{"id":4737,"date":"2024-10-06T13:05:17","date_gmt":"2024-10-06T13:05:17","guid":{"rendered":"https:\/\/workhouse.sweetdishy.com\/?p=4737"},"modified":"2024-10-06T13:05:18","modified_gmt":"2024-10-06T13:05:18","slug":"introducing-the-effects-of-mixing-into-the-energy-balance","status":"publish","type":"post","link":"https:\/\/workhouse.sweetdishy.com\/index.php\/2024\/10\/06\/introducing-the-effects-of-mixing-into-the-energy-balance\/","title":{"rendered":"Introducing the Effects of Mixing into the Energy Balance"},"content":{"rendered":"\n

In Section 12.1 we restricted the discussion and examples to the standard state (25\u00b0C and 1 atm). In this section we proceed with what happens when the temperatures of the inlet and outlet streams differ from 25\u00b0C for a binary\u00a0mixture in an open, steady-state process. (For a closed system the initial and final states of the internal energy would be involved rather than the stream flows.) You can treat problems involving the heat of solution\/mixing in exactly the same way that you can treat problems involving reaction. The heat of solution\/mixing is analogous to the heat of reaction in the energy balance. You can carry out the needed calculations by<\/p>\n\n\n\n

    \n
  1. a.<\/strong> Associating heats of formation of the compounds and solutions with each of the respective compounds and solutions, or<\/li>\n\n\n\n
  2. b.<\/strong>\u00a0Computing the overall lumped heat of solution at the reference state, and for either option calculating the sensible heats (and phase change effects) for the compounds and solutions from the reference state.<\/li>\n<\/ol>\n\n\n\n

    The next example shows the detailed procedure.<\/p>\n\n\n\n

    Example 12.2 Application of Heat of Solution Data<\/strong><\/p>\n\n\n\n

    Hydrochloric acid is an important industrial chemical. To make aqueous solutions of it in a commercial grade (known as\u00a0muriatic acid<\/em>), purified HCl(g) is absorbed in water in a tantalum absorber in a steady-state continuous process. How much heat must be removed from the absorber by the cooling water per 100 kg of product if hot HCl(g) at 120\u00b0C is fed into water in the absorber as shown in\u00a0Figure E12.2? The feed water can be assumed to be at 25\u00b0C, and the exit product HCl(aq) is 25% HCl (by weight) at 35\u00b0C. The cooling water does not mix with the HCl solution.<\/p>\n\n\n\n

    \"A
    Figure E12.2<\/strong><\/figcaption><\/figure>\n\n\n\n

    Solution<\/strong><\/p>\n\n\n\n

    Steps 1\u20134<\/strong><\/p>\n\n\n\n

    You need to convert the process data to moles of HCl to be able to use the data in\u00a0Table 12.1. Consequently, we will first convert the product into moles of HCl and moles of H2<\/sub>O.<\/p>\n\n\n\n

    Table E12.2a<\/strong><\/p>\n\n\n\n

    Component<\/strong><\/th>kg<\/strong><\/th>Mol. wt.<\/strong><\/th>kg mol<\/strong><\/th>Mole Fraction<\/strong><\/th><\/tr><\/thead>
    HCl<\/td>25<\/td>36.37<\/td>0.685<\/td>0.141<\/td><\/tr>
    H2<\/sub>O<\/td>75<\/td>18.02<\/td>4.163<\/td>0.859<\/td><\/tr>
    <\/td>Total<\/td>100<\/td><\/td>4.848<\/td>1.000<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

    The mole ratio of H2<\/sub>O to HCl is 4.163\/0.685 =6.077.<\/p>\n\n\n\n

    Step 5<\/strong><\/p>\n\n\n\n

    The system will be the HCl and water (not including the cooling water).<\/p>\n\n\n\n

    Basis: 100 kg of product<\/p>\n\n\n\n

    Ref. temperature: 25\u00b0C<\/p>\n\n\n\n

    Steps 6 and 7<\/strong><\/p>\n\n\n\n

    The energy balance reduces to\u00a0Q<\/em>\u00a0= \u0394H,<\/em>\u00a0and both the initial and final enthalpies of all of the streams are known or can be calculated directly; hence the problem has zero degrees of freedom. From simple material balances the kilograms and moles of HCl in and out, and the water in and out, are as listed in\u00a0Table E12.2a\u00a0above.<\/p>\n\n\n\n

    Step 3 (continued)<\/strong><\/p>\n\n\n\n

    Next, you have to determine the enthalpy values for the streams. Data are:\u00a0C<\/em>p<\/sub>\u00a0for the HCl(g) is from\u00a0Table G.1;\u00a0C<\/em>p<\/sub>\u00a0for the product is approximately 2.7 J\/(g)(\u00b0C) equivalent to 55.6 J\/(g mol) (\u00b0C);\u00a0\u0394H^Fo\u00a0for HCl \u00b7 6.077 H2<\/sub>O \u224c \u2212157,753 J\/g mol HCl. We will use\u00a0\u0394H^Fo\u00a0values for each stream in the calculation of \u0394H.<\/em><\/p>\n\n\n\n

    Steps 8 and 9<\/strong><\/p>\n\n\n\n

    Table E12.2b<\/strong><\/p>\n\n\n\n

    Stream<\/strong><\/th>g mol<\/strong><\/th>T<\/em>(\u00b0C)<\/strong><\/th>\u0394H^fo(J\/g molHCl)<\/th>\u0394H^sensible(J\/gmol)<\/th><\/tr><\/thead>
    OUT<\/em><\/td><\/td><\/td><\/td><\/td><\/tr>
    <\/td>HCl (aq)<\/td>4.848*<\/a><\/sup><\/td>35<\/td>\u2212157,753<\/td>\u222b25\u00b0C35\u00b0C(2.7)dT<\/td><\/tr>
    IN<\/em><\/td><\/td><\/td><\/td><\/td><\/tr>
    <\/td>H2<\/sub>O(l<\/em>)<\/td>4.163<\/td>25<\/td>\u2212<\/td> <\/td><\/tr>
    <\/td>HCl(g)<\/td>0.685<\/td>120<\/td>\u221292,311<\/td>\u222b25\u00b0C120\u00b0C(29.13\u22120.134\u00d710\u22122T)dT\u2009\u2009\u2009=2758<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

    *<\/sup>HCl=0.685Q=\u0394Hout\u2212\u0394HinOut\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009In=[0.685(\u2212157,753)+4.848(27)]\u2212[0+0.685(\u221292,311)+0.685(2758)]=\u221246,586J<\/p>\n\n\n\n

    If you use heat of solution values, the calculation is (from\u00a0Table 12.1\u00a0the heat of solution is \u221265,442 J\/g mol HCl for the ratio of HCl\/H2<\/sub>O =6.077)Q=\u0394Hout,sensible\u2212\u0394Hin,sensible+\u0394Hsensible=(4.848)(27)\u2212[0.685(2753)+0]+(0.685)(\u221265,442)=\u221246,586\u2062\u2009J \u00a0\u00a0\u00a0\u00a0\u00a0as expected<\/p>\n\n\n\n

    In a process simulation code, table lookup or equations would be used to calculate the heats of formation at various temperatures (and pressures) other than 25\u00b0C (and 1 atm). The details would be buried in the computer code. You can better understand what the calculations for an energy balance involve if you use a graph\u2014at the expense of some accuracy\u2014instead of equations.<\/p>\n\n\n\n

    A convenient graphical way to represent enthalpy data for binary solutions is via an\u00a0enthalpy-concentration diagram.<\/strong>\u00a0Enthalpy-concentration diagrams (H-x<\/em>) are plots of specific enthalpy versus concentration (usually mass or mole fraction) with temperature as a parameter.\u00a0Figure 12.3\u00a0illustrates one such plot. If available,1<\/sup>\u00a0such charts are useful in making combined material\u00a0and energy balance calculations in distillation, crystallization, and all sorts of mixing and separation problems. You will find a few examples of enthalpy-concentration charts in\u00a0Appendix J.<\/p>\n\n\n\n

    1<\/sup>For a literature survey as of 1957, see Robert Lemlich, Chad Gottschlich, and Ronald Hoke,\u00a0Chem. Eng. Data Ser.,<\/em>\u00a02<\/strong>\u00a0, 32 (1957). Additional references: for CC14<\/sub>, see M. M. Krishniah et al,\u00a0J. Chem. Eng. Data,<\/em>\u00a010<\/strong>\u00a0, 117 (1965); for EtOH-EtAc, see Robert Lemlich, Chad Gottschlich, and Ronald Hoke,\u00a0Br. Chem. Eng.,<\/em>\u00a010<\/strong>\u00a0, 703 (1965); for methanol-toluene, see C. A. Plank and D. E. Burke,\u00a0Hydrocarbon Process,<\/em>\u00a045<\/strong>\u00a0, No. 8, 167 (1966); for acetone-isopropanol, see S. N. Balasubramanian,\u00a0Br. Chem. Eng.,<\/em>\u00a012<\/strong>\u00a0, 1231 (1967); for acetonitrile-water-ethanol, see Reddy and Murti, ibid.,\u00a013<\/strong>\u00a0, 1443 (1968); for alcohol-aliphatics, see Reddy and Murti, ibid.,\u00a016<\/strong>\u00a0, 1036 (1971); and for H2<\/sub>SO4<\/sub>, see D. D. Huxtable and D. R. Poole,\u00a0Proc. Int. Solar Energy Soc.,<\/em>\u00a0Winnipeg, August 15, 1976,\u00a08<\/strong>\u00a0, 178 (1977). For more recent sources search the Internet.<\/p>\n\n\n\n

    \"A
    Figure 12.3<\/strong>\u00a0Enthalpy concentration of\u00a0n<\/em>-butane-n<\/em>-heptane at 100 psia. Curve\u00a0DFHC<\/em>\u00a0is the saturated vapor; curve\u00a0BEGA<\/em>\u00a0is the saturated liquid. The dashed lines are equilibrium tie lines connecting\u00a0y,<\/em>\u00a0the mole fraction of C4<\/sub>\u00a0in the vapor, and\u00a0x,<\/em>\u00a0the mole fraction of C4<\/sub>\u00a0in the liquid, at the same temperature.<\/figcaption><\/figure>\n\n\n\n

    As you might expect, the preparation of an enthalpy-concentration chart requires numerous calculations and valid enthalpy or heat capacity data for solutions of various concentrations. Refer to\u00a0Unit Operations of Chemical Engineering<\/em>\u00a0[W. L. McCabe and J. C. Smith, 3rd ed., McGraw-Hill, New York (1976)] for instructions if you have to prepare such a chart. In the next example we show how to use an\u00a0H-x<\/em>\u00a0chart.<\/p>\n\n\n\n

    Example 12.3 Application of an Enthalpy-Concentration Chart<\/strong><\/p>\n\n\n\n

    Six hundred pounds of 10% NaOH per hour at 200\u00b0F are added to 400 lb\/hr of 50% NaOH at the boiling point in an insulated vessel. Calculate the following:<\/p>\n\n\n\n

      \n
    1. a.<\/strong> The final temperature of the exit solution<\/li>\n\n\n\n
    2. b.<\/strong> The final concentration of the exit solution<\/li>\n\n\n\n
    3. c.<\/strong> The pounds of water evaporated per hour during the process<\/li>\n<\/ol>\n\n\n\n

      Solution<\/strong><\/p>\n\n\n\n

      You can use the steam tables and the NaOH-H2<\/sub>O enthalpy-concentration chart in\u00a0Appendix J\u00a0as your sources of data. What are the reference conditions for the chart? The reference conditions for the latter chart are\u00a0\u0394H^Fo\u00a0= 0 at 32\u00b0F for pure liquid water, an infinitely dilute solution of NaOH. Pure caustic has an enthalpy at 68\u00b0F of 455 Btu\/lb above this datum. Treat the process as a flow process even if is not. The energy balance reduces to \u0394H<\/em>=0. Basis: 1000 lb of final solution =1 hr.<\/p>\n\n\n\n

      You can write the following material balances:<\/p>\n\n\n\n

      Component<\/u><\/em><\/td>10% solution<\/u><\/em><\/td>+<\/td>50% solution<\/u><\/em><\/td>=<\/td>Final solution<\/u><\/em><\/td>wt%<\/u><\/em><\/td><\/tr><\/thead>
      NaOH<\/td>60<\/td><\/td>200<\/td><\/td>260<\/td>26<\/td><\/tr>
      H2<\/sub>O<\/td>540<\/td><\/td>200<\/td><\/td>740<\/td>74<\/td><\/tr>
      Total<\/td>600<\/td><\/td>400<\/td><\/td>1000<\/td>100<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

      Enthalpy data from the H-x<\/em> chart<\/p>\n\n\n\n

      \u0394H^(Btu\/lb):\u200310% solution 152\u200350% solution 290<\/p>\n\n\n\n

      The energy balance is<\/p>\n\n\n\n

      10% solution<\/u><\/em><\/td><\/td>50% solution<\/u><\/em><\/td><\/td>Final solution<\/u><\/em><\/td><\/tr>
      600(152)<\/td>+<\/td>400(290)<\/td>=<\/td>\u0394H<\/em><\/td><\/tr>
      91,200<\/td>+<\/td>116,000<\/td>=<\/td>207,200<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

      Note that the enthalpy of the 50% solution at its boiling point is taken from the bubble point at \u03c9NaOH<\/sub> = 0.50. The enthalpy per pound of the final solution is207,200Btu1000lb=207 Btu\/lb<\/p>\n\n\n\n

      \"A
      Figure E12.3<\/strong><\/figcaption><\/figure>\n\n\n\n

      On the enthalpy-concentration chart for NaOH-H2<\/sub>O, for a 26% NaOH solution with an enthalpy of 207 Btu\/lb, you would find that only a two-phase mixture of (1) saturated H2<\/sub>O vapor and (2) NaOH-H2<\/sub>O solution at the boiling point could exist. To get the fraction H2<\/sub>O vapor, you have to make an additional energy (enthalpy) balance. By interpolation, draw the tie line through the point\u00a0x<\/em>=0.26,\u00a0H<\/em>=207 (make it parallel to the 220\u00b0F and 250\u00b0F tie lines). The final temperature of the tie line appears from\u00a0Figure E12.3\u00a0to be 232\u00b0F;\u00a0the enthalpy of the liquid at the bubble point at this temperature is about 175 Btu\/lb. The enthalpy of the saturated water vapor (no NaOH is in the vapor phase) from the steam tables at 232\u00b0F is 1158 Btu\/lb. Let\u00a0x<\/em>=pounds of H2<\/sub>O evaporated.<\/p>\n\n\n\n

      Basis: 1000 lb of final solution<\/p>\n\n\n\n

      x<\/em>(1158) +(1000 \u2212x<\/em>) 175 = 1000 (207.2)<\/p>\n\n\n\n

      x<\/em> = 32.8 of H2<\/sub>O evaporated\/hr<\/p>\n","protected":false},"excerpt":{"rendered":"

      In Section 12.1 we restricted the discussion and examples to the standard state (25\u00b0C and 1 atm). In this section we proceed with what happens when the temperatures of the inlet and outlet streams differ from 25\u00b0C for a binary\u00a0mixture in an open, steady-state process. (For a closed system the initial and final states of […]<\/p>\n","protected":false},"author":1,"featured_media":4607,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[569],"tags":[],"class_list":["post-4737","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-heats-of-solution-and-mixing"],"jetpack_featured_media_url":"https:\/\/workhouse.sweetdishy.com\/wp-content\/uploads\/2024\/09\/heat.png","_links":{"self":[{"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/4737","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/comments?post=4737"}],"version-history":[{"count":1,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/4737\/revisions"}],"predecessor-version":[{"id":4738,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/4737\/revisions\/4738"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/media\/4607"}],"wp:attachment":[{"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/media?parent=4737"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/categories?post=4737"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/tags?post=4737"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}