Equilibrium - TEST Tutorial

( Visit the Closed-Process and Open-Steady pages first. )

States	Closed Process	Closed Steady	Open Steady
Open Process	Closed Cycle	Open Cycles	States-II
HVAC	Combustion		Gas Dynamics

(Each section above is divided into two sub-sections - Manual and Applications.)
Manual

(This page builds upon the Combustion pages.)

Daemons>Equilibrium> Applications

EXAMPLE-1

A 1 kg block of aluminum at 600 K is brought in thermal contact with another identical block at 300 K in an isolated chamber. Show that entropy of the combined system is maximized at equilibrium. What-if Scenario: How would the answers change if the second block is made out of copper?

Time Saver: To reproduce the visual solution, copy and paste this TEST-code on the I/O panel of the appropriate daemon, click the Load button followed by the Super-Calculate button.

# HOME>Daemons>Systems>Closed>Process>Generic>NonUniform>
Nonmixing>SL/SL;

States    {
    State-1: Aluminum(Al), Aluminum(Al);
    Given:       { T1= 600.0 K;   Vel1= 0.0 m/s;   z1= 0.0 m;   m1= 1.0 kg;   Model1= 1.0 UnitLess;   }

    State-2: Aluminum(Al), Aluminum(Al);
    Given:       { T2= 300.0 K;   Vel2= 0.0 m/s;   z2= 0.0 m;   m2= 1.0 kg;   Model2= 2.0 UnitLess;   }

    State-3: Aluminum(Al), Aluminum(Al);
    Given:       { T3= 400.0 K;   Vel3= 0.0 m/s;   z3= 0.0 m;   m3= "m1" kg;   Model3= 1.0 UnitLess;   }

    State-4: Aluminum(Al), Aluminum(Al);
    Given:       { Vel4= 0.0 m/s;   z4= 0.0 m;   m4= "m2" kg;   Model4= 2.0 UnitLess;   }
    }

Analysis    {
    Process-A: b-State = State-1, State-2; f-State = State-3, State-4;
    Given: { Q= 0.0 kJ;   W= 0.0 kJ;   T_B= 25.0 deg-C;   }
    }

Step 1 Launch the Non-Mixing Process Daemon with SL/SL model .

Step 2: Evaluate 4 states, two for bA and bB and two for fA and fB.

Step 3: Set up the Process Panel for Q=W=0.

Step 4: Guess T3 and Super-Calculate T4 and Delta_S.

Step 5: To change working substance, select an appropriate state, change the working substance and Super-Calculate.

Solution

The problem describes a closed process involving a composite system made of two blocks of solids. Threfore, launch the Non-Mixing, Generic, Closed Process Daemon with the SL/SL model. Evaluate the four states as described in the TEST-Code. Notice that State-1 and State-2 describe the bA and bB states. Although the working substances for both sub-systems are identical, for sub-system A select the material from the left menu and for sub-system B use the right menu. Also assume an arbtrary equilibrium temperature for system A as T3=400 K.

On the Process Panel, load the bA, bB, fA and fB states. Enter Q=W=0. Super-Calculate to find T4=500 K and Delta_S=0.095 kJ/K. Now change the guess for T3 to 425 K, Super-Calculate and find T4=475 K and Delta_S to be 0.103 kJ/K. Similarly, a T3= 460 K produces T4=440 K and Delta_S to be 0.105 kJ/K. Finally, a choice of T3=450 K produces T4=450 K and Delta_S=0.106 kJ/K . Entropy of the system is, thus, maximized when temperatures of the two sub-stystems become equa.

For the what-if study, change the second working substance to copper in State-2 and Super-Calculate. Change T3 and Super-Calculate T4 and Delta_S. Entropy is maximized when T3=T4= 510 K. Note that the second working fluid can be changed only for State-2 or State-4 since a change of working fluid automatically recalculates the current state using the selected working substance .

EXAMPLE-2

Determine the more likely of the two compositions - 1 kmol of CO2 versus 0.95CO2+.05CO+.05O - at (a) 100 kPa and 300 K, and (b) 100 kPa and 3000 K.

Solution

The question can be answered by evaluating the Gibb's free energy for the two mixtures. Launch the Reaction Daemon found in the Closed, Process, Specific, Combustion branch. Set up the reaction by entering CO2 as the reactant and CO2, CO and O as the products. Enter the amounts, 1 kmol for CO2 in the reactants block, 0.95 kmol for CO2 in the products, and select Balance Reaction to set up the reaction.

Time Saver: To reproduce the visual solution, copy and paste this TEST-code on the I/O panel of the appropriate daemon, set the reaction manually, and click the Load button followed by the Super-Calculate button.

# HOME>Daemons>Systems>Closed>Process>Specific>Combustion>
Reaction;

States    {
    State-1: Reactants > mixture;
    Given:       { p1= 100.0 kPa;   T1= 26.85 deg-C;   Vel1= 0.0 m/s;   z1= 0.0 m;   Model1= 1.0 UnitLess;   }

    State-2: Products > mixture;
    Given:       { p2= "p1" kPa;   T2= "T1" deg-C;   Vel2= 0.0 m/s;   z2= 0.0 m;   Model2= 3.0 UnitLess;   }
    }

Fig. 2.1 Image of the products state in Ex. 2

Step 1: Set up the reaction.

Step 2: Evaluate the reactants and products states.

Step 3: Compare the Gibb's free energy for the two mixtures.

On the States Panel, evaluate the states as described in the TEST-Code below.

Compare g1=-10400 kJ/kg with g2=-9867 kJ/kg. Clearly, State-1, the reactants, is a more stable state since its Gibb's free energy is lower. Now change T1 to 3000 K and Super-Calculate . The new values for g1 and g2 are -28255 kJ/kg and -28388 kJ/kg respectively. The products mixture is, therefore, a more likely state at 3000 K.

EXAMPLE-3

Evaluate the equilibrium constant, expressed as log₁₀ K for the reaction CO+0.5O2<-->CO2 at (a) 298 K, and (b) 1000 K.

# HOME>Daemons>Systems>Closed>Process>Specific>Combustion>
Reaction;

States    {
    State-1: Reactants > mixture;
    Given:       { p1= 100.0 kPa;   T1= 24.85 deg-C;   Vel1= 0.0 m/s;   z1= 0.0 m;   Model1= 1.0 UnitLess;   }

    State-2: Products > mixture;
    Given:       { p2= "p1" kPa;   T2= "T1" deg-C;   Vel2= 0.0 m/s;   z2= 0.0 m;   Model2= 3.0 UnitLess;   }
    }

Fig. 3.1 Image of the Reaction Panel. To balance the reaction only one of the species amount
needs to be supplied.

Step 1: Set up the reaction.

Step 2: Evaluate the Reactants and Products states.

Step 3: Calculate log(K) in the I/O Panel .

Step 4: Change T1, Super-Calculate, and evaluate log(K) on the I/O Panel again.

Solution

Launch the Reaction Daemon found in the Closed, Process, Specific, Combustion branch. Set up the reaction by entering CO2 and O2 as the reactant and CO2 as the products. Enter 1 kmol for CO2 and select Balance Reaction to set up the reaction.

In the States panel, evaluate the Reactants and Products states at 100 kPa and 298 K as described in the TEST-Code.

Evaluate ln(K) in the I/O Panel as
' =-m1*(g2-g1)/(8.314*T1)' =102.9
Therefore, log(K)=ln(K)/ln(10) '=102.9/ln(10)'= 44.67 .

Change T1 to 1000 K, Super-Calculate and re-evaluate log(K) as 9.80.


Manual