The maximum reversible work in thermodynamics (2)

  • #1
tracker890 Source h
90
11
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
1712596768815.png

referencesol
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 ## :
$$ eneger\ balance:\ \ Q_{in,net}^{\nearrow ^{\sum{Q_k}}}-W_{out,net}^{\nearrow ^W}+m\left( h_i-h_e \right) =\cancel{\bigtriangleup E_{sys}} $$
$$ \therefore W=\sum{Q_k}+m\left( h_i-h_e \right) $$
$$ entropy\ balance:\ \cancel{\bigtriangleup S_{sys}}=\sum{\frac{Q_k}{T_k}}+m\left( s_i-s_e \right) +Sgen $$
$$\therefore T_0Sgen=-\sum{\frac{\ T_0}{T_k}}Q_k-T0\cdot m\left( s_i-s_e \right) $$
$$ \therefore W_{rev}=\sum{\text{(}1-\frac{\,\,T_0}{T_k}\text{)}}Q_k+m\left[ \left( h_i-h_e \right) -T_0\left( s_i-s_e \right) \right] $$
$$ \therefore w_{rev}=\sum{\text{(}1-\frac{\,\,T_0}{T_k}\text{)}}q_k+\left[ \left( h_i-h_e \right) -T_0\left( s_i-s_e \right) \right] =\text{(}1-\frac{\,\,T_0}{T_1}\text{)}q_1+\text{(}1-\frac{\,\,T_0}{T_2}\text{)}q_2+\left[ \left( h_i-h_e \right) -T_0\left( s_i-s_e \right) \right]…. (Ans:w_{rev})$$
$$ i=w_{rev}-w_u^{\nearrow ^0}=w_{rev}............\left( Ans:i \right) $$
 
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  • #2
I agree with the result in the reference you gave.
 
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