Steam Power Engine

The design project presented is about a complex steam power cycle. The purpose of this project was to find the values of the intermediate exit pressures from the turbine, P12, P13, P14, and P15 that give the maximum efficiency for the cycle. In order for us to do that we had to calculate the efficiency of this cycle by using the thermodynamics property toolbox in excel. The toolbox actually helped us to find every single enthalpy faster. In addition, we used the equations given by Dr. Caretto to analyze the problem. The cycle is as below.

Dr. Caretto, Yunus A. Зengel and Michael A. Boles, Thermodynamics, an Engineering Approach, (fourth edition) McGraw-Hill, 2003, Figure 9-17, page 532,

The data given for the project were, condenser pressure 17.4 kPa, steam generator exit pressure for main flow 21.7 MPa, steam generator exit temperature for main flow 473̊̊C, stream generator exit temperature for reheat stream 443̊C, the isentropic efficiency of the first and second pump are respectively 90% and 84%, the isentropic efficiency of the high pressure turbine is 87%, the isentropic efficiency for low pressure turbine at exit 13 is 91%, at exit 14 is 87%, at exit 15 is 85%, at exit 16 is 82%, and finally the heat transfer effectiveness for each closed feedwater heater is 86%.

So basically we had to use all these data plug them into an excel spreadsheet and calculate the enthalpy of each inlet and outlet. We first used the pressure given by Dr. Caretto to get our efficiency for this cycle. After doing that we had to use a solver that helped us get our maximum efficiency form the guesses that we made. Few constraint were to follow such as P11 ≥P12 ≥P13 ≥P14 ≥P15 ≥P16 and mass flow rates are all positive. The solver gave us different new pressures with maximum efficiency such as in the table below.

Row item Four cases with different initial guesses for pressures and same choices for estimates, derivatives and search Different choices for estimates, derivatives or search using guess pressures that gave best efficiency in cases 1 to 4

Case 1 2 3 4 5 6 7

P12 Guess 4845142.81891964 217000 100000 186987 100000 100000 100000

P13 Guess 1046288.22120056 108500 80000 124915 80000 80000 80000

P14 Guess 470638.431407212 54250 40000 23720 40000 40000 40000