{"id":16910,"date":"2018-02-27T19:08:26","date_gmt":"2018-02-27T19:08:26","guid":{"rendered":"http:\/\/sitepourvtc.com\/?page_id=16910"},"modified":"2022-11-10T06:28:39","modified_gmt":"2022-11-10T06:28:39","slug":"work-in-thermodynamics","status":"publish","type":"page","link":"https:\/\/sitepourvtc.com\/nuclear-engineering\/thermodynamics\/laws-of-thermodynamics\/first-law-of-thermodynamics\/work-in-thermodynamics\/","title":{"rendered":"Work in Thermodynamics"},"content":{"rendered":"
In thermodynamics, work<\/strong> performed by a system is the energy transferred by the system to its surroundings. Work is a form of energy<\/strong>, but it is energy in transit<\/strong>. A system contains no work, and work is a process done by or on a system. In general, work is defined for mechanical systems as the action of a force on an object through a distance.<\/p>\n

W = F . d<\/em><\/strong><\/p>\n

where:<\/p>\n

W = work (J)<\/p>\n

F = force (N)<\/p>\n

d = displacement (m)<\/p>\n<\/div><\/div>\n

\n

p\u0394V Work<\/h2>\n

Pressure-volume work (or p\u0394V <\/em><\/strong>Work<\/strong>) occurs when the volume V<\/em> of a system changes. The p\u0394V <\/em><\/strong>Work<\/strong> is equal to the area under the process curve plotted on the pressure-volume diagram. It is also known as boundary work<\/strong>.\u00a0\u00a0<\/strong>Boundary work<\/strong> occurs because the mass of the substance within the system boundary causes a force, the pressure times the surface area, to act on the boundary surface and move it. Boundary work<\/strong>\u00a0(or\u00a0p\u0394V<\/strong><\/em>\u00a0<\/em><\/strong>Work<\/strong>) occurs when the\u00a0volume<\/strong>\u00a0<\/strong>V<\/strong><\/em>\u00a0<\/strong>of a system changes<\/strong>. It is used for calculating piston displacement work in a\u00a0closed system<\/strong>. This happens when steam<\/strong><\/a>\u00a0or gas contained in a piston-cylinder device expands against the piston and forces the piston to move.<\/p>\n

\"pdV<\/a>
p\u0394V Work is equal to the area under the process curve plotted on the pressure-volume diagram.<\/figcaption><\/figure>\n

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

Consider a frictionless piston that is used to provide a constant pressure of 500 kPa<\/strong> in a cylinder containing steam (superheated steam<\/a>) of a volume of 2 m<\/strong>3 <\/sup><\/strong>\u00a0at 500 K<\/strong>.<\/p>\n

Calculate the final temperature if 3000 kJ<\/strong> of heat<\/strong> is added.<\/p>\n

\"Enthalpy<\/a>
Calculate the final temperature if 3000 kJ of heat is added.<\/figcaption><\/figure>\n

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

Using steam tables<\/a> we know, that the specific enthalpy<\/strong> of such steam (500 kPa; 500 K) is about 2912 kJ\/kg<\/strong>. Since the steam has a density of 2.2 kg\/m3<\/sup>, we know about 4.4 kg of steam<\/strong> in the piston at an enthalpy of 2912 kJ\/kg x 4.4 kg = 12812 kJ<\/strong>.<\/p>\n

When we use simply Q = H<\/strong>2<\/sub><\/strong> \u2212 H<\/strong>1<\/sub><\/strong>, then the resulting enthalpy of steam will be:<\/p>\n

H2<\/sub> = H1<\/sub> + Q = 15812 kJ<\/strong><\/p>\n

From steam tables<\/strong>, such superheated steam (15812\/4.4 = 3593 kJ\/kg) will have a temperature of 828 K (555\u00b0C)<\/strong>. Since at this enthalpy, the steam has a density of 1.31 kg\/m3<\/sup>, it is obvious that it has expanded by about 2.2\/1.31 = 1.67 (+67%). Therefore the resulting volume is 2 m3<\/sup> x 1.67 = 3.34 m3<\/sup> and \u2206V = 3.34 m3<\/sup> \u2013 2 m3<\/sup> = 1.34 m3<\/sup>.<\/p>\n

The p\u2206V <\/strong>part of enthalpy, i.e., the work done is:<\/p>\n

W = p\u2206V = 500 000 Pa x 1.34 m<\/strong>3<\/sup><\/strong> = 670 kJ<\/strong><\/p>\n

———–<\/p>\n

During the volume change<\/strong>, the pressure<\/strong> and temperature<\/strong> may also change. To calculate such processes, we would need to know how pressure varies with volume for the actual process by which the system changes from state i to state f<\/strong>. The first law of thermodynamics<\/strong> and the work can then be expressed as:<\/p>\n

\"Work<\/a><\/p>\n

\"Work<\/a>
The work done by the system depends not only on the initial and final states but also on the intermediate states\u2014that is, on the path.<\/figcaption><\/figure>\n

When a thermodynamic system changes from an initial state<\/strong> to a final state<\/strong>, it passes through a series of intermediate states<\/strong>. We call this series of states a path<\/strong>. There are always infinitely many different possibilities for these intermediate states. When they are all equilibrium states, the path can be plotted on a pV-diagram<\/strong>. One of the most important conclusions is that:<\/p>\n

The work done by the system depends not only on the initial and final states but also on the intermediate states\u2014that is, on the path.<\/em><\/p>\n

Q and W are path-dependent, whereas \u0394E<\/strong>int<\/sub><\/strong> is path-independent. <\/strong>As can be seen from the picture (p-V diagram), work is a path-dependent variable. The blue area represents the p\u0394V <\/em>Work<\/strong> done by a system from an initial state i to a final state f. Work W is positive because the system\u2019s volume increases. The second process shows that work is greater, and that depends on the path of the process.<\/p>\n

Moreover, we can take the system through a series of states forming a closed loop<\/strong>, such i \u21d2 f \u21d2 i<\/strong>. In this case, the final state is the same as the initial state<\/strong>, but the total work done<\/strong> by the system is not zero<\/strong>. A positive value for work indicates that work is done by the system in its surroundings. A negative value indicates that work is done on the system by its surroundings.<\/p>\n

<\/span>Cyclic Process<\/div>
\n
\"Cyclic<\/a>
A process that eventually returns a system to its initial state is called a cyclic process.<\/figcaption><\/figure>\n<\/div><\/div><\/div>\n

Example: Turbine Specific Work<\/h2>\n
\"engineering<\/a>
Rankine Cycle – Thermodynamics as Energy Conversion Science<\/figcaption><\/figure>\n

A high-pressure stage<\/strong> of steam turbine<\/strong> operates at a steady state with inlet conditions of \u00a06 MPa<\/strong>, t = 275.6\u00b0C<\/strong>, x = 1 (point C). Steam leaves this turbine stage at a pressure of 1.15 MPa<\/strong>, 186\u00b0C,<\/strong> and x = 0.87<\/strong> (point D). Calculate the enthalpy difference between these two states. Determine the specific work transfer.<\/p>\n

The enthalpy<\/a> for the state C can be picked directly from steam tables<\/strong><\/a>, whereas the enthalpy for the state D must be calculated using vapor quality<\/a>:<\/p>\n

h<\/em><\/strong>1, wet<\/sub><\/em><\/strong> = <\/em><\/strong>2785 kJ\/kg<\/strong><\/p>\n

h<\/em><\/strong>2, wet<\/sub><\/em><\/strong> = h<\/em><\/strong>2,s<\/sub><\/em><\/strong> x + (1 \u2013 x ) h<\/em><\/strong>2,l<\/sub><\/em><\/strong> \u00a0= 2782 . 0.87 + (1 \u2013 0.87) . 790 = 2420 + 103 = 2523 kJ\/kg<\/strong><\/p>\n

\u0394h = 262 kJ\/kg<\/strong><\/p>\n

Since in adiabatic process dh = dw<\/strong>, \u0394h = 262 kJ\/kg is <\/strong>the turbine-specific work<\/strong>.<\/p>\n

First Law in Terms of Enthalpy dH = dQ + Vdp<\/h2>\n

The enthalpy<\/strong><\/a> is defined to be the sum of the internal energy<\/a> E plus the product of the pressure p<\/a> and volume V<\/a>. In many thermodynamic analyses, the sum of the internal energy U and the product of pressure p and volume V appears. Therefore it is convenient to give the combination a name, enthalpy<\/strong>, and a distinct symbol, H.<\/p>\n

H = U + pV<\/span><\/strong><\/em><\/p>\n

See also: Enthalpy<\/a><\/p>\n

The first law of thermodynamics<\/strong> in terms of enthalpy<\/strong> shows us why engineers use the enthalpy in thermodynamic cycles (e.g.,, Brayton cycle<\/strong> or Rankine cycle<\/strong>).<\/p>\n

The classical form of the law is the following equation:<\/p>\n

dU = dQ \u00a0\u2013 dW<\/strong><\/em><\/p>\n

In this equation, dW<\/strong> is equal to dW = pdV<\/strong> and is known as the boundary work<\/strong>.<\/p>\n

<\/span>Boundary Work - pdV Work<\/div>
Boundary work<\/strong> occurs because the mass of the substance contained within the system boundary causes a force, the pressure times the surface area, to act on the boundary surface and make it move. Boundary work<\/strong> (or p\u0394V <\/em><\/strong>Work<\/strong>) occurs when the volume V<\/em> of a system changes<\/strong>. It is used for calculating piston displacement work in a closed system<\/strong>. This happens when steam<\/strong><\/a>\u00a0or gas contained in a piston-cylinder device expands against the piston and forces the piston to move.<\/div><\/div><\/div>\n

Since H = U + pV<\/em><\/strong>, therefore dH = dU + pdV + Vdp<\/strong><\/em> and we substitute dU = dH – pdV – Vdp<\/strong><\/em> into the classical form of the law:<\/p>\n

dH – pdV – Vdp = dQ – pdV<\/strong><\/em><\/p>\n

We obtain the law in terms of enthalpy:<\/p>\n

dH = dQ + Vdp<\/em><\/span><\/strong><\/p>\n

or<\/p>\n

dH = TdS + Vdp<\/em><\/span><\/strong><\/p>\n

In this equation, the term Vdp<\/strong><\/em> is a flow process work. <\/strong>This work, \u00a0Vdp<\/strong><\/em>, is used for open flow systems<\/strong> like a turbine<\/strong> or a pump<\/strong> in which there is a \u201cdp\u201d<\/strong>, i.e., change in pressure. There are no changes in the control volume<\/a>. As can be seen, this form of the law simplifies the description of energy transfer<\/strong>. At constant pressure<\/strong>, the enthalpy change<\/strong> equals the energy<\/strong> transferred from the environment through heating:<\/p>\n

Isobaric process (Vdp = 0): <\/strong><\/p>\n

dH = dQ \u00a0 \u00a0\u00a0\u2192<\/span>\u00a0 \u00a0 \u00a0Q = H<\/strong>2<\/sub><\/strong> \u2013 H<\/strong>1<\/sub><\/strong><\/p>\n

At constant entropy<\/strong>, i.e., in isentropic process, the enthalpy change<\/strong> equals the flow process work<\/strong> done on or by the system:<\/p>\n

Isentropic process (dQ = 0): <\/strong><\/p>\n

dH = Vdp\u00a0 \u00a0 \u00a0\u2192<\/span>\u00a0 \u00a0 \u00a0W = H<\/strong>2<\/sub><\/strong> \u2013 H<\/strong>1<\/sub><\/strong><\/p>\n

It is obvious, and it will be very useful in the analysis of both thermodynamic cycles used in power engineering, i.e., in the Brayton and Rankine cycles.<\/p>\n

Example: First Law of Thermodynamics and Brayton Cycle<\/h2>\n

Let assume the ideal Brayton cycle<\/strong> that describes the workings of a constant pressure<\/strong> heat engine<\/strong>. Modern gas turbine<\/strong> engines and airbreathing jet engines<\/strong> also follow the Brayton cycle. This cycle consist of four thermodynamic processes:<\/p>\n

    \n
  1. \n
    \"first<\/a>
    The ideal Brayton cycle consists of four thermodynamic processes. Two isentropic processes and two isobaric processes.<\/figcaption><\/figure>\n

    Isentropic compression<\/strong> \u2013 ambient air is drawn into the compressor, pressurized (1 \u2192 2). The work required for the compressor is given by W<\/strong>C<\/sub><\/strong> = H<\/strong>2<\/sub><\/strong> \u2013 H<\/strong>1<\/sub><\/strong>.<\/strong><\/li>\n

  2. Isobaric heat addition<\/strong> \u2013 the compressed air then runs through a combustion chamber, burning fuel, and air or another medium is heated (2 \u2192 3). It is a constant-pressure process since the chamber is open to flow in and out. The net heat added is given by Q<\/strong>add<\/sub><\/strong> = H<\/strong>3 <\/sub><\/strong>– H<\/strong>2<\/sub><\/strong><\/li>\n
  3. Isentropic expansion<\/strong> \u2013 the heated, pressurized air then expands on the turbine, gives up its energy. The work done by the turbine is given by W<\/strong>T<\/sub><\/strong> = H<\/strong>4<\/sub><\/strong> \u2013 H<\/strong>3<\/sub><\/strong><\/li>\n
  4. Isobaric heat rejection<\/strong> – the residual heat must be rejected to close the cycle. The net heat rejected is given by Q<\/strong>re<\/sub><\/strong> = H<\/strong>4 <\/sub><\/strong>– H<\/strong>1<\/sub><\/strong><\/li>\n<\/ol>\n

    As can be seen, we can describe and calculate (e.g.,, thermodynamic efficiency) such cycles (similarly for Rankine cycle<\/strong>) using enthalpies<\/a>.<\/p>\n<\/div>\n

    <\/div>\n
    \u00a0
    <\/span>References:<\/div>
    Nuclear and Reactor Physics:<\/strong>\n
      \n
    1. J. R. Lamarsh, Introduction to Nuclear Reactor Theory, 2nd ed., Addison-Wesley, Reading,\u00a0MA (1983).<\/li>\n
    2. J. R. Lamarsh, A. J. Baratta, Introduction to Nuclear Engineering, 3d ed., Prentice-Hall, 2001, ISBN: 0-201-82498-1.<\/li>\n
    3. W. M. Stacey, Nuclear Reactor Physics, John Wiley & Sons, 2001, ISBN: 0- 471-39127-1.<\/li>\n
    4. Glasstone, Sesonske. Nuclear Reactor Engineering: Reactor Systems Engineering,\u00a0Springer; 4th edition, 1994, ISBN:\u00a0978-0412985317<\/li>\n
    5. W.S.C. Williams. Nuclear and Particle Physics.\u00a0Clarendon Press; 1 edition, 1991, ISBN:\u00a0978-0198520467<\/li>\n
    6. Kenneth S. Krane. Introductory Nuclear Physics, 3rd Edition, Wiley, 1987,\u00a0ISBN:\u00a0978-0471805533<\/li>\n
    7. G.R.Keepin. Physics of Nuclear Kinetics.\u00a0Addison-Wesley Pub. Co; 1st edition, 1965<\/li>\n
    8. Robert Reed Burn, Introduction to Nuclear Reactor Operation, 1988.<\/li>\n
    9. U.S. Department of Energy, Nuclear Physics and Reactor Theory.\u00a0DOE Fundamentals Handbook,\u00a0Volume 1 and 2.\u00a0January\u00a01993.<\/li>\n<\/ol>\n

      <\/strong>Advanced Reactor Physics:<\/strong><\/p>\n

        \n
      1. K. O. Ott, W. A. Bezella, Introductory Nuclear Reactor Statics, American Nuclear Society, Revised edition (1989), 1989, ISBN: 0-894-48033-2.<\/li>\n
      2. K. O. Ott, R. J. Neuhold, Introductory Nuclear Reactor Dynamics, American Nuclear Society, 1985, ISBN: 0-894-48029-4.<\/li>\n
      3. D. L. Hetrick, Dynamics of Nuclear Reactors, American Nuclear Society, 1993, ISBN: 0-894-48453-2.\u00a0<\/span><\/li>\n
      4. E. E. Lewis, W. F. Miller, Computational Methods of Neutron Transport, American Nuclear Society, 1993, ISBN: 0-894-48452-4.<\/li>\n<\/ol>\n<\/div><\/div><\/div>
        <\/div><\/div><\/div>
        <\/div>
        <\/div><\/div>
        \n

        See above:<\/h2>\n

        First Law<\/i> <\/span><\/a><\/p><\/div><\/div>

        <\/div><\/div>\n","protected":false},"excerpt":{"rendered":"

        p\u0394V Work Pressure-volume work (or p\u0394V Work) occurs when the volume V of a system changes. The p\u0394V Work is equal to the area under the process curve plotted on the pressure-volume diagram. It is also known as boundary work.\u00a0\u00a0Boundary work occurs because the mass of the substance within the system boundary causes a force, … Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"parent":16837,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"generate_page_header":""},"_links":{"self":[{"href":"https:\/\/sitepourvtc.com\/wp-json\/wp\/v2\/pages\/16910"}],"collection":[{"href":"https:\/\/sitepourvtc.com\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/sitepourvtc.com\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/sitepourvtc.com\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/sitepourvtc.com\/wp-json\/wp\/v2\/comments?post=16910"}],"version-history":[{"count":3,"href":"https:\/\/sitepourvtc.com\/wp-json\/wp\/v2\/pages\/16910\/revisions"}],"predecessor-version":[{"id":34954,"href":"https:\/\/sitepourvtc.com\/wp-json\/wp\/v2\/pages\/16910\/revisions\/34954"}],"up":[{"embeddable":true,"href":"https:\/\/sitepourvtc.com\/wp-json\/wp\/v2\/pages\/16837"}],"wp:attachment":[{"href":"https:\/\/sitepourvtc.com\/wp-json\/wp\/v2\/media?parent=16910"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}