/*! 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 108 Copper tubing is joined to the a... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Copper tubing is joined to the absorber of a flat-plate solar collector as shown. The aluminum alloy (2024-T6) absorber plate is \(6 \mathrm{~mm}\) thick and well insulated on its bottom. The top surface of the plate is separated from a transparent cover plate by an evacuated space. The tubes are spaced a distance \(L\) of \(0.20 \mathrm{~m}\) from each other, and water is circulated through the tubes to remove the collected energy. The water may be assumed to be at a uniform temperature of \(T_{w}=60^{\circ} \mathrm{C}\). Under steady-state operating conditions for which the net radiation heat flux to the surface is \(q_{\text {rad }}^{\prime \prime}=\) \(800 \mathrm{~W} / \mathrm{m}^{2}\), what is the maximum temperature on the plate and the heat transfer rate per unit length of tube? Note that \(q_{\text {rad }}^{\prime \prime}\) represents the net effect of solar radiation absorption by the absorber plate and radiation exchange between the absorber and cover plates. You may assume the temperature of the absorber plate directly above a tube to be equal to that of the water.

Short Answer

Expert verified
The solution to this problem involves finding the temperature distribution within the absorber plate, given the net radiation heat flux and water temperature. Using an energy balance on a control volume within the plate, we obtain a first-order ordinary differential equation for the temperature distribution: \[\frac{dT}{dx} = \frac{h(T - T_w)}{k} - \frac{q_{\text{rad}}^{\prime \prime}}{k}\] The temperature distribution is given by: \[T(x) = T_w + \frac{q_{\text{rad}}^{\prime \prime}}{h}\left(1 - \exp(-\frac{h}{k}x)\right)\] The maximum temperature in the absorber plate is then: \[T_{max} = T_w + \frac{q_{\text{rad}}^{\prime \prime}}{h}\left(1 - \exp(-\frac{h}{k}L)\right)\] To find the heat transfer rate per unit length of tube, we use the heat conduction equation: \[q_{total}^{\prime} = -k\frac{dT}{dx}A\] With the given values, we can calculate the maximum temperature on the plate and the heat transfer rate per unit length of tube.

Step by step solution

01

Control volume and Energy balance

First, consider a control volume within the absorber plate whose width is infinitesimally small, say \(\Delta x\). The heat entering this small control volume is due to conduction from the left end and net radiation heat flux. The heat leaving this small control volume is due to conduction through the right end and convection to the water. By applying the energy balance equation, we can find the temperature distribution within the absorber plate. Energy balance equation: \(q_{in} = q_{out}\)
02

Satisfying energy balance equation for the control volume

Now, by considering the heat transfer processes in this control volume, we can rewrite the energy balance equation as follows: Heat entering the control volume due to conduction: \(-k \frac{dT}{dx}_{left}\) Heat entering the control volume due to net radiation heat flux: \(q_{\text{rad}}^{\prime \prime}\) Heat leaving the control volume due to conduction: \(-k \frac{dT}{dx}_{right}\) Heat leaving the control volume due to convection: \(h(T - T_w)\) Now, let's satisfy the energy balance equation by using these expressions: \[-k \frac{dT}{dx}_{left} + q_{\text{rad}}^{\prime \prime} = -k \frac{dT}{dx}_{right} + h(T - T_w)\]
03

Rewrite the energy balance equation in terms of temperature gradient

Since the width of the control volume \(\Delta x\) is infinitesimally small, we have: \[\frac{dT}{dx}_{left} = \frac{dT}{dx}_{right}\] Now rewrite the energy balance equation in terms of temperature gradient: \[\frac{dT}{dx} = \frac{h(T - T_w)}{k} - \frac{q_{\text{rad}}^{\prime \prime}}{k}\] This is a first-order ordinary differential equation for the temperature distribution within the absorber plate.
04

Solve the differential equation for temperature distribution

Given the boundary conditions and the initial conditions, we can solve this difficulty equation using standard methods (separation of variables or integrating factor method). Here, we assume the solution is given by: \[T(x) = T_w + \frac{q_{\text{rad}}^{\prime \prime}}{h}\left(1 - \exp(-\frac{h}{k}x)\right)\] Now, we can find the maximum temperature in the plate by calculating the maximum value of \(T(x)\).
05

Find the maximum temperature and heat transfer rate

The maximum temperature in the absorber plate is: \[T_{max} = T_w + \frac{q_{\text{rad}}^{\prime \prime}}{h}\left(1 - \exp(-\frac{h}{k}L)\right)\] To find the heat transfer rate per unit length of tube, use the heat conduction equation at the right end (just above the tube): \[q_{total}^{\prime} = -k\frac{dT}{dx}A\] where, \(A\) is the cross-sectional area of the absorber plate. Now, using the given values for \(\:k\), \(\:h\), \(\:q_{\text{rad}}^{\prime\prime}\), \(\:T_w\), and the diameter and length of the tubes, we can calculate the maximum temperature on the plate and the heat transfer rate per unit length of tube.

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.

Solar Collectors
Solar collectors are devices designed to capture and convert sunlight into usable thermal energy. They play a crucial role in solar energy systems and are often used in residential and industrial settings. The most common type is the flat-plate solar collector, which consists of an absorber plate, usually made of metals like aluminum or copper, and is coated in a substance that improves its ability to absorb solar energy. The collector is topped with a transparent cover to minimize heat loss and separated by an evacuated space to further reduce heat transfer to the outside environment.

A crucial feature of flat-plate collectors is their simplicity and durability. They are designed to work efficiently in various weather conditions. The absorber plate's primary responsibility is capturing solar radiation, which is then transferred to a fluid, typically water or a glycol-water mixture, circulating through the system. This fluid carries the heat away for further use or storage. Understanding the mechanics of solar collectors is essential for optimizing their efficiency, especially when calculating heat transfer rates and balance equations.
Thermal Conductivity
Thermal conductivity is a material's ability to conduct heat. In the context of solar collectors, it is a vital factor when considering the performance of the absorber plate. Materials with high thermal conductivity, like copper and aluminum, are favored because they transfer heat efficiently across the surface.

The concept can be described mathematically by Fourier’s Law of Heat Conduction, which relates the heat transfer through a material to its thermal conductivity, surface area, and the temperature gradient. The law is expressed as:
  • \[ q = -k A \frac{dT}{dx} \]
where \( q \) is the heat transfer rate, \( k \) is the thermal conductivity, \( A \) is the area through which heat is being transferred, and \( \frac{dT}{dx} \) is the temperature gradient.

In solar collectors, understanding and optimizing the thermal conductivity of the absorber plate materials can significantly impact the efficiency and heat transfer capabilities of the entire system.
Energy Balance
Energy balance is a fundamental concept crucial for understanding how solar collectors operate effectively. Under steady-state conditions, the energy coming into the system must equal the energy going out. This principle ensures that the absorber plate does not overheat and that the maximum amount of solar energy is used efficiently.

In this context, the energy balance equation helps us predict the temperature distribution in the absorber plate. It incorporates the energy gains from solar radiation and losses through conduction and convection. The equation used is:
  • Incoming energy:
    1. Net radiation: \( q_{\text{rad}}^{\prime \prime} \)
  • Outgoing energy:
    1. Conduction losses: \(-k \frac{dT}{dx} \)
    2. Convection to water: \(h(T - T_w)\)
By maintaining this balance, engineers can ensure that the solar collector works optimally and safely. Calculating the balance requires knowing the material properties and the conditions of operation, such as the fluid's temperature inside the tubes.
Differential Equations
Differential equations are mathematical tools used to describe how a quantity changes with respect to another. In the design of solar collectors, they help in modeling the temperature distribution across the absorber plate under steady-state conditions.

The primary differential equation arises from the energy balance within a control volume. It is a common technique used to solve problems in heat transfer, fluid dynamics, and other engineering domains. For the solar collector, it is expressed as:
  • \[ \frac{dT}{dx} = \frac{h(T - T_w)}{k} - \frac{q_{\text{rad}}^{\prime \prime}}{k} \]
This equation considers both the convection to the coolant and the absorption of solar energy. Solving the differential equation provides insights into the thermal performance of the collector, including the maximum temperature and heat transfer rates.

Advanced mathematical methods, like separation of variables or integrating factor methods, are often applied to solve these equations. This allows for precise predictions of temperature profiles, critical for designing efficient solar thermal systems.

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

Superheated steam at \(575^{\circ} \mathrm{C}\) is routed from a boiler to the turbine of an electric power plant through steel tubes \((k=35 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) of \(300-\mathrm{mm}\) inner diameter and \(30-\mathrm{mm}\) wall thickness. To reduce heat loss to the surroundings and to maintain a safe-to-touch outer surface temperature, a layer of calcium silicate insulation \((k=0.10 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is applied to the tubes, while degradation of the insulation is reduced by wrapping it in a thin sheet of aluminum having an emissivity of \(\varepsilon=0.20\). The air and wall temperatures of the power plant are \(27^{\circ} \mathrm{C}\). (a) Assuming that the inner surface temperature of a steel tube corresponds to that of the steam and the convection coefficient outside the aluminum sheet is \(6 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), what is the minimum insulation thickness needed to ensure that the temperature of the aluminum does not exceed \(50^{\circ} \mathrm{C}\) ? What is the corresponding heat loss(b) Explore the effect of the insulation thickness on the temperature of the aluminum and the heat loss per unit tube length. per meter of tube length?

A wire of diameter \(D=2 \mathrm{~mm}\) and uniform temperature \(T\) has an electrical resistance of \(0.01 \Omega / \mathrm{m}\) and a current flow of \(20 \mathrm{~A}\). (a) What is the rate at which heat is dissipated per unit length of wire? What is the heat dissipation per unit volume within the wire? (b) If the wire is not insulated and is in ambient air and large surroundings for which \(T_{\infty}=T_{\text {sur }}=20^{\circ} \mathrm{C}\), what is the temperature \(T\) of the wire? The wire has an emissivity of \(0.3\), and the coefficient associated with heat transfer by natural convection may be approximated by an expression of the form, \(h=C\left[\left(T-T_{\infty}\right) / D\right]^{1 / 4}, \quad\) where \(C=1.25\) \(\mathrm{W} / \mathrm{m}^{7 / 4} \cdot \mathrm{K}^{5 / 4}\). (c) If the wire is coated with plastic insulation of 2-mm thickness and a thermal conductivity of \(0.25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), what are the inner and outer surface temperatures of the insulation? The insulation has an emissivity of \(0.9\), and the convection coefficient is given by the expression of part (b). Explore the effect of the insulation thickness on the surface temperatures.

Consider two long, slender rods of the same diameter but different materials. One end of each rod is attached to a base surface maintained at \(100^{\circ} \mathrm{C}\), while the surfaces of the rods are exposed to ambient air at \(20^{\circ} \mathrm{C}\). By traversing the length of each rod with a thermocouple, it was observed that the temperatures of the rods were equal at the positions \(x_{\mathrm{A}}=0.15 \mathrm{~m}\) and \(x_{\mathrm{B}}=0.075 \mathrm{~m}\), where \(x\) is measured from the base surface. If the thermal conductivity of rod \(\mathrm{A}\) is known to be \(k_{\mathrm{A}}=70 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), determine the value of \(k_{\mathrm{B}}\) for rod B.

A steam pipe of \(0.12-\mathrm{m}\) outside diameter is insulated with a layer of calcium silicate. (a) If the insulation is \(20 \mathrm{~mm}\) thick and its inner and outer surfaces are maintained at \(T_{s, 1}=800 \mathrm{~K}\) and \(T_{s, 2}=490 \mathrm{~K}\), respectively, what is the heat loss per unit length \(\left(q^{\prime}\right)\) of the pipe? (b) We wish to explore the effect of insulation thickness on the heat loss \(q^{\prime}\) and outer surface temperature \(T_{s, 2}\), with the inner surface temperature fixed at \(T_{s, 1}=\) \(800 \mathrm{~K}\). The outer surface is exposed to an airflow \(\left(T_{\infty}=25^{\circ} \mathrm{C}\right)\) that maintains a convection coefficient of \(h=25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and to large surroundings for which \(T_{\text {sur }}=T_{\infty}=25^{\circ} \mathrm{C}\). The surface emissivity of calcium silicate is approximately \(0.8\). Compute and plot the temperature distribution in the insulation as a function of the dimensionless radial coordinate, \(\left(r-r_{1}\right) /\left(r_{2}-r_{1}\right)\), where \(r_{1}=0.06 \mathrm{~m}\) and \(r_{2}\) is a variable \(\left(0.06

A commercial grade cubical freezer, \(3 \mathrm{~m}\) on a side, has a composite wall consisting of an exterior sheet of \(6.35-\mathrm{mm}\)-thick plain carbon steel, an intermediate layer of \(100-\mathrm{mm}\)-thick cork insulation, and an inner sheet of \(6.35\)-mm-thick aluminum alloy (2024). Adhesive interfaces between the insulation and the metallic strips are each characterized by a thermal contact resistance of \(R_{t, c}^{\prime \prime}=2.5 \times 10^{-4} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). What is the steady-state cooling load that must be maintained by the refrigerator under conditions for which the outer and inner surface temperatures are \(22^{\circ} \mathrm{C}\) and \(-6^{\circ} \mathrm{C}\), respectively?

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.