Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 2: Problem 63

A flat-plate solar collector is used to heat water by having water flow through tubes attached at the back of the thin solar absorber plate. The absorber plate has an emissivity and an absorptivity of \(0.9\). The top surface \((x=0)\) temperature of the absorber is \(T_{0}=35^{\circ} \mathrm{C}\), and solar radiation is incident on the absorber at \(500 \mathrm{~W} / \mathrm{m}^{2}\) with a surrounding temperature of \(0^{\circ} \mathrm{C}\). Convection heat transfer coefficient at the absorber surface is \(5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), while the ambient temperature is \(25^{\circ} \mathrm{C}\). Show that the variation of temperature in the absorber plate can be expressed as \(T(x)=-\left(\dot{q}_{0} / k\right) x+T_{0}\), and determine net heat flux \(\dot{q}_{0}\) absorbed by the solar collector.

Short Answer

Expert verified
Question: Show that the temperature variation in the absorber plate can be expressed as \(T(x)=-\left(\dot{q}_{0} / k\right) x+T_{0}\) and determine the net heat flux \(\dot{q}_{0}\) absorbed by the solar collector. Answer: The temperature variation in the absorber plate can be expressed as \(T(x)=-\left(\dot{q}_{0} / k\right) x+T_{0}\). The net heat flux absorbed \(\dot{q}_{0}\) can be found by calculating the incoming solar radiation absorbed by the absorber plate and the heat loss due to convection to the ambient air using the formula \(\dot{q}_0 = q_{in} - h(T_0 - T_a)\).

Step by step solution

01

Calculate absorber's solar radiation absorptance

First, calculate the amount of solar radiation absorbed by the absorber plate. This can be found by multiplying the incoming solar radiation by the absorber plate's absorptivity. $$ q_{in} = \alpha \times I $$ where, \(\alpha\) is the absorptivity of the absorber plate, \(I\) is the incident solar radiation.
02

Calculate the total heat absorbed

Next, we need to calculate the total heat absorbed by the plate by summing the incoming absorbed solar radiation and the heat loss due to convection to the ambient air: $$ \dot{q}_0 = q_{in} - h(T_0 - T_a) $$ where, \(h\) is the convective heat transfer coefficient, \(T_0\) is the top surface temperature of the absorber plate, and \(T_a\) is the ambient temperature.
03

Calculate the temperature profile

Now, plug in the provided values and solve for the net heat flux: $$ \dot{q}_0 = 0.9 \times 500 \mathrm{~W} / \mathrm{m}^{2} - 5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K} \times (35^{\circ} \mathrm{C} - 0^{\circ} \mathrm{C}) $$ This will give the value for \(\dot{q}_{0}\).
04

Express the temperature variation in the absorber plate

By plugging the net heat flux \(\dot{q}_{0}\) into the following equation, we can express the temperature variation in the absorber plate as a function of \(x\). $$ T(x) = -\left(\frac{\dot{q}_{0}}{k}\right) x+T_{0}, $$ where \(k\) is the thermal conductivity of the absorber plate material. Note that \(k\) is not given in the problem statement, so we cannot find an exact expression for \(T(x)\) in this case.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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 a large plate of thickness \(L\) and thermal conductivity \(k\) in which heat is generated uniformly at a rate of \(\dot{e}_{\text {gen. }}\) One side of the plate is insulated while the other side is exposed to an environment at \(T_{\infty}\) with a heat transfer coefficient of \(h\). \((a)\) Express the differential equation and the boundary conditions for steady one-dimensional heat conduction through the plate, \((b)\) determine the variation of temperature in the plate, and \((c)\) obtain relations for the temperatures on both surfaces and the maximum temperature rise in the plate in terms of given parameters.

What is the difference between the degree and the order of a derivative?

A spherical container, with an inner radius \(r_{1}=1 \mathrm{~m}\) and an outer radius \(r_{2}=1.05 \mathrm{~m}\), has its inner surface subjected to a uniform heat flux of \(\dot{q}_{1}=7 \mathrm{~kW} / \mathrm{m}^{2}\). The outer surface of the container has a temperature \(T_{2}=25^{\circ} \mathrm{C}\), and the container wall thermal conductivity is \(k=1.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Show that the variation of temperature in the container wall can be expressed as \(T(r)=\left(\dot{q}_{1} r_{1}^{2} / k\right)\left(1 / r-1 / r_{2}\right)+T_{2}\) and determine the temperature of the inner surface of the container at \(r=r_{1}\).

Consider a large 5-cm-thick brass plate \((k=\) \(111 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) ) in which heat is generated uniformly at a rate of \(2 \times 10^{5} \mathrm{~W} / \mathrm{m}^{3}\). One side of the plate is insulated while the other side is exposed to an environment at \(25^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(44 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Explain where in the plate the highest and the lowest temperatures will occur, and determine their values.

Consider a \(1.5\)-m-high and \(0.6-\mathrm{m}\)-wide plate whose thickness is \(0.15 \mathrm{~m}\). One side of the plate is maintained at a constant temperature of \(500 \mathrm{~K}\) while the other side is maintained at \(350 \mathrm{~K}\). The thermal conductivity of the plate can be assumed to vary linearly in that temperature range as \(k(T)=\) \(k_{0}(1+\beta T)\) where \(k_{0}=18 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\beta=8.7 \times 10^{-4} \mathrm{~K}^{-1}\). Disregarding the edge effects and assuming steady onedimensional heat transfer, determine the rate of heat conduction through the plate. Answer: \(22.2 \mathrm{~kW}\)
See all solutions

Recommended explanations on Physics Textbooks

Conservation of Energy and Momentum

Read Explanation

Waves Physics

Read Explanation

Electrostatics

Read Explanation

Magnetism

Read Explanation

Famous Physicists

Read Explanation

Dynamics

Read Explanation
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.

Sign-up for free