- #1
tracker890 Source h
- 90
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- Homework Statement
- Open system, maximum reversible work relative to the dead state in a thermodynamic process. I will share the solution process with everyone, and please correct me if there are any mistakes.
- Relevant Equations
- Exergy balance
Energy balance
Entropy balance
reference；sol
The maximum reversible work in thermodynamics
Below is the process of determining the "Available energy" for an open system, shared with everyone as a reference for learning about exergy. If there are any errors in the content, please feel free to correct them.
$$ W_{rev}=W_u^{\nearrow W-P0\cancel{\left( V_2-V_1 \right) }=W}+T0\cdot Sgen=W+T0\cdot Sgen $$
Find## W ## and## T_0\cdot Sgen ## ：
$$ eneger\ balance:\ \ Q_{in,net}^{\nearrow ^{\sum{Q_k}}}-W_{out,net}^{\nearrow ^W}+m\left( h_i-h_e \right) =\cancel{\bigtriangleup E_{sys}} $$
$$ \therefore W=\sum{Q_k}+m\left( h_i-h_e \right) $$
$$ entropy\ balance:\ \cancel{\bigtriangleup S_{sys}}=\sum{\frac{Q_k}{T_k}}+m\left( s_i-s_e \right) +Sgen $$
$$\therefore T_0Sgen=-\sum{\frac{\ T_0}{T_k}}Q_k-T0\cdot m\left( s_i-s_e \right) $$
$$ \therefore W_{rev}=\sum{\text{(}1-\frac{\,\,T_0}{T_k}\text{)}}Q_k+m\left[ \left( h_i-h_e \right) -T_0\left( s_i-s_e \right) \right] $$
$$ \therefore w_{rev}=\sum{\text{(}1-\frac{\,\,T_0}{T_k}\text{)}}q_k+\left[ \left( h_i-h_e \right) -T_0\left( s_i-s_e \right) \right] =\text{(}1-\frac{\,\,T_0}{T_1}\text{)}q_1+\text{(}1-\frac{\,\,T_0}{T_2}\text{)}q_2+\left[ \left( h_i-h_e \right) -T_0\left( s_i-s_e \right) \right]…. (Ans:w_{rev})$$
$$ i=w_{rev}-w_u^{\nearrow ^0}=w_{rev}............\left( Ans:i \right) $$