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class="col-md-9" id="main-column"> <h1 class="page_title"> Cyclic processes <a href="#" class="mark-page-favorite pull-right" data-pid="853" title="Mark as favorite" onclick="return false;"><i class="fa fa-star-o"></i></a> </h1> <ol class="breadcrumb"> <li><a href="../../../physics.html"><i class="fa fa-home"></i></a><i class="fa fa-fw fa-chevron-right divider"></i></li><li><a href="../849/engineering.html">Engineering</a><i class="fa fa-fw fa-chevron-right divider"></i></li><li><span class="gray">Cyclic processes</span></li> <span class="pull-right" style="color: #555" title="Suggested study time: 15 minutes"><i class="fa fa-clock-o"></i> 15'</span> </ol> <article id="main-article"> <p><img alt="" src="../../engineering/cycle-1.png" style="float: left; width: 149px; height: 150px;">A <em>cyclic process</em> is a series of transformations that take the gas back to its original state. These form a closed loop on a <span class="math-tex">\(pV\)</span> diagram.</p> <hr class="hidden-separator"> <div class="panel panel-turquoise panel-has-colored-body"> <div class="panel-heading"> <div> <p>Key Concepts</p> </div> </div> <div class="panel-body"> <div> <div class="panel panel-has-colored-body panel-has-border panel-turquoise"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Heat engine</p> </div> </div> <div class="panel-body"> <div> <p>A simple cyclic process involves:</p> <div class="box"> <ol> <li>Isobaric expansion - work is done on the surroundings</li> <li>Isovolumetric cooling</li> <li>Isobaric compression - work is done on the gas</li> <li>Isovolumetric heating</li> </ol> </div> <p>Let's consider the cycle for a heat engine.</p> <div class="video-embed vimeo"><iframe allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen="" mozallowfullscreen="" webkitallowfullscreen="" height="420" width="100%" src="https://player.vimeo.com/video/355175275"></iframe></div> <p style="text-align: center;"><iframe frameborder="0" height="440" scrolling="no" src="https://www.geogebra.org/material/iframe/id/2162493/width/438/height/440/border/888888/rc/false/ai/false/sdz/false/smb/false/stb/false/stbh/true/ld/false/sri/true/at/auto" width="438"></iframe></p> <ol> </ol> <p>Since <span class="math-tex">\(W=p\Delta V\)</span>, the area between a line on the <span class="math-tex">\(pV\)</span> diagram and the <span class="math-tex">\(V\)</span> axis gives work done. The net work done during a cyclic process is given by the enclosed area.</p> <p>We can show the net work done using a Sankey diagram.</p> <p style="text-align: center;"><iframe frameborder="0" height="439" scrolling="no" src="https://www.geogebra.org/material/iframe/id/rV0c3Tdu/width/691/height/439/border/888888/rc/false/ai/false/sdz/false/smb/false/stb/false/stbh/true/ld/false/sri/true/at/auto" width="691"></iframe></p> </div> </div> <div class="panel-footer"> <div> <p> </p> </div> </div> </div> <div class="panel panel-has-colored-body panel-has-border panel-turquoise panel-expandable"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Refrigerator</p> </div> </div> <div class="panel-body"> <div> <p>A refrigerator completes the opposite process.</p> <div class="video-embed vimeo"><iframe allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen="" mozallowfullscreen="" webkitallowfullscreen="" height="420" width="100%" src="https://player.vimeo.com/video/355175160"></iframe></div> </div> </div> <div class="panel-footer"> <div> <p> </p> </div> </div> </div> </div> </div> <div class="panel-footer"> <div> </div> </div> </div> <div class="panel panel-has-colored-body panel-yellow"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Essentials</p> </div> </div> <div class="panel-body"> <div> <div class="panel panel-has-colored-body panel-has-border panel-yellow"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Thermal efficiency</p> </div> </div> <div class="panel-body"> <div> <p>The dimensionless quantity, <em>thermal efficiency</em>, of any mechanical process is defined as the ratio of the useful work done to the total energy input:</p> <p style="text-align: center;"><span class="math-tex">\(\eta={\textrm{useful work done}\over \textrm{energy input}}\)</span></p> <p>For a heat engine:</p> <ul> <li>Energy input is <span class="math-tex">\(Q_\textrm{H}\)</span></li> <li>Work done is <span class="math-tex">\(W_\textrm{out}=Q_\textrm{H}-Q_\textrm{C}\)</span></li> </ul> <p style="text-align: center;"><span class="math-tex">\(\eta={Q_\textrm{H}-Q_\textrm{C}\over Q_\textrm{H}}=1-{Q_\textrm{C}\over Q_\textrm{H}}\)</span></p> </div> </div> <div class="panel-footer"> <div> <p> </p> </div> </div> </div> <div class="panel panel-has-colored-body panel-has-border panel-yellow panel-expandable"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Carnot cycle</p> </div> </div> <div class="panel-body"> <div> <p>The most efficient cyclic process is known as the <i>Carnot cycle</i>. This involves:</p> <div class="box"> <ol> <li><img alt="" src="../../engineering/carnot1.png" style="float: right; width: 400px; height: 418px;">Isothermal expansion - work is done on the surroundings</li> <li>Adiabatic expansion - work is done on the surroundings</li> <li>Isothermal compression - work is done on the gas</li> <li>Adiabatic compression - work is done on the gas </li> </ol> </div> <ul> </ul> <div class="video-embed vimeo"><iframe allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen="" mozallowfullscreen="" webkitallowfullscreen="" height="420" width="100%" src="https://player.vimeo.com/video/355174861"></iframe></div> <p>Note that you may be required to calculate the work done during any part of the process using the 'counting squares' method. This is a little arduous!</p> <ul> <li>Calculate the size represented by one square using the bottom left square on the axes</li> <li>Count the squares enclosed in the area required and multiply</li> </ul> <div class="video-embed vimeo"><iframe allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen="" mozallowfullscreen="" webkitallowfullscreen="" height="420" width="100%" src="https://player.vimeo.com/video/355174887"></iframe></div> <p>To calculate the thermal efficiency we must consider entropy.</p> <div class="box"> <p>The total change in entropy, <span class="math-tex">\(\Delta S=\Delta S_1+\Delta S_2+\Delta S_3+\Delta S_4\)</span></p> <p>Since stages 2 and 4 are adiabatic, <span class="math-tex">\(Q=0\)</span> and so <span class="math-tex">\(\Delta S _2 = \Delta S_4=0\)</span></p> <p>Stages 1 and 3 take place at constant temperature, so <span class="math-tex">\(\Delta S_1 = {Q_\textrm{H} \over T_\textrm{H}}\)</span> and <span class="math-tex">\(\Delta S_3 = {Q_\textrm{C} \over T_\textrm{C}}\)</span></p> <p>Since the Carnot cycle is a reversible process, <span class="math-tex">\(\Delta S = {Q_\textrm{H} \over T_\textrm{H}} - {Q_\textrm{C} \over T_\textrm{C}} = 0\)</span></p> <p style="text-align: center;"><span class="math-tex">\( {Q_\textrm{H} \over T_\textrm{H}} = {Q_\textrm{C} \over T_\textrm{C}} \Rightarrow {Q_\textrm{C} \over Q_\textrm{H}} = {T_\textrm{C} \over T_\textrm{H}}\)</span></p> <p>Substituting into the general equation for thermal efficiency of a heat engine:</p> <p style="text-align: center;"><span class="math-tex">\(\eta_\textrm{Carnot}=1-{T_\textrm{cold}\over T_\textrm{hot}}\)</span></p> </div> <p style="text-align: center;"><iframe frameborder="0" height="499" scrolling="no" src="https://www.geogebra.org/material/iframe/id/499307/width/548/height/499/border/888888/rc/false/ai/false/sdz/false/smb/false/stb/false/stbh/true/ld/false/sri/true/at/auto" width="548"></iframe></p> <p>The equation for thermal efficiency for a Carnot cycle makes clear that we can approach but never attain 100% efficiency. To do so we would need to have a high temperature thermal energy store approaching infinity or a low temperature thermal energy sink approaching absolute zero:</p> <ul> <li>If <span class="math-tex">\(T_\textrm{hot} \rightarrow \infty\)</span>, <span class="math-tex">\(\eta \rightarrow1\)</span></li> <li>If <span class="math-tex">\(T_\textrm{cold}\rightarrow 0 \textrm{K}\)</span>, <span class="math-tex">\(\eta \rightarrow1\)</span></li> </ul> <p>At the operating temperatures provided, the Carnot cycle is the most efficient process possible.</p> </div> </div> <div class="panel-footer"> <div> <p> </p> </div> </div> </div> </div> </div> </div> <div class="panel panel-has-colored-body panel-green"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Test Yourself</p> </div> </div> <div class="panel-body"> <div> <p><em>Use flashcards to practise your recall.</em></p> <div class="tib-flashcard"><a class="show-flashcards btn btn-success btn-xs-block btn-block " data-levels="1" data-mode="Normal" data-topics="850" data-subject-id="6" data-n-flashcards="6" style="text-align:center">Show flashcards</a></div><hr> <p><em>Use quizzes to practise application of theory.</em> </p> <br><a class="btn btn-primary btn-block text-center" data-toggle="modal" href="#e903dd76"><i class="fa fa-play"></i> START QUIZ!</a><div class="modal fade modal-slide-quiz" id="e903dd76"> <div class="modal-dialog" style="width: 95vw; max-width: 960px"> <div class="modal-content"> <div class="modal-header slide-quiz-title"> <h4 class="modal-title" style="width: 100%;"> Cyclic processes <strong class="q-number pull-right"> <span class="counter">1</span>/<span class="total">1</span> </strong> </h4> </div> <div class="modal-body p-xs-3"> <div class="slide-quiz" data-stats="6-303-853" style="opacity: 0"> <div class="exercise shadow-bottom"><div class="q-question"><p>This <span class="math-tex">\(pV\)</span>diagram shows a cyclic process. The amount of gas present is such that <span class="math-tex">\(nR = 10\)</span> kPa cm<sup>3</sup>K<sup>-1</sup> <span class="math-tex">\(=0.01\)</span> Pa m<sup>3</sup>K<sup>-1</sup>.</p><p><img alt="" height="264" src="../../engineering/screenshot-2019-07-31-at-09.09.08.png" width="290"></p><p>Calculate the temperature change during the transition from A to B.</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>4000 K</span></label> </p><p><label class="radio"> <input type="radio"> <span>3000 K</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>2000 K</span></label> </p><p><label class="radio"> <input type="radio"> <span>1000 K</span></label> </p></div><div class="q-explanation"><p>Using the ideal gas law, calculate temperature separately at each point <span class="math-tex">\(T = {PV\over nR}\)</span>:</p><ul><li><span class="math-tex">\(T_A = 2000\text{ K}\)</span></li><li><span class="math-tex">\(T_B = 4000\text{ K}\)</span></li></ul></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>This <span class="math-tex">\(pV\)</span> diagram shows a cyclic process. The amount of gas present is such that <span class="math-tex">\(nR = 10\)</span> kPa cm<sup>3</sup>K<sup>-1</sup> <span class="math-tex">\(=0.01\)</span> Pa m<sup>3</sup>K<sup>-1</sup>.</p><p><img alt="" height="264" src="../../engineering/screenshot-2019-07-31-at-09.09.08.png" width="290"></p><p>Calculate the internal energy change during the transition from A to B.</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>30 kJ</span></label> </p><p><label class="radio"> <input type="radio"> <span>20 J</span></label> </p><p><label class="radio"> <input type="radio"> <span>20 kJ</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>30 J</span></label> </p></div><div class="q-explanation"><p>The difference in temperature (calculated using the ideal gas law for the two points) is 2000 K.</p><p><span class="math-tex">\(ΔU = {3\over 2}nRΔT = {3\over 2} \times 0.01 \times 2000\)</span></p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>This <span class="math-tex">\(pV\)</span> diagram shows a cyclic process. The amount of gas present is such that <span class="math-tex">\(nR = 10\)</span> kPa cm<sup>3</sup>K<sup>-1</sup> <span class="math-tex">\(=0.01\)</span> Pa m<sup>3</sup>K<sup>-1</sup>.</p><p><img alt="" height="264" src="../../engineering/screenshot-2019-07-31-at-09.09.08.png" width="290"></p><p>Calculate the work done by the gas during the transition from A to B.</p></div><div class="q-answer"><p><label class="radio"> <input class="c" type="radio"> <span>20 J</span></label> </p><p><label class="radio"> <input type="radio"> <span>5 J</span></label> </p><p><label class="radio"> <input type="radio"> <span>10 J</span></label> </p><p><label class="radio"> <input type="radio"> <span>40 J</span></label> </p></div><div class="q-explanation"><p>The work done is equal to the area under the graph: <span class="math-tex">\(W = 200 \times 10^3 \times 100 \times 10^{-6}\)</span></p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>This <span class="math-tex">\(pV\)</span> diagram shows a cyclic process. The amount of gas present is such that <span class="math-tex">\(nR = 10\)</span> kPa cm<sup>3</sup>K<sup>-1</sup> <span class="math-tex">\(=0.01\)</span> Pa m<sup>3</sup>K<sup>-1</sup>.</p><p><img alt="" height="264" src="../../engineering/screenshot-2019-07-31-at-09.09.08.png" width="290"></p><p>Calculate the work done by the gas during the transition from B to C.</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>5 J</span></label> </p><p><label class="radio"> <input type="radio"> <span>10 J</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>0 J</span></label> </p><p><label class="radio"> <input type="radio"> <span>40 J</span></label> </p></div><div class="q-explanation"><p>There is no change in volume, which means there has been no displacement. Nor is there any area enclosed under a vertical straight line. Therefore, no work has been done.</p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>This <span class="math-tex">\(pV\)</span> diagram shows a cyclic process. The amount of gas present is such that <span class="math-tex">\(nR = 10\)</span> kPa cm<sup>3</sup>K<sup>-1</sup> <span class="math-tex">\(=0.01\)</span> Pa m<sup>3</sup>K<sup>-1</sup>.</p><p><img alt="" height="264" src="../../engineering/screenshot-2019-07-31-at-09.09.08.png" width="290"></p><p>Calculate the heat removed from the gas during the transition from B to C.</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>10 J</span></label> </p><p><label class="radio"> <input type="radio"> <span>0 J</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>30 J</span></label> </p><p><label class="radio"> <input type="radio"> <span>40 J</span></label> </p></div><div class="q-explanation"><p>Using the ideal gas law, calculate temperature separately at each point <span class="math-tex">\(T = {PV\over nR}\)</span>:</p><ul><li><span class="math-tex">\(T_B = 4000\text{ K}\)</span></li><li><span class="math-tex">\(T_C = 2000\text{ K}\)</span></li></ul><p><span class="math-tex">\(Q = ΔU = {3\over2}nRΔT = {3\over2} \times 0.01 \times (-2000)\)</span></p><p>We are asked for the heat removed <em>from</em> the gas: <span class="math-tex">\(Q=+30\text{ J}\)</span></p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>This <span class="math-tex">\(pV\)</span> diagram shows a cyclic process. The amount of gas present is such that <span class="math-tex">\(nR = 10\)</span> kPa cm<sup>3</sup>K<sup>-1</sup> <span class="math-tex">\(=0.01\)</span> Pa m<sup>3</sup>K<sup>-1</sup>.</p><p><img alt="" height="264" src="../../engineering/screenshot-2019-07-31-at-09.09.08.png" width="290"></p><p>Calculate the work done on the gas during the transition from C to D.</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>0 J</span></label> </p><p><label class="radio"> <input type="radio"> <span>20 J</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>10 J</span></label> </p><p><label class="radio"> <input type="radio"> <span>40 J</span></label> </p></div><div class="q-explanation"><p>The work done is equal to the area under the graph: <span class="math-tex">\(W = 100 \times 10^3 \times (-100) \times 10^{-6}\)</span></p><p>Since we are asked for work done <i>on</i> the gas: <span class="math-tex">\(W=+10\text { J}\)</span></p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>This <span class="math-tex">\(pV\)</span> diagram shows a cyclic process. The amount of gas present is such that <span class="math-tex">\(nR = 10\)</span> kPa cm<sup>3</sup>K<sup>-1</sup> <span class="math-tex">\(=0.01\)</span> Pa m<sup>3</sup>K<sup>-1</sup>.</p><p><img alt="" height="264" src="../../engineering/screenshot-2019-07-31-at-09.09.08.png" width="290"></p><p>The net work done by the gas in the cycle is:</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>20 J</span></label> </p><p><label class="radio"> <input type="radio"> <span>30 J</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>10 J</span></label> </p><p><label class="radio"> <input type="radio"> <span>40 J</span></label> </p></div><div class="q-explanation"><p>Net work done by the gas is the area enclosed (in a clockwise direction):</p><p><span class="math-tex">\(W = 100 \times 10^3 \times 100 \times 10^{-6}\)</span></p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>This <span class="math-tex">\(pV\)</span> diagram shows a Carnot cycle. The amount of gas present is such that <span class="math-tex">\(nR = 10\)</span> kPa cm<sup>3</sup>K<sup>-1</sup> <span class="math-tex">\(=0.01\)</span> Pa m<sup>3</sup>K<sup>-1</sup>.</p><p><img alt="" height="296" src="../../engineering/screenshot-2019-07-31-at-10.45.24.png" width="339"></p><p>The temperature is maximum at:</p></div><div class="q-answer"><p><label class="radio" style=" float: left; margin-right: 40px; "> <input type="radio"> <span>C and D</span></label> </p><p><label class="radio" style=" float: left; margin-right: 40px; "> <input type="radio"> <span>B and C</span></label> </p><p><label class="radio" style=" float: left; margin-right: 40px; "> <input type="radio"> <span>D and A</span></label> </p><p><label class="radio" style=" float: left; margin-right: 40px; "> <input class="c" type="radio"> <span>A and B</span></label> </p></div><div class="q-explanation"><p>The transition from A to B is a high temperature isotherm. This isotherm is always at higher temperature than the third stage because more work needs to be done <em>by</em> the gas than <em>on</em> it.</p><p>NOTE: B and C (and A and D) are at different temperatures as these stages are adiabatic.</p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>This <span class="math-tex">\(pV\)</span> diagram shows a Carnot cycle. The amount of gas present is such that <span class="math-tex">\(nR = 10\)</span> kPa cm<sup>3</sup>K<sup>-1</sup> <span class="math-tex">\(=0.01\)</span> Pa m<sup>3</sup>K<sup>-1</sup>.</p><p><img alt="" height="296" src="../../engineering/screenshot-2019-07-31-at-10.45.24.png" width="339"></p><p>The net work done by the gas is approximately:</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>20 J</span></label> </p><p><label class="radio"> <input type="radio"> <span>30 J</span></label> </p><p><label class="radio"> <input type="radio"> <span>40 J</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>10 J</span></label> </p></div><div class="q-explanation"><p>The area enclosed by the cycle is about one square. Each square represents <span class="math-tex">\(100\times 10^3\times100\times 10^{-6}\)</span>J.</p><p>HINT: It's easiest to calculate the work done equivalent for one square by looking at the data for the square at the origin.</p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>This <span class="math-tex">\(pV\)</span> diagram shows a Carnot cycle. The amount of gas present is such that <span class="math-tex">\(nR = 10\)</span> kPa cm<sup>3</sup>K<sup>-1</sup> <span class="math-tex">\(=0.01\)</span> Pa m<sup>3</sup>K<sup>-1</sup>.</p><p><img alt="" height="296" src="../../engineering/screenshot-2019-07-31-at-10.45.24.png" width="339"></p><p>Most work is done by the gas from...</p></div><div class="q-answer"><p><label class="radio" style=" float: left; margin-right: 40px; "> <input type="radio"> <span>B to C</span></label> </p><p><label class="radio" style=" float: left; margin-right: 40px; "> <input type="radio"> <span>D to A</span></label> </p><p><label class="radio" style=" float: left; margin-right: 40px; "> <input type="radio"> <span>C to D</span></label> </p><p><label class="radio" style=" float: left; margin-right: 40px; "> <input class="c" type="radio"> <span>A to B</span></label> </p></div><div class="q-explanation"><p>The area between the graph and the horiztonal axis is largest under stage A to B.</p></div><div 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