8A-3 : | Entropy Production of Mixing Two Liquids at Different Temperatures | 8 pts |
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An isolated system of total mass m is formed by mixing two equal masses of the same liquid initially at the temperatures T1 and T2. Eventually, the system attains an equilibrium state. Each mass is incompressible with constant specific heat C. | ||||||||||

a.) | Show that the amount of entropy produced is : |
Eqn 1 | ||||||||

b.) | Demonstrate that Sgen must be positive. | |||||||||

Read : | For part (a) Perform an entropy to determine an equation for Sgen. Then perform an energy balance to determine an expression for the final temperature and substitute the expression into Sgen and simplify. | |||||||||

Given : | Initial State: | Final State : | ||||||||

Chamber 1 : | T1 | Chamber 1 : | Teq | |||||||

Chamber 2 : | T2 | Chamber 2 : | Teq | |||||||

Incompressible fluids with CP = CV = C. | ||||||||||

Find : | Part (a) | Show that the amount of entropy generated is: |
Eqn 1 | |||||||

Part (b) | Demonstrate that Sgen must be positive. | |||||||||

Diagram : | ||||||||||

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 system boundary. | |||||||||

Equations | ||||||||||

Part a.) | Let's begin with the defintion of entropy generation: | Eqn 2 | ||||||||

We can solve Eqn 2 for Sgen : | 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 this equation 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 elimnate Tfinal from Eqn 13. We can determine Tfinal in terms of T1 and T2 by applying the 1st Law to this process. | ||||||||||

Eqn 14 | ||||||||||

No work or heat crosses the sytem boundary, so Eqn 14 becomes : | Eqn 15 | |||||||||

Now, use the constant specific heat of the incompressible fluid to determine ΔU : | Eqn 16 | |||||||||

Eqn 17 | ||||||||||

Now, solve Eqn 17 for Tfinal : | Eqn 18 | |||||||||

Now, we can use Eqn 18 to eliminate Tfinal from Eqn 13 : |
Eqn 19 | |||||||||

Simplify Eqn 19 algebraically : | Eqn 20 | |||||||||

Finally : | Eqn 21 | |||||||||

Part b.) | Entropy generation is non-negative when : | Eqn 22 | ||||||||

The values of m and C must be positive so, Sgen is non-negative 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 T1> 0 K and T2> 0 K ) | ||||||||||

Expand the left-hand side of Eqn 25 : | Eqn 26 | |||||||||

Eqn 27 | ||||||||||

Finally, we get : | Eqn 28 | |||||||||

The inequality in Eqn 28 is satisfied for either T1>T2 or T2>T1. | ||||||||||

The equality in Eqn 28 is satisfied only when T1 = T2. | ||||||||||

Verify : | The assumptions made in this solution cannot be verified with the given information. | |||||||||

Answers : | Part a.) | |||||||||

Part b.) | when : | which is ALWAYS true ! |