6.10: The second law of thermodynamics for open systems
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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, , 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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- For steady-state, steady-flow devices, ; therefore,
\[\displaystyle \sum
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- For steady and isentropic flow devices, and ; therefore,
\[\displaystyle\sum
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where
\[\dot{m}\]
\[\dot{Q}_{c.v.}\]
\[S_{c.v.}\]
\[\dfrac
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\[\dot S_{gen}\]
\[s\]
\[T\]
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 ΔKE=ΔPE=0. Are the change in specific enthalpy Δh=he−hi, specific work w, and specific heat transferq positive, zero, or negative values? What is the relation between w and q?
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 diagram; therefore 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 diagram represents the specific heat transfer of the reversible process; therefore is positive, see Figure 6.9.e2.
(2) The same conclusion, 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, = 0, and the fluid is assumed to be always in thermal equilibrium with the system boundary, or ; 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 and 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 and diagrams must be the same.
Query \(\PageIndex{1}\)
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