Combustion Analysis Software - Web-based Analysis of Chemically Reacting Systems

Tutorial (TOC) > Daemons > Combustion > Examples

Ex-1 Methane (CH4 in gaseous form) is burned with dry air. Determine (a) the air-fuel ratio on a mole basis, and (b) the air-fuel ratio on a mass basis for the theoretical reaction. Determine the air-fuel ratio on a mass basis if (c) the reaction is complete with 25% excess air, (d) the reaction is complete with 10% deficient air.

Solution Any combustion daemon with separated (nonpremixed) fuel and oxidizer can be used for balancing a reaction. Launch the open-steady combustion daemon located at TEST. Daemons. Systems. Open. SteadyState. Specific. Combustion. Separated. IdealGas . For balancing a reaction, it does not matter whether you choose the ideal gas or the perfect gas model.

In the fuel block (pink background) choose methane (CH4) as the fuel. From the action menu, select Theoretical Air. The theoretical reaction is displayed for 1 kmol of fuel. The air amount is displayed as 9.524 kmol. Therefore, the AF ratio (on a mole basis) is 9.524 kmol of air/ kmol of fuel. To make the reaction mass based, click on the Mass button and then Normalize Reaction from the action menu. The reaction is now displayed on the basis of 1 kg of fuel. The mass based AF ratio is calculated as 17.2 kg of air/kg of fuel.

For 25% excess air, λ = 1.25. Enter 1.25 as lambda, and select Excess/Deficient Air from the action menu. The new AF ratio is calculated as 21.5 kg of air/kg of fuel. Similarly, for 10% deficient air, λ = 0.9. Change lambda, and select Excess/Deficient Air command to obtain the new value of AF as 15.48 kg of air/kg of fuel. Note that the equivalent ratio is also reported for any given lambda.

Fig. 1.1 Image of the balanced equation. Choose Normalize to express the reaction in terms of 1 kmol of fuel.

Ex-2 Octane (C8H18 in gaseous form) is burned with dry air. The volumetric analysis of the products on a dry basis is 8.86% CO2, 0.662% CO, 7.51% O2, and 82.978% N2. Balance the equation and determine the air-fuel ratio.

Solution Any combustion daemon with separated (nonpremixed) fuel and oxidizer can be used for balancing a reaction. Launch the open-steady combustion daemon located at TEST. Daemons. Systems. Open. SteadyState. Specific. Combustion. Separated. IdealGas. For balancing a reaction, it does not matter whether you choose the ideal gas or the perfect gas model.

In the fuel block (pink background) choose Octane(C8H18) as the fuel. Air is already the default oxidizer (the blue block). Choose CO and O2 in addition to the default products (green block). We will first balance the reaction on the basis of 100 kmol (volumetric analysis is proportional to molar analysis for gaseous mixture) of dry products. Make sure molar and SI buttons (default) are selected at the top-level control panel.

Enter 8.86 kmol for CO2, 0.662 kmol for CO, 7.51 for O2 (you have to click the checkbox first and then enter the amounts in the textfield that turns yellow, when waiting for input). Entering the N2 amount will over-specify the problem (Try it! The daemon will generate appropriate warning as you select the Balance Reaction choice). Select Balance Reaction from the Select an Action menu and the reaction is balanced as shown in figure below.

Normalize the reaction (again, select Normalize from the select an Action) to express the reaction in terms of 1 kmol of fuel. The amount of air used is 88.24 kmol. If you want to see the amount of oxygen and nitrogen that participate in the reaction (instead of air), select Air->O2,N2 in the action menu.

To determine the amount of excess air, choose Theoretical Air from the action menu. Note that the products list is automatically adjusted for complete combustion and the reaction is balanced on the basis of 1 kmol of fuel, producing the theoretical air amount to be 59.52 kmol. The excess air now can be calculated by hand or using the I/O panel: 100*(88.24-59.52)/59.52 = 48.25%.

Fig. 2.1 Image of the balanced equation. Choose Normalize to express the reaction in terms of 1 kmol of fuel.

Ex-3 Methane (CH4) is burned with 200 percent theoretical air during a combustion process. Assuming complete combustion and a total pressure of 150 kPa, determine (a) the air-fuel ratio and (b) the temperature at which water vapor in the products will start condensing. (c) What-if Scenario: How would the answer in part (b) change if air entered the chamber at 25 deg-C with a relative humidity of 80%?

Solution The open, steady combustion daemon with separated entry for fuel and oxidizer should be used for analyzing this problem. Launch the open-steady combustion daemon located at TEST. Daemons. Systems. Open. SteadyState. Specific. Combustion. Separated. IdealGas (the material model does not matter in this problem).

Leaving the default choice of SI and Mole buttons, select methane in the fuel block. 200% theoretical air means λ = 2; enter it in the lambda widget. From the action menu, select Excess/Deficient Air and the reaction is balanced for complete combustion. An air-fuel ratio of 34.4, is displayed on the message panel. If you want, you can select the Mass radio button to convert the reaction into mass basis (16.04 kg of fuel), then Normalize (using the action menu). Expressed in terms of 1 kg of fuel, the amount of air, 34.4 kg, is the air-fuel ratio.

Get back the molar representation by selecting the Mole button and then Normalize from the action menu (the reaction is expressed in terms of 1 kmol of fuel now). The mole fraction of water vapor in the products can now be calculated as 2/(1+2+15.05+2)=0.1. The vapor pressure of water, therefore, must be 150*0.1=15 kPa. On a separate window, launch the system state daemon with PC model (located at Daemons.States.System.PC Model). Select H2O, enter p1=15 kPa and x1=1 (or any value between 0 and 1). The saturation temperature is found as 53.9 deg-C. This must be the dew point temperature of the products mixture.

Fig. 3.1 Convert percent theoretical air into lambda (fraction), enter it, and select Excess/Deficient Air from the action menu.

For the what-if scenario we determine the amount of moisture that accompany 19.048 kmol of air (the balanced reaction displays this amount). The partial pressure of water vapor is (R.H.)(p_sat@25°C)=0.8*3.157 (you can use the system state daemon with PC model and enter T1=25 deg-C and x1=1 and get p1 as p_sat) = 2.53 kPa. This means that the mole fraction of water vapor in the moist air must be 2.53/150=0.0169, that is,

0.0169=n_H2O/(19.048+n_H20)

Solving, n_H2O=0.327 kmol. To add this moisture to the air, initialize the reaction, select CH4 (1 kmol), enter air amount as 19.048 kmol, select H2O in the oxidizer block and enter the amount of 0.0169, and select Balance Reaction from the action menu. The vapor pressure in the products of this new reaction is 2.327*(150)/(1+2.327+15.05+2)=17.13 kPa. The corresponding saturation vapor pressure is T_sat@17.13 kPa=56.7 deg-C.

Fig. 3.2 Calculate vapor amount manually and balance the reaction for moist air.

Ex-4 A coal sample has a mass analysis of 76.39% carbon, 4.2% hydrogen (H₂), 5.32% oxygen (O₂), 1.63% nitrogen (N₂), 1.5% sulfur, and the rest is ash. For complete combustion with 120% of the theoretical air, determine the air-fuel ratio on a mass basis.

Solution Any combustion daemon with separated entry for fuel and oxidizer can be used for balancing a reaction. Launch the open-steady combustion daemon located at TEST. Daemons. Systems. Open. SteadyState. Specific. Combustion. Separated. IdealGas.

Select the SI and Mass mode. On the fuel block, select all fuel components including ash and enter their masses out of a total amount of 100 kg. In the input field for Excess or Deficient air enter the excess air amount, 20%. From the action menu, select Excess/Deficient Air. The products are automatically selected and the reaction is normalized for 1 kg of fuel mixture (coal). The air fuel ratio is calculated as 12.05.

Ex-5 Determine the enthalpy of combustion of liquid propane at 300 K and 100 kPa.

Solution The enthalpy of combustion is the hypothetical heat transfer as liquid dodecane and stoichiometic air react at 300 K and the products leave the steady-state combustion chamber at 300 K. The open, steady combustion daemon with separated entry for fuel and oxidizer is ideal for this problem. Launch the open-steady combustion daemon located at TEST. Daemons. Systems. Open. SteadyState. Specific. Combustion. Separated. IdealGas.

On the fuel block, select liquid dodecane and enter 1 kmol as its amount. On the products panel, de-select H2O and select H2O(L) in its place (water must be in liquid form at 300 K). There will be no excess O2 left in a stoichiometric reaction, so de-select O2 in the products. From the action menu select Balance Reaction and the reaction is balanced.

Switch to the state panel. Evaluate the fuel state (fuel is the default choice for state-1) as state-1 with p1=100 kPa and T1=300 K. Select state-2 and oxidizer, enter p2=p1 and T2=T1, and Calculate the oxidizer state. Likewise, calculate state-3 for products with p3=p1 and T3=T1. Switch to the Device analysis panel. Load state-1 and 2 as the i1 and i2 states and state-3 as the e state. Enter Wdot_ext=0 and Calculate Qdot as -8101021.5 kW. Since the flow rate of fuel is 1 kmol/s, this translates to -8101021.5 kJ/kmol. To obtain the enthalpy of combustion in kJ/kg, divide this by the molar mass of fuel. In the I/O panel type in =-8101021.5/MM1=. -47653.06 kJ/kg or -47.65 MJ/kg.

Fig. 5.1 State of products. Note that the composition is also displayed in each state.

Ex-6 In an adiabatic combustion chamber, CH4 is burned at a rate of 1 kg/s with 1 kg/s of O2 with both the fuel and oxidizer entering the chamber separately at 100 kPa and 300 K. Determine (a) the temperature of the products and (b) the rate of entropy generation.

Solution This problem is almost identical to the previous problem, except here we have an open steady system, where fuel and oxidizer enter separately. The adiabatic temperature is expected to be much lower since there is no compression involved at the end of heat release.

Start the open-steady combustion daemon located at TEST. Daemons. Systems. Open. Steady. Specific. Combustion. Separated. IdealGas.

Balance the reaction by (i) selecting Mass and SI buttons, (ii) selecting the fuel as methane (CH4), (iii) entering an amount of 1 kg, (iv) de-selecting air and selecting O2 as oxidizer, (v) entering the amount of 1 kg, (vi) de-selecting N2 from the products, and (vii) choosing the Balance Reaction item from the action menu. 0.75 kg of CH4 will be left unburned.

Fig. 6.1 Image of the products state completely evaluated.

In the state panel, you will find Fuel, Oxidizer and Products as three different working fluids. Choose Fuel, enter the total pressure p1=100 kPa and T1=300 K. Velocity and z are already initialized to zero. A Calculate produces the complete state. For Oxidizer do exactly the same (see Fig. 4.1) with p2=p1 and T2=T1. For the products enter the pressure as p3=p1. A Calculate fails to produce the final temperature at this point.

Fig. 6.2 Image of device panel.

In the device panel, load the three states as the i1, i2 and e states. Enter Wdot_ext=Qdot=0 and Calculate and SuperCalculate. In the state panel, T3 is found as T=2431 K . Note that it is significantly smaller than the constant-volume adiabatic flame temperature obtained in the previous problem.

In the device panel entropy generation rate is calculated as Sdot_gen=11.53 kW/K .

In the state panel, change the fuel to C2H6 and enter its amount as 1 kg. Super-Calculate. The new answers can be found on State-3 and the device panel as as T=2521 K and Sdot_gen=8.155 kW/K.

# TEST-codes for Ex-6;
# Daemon Path: TEST>Daemons>Systems> Open > Steady > Specific>
# Combustion>Separated>IG Model;

States    {
    State-1: Fuel > 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: Oxidizer > mixture;
    Given:       { p2= "p1" kPa;   T2= "T1" deg-C;   Vel2= 0.0 m/s;   z2= 0.0 m;   Model2= 2.0 UnitLess;   }

    State-3: Products > mixture;
    Given:       { p3= "p1" kPa;   Vel3= 0.0 m/s;   z3= 0.0 m;   Model3= 3.0 UnitLess;   }
    }

Analysis    {
    Device-A: i-State = State-1, State-2; e-State = State-3; Mixing: true;
    Given: { Qdot= 0.0 kW;   Wdot_ext= 0.0 kW;   T_B= 25.0 deg-C;   }
    }

Ex-7 Liquid octane is completely burned with 300% theoretical air in a steady combustion chamber. If the inlet temperature of both fuel and air is 25 deg-C, determine (a) the exit temperature neglecting the exit ke. (b) What-if Scenario: Determine the exit velocity if the exit temperature was 870 deg-C.

Solution Start TEST. Daemons. Systems. Open. Steady. Specific. Combustion. Separated. IdealGas daemon.

In the Reaction panel, select Octane(L) as the fuel, enter lambda as 3, and select Excess/Deficient Air from the action menu. The balance reaction is displayed as shown on the figure below.

In the state panel, evaluate State-1 with T1=25 deg-C for Fuel, State-2 with T2=25 deg-C for Oxidizer, and State-3 with Vel3=0. Note that you have to explicitly select Fuel, Oxidizer or Products in setting up the states.

In the device panel, enter Qdot=0 and Wdot_ext=0. Calculate and Super-Calculate. In State-3 the exit temperature is posted as T3=879.4 deg-C.

For the what-if study, make Vel2 an unknown and Enter T2 as 870 deg-C. Super Calculate to evaluate Vel2 as 153 m/s.

Fig. 7.1 Image of the reaction panel for Ex. 7 .

Fig. 7.2 Image of the products state, showing the adiabatic flame temperature.

# TEST-codes for Ex-7;
# Daemon Path:TEST>Daemons>Systems>Open>Steady>Specific>Combustion>Separated>IG Model

States    {
    State-1: Fuel > mixture;
    Given:       { p1= 100.0 kPa;   T1= 24.999994 deg-C;   Vel1= 0.0 m/s;   z1= 0.0 m;   Model1= 1.0 UnitLess;   }

    State-2: Oxidizer > mixture;
    Given:       { p2= "p1" kPa;   T2= "T1" deg-C;   Vel2= 0.0 m/s;   z2= 0.0 m;   Model2= 2.0 UnitLess;   }

    State-3: Products > mixture;
    Given:       { p3= "p1" kPa;   Vel3= 0.0 m/s;   z3= 0.0 m;   Model3= 3.0 UnitLess;   }
    }

Analysis    {
    Device-A: i-State = State-1, State-2; e-State = State-3; Mixing: true;
    Given: { Qdot= 0.0 kW;   Wdot_ext= 0.0 kW;   T_B= 25.0 deg-C;   }
    }

Ex-8 A fuel mixture composed of 60% CH₄ and 40% C₂H₂ by volume is burned with 10% access air in an adiabatic combustion chamber. The fuel mixture enters the combustion chamber at 500 kPa, 25°C and the air at 500 kPa, 120°C. Combustion is complete and the products, leaving the combustor without any significant pressure drop, enters an isentropic nozzle where it expands to a pressure of 100 kPa at the nozzle exit. Neglecting the KE at the nozzle inlet, determine (a) the nozzle inlet temperature, (b) the exit velocity, and (c) the exit temperature. Use variable c_p. (d) What-if scenario: How would the answer in part (b) change if the combustion products modeled as air?

Solution The fuel mixture and air enter the combustion chamber through separate inlets. Therefore, we select the open steady combustion daemon located at TEST. Daemons. Systems. Open. SteadyState. Specific. Combustion. Separated. IdealGas.

Select the SI and Mole mode (the default). Compose the fuel mixture on the basis of 100 kmol (recall from Chapter 11 that volumetric composition of an ideal gas mixture is identical to its molar composition). For 10% excess air, enter lambda as 1.1 (see Fig. 8.1). From the action menu, select Excess/Deficient Air.

The daemon selects the products automatically and normalizes the balanced reaction for 1 kmol of fuel. If you want to base the reaction on a different amount of fuel, you can multiply the reaction by a suitable factor by entering it in the Excess/Deficient air widget and selecting Multiply Reaction from the action menu.

Fig. 8.1 Setting up the reaction for a fuel mixture.

Switch to the state panel. We will calculate the fuel state first. Notice that Fuel is the default choice in the working fluid selector and its composition is displayed in a box. State-1 is also the default choice. Enter p1=500 kPa and T1=25 deg-C and calculate the fuel state. Select state-2 from the state selector and oxidizer from the fluid selector, enter p2 as '=p1' and T2=120 deg-C, and Calculate the oxidizer (air) state. For state-3, the products state, only p3 is known at this point. Similarly for state-4, the nozzle exit state, p4=100 kPa and all other properties including Vel4 are unknown. The TEST-codes below list exactly what we entered as input (note that Vel4 is made an unknown).

With the states partially computed, switch to the device panel. Set up device-A as the combustion chamber with Qdot=Wdot_ext=0. Select the two inlet states (fuel and air, state-1 and 2) and the exit state (products, state-3). Calculate to solve the energy equation producing j3, which is posted on state-3 (switch to state-3 to check). State-3 now can be completely calculated. But before we do that, let us set up the nozzle as device-B. Select state-3 as the inlet state and state-4 as the exit of the nozzle. Enter Qdot=Wdot_ext=0. Also, Sdot_gen=0 for an isentropic nozzle. Analysis of the nozzle produces j4=j3 and s4=s3 (which could be directly entered in state-4). However, the device analysis posts these properties in state-4 for you. To update all states, press the Super-Calculate button. If some states are not updated (state-4 in this case), you have to individually Calculate those states (there is no Super-Iterate button in the combustion daemons).

T3, T4, and Vel4 are found as 2215.2 deg-C, 1522.2 deg-C, and 1410.4 m/s respectively.

# TEST-codes for Ex-8;
# Daemon Path:TEST>Daemons>Systems>Open>Steady>Specific>Combustion>Separated>IG Model

States {
State-1: Fuel > mixture;
Given: { p1= 500.0 kPa; T1= 24.999994 deg-C; Vel1= 0.0 m/s; z1= 0.0 m; Model1= 1.0 UnitLess; }

State-2: Oxidizer > mixture;
Given: { p2= "p1" kPa; T2= 120.00002 deg-C; Vel2= 0.0 m/s; z2= 0.0 m; Model2= 2.0 UnitLess; }

State-3: Products > mixture;
Given: { p3= "p1" kPa; Vel3= 0.0 m/s; z3= 0.0 m; Model3= 3.0 UnitLess; }

State-4: Products > mixture;
Given: { p4= 100.0 kPa; z4= 0.0 m; Model4= 3.0 UnitLess; }
}

Analysis {
Device-A: i-State = State-1, State-2; e-State = State-3; Mixing: true;
Given: { Qdot= 0.0 kW; Wdot_ext= 0.0 kW; T_B= 25.0 deg-C; }

Device-B: i-State = State-3; e-State = State-4; Mixing: true;
Given: { Qdot= 0.0 kW; Wdot_ext= 0.0 kW; T_B= 25.0 deg-C; Sdot_gen= 0.0 kW/K; }
}

For the what-if study, we calculate two new states, state-5 and 6 with air (oxidizer) as the working fluid. However, when we select oxidizer for, say, state-5, the mass flow rate is picked up from the reaction. You can overwrite the mass flow rate and set mdot5 as '=mdot3'. The states are calculated as described by the codes below.

State-5: Oxidizer > mixture;
Given: { p5= "p3" kPa; T5= "T3" deg-C; Vel5= 0.0 m/s; z5= 0.0 m; mdot5= "mdot3" kg/s; Model5= 2.0 UnitLess; }

State-6: Oxidizer > mixture;
Given: { p6= "p4" kPa; s6= "s5" kJ/kg.K; z6= 0.0 m; j6= "j5" kJ/kg; mdot6= "mdot5" kg/s; Model6= 2.0 UnitLess; }

The exit velocity is found at state-6 as 1384.7 m/s.

Ex-9 1 kg of methane and 1 kg of oxygen are kept at 100 kPa and 300 K in a rigid insulated container. Determine (a) the final pressure and (b) temperature after the mixture is ignited and the products of complete combustion come to equilibrium. Assume variable specific heat.

Solution Premixed fuel and oxidizer go through a combustion process (b-state to f-state) in this problem. Variable specific heat means the IG model must be used instead of the PG model. Launch the closed-process combustion daemon located at TEST. Daemons. Systems. Closed. Process. Specific. Combustion. Premixed. IdealGas.

Set up the Reactants block as a mixture of 1 kg of CH4 and 1 kg of O2 and the products as CO2, H2O and CH4. Use Balance Reaction to set up the reaction (see Fig. 7.1). Evaluate two states, State-1 for fuel and State-2 for products as shown in the TEST-codes below.

States    {
    State-1: Fuel > 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:       { Vel2= 0.0 m/s;   z2= 0.0 m;   Vol2= "Vol1" m^3;   Model2= 3.0 UnitLess;   }
    }

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

After the process panel is set up as shown in the TEST-codes above, Super-Calculate to produce T2=2413 deg-C and p2=895 kPa.

Fig. 9.1 Image of the balanced reaction between methane and oxygen.

Fig. 9.2 Image of the process panel.

Ex-10 Determine the stoichiometric adiabatic flame temperature at a constant pressure for the following fuels: (a) methane, (b) ethane, and (c) propane. Assume the reactants temperature to be 300 K and complete combustion. (d) What-if Scenario: How would the answers change if the combustion is carried out in a constant volume manner?

Solution Start TEST. Daemons. Systems. Open. Steady. Specific. Combustion. Separated. IdealGas daemon.

On the Reaction panel, select methane (CH4) on the reaction block and select Theoretical Air from the action menu. The stoichiometric reaction is immediately displayed.

In the state panel, evaluate State-1 with T1=300 K for Fuel, State-2 with T2=T1 for Oxidizer, and State-3 for products leaving T3 and p3 unknowns.

In the device panel, enter Qdot=Wdot_ext=0. Calculate and Super-Calculate. In State-3 the exit temperature, which is the adiabatic flame temperature, is posted as T3=2330 K. Go back to the states panel, de-select methane and select methane on the fuel block, select Theoretical Air from the action menu, and Super-Calculate. On the states panel, T3 is updated to 2385 K. Likewise, for propane, it is 2397 K.

Fig. 10.1 Stoichiometric adiabatic temperature for complete combustion of propane.

For constant volume process, launch the closed process combustion daemon, located at Daemons.Systems.Closed.Process.Specific.Combustion.Premixed.Idealgas page. It is also a good idea to launch the Separated daemon on a separate window to obtain the balanced reaction for stoichiometric combustion of methane. In the closed-process daemon obtain the balanced reaction in the following way. For the reactants enter methane (1 kmol) and air (9.5238 kmol) (you can balance the stoichiometric reaction in the separated combustion daemon and obtain the air quantity from there) and for products select CO2, H2O, and N2. Now select Balance Reaction from the action menu and the reaction is balanced.

Go to the states panel and calculate the reactants state with T1=300 K and the products state with p1 and T1 unknown. In the process panel, select state-1 and state-2 as the b and f states, enter Q=W=0 and Calculate to solve the energy equation and post e2 back to state-2. Go back to the state panel and Calculate state-2 or Super-Calculate to obtain T2=2824 K. Repeat the same process for other fuels.

Fig. 10.2 Constant volume adiabatic temperature for methane.