/*! 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 96 One scheme for extending the ope... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

One scheme for extending the operation of gas turbine blades to higher temperatures involves applying a ceramic coating to the surfaces of blades fabricated from a superalloy such as inconel. To assess the reliability of such coatings, an apparatus has been developed for testing samples under laboratory conditions. The sample is placed at the bottom of a large vacuum chamber whose walls are cryogenically cooled and which is equipped with a radiation detector at the top surface. The detector has a surface area of \(A_{d}=10^{-5} \mathrm{~m}^{2}\), is located at a distance of \(L_{\text {sl }}=1 \mathrm{~m}\) from the sample, and views radiation originating from a portion of the ceramic surface having an area of \(\Delta A_{c}=10^{-4} \mathrm{~m}^{2}\). An electric heater attached to the bottom of the sample dissipates a uniform heat flux, \(q_{b}^{\prime \prime}\), which is transferred upward through the sample. The bottom of the heater and sides of the sample are well insulated. Consider conditions for which a ceramic coating of thickness \(L_{c}=0.5 \mathrm{~mm}\) and thermal conductivity \(k_{c}=\) \(6 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) has been sprayed on a metal substrate of thickness \(L_{s}=8 \mathrm{~mm}\) and thermal conductivity \(k_{s}=\) \(25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The opaque surface of the ceramic may be approximated as diffuse and gray, with a total, hemispherical emissivity of \(\varepsilon_{c}=0.8\). (a) Consider steady-state conditions for which the bottom surface of the substrate is maintained at \(T_{1}=1500 \mathrm{~K}\), while the chamber walls (including the surface of the radiation detector) are maintained at \(T_{w}=90 \mathrm{~K}\). Assuming negligible thermal contact resistance at the ceramic- substrate interface, determine the ceramic top surface temperature \(T_{2}\) and the heat flux \(q_{b}^{\prime \prime}\). (b) For the prescribed conditions, what is the rate at which radiation emitted by the ceramic is intercepted by the detector?

Short Answer

Expert verified
In summary, for the given conditions, the ceramic top surface temperature \(T_2\) is \(663.87 \text{ K}\) and the heat flux \(q_{b}''\) is \(10,054.4 \text{ W/m}^2\). The rate at which radiation emitted by the ceramic is intercepted by the detector is \(97.55 \text{ W}\).

Step by step solution

01

Heat Transfer through Ceramic Coating

To find the heat transfer through the ceramic coating, we can use Fourier's Law applied to a one-dimensional, steady conduction as follows: \(q_c = k_c\frac{T_1-T_2}{L_c}\) We will use this equation later to link the ceramic coating and the metal substrate.
02

Heat Transfer through Metal Substrate

Similarly, we will find the heat transfer through the metal substrate using Fourier's Law: \(q_s = k_s\frac{T_2-T_w}{L_s}\)
03

Equate the Heat Transfers

Since the heat transfer through the ceramic coating must equal the heat transfer through the metal substrate, we can equate the two equations we derived earlier: \(k_c\frac{T_1-T_2}{L_c} = k_s\frac{T_2-T_w}{L_s}\) We can now solve this equation for T2: \(T_2 = \frac{k_c L_s T_1 + k_s L_c T_w}{k_c L_s + k_s L_c}\) Substitute the given values: - \(k_c = 6 \text{ W/mK}\) - \(L_c = 0.5 \times 10^{-3} \text{ m}\) - \(k_s = 25 \text{ W/mK}\) - \(L_s = 8 \times 10^{-3} \text{ m}\) - \(T_1 = 1500 \text{ K}\) - \(T_w = 90 \text{ K}\) to get the value for \(T_2\): \(T_2 = 663.87 \text{ K}\)
04

Calculate the Heat Flux\(q_{b}''\)

Now to find the heat flux, we can substitute T2 in any of the two heat transfer equations from step 1 or step 2. Using the ceramic heat transfer equation: \(q_{b}'' = q_c = k_c\frac{T_1-T_2}{L_c}\) \(q_{b}'' = 6\frac{1500-663.87}{0.5 \times 10^{-3}}\) \(q_{b}'' = 10,054.4 \text{ W/m}^2\) So the heat flux is \(10,054.4 \text{ W/m}^2\).
05

Radiation Emitted by the Ceramic

For part (b), we will use the following relationship for the rate at which radiation is emitted: \(q_{r} = \varepsilon_{c} \sigma (T_2^4 - T_w^4)\) where: - σ is the Stefan–Boltzmann constant (\(5.67 \times 10^{-8} \text{W/m}^2\text{K}^4\)) - ε is the emissivity of the ceramic (\(0.8\)) Substitute the values to find the radiation emitted by the ceramic: \(q_{r} = 0.8 \times 5.67 \times 10^{-8} (663.87^4 - 90^4)\) \(q_{r} = 975.52 \text{ W/m}^2\)
06

Radiation Intercepted by Detector

Now we can calculate the radiation emitted by the ceramic intercepted by the detector using the following relationship: \(q_{d} = q_r \frac{A_d}{\Delta A_c}\) Substitute the given values and calculated radiation emitted: - \(A_d = 10^{-5}\text{ m}^2\) - \(\Delta A_c = 10^{-4}\text{ m}^2\) - \(q_r = 975.52 \text{ W/m}^2\) to get the value for \(q_d\): \(q_{d} = 975.52 \frac{10^{-5}}{10^{-4}}\) \(q_{d} = 97.55 \text{ W}\) So the rate at which radiation is intercepted by the detector is \(97.55 \text{ W}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Ceramic Coating
A ceramic coating is a thin layer of ceramic material applied to a substrate, typically metal. In the context of gas turbine blades, this coating serves as a protective layer to handle high temperatures. Ceramic materials are chosen for their ability to withstand extreme heat while also providing thermal insulation.

The ceramic coating discussed here is applied to turbine blades made from a superalloy such as Inconel. This helps the blade endure high operational temperatures by reducing the thermal load on the metallic substrate beneath the ceramic layer.

Ceramic coatings have several benefits:
  • Heat Resistance: Ceramics can tolerate high temperatures much better than metals, helping protect the underlying alloy.
  • Thermal Barrier: By acting as a thermal barrier, ceramics reduce the temperature experienced by the blade, enhancing its lifespan.
  • Weight Efficiency: Since they are thin, ceramic coatings add minimal weight to the blade.
Overall, the efficacy of a ceramic coating is heavily influenced by its thermal conductivity, which determines how effectively it can insulate the underlying metal substrate from heat.
Radiation Detector
In the setup for this exercise, a radiation detector plays a crucial role in measuring the effectiveness of the ceramic coating. The detector is placed in a vacuum chamber and is responsible for capturing the radiation emitted by the ceramic surface.

The placement and surface area of the radiation detector are critical for accurate readings. It captures the radiation originating from the surface area of the ceramic, allowing for the determination of heat transfer efficiency and coating performance.
  • Location: Positioned at the top of the vacuum chamber, the radiation detector measures the radiation intercepted within its area.
  • Purpose: Its function is to measure the radiant energy that emits from the ceramic surface and reaches the detector.
  • Calculation Aid: Helps in calculating the rate of radiation, which involves applying concepts of emissivity and the Stefan-Boltzmann law.
The radiation detector's role in assessing radiative heat transfer provides insights into the thermal performance of the coating and can help improve thermal efficiency strategies.
Thermal Conductivity
Thermal conductivity is a measure of a material's ability to conduct heat. This property is vital when considering the application of ceramic coatings in high-temperature environments like turbines.

In our exercise, two different materials are examined for their thermal conductivity: ceramic and metal substrates. The ceramic has a thermal conductivity of 6 W/m·K, while the metal substrate has a higher conductivity of 25 W/m·K. This difference impacts how heat flows through the multilayered structure of coating and substrate.
  • Importance: High thermal conductivity in metals means heat flows quickly, whereas lower values in ceramics slow the heat transfer, providing insulation.
  • Calculations: By applying Fourier's Law, we can calculate the heat transfer through both materials, understanding how each layer affects the thermal flow.
  • Balancing Act: In applications like turbine blades, it's crucial to balance the thermal conductivity to ensure that the metal remains protected by the ceramic layer without excessive heat transfer.
Understanding thermal conductivity is fundamental to designing efficient ceramic coatings, ensuring that they can provide the desired thermal insulation while maintaining structural integrity under operational stresses.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Consider an opaque, gray surface whose directional absorptivity is \(0.8\) for \(0 \leq \theta \leq 60^{\circ}\) and \(0.1\) for \(\theta>60^{\circ}\). The surface is horizontal and exposed to solar irradiation comprised of direct and diffuse components. (a) What is the surface absorptivity to direct solar radiation that is incident at an angle of \(45^{\circ}\) from the normal? What is the absorptivity to diffuse irradiation? (b) Neglecting convection heat transfer between the surface and the surrounding air, what would be the equilibrium temperature of the surface if the direct and diffuse components of the irradiation were 600 and \(100 \mathrm{~W} / \mathrm{m}^{2}\), respectively? The back side of the surface is insulated.

A proposed method for generating electricity from solar irradiation is to concentrate the irradiation into a cavity that is placed within a large container of a salt with a high melting temperature. If all heat losses are neglected, part of the solar irradiation entering the cavity is used to melt the salt while the remainder is used to power a Rankine cycle. (The salt is melted during the day and is resolidified at night in order to generate electricity around the clock.) Consider conditions for which the solar power entering the cavity is \(q_{\mathrm{sal}}=7.50 \mathrm{MW}\) and the time rate of change of energy stored in the salt is \(\dot{E}_{\mathrm{st}}=3.45 \mathrm{MW}\). For a cavity opening of diameter \(D_{s}=1 \mathrm{~m}\), determine the heat transfer to the Rankine cycle, \(q_{R}\). The temperature of the salt is maintained at its melting point, \(T_{\text {salt }}=T_{\text {m }}=1000^{\circ} \mathrm{C}\). Neglect heat loss by convection and irradiation from the surroundings.

A furnace with an aperture of 20 -mm diameter and emissive power of \(3.72 \times 10^{5} \mathrm{~W} / \mathrm{m}^{2}\) is used to calibrate a heat flux gage having a sensitive area of \(1.6 \times 10^{-5} \mathrm{~m}^{2}\). (a) At what distance, measured along a normal from the aperture, should the gage be positioned to receive irradiation of \(1000 \mathrm{~W} / \mathrm{m}^{2}\) ? (b) If the gage is tilted off normal by \(20^{\circ}\), what will be its irradiation? (c) For tilt angles of 0,20 , and \(60^{\circ}\), plot the gage irradiation as a function of the separation distance for values ranging from 100 to \(300 \mathrm{~mm}\).

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, \(T_{x j}\).

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.