8A3 :  Entropy Production of Mixing Two Liquids at Different Temperatures  8 pts 

The initial and final states of a sealed, insulated, rigid tank are shown below.Each side of the tank contains a different incompressible liquid at a different temperature, T_{1} and T_{2}.  


The mass of liquid initally on each side of the tank is the same: m_{1} = m_{2} = m/2. The barrier between the two sides of the tank is removed and the two liquids mix and eventually reach the final equilibrium state.  
Assume each liquid has a constant heat capacity and there are no thermal effects due to the mixing of the fluids. a.) Show that S_{gen} is given by the following equation: 



b.) Show that S_{gen} must be positive.  
Read :  For part (a) Perform an entropy balance to determine an equation for S_{gen}. Then perform an energy balance to determine an expression for the final temperature and substitute the expression into S_{gen} and simplify.  
Given:  Initial State:  Final State :  
Chamber 1 :  T_{1}  Chamber 1 :  T_{eq}  
Chamber 2 :  T_{2}  Chamber 2 :  T_{eq}  
Incompressible fluids with C_{P} = C_{V} = C.  
Find:  Part (a)  Show that the amount of entropy generated is: 

Eqn 1  
Part (b)  Demonstrate that S_{gen} must be positive.  ^{}  
^{}  
Diagram:  The diagram in the problem statement is adequate.  
Assumptions:  1   The system consists of the total mass of liquid in the entire tank.  
2   The system is isolated (adiabatic and closed).  
3   The liquid is incompressible with constant specific heat, C.  
4   No work crosses the sytem boundary.  
Equations / Data / Solve:  
Part a.)  Let's begin with the defintion of entropy generation: 

Eqn 2  
We can solve Eqn 2 for S_{gen} : 

Eqn 3  


Since the system is isolated, there is no heat transferred: 

Eqn 4  
We can use Eqn 4 to simplify Eqn 3, yielding : 

Eqn 5  
The change in the entropy of the system is : 

Eqn 6  

Eqn 7  
We can rearrange Eqn 7 to show that the total change in entropy for the system is the sum of the changes in entropy of each of the two fluids.  

Eqn 8  
The entropy change for an incompressible fluid depends only on temperature. 

Eqn 9  
Because the heat capacity in this problem is a constant, it is relatively easy to integrate Eqn 9 to get: 

Eqn 10  
Next, apply Eqn 10 to determine the entropy change of each fluid in this process and substitute the result into Eqn 8 : 

Eqn 11  
Properties of logarithms let us rearrange Eqn 11 to : 

Eqn 12  
Combining Eqn 12 with Eqn 5 gives us : 

Eqn 13  
To complete this derivation, we must eliminate T_{final} from Eqn 13. We can determine T_{final} in terms of T_{1} and T_{2} by applying the 1st Law to this process.  

Eqn 14  
No work or heat crosses the system boundary, so Eqn 14 becomes : 

Eqn 15  
Now, use the constant specific
heat of the incompressible fluid to determine DU : 

Eqn 16  

Eqn 17  
Now, solve Eqn 17 for T_{final} : 

Eqn 18  
Now, we can use Eqn 18 to eliminate T_{final} from Eqn 13 : 

Eqn 19  
Simplify Eqn 19 algebraically : 

Eqn 20  
Finally : 

Eqn 21  
Part b.)  Entropy generation is nonnegative when : 

Eqn 22  
The values of m and C
must be positive
so, S_{gen} is nonnegative when : 

Eqn 23  
Simplify Eqn 23 by algebraic
manipulation, as follows : 

Eqn 24  
Squaring both sides of Eqn 24 yields : 

Eqn 25  
( This is OK because T_{1} > 0 K and T_{2} > 0 K )  
Expand the lefthand side of Eqn 25 : 

Eqn 26  

Eqn 27  
Finally, we get : 

Eqn 28  
The inequality in Eqn 28 is satisfied for either T_{1} > T_{2} or T_{2} > T_{1}.  
The equality in Eqn 28 is satisfied only when T_{1} = T_{2}.  
Verify:  The assumptions made in this solution cannot be verified with the given information.  
Answers :  Part a.) 


Part b.) 


which is ALWAYS true !  