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TEST-Code |
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| (Each section above is divided into two sub-sections - Manual and Applications.) | |
| Manual | 
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(This page builds upon Closed Process and Open Steady Pages.)
Daemons>Combustion> Applications |
| EXAMPLE-1 | 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 (a) the air-fuel ratio. |
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Step 1: Reaction balancing daemon can be found both in the Closed Process
and Open Steady branches.
Step 2: Balance the reaction. Use the N2 amount as a check.
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Solution
Answering the questions described in the Approach section leads you to the following daemon page: HOME. Daemons. Systems. Open. SteadyState. Specific. Combustion. Reaction . Choose C8H18(g) as fuel. Air is the default oxidizer. Choose CO and O2 in addition to the default products. Because volume is proportional to the mole fraction, we enter 8.86 kmol for CO2, 0.662 kmol for CO, 7.51 for O2. 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 and the reaction is balanced. Normalize the reaction to express the reaction in terms of 1 kmol of fuel. The amount of air used is 88.24 kmol. To determine the excess air, choose Theoretical Air from the action menu. The air amount is found as 59.52 kmol. The excess air now can be easily calculated as 100*(88.24-59.52)/59.52 % |
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| Fig. 1.1 Image
of the balanced equation. Choose Normalize to express the reaction in terms
of 1 kmol of fuel. |
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EXAMPLE-2
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1 kg of methane and 1 kg of oxygen are kept at 100 kPa and 300 K in a
rigid insulated container. Determine the final pressure and temperature after
the mixture is ignited. Assume variable specific heat.
Solution Answering the questions described in the Approach section leads you to the following daemon page: HOME. Daemons. Systems. Closed. Process. Specific. Combustion. IdealGas .
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| Fig. 2.1 Image of the balanced reaction between methane and oxygen. |
| Step 1: Launch the Combustion Process Daemon. Step 2: Balance the reaction. |
Choose CH4 as the fuel. Set O2
as the oxidizer (get rid of default Air by double-clicking its check box
and then choose O2). We cannot assume the proportions of CH4 and O2 to be
theoretical; therefore, either excess O2 or excess CH4 must be left over
as products. If you do not realize that fuel is in excess in this reaction,
try using O2 as the additional product. A Balance Reaction will produce negative
amount of O2 signalling that it is the fuel that is left over in this reaction.
Now un-check N2 and set up the reaction with CH4 on the product side too.
Put CH4(fuel)=1kg, O2(Oxidizer)=1kg and
Balance Reaction. |
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| Fig. 2.2: Image of the States Panel with Fuel as the working fluid. |
| Step 3: Calculate the three states as best as possible.
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On the States panel, evaluate the fuel, oxidizer and products states as much as possible (you have to select the state number as well as the type of mixture, Fuel, Oxidizer, or Products). For the fuel state, T1=300 K. However, p1 is not 100 kPa, instead, the partial pressure of CH4 must be used. On the Reaction Panel click on the Molar button to find out the mole fraction of CH4 to be 0.0623/(0.0623+0.0313)= 0.665 . Therefore, p1=0.666*100. Note that the mass of m1=1 kg is automatically posted. Calculate State-1. For State-2, enter p2=100-p1 (Dalton's model) and T2=300 K. Calculate. Notice that the calculated values of Vol2 and Vol1 are equal as expected. For State-3, enter Vol3=Vol1 and Calculate. p3 and T3 are both unknown at this stage. |
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| Fig. 2.3 Image of the Process Panel after Super-Calculate. |
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Step 4: On the Analysis Panel, load the anchor states, enter the process
variables and Calculate.
Step 5: Super-Calculate. |
On the Process Panel load State-1 and 2 as the bA and bB states and State-3 as the f-State. Enter the process variables, W=Q=0. Press the Enter key and Super-Calculate. The calculated e3 is posted back to State-3 and all states and processes are re-calculated. The entropy generated can be seen on this panel S_gen=10.056 kJ/K. |
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| Fig. 2.4 Image of the product States Panel with all the necessary variables entered. |
| Step 6 Adiabatic temperature can be found in State-3. | On the State Panel (Product)
find p3=895
and T3=2686 K
. Another way to solve the same problem is to set the Fuel 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. Evaluate two states, State-1 for fuel and State-2 for products as shown in the TEST-Code 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-Code above, Super-Calculate to produce T2=2413 deg-C and p2=895 kPa. |
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EXAMPLE-3
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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 at
100 kPa and 300 K. Determine (a) the temperature of the products and (b) the
rate of entropy generation. What-if Scenario:
Solution This problem is almost identical to the previous problem, except here we have an open steady system. The adiabatic temperature is expected to be much lower since there is no compression involved at the end of heat release. Start HOME. Daemons. Systems. Open. Steady. Specific. Combustion. IdealGas daemon. Balance the reaction following the procedure described in the previous example. As before 0.75 kg of CH4 will be left unburned (see Fig 2.1). |
