{"id":1621,"date":"2024-07-25T20:13:26","date_gmt":"2024-07-25T20:13:26","guid":{"rendered":"https:\/\/workhouse.sweetdishy.com\/?p=1621"},"modified":"2024-07-25T20:13:26","modified_gmt":"2024-07-25T20:13:26","slug":"application-of-first-law-of-thermodynamics-in-steady-flow-process-and-variable-flow-process","status":"publish","type":"post","link":"https:\/\/workhouse.sweetdishy.com\/index.php\/2024\/07\/25\/application-of-first-law-of-thermodynamics-in-steady-flow-process-and-variable-flow-process\/","title":{"rendered":"Application of First Law of Thermodynamics in Steady Flow Process and Variable Flow Process"},"content":{"rendered":"\n

In a steady flow process, thermodynamic properties at any section remain constant with respect to time; it can vary only with respect to space. A schematic diagram of steady flow process is shown in\u00a0Figure 1.7.<\/p>\n\n\n\n

\"Figure<\/figure>\n\n\n\n

Figure 1.7<\/strong> Schematic Diagram of Steady Flow Process<\/p>\n\n\n\n

\"Equation\"\/<\/figure>\n\n\n\n

<\/a>From continuity equation:<\/p>\n\n\n\n

\"Equation\"\/<\/figure>\n\n\n\n

Energy balance equation:<\/p>\n\n\n\n

\"Equation\"\/<\/figure>\n\n\n\n

This is known as steady flow energy equation (SFEE) for single stream.<\/p>\n\n\n\n

Variable Flow Process<\/h4>\n\n\n\n

In some flow process, mass flow rate is not steady but varies with respect to time. In such a case, the difference in energy flow is stored in system as \u0394E<\/em>v<\/sub>.<\/p>\n\n\n\n

\"Equation\"\/<\/figure>\n\n\n\n

Rate of energy increase = Rate of energy inflow \u2212 Rate of energy outflow<\/p>\n\n\n\n

\"Equation\"\/<\/figure>\n\n\n\n

Example 1.14:<\/strong>\u00a0An air conditioning system, as shown in\u00a0Figure 1.8,\u00a0handling 1 kg\/s of air at 37\u00b0C and consumes a power of 20 kW and rejects heat of 38 kW. The inlet and outlet velocities of air are 50 and 80 m\/s, respectively. Find the exit air temperature, assuming adiabatic conditions. Take\u00a0Cp<\/sub><\/em>\u00a0of air as 1.005 kJ\/kg.<\/p>\n\n\n\n

\"Figure<\/figure>\n\n\n\n

Figure 1.8<\/strong> Air Conditioning System<\/p>\n\n\n\n

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

\"Equation\"\/<\/figure>\n\n\n\n

Example 1.15:<\/strong>\u00a0In a cooling tower of a power plant (Figure 1.9), air enters at a height of 1 m above the ground and leaves at 8 m. The inlet and outlet velocities are 18 and 27 m\/s, respectively. Water enters at a height of 10 m and leaves at a height of 0.5 m. The velocity of water at entry and exit are 5 and 1.5 m\/s, respectively. Water temperatures are 85 and 42\u00b0C at inlet and exit, respectively. Air temperatures are 27 and 70\u00b0C at entry and exit, respectively. The cooling tower is fully insulated and a fan of 2.5 kW drives air through the cooler. Find the air per second required for 1 kg\/s of water flow. The values of\u00a0Cp<\/sub><\/em>\u00a0of air and water are 1.005 and 4.18 kJ\/kg K, respectively.<\/p>\n\n\n\n

\"Figure<\/figure>\n\n\n\n

Figure 1.9<\/strong> Cooling Tower<\/p>\n\n\n\n

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

\"Equation\"\/<\/figure>\n\n\n\n

Example 1.16:<\/strong> In a centrifugal air compressor, initial pressure, and specific volume are 1 bar and 1 m3<\/sup>\/kg, respectively. The air flow rate is 30 kg\/min. Heat liberated to atmosphere from compressor is 100 kW and inlet velocity of air = 10 m\/s, outlet velocity of air = 5 m\/s. Find<\/p>\n\n\n\n

    \n
  1. Compressor work.<\/li>\n\n\n\n
  2. Ratio of inlet and outlet area, if internal energy at outlet is 100 kJ more than that of inlet. Solve the problem using SFEE.<\/li>\n<\/ol>\n\n\n\n

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

      \n
    1.   \"equation\"<\/li>\n\n\n\n
    2.   \"equation\"<\/li>\n<\/ol>\n\n\n\n

      Example 1.17:<\/strong> Air enters in a compressor at the rate of 0.5 kg\/s, at 8 m\/s with a pressure of 1 bar and a specific volume of 0.85 m3<\/sup>\/kg, and leaving at 5 m\/s with a pressure of 6 bar and a specific volume of 0.2 m3<\/sup>\/kg (Figure 1.10<\/a>). The internal energy of the air leaving is 80 kJ\/kg greater than that of the air entering. Cooling water in a jacket surrounding the cylinder absorbs heat from the air at the rate of 70 W. Calculate the power required to drive the compressor and the inlet, and outlet cross-sectional areas.<\/p>\n\n\n\n

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

      \"Figure<\/figure>\n\n\n\n

      Figure 1.10<\/strong> Compressor<\/p>\n\n\n\n

      \"Equation\"\/<\/figure>\n\n\n\n

      <\/a>1.4.7  Limitations of First Law of Thermodynamics<\/h4>\n\n\n\n

      First law of thermodynamics does not tell about the following:<\/p>\n\n\n\n

        \n
      • How much of the given quantity of heat is changed into work?<\/li>\n\n\n\n
      • In which direction is changing take place (heat to work or work to heat)?<\/li>\n\n\n\n
      • Under which condition the changing will take place?<\/li>\n<\/ul>\n","protected":false},"excerpt":{"rendered":"

        In a steady flow process, thermodynamic properties at any section remain constant with respect to time; it can vary only with respect to space. A schematic diagram of steady flow process is shown in\u00a0Figure 1.7. Figure 1.7 Schematic Diagram of Steady Flow Process From continuity equation: Energy balance equation: This is known as steady flow energy […]<\/p>\n","protected":false},"author":1,"featured_media":1601,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-1621","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-blog"],"jetpack_featured_media_url":"https:\/\/workhouse.sweetdishy.com\/wp-content\/uploads\/2024\/07\/thermodynamics-1-3.png","_links":{"self":[{"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/1621","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=1621"}],"version-history":[{"count":1,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/1621\/revisions"}],"predecessor-version":[{"id":1622,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/1621\/revisions\/1622"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/media\/1601"}],"wp:attachment":[{"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/media?parent=1621"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/categories?post=1621"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/tags?post=1621"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}