91Ó°ÊÓ

An opaque, horizontal plate has a thickness of \(L=21 \mathrm{~mm}\) and thermal conductivity \(k=25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Water flows adjacent to the bottom of the plate and is at a temperature of \(T_{x, w}=25^{\circ} \mathrm{C}\). Air flows above the plate at \(T_{x, a}=260^{\circ} \mathrm{C}\) with \(h_{a}=40 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The top of the plate is diffuse and is irradiated with \(G=1450 \mathrm{~W} / \mathrm{m}^{2}\), of which \(435 \mathrm{~W} / \mathrm{m}^{2}\) is reflected. The steady-state top and bottom plate temperatures are \(T_{t}=43^{\circ} \mathrm{C}\) and \(T_{b}=35^{\circ} \mathrm{C}\), respectively. Determine the transmissivity, reflectivity, absorptivity, and emissivity of the plate. Is the plate gray? What is the radiosity associated with the top of the plate? What is the convection heat transfer coefficient associated with the water flow?

Step by step solution

01

Analyze the energy balance of the plate bottom

At the bottom of the plate, the energy balance involves convective heat transfer from the water flow. The energy balance equation can be written as: \(q_{conv} = h_w (T_{b} - T_{x, w})\) Where \(h_w\) is the convection heat transfer coefficient associated with the water flow.

02

Analyze the energy balance of the plate top

At the top of the plate, the energy balance involves convective heat transfer from the air flow and radiative heat transfer from the irradiation. The energy balance equation for the top of the plate can be written as: \(q_{conv} + q_{rad} = h_a (T_{x, a} - T_t) + G - q_{reflected}\) Where \(q_{rad}\) is the net radiative heat transfer and \(q_{reflected}\) is the reflected irradiation (absorptivity = \(G - q_{reflected} = G(1 - \rho)\), where \(\rho\) is reflectivity).

03

Determine the heat transfer properties of the plate

Since the steady-state temperatures of the plate's surfaces are given, we can establish the conductive heat transfer through the plate \(q_{cond}\). The formula for conductive heat transfer is: \(q_{cond} = \frac{k(T_t - T_b)}{L}\) With the given values for \(T_t\), \(T_b\), \(L\), and \(k\), we can calculate the conductive heat transfer.

04

Calculate the heat transfer coefficients

Using the energy balances from Steps 1 and 2, and the results from Step 3, we can calculate the convection heat transfer coefficient associated with the water flow (\(h_w\)): \(h_w = \frac{q_{cond}}{T_b - T_{x, w}}\) And the net radiative heat transfer (\(q_{rad}\)): \(q_{rad} = q_{cond} - h_a (T_{x, a} - T_t)\)

05

Determine the transmissivity, reflectivity, absorptivity, and emissivity of the plate

Since the plate is opaque, its transmissivity is zero: Transmissivity = 0 From the net radiative heat transfer calculation, we can determine the reflectivity of the plate: Reflectivity (\(\rho\)) = \(1 - \frac{q_{reflected}}{G}\) And absorptivity: Absorptivity (\(\alpha\)) = \(1 - \rho\) Since the plate is in radiative equilibrium, its absorptivity must equal its emissivity (\(\boldsymbol{\epsilon}\)): Emissivity (\(\epsilon\)) = Absorptivity (\(\alpha\))

06

Check if the plate is gray

A plate is considered gray if its emissivity and absorptivity are wavelength-independent. Since we cannot derive this information from the given data, we cannot confirm if the plate is gray.

07

Calculate the radiosity associated with the top of the plate

The radiosity (\(J_t\)) associated with the top of the plate is the sum of the emitted and reflected radiation: \(J_t = \epsilon \sigma T_t^4 + \rho G\) Where \(\sigma\) is the Stefan-Boltzmann constant. With the reflectivity and emissivity values obtained in Step 5 and the given values for top surface temperature and irradiation, we can calculate the radiosity. Now we have calculated all the properties requested in the exercise.

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.

Heat Transfer Coefficient

The heat transfer coefficient is a crucial factor in thermodynamics, as it quantifies the thermal energy exchanged between a solid surface and a fluid per unit surface area, per unit temperature difference. This measurement is pivotal in determining how effectively heat is conveyed through mediums like metal plates or boundary layers surrounding fluids.
A greater heat transfer coefficient indicates more proficient heat transfer, which is desirable in various applications like enhancing cooling systems or improving energy efficiency.

• The equation for calculating the convective heat transfer at the bottom of the plate is:\[q_{conv} = h_w (T_b - T_{x, w})\]
• Where \(h_w\) is the convection heat transfer coefficient associated with the water flow.

This equation assists in understanding how heat moves from the plate toward the fluid and plays a key role in energy balance calculations.

Radiative Heat Transfer

Radiative heat transfer concerns the emission of thermal energy through electromagnetic waves. It is computed based on surface temperatures and emissivity, which describe how much radiation is emitted compared to that of a perfect blackbody.
In the given exercise, radiative heat transfer at the top of the plate is affected by irradiation, a process by which energy, in the form of waves or particles, is emitted and propagated through a space or medium.

• Key aspects addressed include net radiative heat transfer \(q_{rad}\), determined from reflected irradiation and absorptivity.
• The equation for the net radiative heat exchange is:\[q_{rad} = q_{cond} - h_a (T_{x, a} - T_t)\]

This concept is instrumental in augmenting or reducing temperatures of objects not in direct contact, emphasizing its importance in thermal analysis.

Convective Heat Transfer

Convective heat transfer refers to the thermal energy transfer between a surface and a fluid flowing over it, depending on fluid properties, flow characteristics, and surface conditions.
This process is essential for systems involving heating or cooling, where fluid like air or water carries heat away from or toward a surface.

• The convective heat transfer at the top of the plate involves air passing over the plate with a convective coefficient \(h_a = 40 \, \mathrm{W} / \mathrm{m}^2 \, \cdot \, \mathrm{K}\).
• Convective heat exchange is dictated by the following equation:\[q_{conv} = h_a (T_{x, a} - T_t)\]

This process underlines how moving fluids significantly influence temperature control and regulation.

Energy Balance

An energy balance analyzes the conservation of energy within a system by equating energy input with energy output, considering various heat transfer mechanisms like conduction, convection, and radiation.
This method allows engineers to predict temperature distributions and energy consumption accurately.
For the given exercise, energy balance involved:

• Calculating conductive heat transfer through the thickness of the plate using:\[q_{cond} = \frac{k(T_t - T_b)}{L}\]
• Evaluating convective and radiative exchanges to ensure all heat inputs and outputs align.

This comprehensive analysis helps in understanding how different heat transfer forms contribute to the system's overall thermal management.

Emissivity

Emissivity is a measure of a surface's ability to emit thermal radiation, compared to that of an ideal blackbody, which has an emissivity of 1.
This property is crucial in calculating radiative heat transfer and plays a significant role in thermal equilibrium.
In the exercise, the plate's emissivity determined by:

• It equates absorptivity when radiative equilibrium is achieved, meaning emissivity equals absorptivity since \(1 - \rho\).
• The radiosity equation derived from emissivity provides the net energy leaving the surface:\[J_t = \epsilon \sigma T_t^4 + \rho G\]

Understanding emissivity helps tailor materials and designs to facilitate or hinder the release of heat as needed, highlighting its importance in thermal science.