/*! 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 27 A salt-gradient solar pond is a ... [FREE SOLUTION] | 91Ó°ÊÓ

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A salt-gradient solar pond is a shallow body of water that consists of three distinct fluid layers and is used to collect solar encrgy. The upper- and lower-most layers are well mixed and serve to maintain the upper and lower surfaces of the central layer at uniform temperatures \(T_{1}\) and \(T_{2}\). where \(T_{2}>T_{1}\). Although there is bulk fluid motion in the mixed layers, there is no such motion in the central layer. Consider conditions for which solar radiation absorption in the central layer provides nonuniform heat generation of the form \(\dot{q}=A e^{-u ;}\), and the lemperature distribution in the central layer is $$ T(x)=-\frac{A}{k a^{2}} e^{-a x}+B x+C $$ The quantities \(A\left(\mathrm{~W} / \mathrm{m}^{3}\right), a(1 / \mathrm{m}), B(\mathrm{~K} / \mathrm{m})\), and \(C(\mathrm{~K})\) are known constants having the prescribed units, and \(k\) is the thermal conductivity, which is also constant. (a) Obtain expressions for the rate at which heat is transferred per unit area from the lower mixed layer to the central layer and from the central layer to the upper mixed layer. (b) Determine whether cunditions are steady or transicnt. (c) Obtain an expression for the rate at which thermal cnergy is generated in the entire central layer, per unit surface area.

Short Answer

Expert verified
(a) Heat transfer rates: \( q''_{12} = -A - kB \), \( q''_{21} = -A e^{-a L} - kB \). (b) Conditions are steady if these rates are equal. (c) Heat generation rate: \( \frac{A}{a}(1-e^{-aL}) \).

Step by step solution

01

Understand the Heat Transfer Problem

We have a central layer in a solar pond where heat is generated uniformly. The task is to find heat transfer rates from lower and upper mixed layers to this central layer, check if conditions are steady or transient, and compute heat generation per unit surface area in the central layer. We must use given equations and understand their roles in this physical scenario.
02

Apply Fourier's Law for Heat Transfer

To compute heat transfer from one layer to another, use Fourier's law for heat conduction: \[ q'' = -k \frac{dT}{dx} \].Evaluate this at the interfaces \( x = 0 \) and \( x = L \) for heat transfer rates \( q''_{12} \) and \( q''_{21} \) from lower to central layer and from central to upper layer respectively.
03

Compute Temperature Gradient

Find \( \frac{dT}{dx} \) from the temperature distribution \( T(x) = -\frac{A}{k a^{2}} e^{-a x} + B x + C \). The derivative becomes:\[ \frac{dT}{dx} = \frac{A}{k} e^{-a x} + B \].
04

Calculate Heat Transfer from Lower Layer to Central Layer

At \( x = 0 \), substitute into Fourier's law:\[ q''_{12} = -k \left( \frac{A}{k} e^{-a(0)} + B \right) = -A - kB \].
05

Calculate Heat Transfer from Central Layer to Upper Layer

Similarly, at \( x = L \), substitute into Fourier's law:\[ q''_{21} = -k \left( \frac{A}{k} e^{-a L} + B \right) = -A e^{-a L} - kB \].
06

Determine Steady or Transient Conditions

In steady conditions, heat transferred into the central layer equals heat transferred out. Check if \( q''_{12} = q''_{21} \).Compare results from Steps 4 and 5, if equal then it is steady, else transient.
07

Compute Thermal Energy Generation Rate

Total thermal energy generated per unit surface area in the central layer is: \[ q''_g = \int_0^L \dot{q} \, dx = \int_0^L A e^{-a x} \, dx = \left[-\frac{A}{a} e^{-ax} \right]_0^L = \frac{A}{a}(1 - e^{-aL})\].

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Key Concepts

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

Salt-Gradient Solar Pond
A salt-gradient solar pond is an innovative way to harness solar power. It consists of three layers. The top is mixed, maintaining a cooler temperature, while the bottom is also mixed but warmer.
The middle layer is key; it's stable, without fluid motion, making it ideal for contained heat generation. Sunlight enters, heating this central layer differently, creating a unique temperature and energy profile.
In this pond, as solar radiation is absorbed, it maintains a thermal gradient using salt. This allows for significant heat storage. The absence of motion in the middle layer prevents heat escape, making the pond efficient for energy capture.
  • Top and bottom: mixed layers with distinct temperatures.
  • Middle: stationary, creating a temperature gradient.
  • Stability from salt maintains solar heat efficiently.
Fourier's Law
Fourier's Law is essential for understanding heat transfer in a salt-gradient solar pond. It helps measure how heat moves through layers in the pond. The law states that the rate of heat transfer through a medium is proportional to the negative gradient of temperatures and the area through which the heat is conducted.
Mathematically, it is expressed as: \[ q'' = -k \frac{dT}{dx} \]
Here, \( q'' \) is the heat flux, \( k \) is thermal conductivity, and \( \frac{dT}{dx} \) is the temperature gradient. This equation helps in computing how energy dissipates between layers.
  • Crucial for calculating interface heat transfer.
  • Uses known conductivity and temperature changes.
  • Allows assessment of pond's thermal efficiency.
Steady and Transient Conditions
In our solar pond, whether conditions are steady or transient greatly affects energy efficiency. **Steady conditions** mean the heat entering a layer equals the heat leaving, creating a stable system. If the heat balance isn’t achieved, we experience **transient conditions** where temperature changes over time.
To identify these, compare the calculated heat transfers from the middle layer upwards and downwards. If they match, steady conditions prevail; otherwise, the system is transient. By ensuring steady conditions, energy generation and retention are maximized.
  • **Steady:** Heat flow balance, stable temperatures.
  • **Transient:** Unbalanced flow, varying temperatures.
  • Essential for predicting system performance.
Thermal Energy Generation
Thermal energy generation in a salt-gradient solar pond is crucial for energy storage and utilization. Here, the solar energy is absorbed primarily in the central layer, converting it into usable thermal energy.
In the exercise, the energy generation rate is derived by integrating the heat generation expression across the layer's thickness. This quantifies total energy produced within a set surface area. It shows how efficiently the solar pond converts sunlight into heat.
The formula used is:\[ q''_g = \int_0^L A e^{-a x} \, dx = \frac{A}{a}(1 - e^{-aL}) \]
This describes how the environment of the pond harnesses sunlight, a critical step in developing efficient renewable energy systems.
  • Solar ponds as effective thermal energy sources.
  • Calculates energy efficiency through heat generation rate.
  • Provides insights into sustainability and energy production.

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Most popular questions from this chapter

An apparatus for measuring thermal conductivity cmploys an electrical heater sandwiched between two identical samples of diameter \(30 \mathrm{~mm}\) and length \(60 \mathrm{~mm}\), which are presced between plates maintained at a uniform tempenature \(T_{n}=77^{\circ} \mathrm{C}\) by a circulating fluid. \(\mathrm{A}\) conducting grease is placed between all the surfaces to ensure good thermal contact. Differential thermocouples are imbedded in the samples with a spocing of \(15 \mathrm{~mm}\). The lateral sides of the samples are insulated to ensure cee-dimensional heat transfer thrueugh the samples. (a) With two saruples of SS316 in the apparatus, the heater draws \(0.353 \mathrm{~A}\) at \(100 \mathrm{~V}\) and the differential thermocouples indicate \(\Delta T_{1}=\Delta T_{2}=25.0^{\circ} \mathrm{C}\). What is the thermal conductivity of the stainless steel sample material? What is the average tempenture of the samples? Compare your result with the therrmal conductivity value reported for this material in Table A.1. (b) By mistake, an Armeo iron sample is placed in the lower position of the apparatus with one of the SS 316 samples from part (a) in the upper portion. For this situation, the heater draws \(0.601 \mathrm{~A}\) at \(100 \mathrm{~V}\) and the differential thermocouples indicate \(\Delta T_{1}=\) \(\Delta T_{2}=15.0^{\circ} \mathrm{C}\). What are the thermal conductivity and average temperature of the Armco iron sample? (c) What is the advantage in constructing the apparatus with two identical samples sandwiching the heuter rather than with a single heater-sample combination? When would heat leakage out of the lateral surfaces of the samples become significant? Under what conditions would you cxpect \(\Delta T_{1} \neq \Delta T_{2}\) ?

Asume steady-state, one-dimensional heat conduction through the symmetric shape shown. Assuming that there is no internal heat generation, derive an expresion for the thermal conductivity \(k(x)\) for these conditions \(A(x)=(1-x), T(x)=300\left(1-2 x-x^{3}\right)\), and \(q=6000 \mathrm{~W}\), where \(A\) is in square meters, \(T\) in kelvins, and \(x\) in meters.

A spherical shell with inner radius \(r_{1}\) and outer radius \(r_{2}\) has surface temperatures \(T_{1}\) and \(T_{2}\), respectively, where \(T_{1}>T_{2}\). Sketch the temperature distribution on \(T-r\) coontinates assuming steady- state, one-timensional concuction with constant properties. Briefly justify the shape of your curve.

At a given instant of time the temperature distribution within an infinite homogeneous body is given by the function $$ T(x, y, z)=x^{2}-2 y^{2}+z^{2}-x y+2 y t $$ Assuming constant propertics and no internal heat generation, determine the regions where the temperature changes with time.

Sections of the trans-Alaska pipeline run above the ground and ate supported by vertical steel shafts \((k=25\) W/m \(-\mathrm{K})\) that are \(1 \mathrm{~m}\) long and have a cross-sectional area of \(0.005 \mathrm{~m}^{2}\). Under normal cperating conditions, the temperature variation along the length of a shaft is known to be governed by an expression of the form $$ T=100-150 x+10 x^{2} $$ where \(T\) and \(x\) have units of \({ }^{\circ} \mathrm{C}\) and meters, respectively. Temperature variations are small over the shaft cross section. Evaluate the temperature and conduction heat rate at the shaft-pipeline joint \((x=0)\) and at the shaft-ground interface \((x=1 \mathrm{~m})\). Explain the difference in the heat rates.

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