Example 8A - 3: Entropy Production of Mixing Two Liquids at Different Temperatures

8A-3 :	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, T₁ and T₂.

The mass of liquid initally on each side of the tank is the same: m₁ = m₂ = 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 :

T₁

Chamber 1 :

T_eq

Chamber 2 :

T₂

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 T₁ and T₂ by applying the 1st Law to this process.