The maximum reversible work in thermodynamics (2)

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The discussion focuses on calculating the maximum reversible work in thermodynamics for an open system, emphasizing the concept of available energy and exergy. The equations provided outline the relationships between work, heat transfer, and entropy generation, leading to the formulation of reversible work. Key equations illustrate how to derive the expressions for work and entropy balance, ultimately defining reversible work in terms of heat transfers and enthalpy changes. Participants are encouraged to correct any errors in the shared content for clarity and accuracy. The discussion reinforces the importance of understanding these thermodynamic principles for effective energy management.
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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
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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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I agree with the result in the reference you gave.
 
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