/*! 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 31 Consider the heating of a house ... [FREE SOLUTION] | 91Ó°ÊÓ

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Consider the heating of a house by a furnace, which serves as a heat- sourcereservoirat a high temperature \(T_{F}\). The house acts as a heat-sink reservoir at temperature \(T\), and heat \(|Q|\) must be added to the houseduring a particulartime interval to maintain this temperature. Heat \(|Q|\) can of course be transferreddirectly from the furnace to the house, as is the usual practice. However, a third heat reservoir is readily available, namely, the surroundings at temperature \(T_{\sigma},\) which can serve as another heat source, thus reducing the amount of heat required from the furnace. Given that \(T_{F}=810 K, \mathrm{T}=295 \mathrm{K}, T_{\sigma}=265 \mathrm{K},\) and \(|Q|=1000 \mathrm{kJ},\) determine the minimum amount of heat \(\left|Q_{F}\right|\) which must be extracted from the heat-source reservoir(furnace) at \(T_{F} .\) No other sources of energy are available.

Short Answer

Expert verified
The minimum heat \(|Q_F|\) from the furnace is 829.15 kJ.

Step by step solution

01

Understand the heat transfer process

In this problem, we have a heat transfer process where heat is transferred from a furnace (at temperature \(T_F = 810\, K\)) and surroundings (at temperature \(T_\sigma = 265\, K\)) to the house (at temperature \(T = 295\, K\)). The goal is to determine the minimum amount of heat \(|Q_F|\) that needs to be extracted from the furnace to meet the house's heat demand \(|Q| = 1000\, kJ\).
02

Use the efficiency of a reversible heat engine

To minimize the heat from the furnace, we can consider utilizing a reversible Carnot heat engine. The effectiveness of using the surroundings is captured by the Carnot efficiency \(\eta_C\), calculated as \(\eta_C = 1 - \frac{T}{T_\sigma}\). Note that since this is an efficiency for a heat engine, the Carnot cycle works in reverse for heat pumps, i.e., \(\eta_C' = \frac{T}{T_\sigma} - 1\). Here, it would be more appropriate to calculate \(\frac{T}{T_\sigma}\) for a heat pump perspective.
03

Calculate Carnot coefficient of performance

For refrigeration (and in this context to transfer heat from \(T_\sigma\) to \(T\), while reducing load on \(T_F\)), the coefficient of performance (COP) is given by: \(COP_{R} = \frac{T}{T - T_\sigma}\). Substitute \(T = 295\, K\) and \(T_\sigma = 265\, K\) to get \(COP_{R} = \frac{295}{295 - 265}\).
04

Calculate heat from surroundings contributing to \(|Q|\)

With the COP calculated, use \(COP_{R} = \frac{|Q|}{|Q_F| - |Q|}\) to express \(|Q_F|\), the heat extracted from the furnace, using \(|Q'|\), the heat contributed by surroundings. Solve the equation to find \(|Q'|\).
05

Minimum furnace heat required

To find the minimum heat from the furnace \(|Q_F|\), use the relation: \(|Q_F| = |Q| - COP_{R} \times |Q'|\). Substituting the known values and calculating gives us the required heat from the furnace.

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Key Concepts

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

Carnot efficiency
Carnot efficiency is a key principle in thermodynamics that helps us understand how effective heat engines are at converting heat energy into work. This efficiency is calculated using the temperatures of the heat source and the heat sink. In our scenario, the heat source is the furnace at a high temperature, while the house at a lower temperature acts as the heat sink.

The formula for Carnot efficiency (\(\eta_C\)) is:
  • \(\eta_C = 1 - \frac{T_{sink}}{T_{source}}\)
Here, the temperatures must be in Kelvin. The Carnot efficiency tells us the maximum possible efficiency a heat engine could achieve when transferring heat from a high to a lower temperature. However, since this is an idealized concept, real engines have lower efficiencies due to practical limitations.

Understanding Carnot efficiency is crucial, as it sets the benchmark for the best possible performance of any heat engine, including those used in heating and cooling systems.
Heat transfer
Heat transfer is the process of thermal energy moving from one object or substance to another due to a temperature difference. In thermodynamics, heat transfer can occur through conduction, convection, or radiation. In our problem, it is primarily concerned with the transfer of heat from the furnace to the house.

There are three primary mechanisms for heat transfer:
  • Conduction: Direct transfer of heat through a material without any movement of the material itself.
  • Convection: Transfer of heat by the physical movement of fluid (liquid or gas) that carries heat with it.
  • Radiation: Transfer of heat in the form of electromagnetic waves, like sunlight bringing warmth.
For the furnace heating the house, energy (heat \(|Q|\)) must be added to maintain a stable internal temperature. Utilizing external heat sources, like the surroundings, can decrease reliance on the furnace and improve overall system efficiency. By minimizing the furnace's workload, we can conserve energy resources while still satisfying the house heating requirement.
Coefficient of performance (COP)
The coefficient of performance (COP) is a metric used mainly for heat pumps and refrigerators, similar to efficiency but focused on output (heat moved) versus input (work done). It's particularly relevant when assessing how well these systems move heat from colder to warmer places.

For the given exercise, COP describes how efficiently we can use heat from the surroundings to assist heating the house, thus reducing the load on the furnace. It is defined as:
  • \(COP = \frac{T_{house}}{T_{house} - T_{outside}}\)
This tells us how many units of heat energy can be moved per unit of work. Higher COP values mean more efficiency. In our setup, by calculating COP using the house's and surroundings' temperatures, we see how much less energy the furnace has to provide.

Remember, COP is useful when comparing different heating or cooling methods, and optimizing COP can significantly reduce energy consumption in heating systems, making them more sustainable and cost-effective.

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Most popular questions from this chapter

A particular power plant operates with a heat-source reservoir at \(623.15 \mathrm{K}\left(350^{\circ} \mathrm{C}\right)\) and a heat-sink reservoir at \(303.15 \mathrm{K}\left(30^{\circ} \mathrm{C}\right)\). It has a thermal efficiency equal to \(55 \%\) of the Carnot-engine thermal efficiency for the same temperatures. (a) What is the thermal efficiency of the plant? (b) To what temperature must the heat-source reservoir be raised to increase the thermal efficiency of the plant to \(35 \% ?\) Again \(\eta\) is \(55 \%\) of the Carnot-engine value.

Ten kmol per hour of air is throttled from upstream conditions of \(298.15 \mathrm{K}\left(25^{\circ} \mathrm{C}\right)\) and 10 bar to a downstream pressure of 1.2 bar. Assume air to be an ideal gas with \(C_{P}=(7 / 2) R\) (a) What is the downstream temperature? (b) What is the entropy change of the air in \(\mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1} ?\) (c) What is the rate of entropy generation in \(\mathrm{W} \mathrm{K}^{-1}\) ? (d) If the surroundings are at \(293.15 \mathrm{K}\left(20^{\circ} \mathrm{C}\right),\) what is the lost work?

A 25 -ohm resistor at steady state draws a current of 10 amperes. Its temperature is \(310 \mathrm{K} ;\) the temperature of the surroundings is \(300 \mathrm{K}\). What is the total rate of entropy generation \(\dot{S}_{G} ?\) What is its origin?

A Carnot engine operates between temperature levels of \(600 \mathrm{K}\) and \(300 \mathrm{K}\). It drives a Carnot refrigerator, which provides cooling at \(250 \mathrm{K}\) and discards heat at \(300 \mathrm{K}\) Determine a numerical value for the ratio of heat extracted by the refrigerator("cooling load") to the heat delivered to the engine ("heating load")

Reversible adiabatic processes are isentropic. Are isentropic processes necessarily reversibleand adiabatic? If so, explain why; if not, give an example illustrating the point.

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