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6.10: The second law of thermodynamics for open systems

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    88860
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    Entropy can be transferred to a system via two mechanisms: (1) heat transfer and (2) mass transfer. For open systems, the second law of thermodynamics is often written in the rate form; therefore, we are interested in the time rate of entropy transfer due to heat transfer and mass transfer.

    \[\dot{S}_{heat} =\dfrac{dS_{heat}}{dt} \cong \displaystyle\sum\dfrac{\dot{Q}_k}{T_k}\]

    \[\dot{S}_{mass} = \displaystyle\sum\dfrac{dS_{mass}}{dt} = \displaystyle \sum \dot{m}_{k}s_{k}\]

    where

    \[\dot{m}\]

    \[\dot{Q}_k\]

    \[\dot{S}_{heat}\]

    \[\dot{S}_{mass}\]

    \[s_k\]

    Applying the entropy balance equation, \Delta \rm {entropy= + in - out + gen}, to a control volume, see Figure 6.9.1, we can write the following equations:

    • General equation for both steady and transient flow devices

    \[\dfrac

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    =\displaystyle\left(\sum
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    +\displaystyle\sum\frac
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    \right)-\displaystyle\left(\sum
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    \right)+\displaystyle{\dot
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    }_
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    \ \ \ \ \ \ ({\dot
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    }_
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    \ge 0)\]

    • For steady-state, steady-flow devices, \dfrac{{dS}_{c.v.}}{dt}=0; therefore,

    \[\displaystyle \sum

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    -\sum
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    =\displaystyle \sum\dfrac
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    +{\dot
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    }_
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    \ \ \ \ \ \ ({\dot
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    \ge 0)\]

    • For steady and isentropic flow devices, \dot{Q}_{c.v.}=0 and \dot S_{gen}=0; therefore,

    \[\displaystyle\sum

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    =\displaystyle\sum
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    \]

    where

    \[\dot{m}\]

    \[\dot{Q}_{c.v.}\]

    \[S_{c.v.}\]

    \[\dfrac

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    {dt}\]

    \[\dot S_{gen}\]

    \[s\]

    \[T\]

    Flow through a control volume, showing the entropy transfers and entropy generation
    Figure 6.9.1 Flow through a control volume, showing the entropy transfers and entropy generation

    Example 1

    The diagrams in Figure 6.9.e1 show a reversible process in a steady-state, single flow of air. The letters i and e represent the initial and final states, respectively. Treat air as an ideal gas and assume ΔKEPE=0. Are the change in specific enthalpy Δh=hehi, specific work w, and specific heat transferq positive, zero, or negative values? What is the relation between w and q?

    T-s and P-v diagrams of a reversible process
    Figure 6.9.e1 T-s and P-v diagrams of a reversible process for an ideal gas

    Solution:

    The specific work can be evaluated mathematically and graphically.

    (1) Mathematically,

    \[\because v_{e} > v_{i}\]

    \[\therefore w = \displaystyle\int_{i}^{e}{Pd{v}\ } >\ 0\]

    (2) Graphically, the specific work is the area under the process curve in the P-v diagram; therefore w is positive, see Figure 6.9.e2.

    In a similar fashion, the specific heat transfer can also be evaluated graphically and mathematically.

    (1) Graphically,

    \[\because ds=\left(\displaystyle\frac{\delta q}{T}\right)_{rev}\]

    \[\therefore q_{rev} = \displaystyle \int_{i}^{e}{Tds} = T(s_e-s_i)\ >\ 0\]

    For a reversible process, the area under the process curve in the T-s diagram represents the specific heat transfer of the reversible process; therefore q=q_{rev} is positive, see Figure 6.9.e2.

    (2) The same conclusion, q_{rev}0" class="latex mathjax" title="q_{rev}>0" src="/@api/deki/files/59236/d00f283ba44c47860e35c0b010cd6fb7.png">, can also be derived from the second law of thermodynamics mathematically, as follows.

    \[\dot{m}(s_e-s_i)=\displaystyle\sum\frac{\dot{Q}}{T_{surr}}+\dot{S}_{gen}\]

    For a reversible process, \dot{S}_{gen}= 0, and the fluid is assumed to be always in thermal equilibrium with the system boundary, or T = T_{surr}; therefore,

    \[q_{rev} = \dfrac{\dot{Q}}{\dot{m}} = T(s_e-s_i) > 0\]

    The change in specific enthalpy can then be evaluated. For an ideal gas,

    \[\Delta h = h_e - h_i = C_p(T_e - T_i)\]

    \[\because T_e = T_i\]

    \[\therefore h_e = h_i\]

    Now, we can determine the relation between w and q_{rev} from the first law of thermodynamics for control volumes.

    \[\because \dot{m}( h_e - h_i ) = \dot{Q}_{rev} - \dot{W} = 0\]

    \[\therefore \dot{Q}_{rev} = \dot{W}\]

    \[\therefore q_{rev} = w\]

    In this reversible process, the specific heat transfer and specific work must be the same. Graphically, the two areas under the P-v and T-s diagrams must be the same.

    T-s and P-v diagrams, showing the solutions for a reversible process for an ideal gas
    Figure 6.9.e2 T-s and P-v diagrams, showing the solutions for a reversible process of an ideal gas

    Query \(\PageIndex{1}\)

    Media Attributions

    • Entropy transfers and entropy generation through a C.V. © Pbroks13 is licensed under a Public Domain license

    This page titled 6.10: The second law of thermodynamics for open systems is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Claire Yu Yan (BC Campus) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.