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Tutorial (TOC) > Daemons > Equilibrium > Examples |
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Ex-1 Pure oxygen (O2) is heated at a constant pressure of 100 kPa to a temperautre of 2500 deg-C. (a) Determine the (a) mass and (b) mole fraction of oxygen atom in the dissociated mixture. (c) What-if scenario: How would the answers change if the pressure was reduced to 50 kPa? |
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Solution Oxygen undergoes a constant-pressure dissociation process. We can use either the steady-flow or the closed-process equilibrium daemon. Launch the closed process chemical equilibrium daemon located at Closed, Process, Specific, Combustion, Equilibrium branch.
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Switch to the state panel, select Products radio button, enter p1 and T1, and Calculate. The state including its composition is calculated. While the state properties are displayed on the state panel, the composition can be found on the composition panel as shown below. The mass and mole fractions of O are calculated as 0.0244 and 0.0477 respectively.
For the what-if scenario, change p1 to 50 kPa, press the Enter key, and Super-Calculate to produce x_O as 0.0345 and y_O as 0.0667. |
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Ex-2 A mixture of 1 kmol of H2O and 3 kmol of O2 is heated to 3500 K at a pressure of 1 atm. Determine the equilibrium composition of this mixture, assuming the products to contain H2O, OH, O2 and H2 only. |
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Solution Launch the closed process chemical equilibrium daemon located at Closed, Process, Specific, Combustion, Equilibrium branch.
Now select the Products radio button in the state panel, enter p1=1 atm, T1=2500 K, and Calculate. The state is displayed as shown below.
The molar composition can be found in the composition panel (see image below) as 0.4783H2O + 0.1123H2 + 0.8188OH + 2.8515O2.
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Ex-3 Determine the stoichiometric equlibrium flame temperature of methane burning with air at a constant pressure of 100 kPa. |
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Solution Using the open, steady, separated combustion daemon obtain the stoichiometric composition of the reactants as CH4 + 2O2 + 7.5238N2. (Start the combustion daemon at open.steady.specific.combustion.separated.IGmixture page, select methane as the fuel and Theoretical Air from the action menu. Now select Air->O2,N2 from the action menu to obtain the balanced reaction). Launch the open steady equilibrium daemon located at Open, Steady, Specific, Combustion, Equilibrium branch. On the composition panel, set up the reactants by adding CH4, O2 and N2 in known molar amounts. Also select relevant species for the products (you can select "standard combustion products" as a shortcut). Calculate state-1 as the reactants state (with the Reactants button selected) with p1=100 kPa and T1=25 deg-C, and an assumed mass flow rate mdot1= 1 kg/s. Note the enthalpy calculated as h1=-985.02 kJ/kg of reactants mixture. In the state panel calculate State-2 with p2=p1. In the device panel, select state-1 as the i-state and state-2 as the e-state. Enter Wdot_ext=0. Go back to the state-panel and use a guessed value for T2, say, 1900 deg-C. Press the Enter key and then Super-Calculate. This will update the device panel. Qdot is negative, indicating that the adiabatic temperature must be higher. Increase T2 and Super-Calculate again. Continue until yo uget Qdot is close to zero. At that point T2 is 1965 deg-C. Therefore, the equilibrium temperature (for the chosen products) must be quite close to 1965 deg-C. In the near future, this daemon will be able to handle input such as h2=h1. By entering Qdot=0, or h2=h1, T2 can be directly calculated without any need for iterations.
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Ex-4 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 was made of copper? |
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Solution The problem describes a closed process involving
a composite system made of two blocks of solids. Therefore, launch the
Non-Mixing, Generic, Closed Process
Daemon with the SL/SL model. Evaluate the
four states as described in the TEST-codes. 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 arbitrary 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 equal. 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 . |
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Ex-5 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. We can use several different daemons - system or flow state daemons with IG mixture, closed or open combustion daemons with IG or PG model - to evaluate g of a gas mixture. |
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| Fig. 5.1 Image of the products state in Ex. 2 |
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Launch the premixed closed process combustion daemon located at Closed, Process, Specific, Combustion, Premixed, IdealGas 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. On the States Panel, evaluate the states (reactants and products) as described in the TEST-codes above. 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. |
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Ex-6 Evaluate the equilibrium constant, expressed as log10 K for the reaction CO+0.5O2<----->CO2 at (a) 298 K, and (b) 1000 K. |
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| Fig. 6.1 Image of the Reaction Panel. To balance the reaction
only one of the species amount needs to be supplied. |
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Solution Launch the closed process premixed combustion daemon found in the Closed, Process, Specific, Combustion, Premixed, IG Model branch. Set up the reaction by entering CO2 and O2 as reactants 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-codes.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. |
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