted from the opening?

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