6.9 The second law of thermodynamics for open systems

Claire Yu Yan

6. Entropy and the Second Law of Thermodynamics

6.9 The second law of thermodynamics for open systems

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

$\dot{m}$ : rate of mass transfer

$\dot{Q}_k$ : rate of heat transfer via the location $k$ of the system boundary, which is at a temperature of $T_k$ in Kelvin

$\dot{S}_{heat}$ : time rate of entropy transfer due to heat transfer

$\dot{S}_{mass}$ : time rate of entropy transfer that accompanies the mass transfer into or out of a control volume

$s_k$ : specific entropy of the fluid

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

$\dfrac{{{dS}}_{{c}.{v}.}}{{dt}}=\displaystyle\left(\sum{{\dot{{m}}}_{i}{s}_{i}}+\displaystyle\sum\frac{{\dot{{Q}}}_{{c}.{v}.}}{{T}}\right)-\displaystyle\left(\sum{{\dot{{m}}}_{e}{s}_{e}}\right)+\displaystyle{\dot{{S}}}_{{gen}}\ \ \ \ \ \ ({\dot{{S}}}_{{gen}} \ge 0)$

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

$\displaystyle \sum{{\dot{{m}}}_{e}{s}_{e}}-\sum{{\dot{{m}}}_{i}{s}_{i}}=\displaystyle \sum\dfrac{{\dot{{Q}}}_{{c}.{v}.}}{{T}}+{\dot{{S}}}_{{gen}}\ \ \ \ \ \ ({\dot{{S}}}_{{gen}} \ge 0)$

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

$\displaystyle\sum{{\dot{{m}}}_{e}{s}_{e}}=\displaystyle\sum{{\dot{{m}}}_{i}{s}_{i}}$

where

$\dot{m}$ : rate of mass transfer of the fluid entering or leaving the control volume via the inlet $i$ or exit $e$ , in kg/s

$\dot{Q}_{c.v.}$ : rate of heat transfer into the control volume via the system boundary (at a constant $T$ ), in kW

$S_{c.v.}$ : entropy in the control volume, in kJ/K

$\dfrac{{dS}_{c.v.}}{dt}$ : time rate of change of entropy in the control volume, in kW/K

$\dot S_{gen}$ : time rate of entropy generation in the process, in kW/K

$s$ : specific entropy of the fluid entering or leaving the control volume via the inlet $i$ or exit $e$ , in kJ/kgK

$T$ : absolute temperature of the system boundary, in Kelvin

**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 ΔKE=ΔPE=0. Are the change in specific enthalpy Δh=h_e−h_i, specific work w, and specific heat transfer q positive, zero, or negative values? What is the relation between w and q?

**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, $q_{rev}>0$ , 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,

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

$\because T_e = T_i$

$\therefore h_e = h_i$ and $\Delta h =0$

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

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