{"id":1831,"date":"2021-07-27T20:21:56","date_gmt":"2021-07-28T00:21:56","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/thermo1\/chapter\/6-1-heat-engine\/"},"modified":"2022-08-11T17:24:24","modified_gmt":"2022-08-11T21:24:24","slug":"6-1-heat-engine","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/thermo1\/chapter\/6-1-heat-engine\/","title":{"raw":"6.1 Heat engine","rendered":"6.1 Heat engine"},"content":{"raw":"
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A heat engine is a continuously operating device that produces work by transferring heat from a [pb_glossary id=\"2427\"]heat source[\/pb_glossary] (high-temperature body) to a [pb_glossary id=\"2428\"]heat sink[\/pb_glossary] (low-temperature body) using a working fluid. In a heat engine cycle, a working fluid may remain as a single-phase fluid or experience phase changes.<\/p>\r\n \r\n

A steam engine is a type of heat engine commonly used in steam power generating plants. It operates on Rankine cycles and uses water as the working fluid. We will use a steam engine to illustrate how heat is converted to work in heat engines. A typical steam engine consists of four main equipment: boiler, turbine, condenser, and pump, as shown in Figure 6.1.1<\/a>. The T-s<\/em> diagram in Figure 6.1.2<\/a> illustrates the four processes in a Rankine cycle:<\/p>\r\n\r\n

    \r\n \t
  1. Water at a low pressure and a low temperature (state 1) is pumped to a boiler.\u00a0 The pump consumes power, [latex]\\dot{W}_{pump}[\/latex]<\/span>, in order to maintain a continuous supply of water to the boiler while increasing the pressure of the water entering the boiler (state 2). Process 1-2 may be assumed adiabatic.<\/li>\r\n \t
  2. In the boiler, the liquid water absorbs heat, [latex]\\dot{Q}_{H}[\/latex], <\/span>from an external heat source and changes into high-temperature, high-pressure steam (state 3). The pressure drop in the boiler is usually negligible; therefore, process 2-3 may be assumed isobaric.<\/li>\r\n \t
  3. The high-temperature, high-pressure steam then expands in the turbine, making the turbine rotate continuously, and thus generating mechanical power, [latex]\\dot{W}_{turbine}[\/latex]<\/span>. During the expansion process, the temperature and pressure of the steam decrease. Consequently, the steam leaving the turbine (state 4) becomes a low-temperature, low-pressure, two-phase mixture. Process 3-4 may be assumed adiabatic.<\/li>\r\n \t
  4. The steam leaving the turbine then enters a condenser and is condensed to a saturated or compressed liquid (state 1). During this process, heat, [latex]\\dot{Q}_{L}[\/latex], is removed <\/span>from the steam. The pressure drop in the condenser is usually negligible; <\/a> therefore, process 4-1 may be assumed isobaric.<\/li>\r\n<\/ol>\r\n \r\n\r\n[caption id=\"attachment_2215\" align=\"aligncenter\" width=\"500\"]\"Rankine<\/a> <\/a>Figure 6.1.1<\/strong> Rankine cycle<\/em>[\/caption]\r\n\r\n
    \r\n\r\n[caption id=\"attachment_2216\" align=\"aligncenter\" width=\"500\"]\"T-s<\/a> Figure 6.1.2<\/strong>\u00a0T-s diagram of a Rankine cycle<\/em>[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n

    Figure 6.1.3<\/a> is a simplified schematic for analyzing the energy conservation in heat engines. Applying the first law of thermodynamics to the cycle, we can write<\/p>\r\n \r\n

    [latex]\\dot{Q}_{H} - \\dot{Q}_{L} = \\dot{W}_{turbine} - \\dot{W}_{pump} = \\dot{W}_{net, out} \u00a0[\/latex]<\/p>\r\n \r\n

    Clearly, the heat removed by the condenser, [latex]\\dot{Q}_L[\/latex], cannot be converted to useful work. It is waste<\/em>d<\/em> in order to complete the cycle. In other words, a heat engine cannot convert all<\/em> the heat supplied by the heat source (e.g., boiler) to useful work, even under ideal conditions. Thermal efficiency is a dimensionless parameter used to measure the performance of a heat engine.<\/p>\r\n \r\n

    [latex]\\eta_{th}=\\displaystyle\\frac{desired\\ output}{required\\ input}=\\frac{{\\dot{W}}_{net,\\ out}}{{\\dot{Q}}_H}=1-\\frac{{\\dot{Q}}_L}{{\\dot{Q}}_H}[\/latex]<\/p>\r\n \r\n\r\nwhere\r\n

    [latex]\\dot{Q}_H[\/latex]: heat absorbed from the heat source, in kW<\/p>\r\n

    [latex]\\dot{Q}_L[\/latex]: heat rejected to the heat sink, in kW<\/p>\r\n

    [latex]\\dot{W}_{net,\\ out}[\/latex]: net work output from the heat engine, in kW<\/p>\r\n

    [latex]\\eta_{th}[\/latex]: thermal efficiency of the heat engine, <\/a>dimensionless<\/p>\r\n \r\n\r\n[caption id=\"attachment_3512\" align=\"aligncenter\" width=\"276\"]\"Schematic<\/a> Figure 6.1.3<\/strong> Schematic of a heat engine<\/em>[\/caption]\r\n\r\n

    \r\n
    \r\n

    Example 1<\/p>\r\n\r\n<\/header>\r\n

    \r\n\r\nFigures 6.1.e1<\/a> and 6.1.e2<\/a> illustrate a Rankine cycle consisting of a two-stage steam engine and a feedwater heater. The steam engine is enclosed in the red outlines in Figure 6.1.e1<\/a>. The two stages of the turbine are labelled as HE1 and HE2, respectively. In stage 1, the steam absorbs heat,\u00a0 [latex]\\dot{Q}_H[\/latex], from the boiler and generates a power, [latex]\\dot{W}_{1}[\/latex]. A portion of the exhaust steam from stage 1 then enters stage 2, further generating a power, [latex]\\dot{W}_{2}[\/latex]. The remaining exhaust steam from stage 1 is used to preheat the feed water. If the thermal efficiencies of the two turbine stages are [latex]\\eta_{th,1}[\/latex] and [latex]\\eta_{th,2}[\/latex], what is the overall thermal efficiency of the cycle as a function of [latex]\\eta_{th,1}[\/latex] and [latex]\\eta_{th,2}[\/latex]? Assume 90% of the exhaust steam exiting from stage 1 enters stage 2 and generates the power, <\/a>[latex]\\dot{W}_{2}[\/latex].\r\n\r\n \r\n\r\n[caption id=\"attachment_2235\" align=\"aligncenter\" width=\"500\"]\"Two-stage<\/a> <\/a>Figure 6.1.e1 <\/strong>Two-stage steam turbine with a feed-water heater<\/em>[\/caption]\r\n\r\n[caption id=\"attachment_2237\" align=\"aligncenter\" width=\"500\"]\"Schematic<\/a> Figure 6.1.e2<\/strong>\u00a0Schematic of the two-stage heat engine<\/em>[\/caption]\r\n\r\nSolution:<\/em><\/span>\r\n\r\nThe thermal efficiency of the first and second stages of the steam turbine can be written as\r\n

    [latex]\\eta_{th,1} = \\dfrac{\\dot{W}_{1}}{\\dot{Q}_{H}}[\/latex]\u00a0\u00a0 \u00a0 \u00a0\u00a0 \u00a0\u00a0 [latex]\\eta_{th,2} = \\dfrac{\\dot{W}_{2}}{\\dot{Q}_{M}}[\/latex]<\/p>\r\nThe desired output of the cycle is the total power generated by the turbine and the required energy input comes from the boiler; therefore,\r\n

    [latex]\\eta_{th} = \\dfrac{\\dot{W}_{tot}}{\\dot{Q}_{H}}=\\dfrac{\\dot{W}_{1} + \\dot{W}_{2}}{\\dot{Q}_{H}} [\/latex]<\/p>\r\nApply the first law of thermodynamics to the first stage, HE1. Note that 90% of the exhaust steam from stage 1 enters stage 2; therefore,\r\n

    [latex] \\dot{Q}_{M} + \\dot{Q}_{Heater} = \\dot{Q}_{H} - \\dot{W}_{1}[\/latex]<\/p>\r\n

    and<\/p>\r\n

    [latex]\\dot{Q}_{M} = 0.9 (\\dot{Q}_{H} - \\dot{W}_{1})[\/latex]<\/p>\r\nCombine the above equations and rearrange,\r\n

    [latex]\\begin{align*} \\because\\eta_{th} &= \\dfrac{\\dot{W}_{1}}{\\dot{Q}_{H}} + \\dfrac{\\dot{W}_{2}}{\\dot{Q}_{H}} \\\\&=\\eta_{th,1} + \\dfrac{\\eta_{th,2}\\dot{Q}_{M}}{\\dot{Q}_{H}} \\\\&=\\eta_{th,1} + \\dfrac{\\eta_{th,2} \\times 0.9(\\dot{Q}_{H}-\\dot{W}_{1})}{\\dot{Q}_{H}} \\end{align*}[\/latex]<\/p>\r\n

    [latex] \\therefore \\eta_{th} = \\eta_{th,1} + 0.9 \\eta_{th,2}(1-\\eta_{th,1}) [\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

    \r\n

    Practice Problems<\/p>\r\n\r\n<\/header>\r\n

    \r\n\r\n[h5p id=\"40\"]\r\n\r\n<\/div>\r\n<\/div>","rendered":"
    \n

    A heat engine is a continuously operating device that produces work by transferring heat from a heat source<\/a> (high-temperature body) to a heat sink<\/a> (low-temperature body) using a working fluid. In a heat engine cycle, a working fluid may remain as a single-phase fluid or experience phase changes.<\/p>\n

     <\/p>\n

    A steam engine is a type of heat engine commonly used in steam power generating plants. It operates on Rankine cycles and uses water as the working fluid. We will use a steam engine to illustrate how heat is converted to work in heat engines. A typical steam engine consists of four main equipment: boiler, turbine, condenser, and pump, as shown in Figure 6.1.1<\/a>. The T-s<\/em> diagram in Figure 6.1.2<\/a> illustrates the four processes in a Rankine cycle:<\/p>\n

      \n
    1. Water at a low pressure and a low temperature (state 1) is pumped to a boiler.\u00a0 The pump consumes power, [latex]\\dot{W}_{pump}[\/latex]<\/span>, in order to maintain a continuous supply of water to the boiler while increasing the pressure of the water entering the boiler (state 2). Process 1-2 may be assumed adiabatic.<\/li>\n
    2. In the boiler, the liquid water absorbs heat, [latex]\\dot{Q}_{H}[\/latex], <\/span>from an external heat source and changes into high-temperature, high-pressure steam (state 3). The pressure drop in the boiler is usually negligible; therefore, process 2-3 may be assumed isobaric.<\/li>\n
    3. The high-temperature, high-pressure steam then expands in the turbine, making the turbine rotate continuously, and thus generating mechanical power, [latex]\\dot{W}_{turbine}[\/latex]<\/span>. During the expansion process, the temperature and pressure of the steam decrease. Consequently, the steam leaving the turbine (state 4) becomes a low-temperature, low-pressure, two-phase mixture. Process 3-4 may be assumed adiabatic.<\/li>\n
    4. The steam leaving the turbine then enters a condenser and is condensed to a saturated or compressed liquid (state 1). During this process, heat, [latex]\\dot{Q}_{L}[\/latex], is removed <\/span>from the steam. The pressure drop in the condenser is usually negligible; <\/a> therefore, process 4-1 may be assumed isobaric.<\/li>\n<\/ol>\n

       <\/p>\n

      \"Rankine<\/a>
      <\/a>Figure 6.1.1<\/strong> Rankine cycle<\/em><\/figcaption><\/figure>\n
      \n
      \"T-s<\/a>
      Figure 6.1.2<\/strong>\u00a0T-s diagram of a Rankine cycle<\/em><\/figcaption><\/figure>\n<\/div>\n<\/div>\n

      Figure 6.1.3<\/a> is a simplified schematic for analyzing the energy conservation in heat engines. Applying the first law of thermodynamics to the cycle, we can write<\/p>\n

       <\/p>\n

      [latex]\\dot{Q}_{H} - \\dot{Q}_{L} = \\dot{W}_{turbine} - \\dot{W}_{pump} = \\dot{W}_{net, out} \u00a0[\/latex]<\/p>\n

       <\/p>\n

      Clearly, the heat removed by the condenser, [latex]\\dot{Q}_L[\/latex], cannot be converted to useful work. It is waste<\/em>d<\/em> in order to complete the cycle. In other words, a heat engine cannot convert all<\/em> the heat supplied by the heat source (e.g., boiler) to useful work, even under ideal conditions. Thermal efficiency is a dimensionless parameter used to measure the performance of a heat engine.<\/p>\n

       <\/p>\n

      [latex]\\eta_{th}=\\displaystyle\\frac{desired\\ output}{required\\ input}=\\frac{{\\dot{W}}_{net,\\ out}}{{\\dot{Q}}_H}=1-\\frac{{\\dot{Q}}_L}{{\\dot{Q}}_H}[\/latex]<\/p>\n

       <\/p>\n

      where<\/p>\n

      [latex]\\dot{Q}_H[\/latex]: heat absorbed from the heat source, in kW<\/p>\n

      [latex]\\dot{Q}_L[\/latex]: heat rejected to the heat sink, in kW<\/p>\n

      [latex]\\dot{W}_{net,\\ out}[\/latex]: net work output from the heat engine, in kW<\/p>\n

      [latex]\\eta_{th}[\/latex]: thermal efficiency of the heat engine, <\/a>dimensionless<\/p>\n

       <\/p>\n

      \"Schematic<\/a>
      Figure 6.1.3<\/strong> Schematic of a heat engine<\/em><\/figcaption><\/figure>\n
      \n
      \n
      \n

      Example 1<\/p>\n<\/header>\n

      \n

      Figures 6.1.e1<\/a> and 6.1.e2<\/a> illustrate a Rankine cycle consisting of a two-stage steam engine and a feedwater heater. The steam engine is enclosed in the red outlines in Figure 6.1.e1<\/a>. The two stages of the turbine are labelled as HE1 and HE2, respectively. In stage 1, the steam absorbs heat,\u00a0 [latex]\\dot{Q}_H[\/latex], from the boiler and generates a power, [latex]\\dot{W}_{1}[\/latex]. A portion of the exhaust steam from stage 1 then enters stage 2, further generating a power, [latex]\\dot{W}_{2}[\/latex]. The remaining exhaust steam from stage 1 is used to preheat the feed water. If the thermal efficiencies of the two turbine stages are [latex]\\eta_{th,1}[\/latex] and [latex]\\eta_{th,2}[\/latex], what is the overall thermal efficiency of the cycle as a function of [latex]\\eta_{th,1}[\/latex] and [latex]\\eta_{th,2}[\/latex]? Assume 90% of the exhaust steam exiting from stage 1 enters stage 2 and generates the power, <\/a>[latex]\\dot{W}_{2}[\/latex].<\/p>\n

       <\/p>\n

      \"Two-stage<\/a>
      <\/a>Figure 6.1.e1 <\/strong>Two-stage steam turbine with a feed-water heater<\/em><\/figcaption><\/figure>\n
      \"Schematic<\/a>
      Figure 6.1.e2<\/strong>\u00a0Schematic of the two-stage heat engine<\/em><\/figcaption><\/figure>\n

      Solution:<\/em><\/span><\/p>\n

      The thermal efficiency of the first and second stages of the steam turbine can be written as<\/p>\n

      [latex]\\eta_{th,1} = \\dfrac{\\dot{W}_{1}}{\\dot{Q}_{H}}[\/latex]\u00a0\u00a0 \u00a0 \u00a0\u00a0 \u00a0\u00a0 [latex]\\eta_{th,2} = \\dfrac{\\dot{W}_{2}}{\\dot{Q}_{M}}[\/latex]<\/p>\n

      The desired output of the cycle is the total power generated by the turbine and the required energy input comes from the boiler; therefore,<\/p>\n

      [latex]\\eta_{th} = \\dfrac{\\dot{W}_{tot}}{\\dot{Q}_{H}}=\\dfrac{\\dot{W}_{1} + \\dot{W}_{2}}{\\dot{Q}_{H}}[\/latex]<\/p>\n

      Apply the first law of thermodynamics to the first stage, HE1. Note that 90% of the exhaust steam from stage 1 enters stage 2; therefore,<\/p>\n

      [latex]\\dot{Q}_{M} + \\dot{Q}_{Heater} = \\dot{Q}_{H} - \\dot{W}_{1}[\/latex]<\/p>\n

      and<\/p>\n

      [latex]\\dot{Q}_{M} = 0.9 (\\dot{Q}_{H} - \\dot{W}_{1})[\/latex]<\/p>\n

      Combine the above equations and rearrange,<\/p>\n

      [latex]\\begin{align*} \\because\\eta_{th} &= \\dfrac{\\dot{W}_{1}}{\\dot{Q}_{H}} + \\dfrac{\\dot{W}_{2}}{\\dot{Q}_{H}} \\\\&=\\eta_{th,1} + \\dfrac{\\eta_{th,2}\\dot{Q}_{M}}{\\dot{Q}_{H}} \\\\&=\\eta_{th,1} + \\dfrac{\\eta_{th,2} \\times 0.9(\\dot{Q}_{H}-\\dot{W}_{1})}{\\dot{Q}_{H}} \\end{align*}[\/latex]<\/p>\n

      [latex]\\therefore \\eta_{th} = \\eta_{th,1} + 0.9 \\eta_{th,2}(1-\\eta_{th,1})[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n

      \n
      \n

      Practice Problems<\/p>\n<\/header>\n

      \n
      \n