MEMS 1051: APPLIED THERMODYNAMICS Group Design Project. December 3, 2014

Transcription

1 MEMS 1051: APPLIED THERMODYNAMICS Group Design Project December 3, 2014 Team Superheat Matt Adams Kyle Fragassi Rachel Rohr Steven Terrana Michelle Underwood

2 ABSTRACT Problem Statement We have been tasked to build a facility that can produce 80 MW of electrical power at minimum cost. Three different cycle configurations were analyzed due to space considerations: a Brayton cycle with regeneration, a single-shaft Brayton cycle combined with a Rankine cycle, and a splitshaft Brayton cycle combined with a Rankine cycle. The optimum configuration was then analyzed to see if it could provide power at a competitive price. Overview of Analysis Using the air-standard assumption, the state points of all three configurations were found using a set of assumptions and known relationships. A cost equation was developed based on machinery initial costs, maintenance costs, and fuel costs. This cost was then determined for various pressure ratios using MATLAB and plotted to view the results. This was done for each cycle independently such that a unique optimum compressor pressure ratio was found corresponding to the minimum cost for each cycle. Results/Recommendations As a result of our analysis, we recommend the second configuration of a single-shaft Brayton cycle combined with a Rankine cycle. The single-shaft Brayton cycle and Rankine cycle has the cheapest 10-year cycle (it is $4,110, less than the next cheapest cycle) while simultaneously boasting the highest overall cycle efficiency. While this configuration uses a 49.29% higher compressor pressure ratio than the configuration with a split-shaft Brayton cycle, the maximum pressure is only kpa, well within the safe pressure limits of the machinery. This configuration will be able to provide power at a competitive price. At the optimum compressor pressure ratio, the second configuration can provide power at $ /kW hr. Risks or Downsides While this analysis provided a reasonable comparison among the three configurations, in reality, all three cost analyses are idealized and neglect a number of additional factors. First, by eliminating the use of the air standard assumption, the cost per kw could be more accurately predicted. Since atmospheric air is primarily made of ideal gasses, the air standard assumption could be used to find enthalpy and entropy values. However, air also contains water vapor, which is not accounted for in the assumption. Next, the cost analysis neglected the cost for piping. It is also assumed that all machinery will maintain its initial efficient operation, which is not realistic over a 10-year period. Last, inflation and other economic factors were neglected. The cost analysis was also limited by the machinery available. While higher power machinery will be more expensive, it could ultimately allow for a cheaper configuration due to increased efficiency. Page 2 of 32

3 PROBLEM DESCRIPTION The facility being designed must be able to produce 80 MW of electrical power. Three different cycle configurations were analyzed due to space considerations: a Brayton cycle with regeneration, a single-shaft Brayton cycle combined with a Rankine cycle, and a split-shaft Brayton cycle combined with a Rankine cycle. By combining the Rankine cycle with the Brayton cycle, the large amount of energy that would have been lost from the Brayton cycle can power the Rankine cycle and increase the work output, increasing the efficiency. We first analyzed a Brayton cycle alone that utilizes two compressor stages, intercooling, two turbine stages, two combustion chambers, and a regenerator. Figure 1 First Brayton Cycle Configuration The next two configurations combine both the Rankine and Brayton cycles. The second configuration uses a single-shaft Brayton cycle with one stage of reheat, no intercooling, and no regeneration, with a waste heat steam generator to allow for the Brayton cycle to supply the total heating required for the Rankine cycle. In the single-shaft configuration, the work from each turbine stage is identical. Page 3 of 32

4 Figure 2 Second Configuration Includes a Single-Shaft Brayton Cycle Combined with a Rankine Cycle The third configuration uses a split-shaft Brayton cycle with one stage of reheat, no intercooling, and no regeneration, with a waste heat steam to heat the Rankine cycle. In the split-shaft configuration, the work produced from the first turbine is equal to the work required by the compressor. Figure 3 Third Configuration Includes a Split-Shaft Brayton Cycle Combined with a Rankine Cycle Page 4 of 32

5 The power plant was designed to operate 24 hours a day for 10 years. Each piece of equipment would require maintenance throughout its life time; each year, the maintenance cost was approximately 10% of the original cost of the machinery. The cost of each piece of equipment was related to the cost of operation individually, and fuel costs encompass the whole cycle. The given values for the cycle state points and machinery operation include: Rankine Cycle Turbine Inlet Pressure = 2000 kpa Turbine Inlet Temperature = 370 C Turbine Outlet Pressure = 14 kpa Condenser Outlet State = Sat. Liquid Expansion Efficiency = 85% Pump Efficiency = 65% Brayton Cycle Compressor Inlet Pressure = 101 kpa (All) Compressor Inlet Temperature = 25 C Combustion/Reheat Exit Temp. = 1000 C Expansion Efficiency (all turbines) = 87% Compression Efficiency = 85% Regenerator Efficiency = 80% The given values for the costs include: Initial Costs: Gas Compressor: $50/kW of required power input Gas Turbine: $45/kW of gross power output Gas Combustor: $2 per lbm/hr of air flow (note units here!) Gas Reheat Combustor: $2 per lbm/hr of air flow Gas Regenerator: $2 per lbm/hr of air flow being heated Waste Heat Steam Generator: $5 per lbm/hr of steam flow through the generator Steam Turbine: $40/kW of gross power output Steam Condenser: $6 per lbm/hr of steam flow Pump: $150/kW of required power input Operating Costs: The power plant will operate 24 hours/day for 10 years. To take leap years into account, assume that there are days/year. Maintenance Costs: Each year, each piece of equipment will require approximately 10% its initial cost for parts and labor. Fuel Costs: Fuel costs $4/million Btus of heat transferred into the boilers in the Brayton cycle. Page 5 of 32

6 DESCRIPTION OF ANALYSIS To complete the analysis, the air-standard assumption was used. Air was treated as an ideal gas without constant specific heats. For all three Brayton configurations, the known state points were first established and the enthalpy (h) values and standard entropy (s 0 ) values were found using the Ideal Gas Properties of Air Table from Fundamentals of Engineering Thermodynamics. For the Rankine cycles in the second and third configuration, the known state points were established and the h and entropy (s) values were found from the International Association for the Properties of Water and Steam. For all three configurations, the following assumptions and equations were then used to find h and s 0 or s values for the unknown state points in each configuration: 1. Pressure stays constant across an intercooler, combustor, regenerator, condenser, and waste heat steam generator. 2. η t = h a h b h a h bi Where: η t = turbine/expansion efficiency h a = enthalpy of the state point before the turbine h bi = ideal enthalpy of the state point after the turbine h b = actual enthalpy of the state point after the turbine 3. η c = h a h bi h a h b Where: η c = pump/compression efficiency 4. For the Brayton cycle (air), ideal enthalpy is solely dependent on ideal standard entropy. Ideal standard entropy was found by treating the compressors and turbines as ideal and isentropic. Isentropic ideal gases adhere to the following equation: 0 = s b 0 s a 0 Rln ( P b P a ) Where: s b 0 = ideal standard entropy of the state point after the compressor or turbine s a 0 = standard entropy of the state point before the compressor or turbine R = ideal gas constant for air P b = pressure of the state point after the compressor or turbine P a = pressure of the state point before the compressor or turbine Page 6 of 32

7 For the compressor, P b P a = r c Where: r c = the compressor pressure ratio For the turbine, P b P a = 1 r t Where: r t = the turbine pressure ratio Once s b 0 was calculated, h bi could be found from the ideal gas properties air table. 5. For the Rankine cycle (steam), ideal enthalpy is dependent on two known characteristics of a state point: ideal entropy and temperature, pressure, or quality. Ideal entropy is found by treating the compressors and turbines as ideal and isentropic: s bi = s a Where: s bi = ideal entropy of the state point after the compressor or turbine s a = entropy of the state point before the compressor turbine Based on s bi and another known value, h bi could be found from the steam tables. For the first configuration, the following additional assumptions and equations were used: 1. The pressure ratio across each device was identical: r c1 = r c2 = r t1 = r t2 = r 2. The mass flow rate of the air (m b) could be determined by the desired electrical power production of the facility. W net = 80,000 kw W c1 + W c2 + W t1 + W t2 = 80,000 m b[(h 1 h 2 ) + (h 3 h 4 ) + (h 6 h 7 ) + (h 8 h 9 )] = 80,000 80,000 m b = (h 1 h 2 ) + (h 3 h 4 ) + (h 6 h 7 ) + (h 8 h 9 ) For the second configuration, the following set of assumptions and equations were used: 1. The work from the first turbine must equal the work from the second turbine. m brayton(h 3 h 4 ) = m brayton(h 5 h 6 ) h 3 h 4 = h 5 h 6 Page 7 of 32

8 Because all combustors and reheaters have the same exit temperature, the turbine inlet temperatures are identical and therefore have the same enthalpy and entropy. h 3 = h 5 h 3 h 4 = h 3 h 6 h 4 = h 6 If the exit enthalpies are identical, the turbine outlet temperatures must be the same as well. Therefore, the standard entropies of the outlets must be the same. s 0 4 s 0 3 Rln ( 1 ) = s 0 r 6 s 0 5 Rln ( 1 ) t1 r t2 s = s 5 s = s 6 r t1 = r t2 = r t 2. Because P 1 = P 7 = 101 kpa, the ratio by which the air is compressed must be equal to the ratio by which the air is expanded. r c = r turbine total = r t 2 r t = r c Therefore, the second configuration only had one variable that could be altered: r c. 3. The mass flow rate of the steam (m r) could be determined using the mass flow rate of the air by performing an energy balance of the waste heat steam generator. Q in Rankine = Q out Brayton m r(h 8 h 11 ) = m b(h 6 h 7 ) m r = h 6 h 7 h 8 h 11 m b 4. The mass flow rate of the air could be determined by the desired electrical power production of the facility. W net = 80,000 kw W c + W t1 + W t2 + W p + W t3 = 80,000 m b[(h 1 h 2 ) + (h 3 h 4 ) + (h 5 h 6 )] + m r[(h 8 h 9 ) + (h 10 h 11 )] = 80,000 m b [(h 1 h 2 ) + (h 3 h 4 ) + (h 5 h 6 ) + ( h 6 h 7 ) [(h h 8 h 8 h 9 ) + (h 10 h 11 )]] 11 = 80,000 80,000 m b = (h 1 h 2 ) + (h 3 h 4 ) + (h 5 h 6 ) + ( h 6 h 7 h 8 h 11 ) [(h 8 h 9 ) + (h 10 h 11 )] Once m b was found, m r was calculated. Page 8 of 32

9 For the third configuration, the following set of assumptions and equations were used: 1. The work from the first turbine must equal the work required by the compressor. m brayton(h 3 h 4 ) = m brayton(h 2 h 1 ) h 3 h 4 = h 2 h 1 h 4 = h 1 h 2 + h 3 While h 1 and h 3 were predetermined from known temperatures, h 2 varied with r c. Once h 4 was calculated, h 4i could be found using the turbine efficiency equation. Then, s 4 0 was found using the ideal air table and h 4i. Using the ideal gas isentropic equation and knowing s 3 0 and s 4 0, it was established that r t1 was a set value that could not vary independently: h 4i = h 3 h 3 h 4 η t h 4i s 4i 0 = s 0 4 s 0 3 Rln ( 1 ) r r t1 = e (s 4i s 3 ) R t1 2. Like the second configuration, P 1 = P 7 = 101 kpa and the ratio by which the air is compressed must be equal to the ratio by which the air is expanded. r c = r turbine total = r t1 r t2 r t2 = r c r c = r t1 e (s 4i s 3 ) R Therefore, the third configuration had only one variable that could be altered: r c. 3. The mass flow rate of the steam (m r) could be determined using the mass flow rate of the air by performing an energy balance of the waste heat steam generator. Q in Rankine = Q out Brayton m r(h 8 h 11 ) = m b(h 6 h 7 ) m r = h 6 h 7 h 8 h 11 m b 4. The mass flow rate of the air could be determined by the desired electrical power production of the facility. W net = 80,000 kw W c + W t1 + W t2 + W p + W t3 = 80,000 m b[(h 1 h 2 ) + (h 3 h 4 ) + (h 5 h 6 )] + m r[(h 8 h 9 ) + (h 10 h 11 )] = 80,000 Page 9 of 32

10 m b [(h 1 h 2 ) + (h 3 h 4 ) + (h 5 h 6 ) + ( h 6 h 7 ) [(h h 8 h 8 h 9 ) + (h 10 h 11 )]] 11 = 80,000 80,000 m b = (h 1 h 2 ) + (h 3 h 4 ) + (h 5 h 6 ) + ( h 6 h 7 h 8 h 11 ) [(h 8 h 9 ) + (h 10 h 11 )] Once m b was found, m r was calculated. For the first configuration, the following equations were used to calculate the total cost over 10 years as well as the total cost per kw produced. First, the initial costs of each piece of equipment were calculated. Note that absolute values of the change in enthalpies were used because a loss in enthalpy will still correspond to a positive cost. Initial Cost gas compressors = ( $50 kw ) (m kg b s ) ( h 1 h 2 + h 3 h 4 kj kg ) Initial Cost gas turbines = ( $45 kw ) (m b Initial Cost gas combustors = (2 combustors) ( $2 ) (m b lbm hr Initial Cost regenerator = ( $2 ) (m b lbm hr kg s ) ( h 6 h 7 + h 8 h 9 kj kg ) kg s kg s ) (3600 s hr ) (3600 s hr lbm ) ( ) kg lbm ) ( ) kg Total Initial Cost = Initial Costs = Initial Cost gas compressors + Initial Cost gas turbines + Initial Cost gas combustors + Initial Cost regenerator Next, the yearly maintenance costs were calculated: Yearly Maintenance Costs = (Total Initial Cost)(10%) Then, the yearly fuel costs were calculated: Yearly Fuel Costs = $4 ( 10 6 Btus ) ( 1 Btu kj ) (m kg b s ) ( h 6 h 5 + h 8 h 7 kj s 24 hr day ) (3600 ) ( ) ( ) kg hr day year Page 10 of 32

11 The total cost of the first configuration over the 10-year period could then be calculated with the following equation: Total Cost = (Total Initial Cost) + (10 years)(yearly Maintenance Costs + Yearly Fuel Costs) For the second and third configurations, the following equations were used to calculate the total cost over 10 years as well as the total cost per kw produced. First, the initial costs of each piece of equipment were calculated. Initial Cost gas compressor = ( $50 kw ) (m kg b s ) ( h 1 h 2 kj kg ) Initial Cost gas turbines = ( $45 kw ) (m b Initial Cost gas combustors = (2 combustors) ( $2 ) (m b lbm hr Initial Cost steam turbine = ( $40 kw ) (m r Initial Cost pump = ( $150 kw ) (m r Initial Cost steam condenser = ( $6 ) (m r lbm hr kg s ) ( h 3 h 4 + h 5 h 6 kj kg ) kg s ) (3600 s hr kg s ) ( h 8 h 9 kj kg ) kg s ) ( h 10 h 11 kj kg ) kg s ) (3600 s hr Initial Cost waste heat steam generator = ( $5 ) (m r lbm hr kg s lbm ) ( ) kg lbm ) ( ) kg ) (3600 s hr Total Initial Cost = Initial Costs = Initial Cost gas compressor + Initial Cost gas turbines + Initial Cost gas combustors + Initial Cost steam turbine + Initial Cost pump + Initial Cost steam condenser + Initial Cost waste heat steam generator Next, the yearly maintenance costs were calculated: Yearly Maintenance Costs = (Total Initial Cost)(10%) lbm ) ( ) kg Page 11 of 32

12 Then, the yearly fuel costs were calculated: Yearly Fuel Costs = $4 ( 10 6 Btus ) ( 1 Btu kj ) (m kg b s ) ( h 3 h 2 + h 5 h 4 kj s 24 hr day ) (3600 ) ( ) ( ) kg hr day year The total cost of the first configuration over the 10-year period could then be calculated with the following equation: Total Cost = (Total Initial Cost) + (10 years)(yearly Maintenance Costs + Yearly Fuel Costs) Using MATLAB, the total cost of each configuration was plotted as a function of the changing pressure ratio to find the minimum total cost per kw of produced power and the optimum compressor pressure ratio that resulted in the minimum total cost. RESULTS AND RECOMMENDATIONS For the first configuration with only a Brayton cycle, both the 10-year cost divided by the desired 80,000 kw and the price per kw hour were plotted against an increasing pressure ratio for each device. Figure 4 Pressure Ratio vs. Cost ($/kw) for the First Cycle Configuration Page 12 of 32

13 Figure 5 Pressure Ratio vs. Price per kw hr ($/(kw hr)) for the First Cycle Configuration The optimum device compressor pressure ratio was found to be 3.40 for a total pressure ratio of Table 1 Optimum Values for the First Cycle Configuration Results for Configuration 1 Pressure Ratio: 3.40 Mass Flow Rate: (kg/s) Cycle Efficiency: 47.85% Optimized Cost: $239,301, Optimized Cost per KW: $2, Optimized Cost per KW HR: $ For the second configuration with a single-shaft Brayton cycle combined with a Rankine cycle, both the 10-year cost divided by the desired 80,000 kw and the price per kw hour were plotted against an increasing pressure ratio for each device. Page 13 of 32

14 Figure 6 Pressure Ratio vs. Cost ($/kw) for the Second Cycle Configuration Figure 7 Pressure Ratio vs. Price per kw hr ($/(kw hr)) for the Second Cycle Configuration Page 14 of 32

15 The optimum compressor pressure ratio was found to be Table 2 Optimum Values for the Second Cycle Configuration Results for Configuration 2 Compressor Pressure Ratio: Turbine Pressure Ratio: _ Cycle Efficiency: Brayton Efficiency: Rankine Efficiency: Carnot Efficiency: Brayton Mass Flow Rate: Rankine Mass Flow Rate: (kg/s) (kg/s) Optimized Cost: $235,190, Optimized Cost Per Kilowatt: $2, Optimized Cost Per KW HR: $ Net Power From Brayton Cycle: KW (67.63%) Net Power From Rankine Cycle: KW (32.37%) For the third configuration with a split-shaft Brayton cycle combined with a Rankine cycle, both the 10-year cost divided by the desired 80,000 kw and the price per kw hour were plotted against an increasing pressure ratio for each device. Page 15 of 32

16 Figure 8 Pressure Ratio vs. Cost ($/kw) for the Third Cycle Configuration Figure 9 Pressure Ratio vs. Price per kw hr ($/(kw hr)) for the Third Cycle Configuration Page 16 of 32

17 The optimum compressor pressure ratio was found to be Table 3 Optimum Values for the Third Cycle Configuration Results for Configuration 3 Compressor Pressure Ratio: Turbine Pressure Ratio: _ Cycle Efficiency: Brayton Efficiency: Rankine Efficiency: Carnot Efficiency: Brayton Mass Flow Rate: Rankine Mass Flow Rate: (kg/s) (kg/s) Optimized Cost: $239,732, Optimized Cost Per KW HR: $ Optimized Cost per KW: $2, Net Power From Brayton Cycle: KW (59.93%) Net Power From Rankine Cycle: KW (40.07%) As a result of these findings, we recommend the second configuration of a single-shaft Brayton cycle combined with a Rankine cycle. Configuration two has the cheapest 10-year cycle (it is $4,110, less than the next cheapest cycle) while simultaneously boasting the highest overall cycle efficiency. While configuration two uses a 49.29% higher compressor pressure ratio than configuration three with a split-shaft Brayton cycle, the maximum pressure is only kpa, well within the safe pressure limits of the machinery. This configuration will be able to provide power at a competitive price. At the optimum compressor pressure ratio, the second configuration can provide power at $ /kW hr. According to the U.S. Energy Information Administration s most recent Electricity Monthly Update from September 2014, the average national retail cost was $0.1080/kW hr. The average national residential cost was $0.1294/kW hr. In Pennsylvania, the average residential cost was even higher at $0.1349/kW hr. Page 17 of 32

18 CONCLUSION While this analysis provided a reasonable comparison among the three configurations, in reality, all three cost analyses are idealized and neglect a number of additional factors. First, by eliminating the use of the air standard assumption, the cost per kw could be more accurately predicted. Since atmospheric air is primarily made of ideal gasses, the air standard assumption could be used to find enthalpy and entropy values. However, air also contains water vapor, which is not accounted for in the assumption. Next, the cost analysis neglected the cost for piping. It also assumed that all machinery will maintain its initial efficient operation, which is not realistic over a 10-year period. Last, inflation and other economic factors were neglected. The cost analysis was also limited by the machinery available. While higher power machinery will be more expensive, it could ultimately allow for a cheaper configuration due to increased efficiency. Additionally, although we successfully found a minimum total cost and optimum pressure ratio for all three configurations, it would be beneficial to investigate the spikes in cost found in the cost vs. pressure ratio graph to understand their origin. Page 18 of 32

19 APPENDIX References "Electricity Monthly Update With Data for September 2014." U.S. Energy Information Administration (EIA). 25 Nov Web. 28 Nov < Holmgen, Magnus (1999) XSteam [MATLAB Function] Available at < (Accessed November 25 th, 2014) Johnson s (Revised) Universal Cyclopedia Scientific and Popular Treasury of Useful Knowledge. (1886). A.J. Johnson & Company. Macquorn, William John (2008). Rankine. Complete Dictionary of Scientific Biography. Miller, Jared (2009) IdealAir [MATLAB Function]. Available at < (Accessed November 25 th, 2014) Page 19 of 32

20 Matlab Code %% Cycle One % These variables determine the start and end % points for the pressure ratio r to be varied. startpoint = 1; endpoint = 10; step = 0.1; % initializes the indexing variable for storing % the data generated by the loop i = 1; for r =startpoint :step: endpoint % Pressures (KPa) P1 = 101; P2 = P1 * r; P2_ideal = P2; P3 = P2; P4 = P3 * r; P4_ideal = P4; P5 = P4; P6 = P4; P7 = P6 / r; P7_ideal = P7; P8 = P7; P9 = P1; P9_ideal = P9; P10 = P1; % Known Temperatures (K) T1 = 298; T3 = T1; T6 = 1273; T8 = T6; % Known Enthalpies and Standard Entropies S1 = IdealAir(T1,'T','so'); H1 = IdealAir(T1,'T','h'); S3 = IdealAir(T3,'T','so'); H3 = IdealAir(T3,'T','h'); S6 = IdealAir(T6,'T','so'); H6 = IdealAir(T6,'T','h'); S8 = IdealAir(T8,'T','so'); H8 = IdealAir(T8,'T','h'); % Finding Standard Entropies of Ideal States R = ; S2_ideal = S1 + R * log(p2/p1); S4_ideal = S3 + R * log(p4/p3); S7_ideal = S6 + R * log(p7/p6); S9_ideal = S8 + R * log(p9/p8); % Finding ideal enthalpies by using IdealAir % to look up thermodynamic properties of air, based on the ideal entropies found previously. h = IdealAir([S2_ideal,S4_ideal,S7_ideal,S9_ideal],'so ','h'); H2_ideal = h(1); H4_ideal = h(2); H7_ideal = h(3); H9_ideal = h(4); % Finding actual enthalpies compressor_efficiency = 0.85; turbine_efficiency = 0.87; regeneration_efficiency = 0.80; % For the statepoints after the % compressors H2 = H1 - (H1 - H2_ideal) / compressor_efficiency; H4 = H3 - (H3 - H4_ideal) / compressor_efficiency; % For the statepoints after the turbines H7 = H6 - turbine_efficiency * (H6 H7_ideal); H9 = H8 - turbine_efficiency * (H8 H9_ideal); % For the statepoint after the regenerator H5 = regeneration_efficiency * (H9-H4) + H4; Page 20 of 32

21 Matlab Code % Finding Mass Flow Rate: % desired_output = m * (Wt1 + Wt2 - Wc1 - % Wc2); desired_output = 80 * 10^3; % KW m = desired_output / (H6 - H7 + H8 - H9 + H1 - H2 + H3 - H4); % kg/s % Finding H10 using a basic energy balance H10 = H4 - H5 + H9; % Populates the Pressure_Ratio matrix Pressure_Ratio(i) = r; % Cost Equations % Initial Costs % Compressor a = 50; Wc1 = m * abs(h1 - H2); % KW into compressor 1 Wc2 = m * abs(h3 - H4); % KW into compressor 2 ic_compressors = a * (Wc1 + Wc2); %Turbine b = 45; Wt1 = abs( m * (H6 - H7) ); % KW out of T1 Wt2 = abs( m * (H8 - H9) ); % KW out of T2 ic_turbines = b * (Wt1 + Wt2); % Air Flow Costs ic_airflow = 3 * ( 2 * (m * 3600 * )); initial_costs = ic_compressors + ic_turbines + ic_airflow; % Maintenance Costs Per Year Maintenance = (0.10) * initial_costs; % Fuel Costs Per Year Qin1 = abs( m * (H5 - H6) ); %KW into C1 Qin2 = abs( m * (H7 - H8) ); %KW into C2 Qin = Qin1 + Qin2; Fuel_Cost = Qin * 1000 * * 24 * 3600 * (1/ ) * (4 / 10 6 ); % Total Cost for 10 years Total_Cost(i) = initial_costs + 10 * (Maintenance + Fuel_Cost); % Finding the rest of the state points, for % documentation T2 = IdealAir(H2,'h','T'); S2 = IdealAir(H2,'h','so'); T4 = IdealAir(H4,'h','T'); S4 = IdealAir(H4,'h','so'); T5 = IdealAir(H5,'h','T'); S5 = IdealAir(H5,'h','so'); T7 = IdealAir(H7,'h','T'); S7 = IdealAir(H7,'h','so'); T9 = IdealAir(H9,'h','T'); S9 = IdealAir(H9,'h','so'); T10 = IdealAir(H10,'h','T'); S10 = IdealAir(H10,'h','so'); % This if statement stores the mass flow rate, % optimized pressure ratio, and statepoint % data for the minimized cost. if Total_Cost(i) == min(total_cost) ideal_m = m; Ideal_Pressure_Ratio = Pressure_Ratio(i); Minimized_Cost = Total_Cost(i); Ideal_Cycle_Efficiency = desired_output / Qin * 100; T = [T1 T2 T3 T4 T5 T6 T7 T8 T9 T10]; H = [H1 H2 H3 H4 H5 H6 H7 H8 H9 H10]; S = [S1 S2 S3 S4 S5 S6 S7 S8 S9 S10]; end Page 21 of 32

22 Matlab Code % These logical constructs show a plot of the % cost per kilowatt over the life of the plant % as they update in real time based on the % progress of the for loop. A horizontal % black dotted line is created at the minimum % cost piont, as well as a stem plot at the % minimum to clearly identify it. These also % update in real time with the loop. if i == 2 % Plotting our results figs(1)=figure('windowstyle','docked','name', 'Cycle 1 Pressure Ratio Versus Cost/KW','NumberTitle','off'); h1= plot(pressure_ratio(2),total_cost(2)/desired_output, 'LineWidth',2); hold on; xlabel('pressure Ratio','FontSize',20) ylabel('total Cost ($/KW)','FontSize',20) title('pressure Ratio Versus Cost','FontSize',22) set(gca,'fontsize',15) axis([ 0 endpoint ]) h2 = stem(ideal_pressure_ratio,minimized_cost/desired_output, 'color','r','linewidt h',2); h3 = line( [ 0 endpoint ], Minimized_Cost/desired_output * ones(1,2),'linestyle','--','color','k'); h4 = plot(pressure_ratio(2),total_cost(2)/desired_outpu t,'*','linewidth',2,'markerfacecolor','y','markere dgecolor','r'); tilefigs elseif i>2 set(h1,'xdata',pressure_ratio(2:end),'ydata', Total_Cost(2:end)/desired_output) set(h2,'xdata',ideal_pressure_ratio,'ydata', Minimized_Cost/desired_output) set(h3,'ydata', Minimized_Cost/desired_output * ones(1,2)) set(h4,'xdata',pressure_ratio(end),'ydata', Total_Cost(end)/desired_output) End pause(0.01) % for lack of better words, without % this pause the script trips over % itself. i = i + 1; end % Finding the minimum cost and ideal pressure ratio [Minimized_Cost,index] = min(total_cost); Ideal_Pressure_Ratio = Pressure_Ratio(index); % Converting the money into a readable format dollars = commas(floor(minimized_cost)); cents = Minimized_Cost - floor(minimized_cost); cents = ceil(cents * 100); money = ['$',dollars,'.',num2str(cents)]; % Converting the price per KW into a readable format perkw = Minimized_Cost / (desired_output); dollars = commas(floor(perkw)); cents = perkw - floor(perkw); cents = ceil(cents * 100); perkw = ['$',dollars,'.',num2str(cents)]; % Finding price per kilowatt hour kwhr = desired_output * 10 * * 24; priceperkwhr = Minimized_Cost / kwhr ; Page 22 of 32

23 Matlab Code % Converting the mass flow rate into a readable format n = commas(floor(ideal_m)); d = ideal_m - floor(ideal_m); dec = ceil(d * 100); flow = [n,'.',num2str(dec)]; % Display the results disp(' ') cprintf('_black',' Results for Cycle 1 \n') cprintf('black',[' Ideal Pressure Ratio: ',num2str(ideal_pressure_ratio),'\n']) cprintf('black',[' Mass Flow Rate: ',flow,' (kg/s)\n']) cprintf('black',[' Ideal Cycle Efficiency: ',num2str(ideal_cycle_efficiency),' %% \n\n']) cprintf('black',[' Optimized Cost: ',money,'\n']) cprintf('black',[' Optimized Cost per KW: ',perkw,'\n']) cprintf('black',['optimized Cost per KW HR: $',num2str(priceperkwhr),'\n']) disp(' ') % Plotting Price Vs. Compressor Pressure Ratio figs(2)=figure('windowstyle','docked','name', 'Cycle 1 - Pressure Ratio vs Cost Per KW-HR','NumberTitle','off'); plot(pressure_ratio,total_cost/(desired_output * 10 * * 24),'LineWidth',2); hold on; ax(2) = gca; xlabel('pressure Ratio','FontSize',20) ylabel('price per KW-HR ($)','FontSize',20) title('pressure Ratio Versus Cost','FontSize',22) set(gca,'fontsize',15) axis([ 0 endpoint 0.05]) stem(ideal_pressure_ratio,minimized_cost/( desired_output * 10 * * 24),'color','r','LineWidth',2); line( [ 0 endpoint],minimized_cost/(desired_output * 10 * * 24) * ones(1,2), 'LineStyle','--','Color','k'); clear all; %% Cycle Two % RANKINE PORTION OF THE CYCLE % Pressures P8 = 20; % BAR P9 =.14; % BAR P10 = P9; % BAR P11 = P8; % BAR % Known Temperatures T8 = 370; % deg C % Enthalpy and Entropy for State 8 as a % function of Pressure and Temperature H8 = XSteam('h_pT',P8,T8); S8 = XSteam('s_pT',P8,T8); % Ideal Entropy for State 9 S9_ideal = S8; % Finding ideal enthalpy at State 9 H9_ideal = XSteam('h_ps',P9,S9_ideal); % Finding H9 and subsequently T9 rankine_turbine_efficiency = 0.85; H9 = H8 - rankine_turbine_efficiency * (H8 - H9_ideal); T9 = XSteam('T_ph',P9,H9); % deg C S9 = XSteam('s_ph',P9,H9); Page 23 of 32

24 Matlab Code % Finding enthalpy and entropy for State 10 and subsequently T10 % At state 10, x = 0; H10 = XSteam('hL_p',P10); S10 = XSteam('sL_p',P10); T10 = XSteam('T_ph',P10,H10); % deg C % Finding ideal entropy for State 11 S11_ideal = S10; % Finding ideal enthalpy for State 11 H11_ideal = XSteam('h_ps',P11,S11_ideal); % Finding actual enthalpy for State 11 and % subsequently T11 rankine_pump_efficiency = 0.65; H11 = H10 + (H11_ideal - H10) / rankine_pump_efficiency; T11 = XSteam('T_ph',P11,H11); % C S11 = XSteam('s_ph',P11,H11); % Determines the start and end points for the % compressor pressure ratio as well as the turbine % pressure ratio and sets the steps to increase % by. Every combination of compressor pressure % ratio and turbine pressure ratio between the % start and end points will be analyzed for a % minimum cost. step =.1; startpointc = 1; endpointc = 30; % initializes the indexing variable to store the % data generated by the for loops. i = 1; for compressor_pressure_ratio = startpointc :step: endpointc % Known Temperatures T1 = 298; % K T3 = 1273; % K T5 = T3; % K T7 = T ;% K % Known Enthalpies and Standard Entropies S1 = IdealAir(T1,'T','so'); H1 = IdealAir(T1,'T','h'); S3 = IdealAir(T3,'T','so'); H3 = IdealAir(T3,'T','h'); S5 = IdealAir(T5,'T','so'); H5 = IdealAir(T5,'T','h'); S7 = IdealAir(T7,'T','so'); H7 = IdealAir(T7,'T','h'); % Define the turbine pressure ratio so P7 = P1; turbine_pressure_ratio = sqrt(compressor_pressure_ratio); % Finding Standard Entropies of Ideal States R = ; S2_ideal = S1 + R * log(compressor_pressure_ratio); S4_ideal = S3 + R * log(1 / turbine_pressure_ratio); S6_ideal = S5 + R * log(1 / turbine_pressure_ratio); % Finding ideal enthalpies h=idealair([s2_ideal,s4_ideal,s6_ideal],'so','h'); H2_ideal = h(1); H4_ideal = h(2); H6_ideal = h(3); % BRAYTON PORTION OF CYCLE Page 24 of 32

25 Matlab Code % Finding actual enthalpies brayton_compressor_efficiency = 0.85; brayton_turbine_efficiency = 0.87; H2 = H1 - (H1 - H2_ideal) / brayton_compressor_efficiency; T2 = IdealAir(H2,'h','T'); S2 = IdealAir(H2,'h','so'); H4 = H3 - brayton_turbine_efficiency * (H3 H4_ideal); T4 = IdealAir(H4,'h','T'); S4 = IdealAir(H4,'h','so'); H6 = H5 - brayton_turbine_efficiency * (H5 H6_ideal); T6 = IdealAir(H6,'h','T'); S6 = IdealAir(H6,'h','so'); % Determining Mass Flow Rates from the net work % equation and the energy balance of the waste % heat generator desired_output = 80 * 10^3; % KW waste_heat_generator_efficiency = 1 ; constant = H1 - H2 + H3 - H4 + H5 - H6 + waste_heat_generator_efficiency * ((H6-H7)/(H8-H11))* (H8-H9+H10-H11); m_brayton = desired_output / constant; m_rankine = ( (H6 - H7)/(H8-H11) ) * m_brayton * waste_heat_generator_efficiency; % Populating the pressure ratio matrices cpr_matrix(i) = compressor_pressure_ratio; % Cost Equation % Initial Costs % Rankine Turbine c = 40; Wt3 = m_rankine * abs(h8 - H9); % KW out of T1 ic_rankine_turbine = c * Wt3; % Rankine Pump d = 150; Wp = m_rankine * abs(h10 - H11); % KW into pump ic_rankine_pump = d * Wp; % Waste Heat Steam Generator Air Flow Cost ic_waste_heat_airflow = 5 * (m_rankine * 3600 * ); % Condenser Air Flow Cost ic_rankine_condenser_airflow = 6 * (m_rankine * 3600 * ); Rankine_Initial_Costs = ic_rankine_turbine + ic_rankine_pump + ic_waste_heat_airflow + ic_rankine_condenser_airflow; Rankine_Power = Wt3 - Wp; % Brayton Compressor a = 50; Wc = m_brayton * abs(h1 - H2); % KW into comp. ic_brayton_compressors = a * Wc; % Brayton Turbines b = 45; Wt1 = m_brayton * abs(h3 - H4); % KW out of T1 Wt2 = m_brayton * abs(h5 - H6); % KW out of T2 ic_brayton_turbines = b * (Wt1 + Wt2); % Brayton Air Flow Costs ic_brayton_airflow = 2 * ( 2 * (m_brayton * 3600 * )); Brayton_Initial_Costs = ic_brayton_compressors + ic_brayton_turbines + ic_brayton_airflow; Brayton_Power = Wt1 + Wt2 - Wc; Page 25 of 32

26 Matlab Code % Determines the total inital costs initial_costs = Rankine_Initial_Costs + Brayton_Initial_Costs; % Maintenance Costs Per Year Maintenance = (0.10) * initial_costs; % Fuel Costs Per Year Qin1 = m_brayton * abs(h2 - H3); %KW into C1 Qin2 = m_brayton * abs(h4 - H5); %KW into C2 Qin = Qin1 + Qin2; Fuel_Cost = Qin * 1000 * * 24 * 3600 * (1/ ) * (4 / 10^6); % Total Cost for 10 years Total_Cost(i) = initial_costs + 10 * (Maintenance + Fuel_Cost); % Calculates the efficiencies of each cycle % overall efficiency cycle_efficiency(i) = desired_output / Qin * 100; % brayton effiency brayton_efficiency(i) = (Wt1 + Wt2 - Wc) / Qin * 100; % rankine efficiency heat_transferred = m_rankine * abs(h11 - H8); rankine_efficiency(i) = (Wt3 - Wp)/ heat_transferred * 100; % carnot efficiency T = [T1 T2 T3 T4 T5 T6 T7 T8 T9 T10 T11]; carnot = (1 - min(t)/max(t)) * 100; if Total_Cost(i) == min(total_cost) ideal_m_brayton = m_brayton; ideal_m_rankine = m_rankine; Ideal_Compressor_Ratio = compressor_pressure_ratio; Ideal_Turbine_Ratio = turbine_pressure_ratio; Ideal_Cycle_Efficiency = cycle_efficiency(i); Ideal_Carnot_Efficiency = carnot; Ideal_Brayton_Efficiency = brayton_efficiency(i); Ideal_Rankine_Efficiency = rankine_efficiency(i); Minimized_Cost = Total_Cost(i); ideal_t = [T1 T2 T3 T4 T5 T6 T7 T8+273 T9+273 T T11+273]; ideal_s = [S1 S2 S3 S4 S5 S6 S7 S8 S9 S10 S11]; ideal_h = [H1 H2 H3 H4 H5 H6 H7 H8 H9 H10 H11]; end % These graphs update in real time as the for % index. if i == 2 % Plotting Price Vs. Compressor Pressure Ratio figs(3)=figure('windowstyle','docked','name', 'Cycle 2 - CPR vs Cost Per KW','NumberTitle','off'); h5 = plot(cpr_matrix(2),total_cost(2)/desired_output, 'LineWidth',2); hold on; ax(2) = gca; xlabel('compressor Pressure Ratio','FontSize',20) ylabel('total Cost ($/KW)','FontSize',20) title('compressor Pressure Ratio Versus Cost','FontSize',22) set(gca,'fontsize',15) axis([ 0 endpointc 0 max(total_cost)/desired_output]) h6 = stem(ideal_compressor_ratio,minimized_cost/ desired_output,'color','r','linewidth',2); Page 26 of 32

27 Matlab Code h7 = line( [ 0 endpointc], Minimized_Cost/desired_output * ones(1,2),'linestyle','--','color','k'); h8 = plot(cpr_matrix(2),total_cost(2)/desired_output, '*','LineWidth',2,'MarkerFaceColor','y', 'MarkerEdgeColor','r'); % makes the figure windows equally sized Tilefigs(figs) elseif i>2 set(h5,'xdata',cpr_matrix,'ydata',total_cost/ desired_output) set(h6,'xdata',ideal_compressor_ratio,'ydata', Minimized_Cost/desired_output) set(h7,'ydata',minimized_cost/desired_output * ones(1,2)) set(h8,'xdata',cpr_matrix(end),'ydata', Total_Cost(end)/desired_output) end pause(.001) i = i + 1; end % Converting the money into a readable format dollars = commas(floor(minimized_cost)); cents = Minimized_Cost - floor(minimized_cost); cents = ceil(cents * 100); money = ['$',dollars,'.',num2str(cents)]; % Converting the price per KW into a readable format perkw = Minimized_Cost / desired_output; dollars = commas(floor(perkw)); cents = perkw - floor(perkw); cents = ceil(cents * 100); perkw = ['$',dollars,'.',num2str(cents)]; % Finding price per kilowatt hour kwhr = desired_output * 10 * * 24; priceperkwhr = Minimized_Cost / kwhr ; % Display the results disp(' ') cprintf('_black',' Results for Cycle 2 \n') cprintf('black',[' Ideal Compressor Pressure Ratio: ', num2str(ideal_compressor_ratio),'\n']) cprintf('black',[' Ideal Turbine Pressure Ratio: ', num2str(ideal_turbine_ratio),'\n\n']) cprintf('black',[' Ideal Cycle Efficiency: ', num2str(ideal_cycle_efficiency),'\n']) cprintf('black',[' Ideal Brayton Efficiency: ', num2str(ideal_brayton_efficiency),'\n']) cprintf('black',[' Ideal Rankine Efficiency: ', num2str(ideal_rankine_efficiency),'\n']) cprintf('black',[' Carnot Efficiency: ', num2str(ideal_carnot_efficiency),'\n\n']) cprintf('black',[' Ideal Brayton Mass Flow Rate: ', num2str(ideal_m_brayton),' (kg/s)\n']) cprintf('black',[' Ideal Rankine Mass Flow Rate: ', num2str(ideal_m_rankine), ' (kg/s)\n\n']) cprintf('black',[' Optimized Cost: ',money,'\n']) cprintf('black',[' Optimized Cost Per Kilowatt: ',perkw,'\n']) cprintf('black',[' Optimized Cost Per KW HR: $', num2str(priceperkwhr),'\n\n']) Page 27 of 32

28 Matlab Code cprintf('black',[' Net Power From Brayton Cycle: ', num2str(brayton_power,'%.2f'),' KW (', num2str(brayton_power/desired_output * 100,'%.2f'),'%%) \n']) cprintf('black',[' Net Power From Rankine Cycle: ', num2str(rankine_power,'%.2f'),' KW (', num2str(rankine_power/desired_output * 100,'%.2f'),'%%) \n']) disp(' ') % Plotting Price Vs. Compressor Pressure Ratio figs(4)=figure('windowstyle','docked','name', 'Cycle 2 - CPR Vs Cost Per KW- HR','NumberTitle','off'); plot(cpr_matrix,total_cost/(desired_output * 10 * * 24),'LineWidth',2); hold on; ax(2) = gca; xlabel('compressor Pressure Ratio','FontSize',20) ylabel('price per KW-HR ($)','FontSize',20) title('compressor Pressure Ratio Versus Cost','FontSize',22) set(gca,'fontsize',15) axis([ 0 endpointc 0.05]) stem(ideal_compressor_ratio,minimized_cost/ (desired_output * 10 * * 24),'color','r','LineWidth',2); line( [ 0 endpointc], Minimized_Cost/ (desired_output * 10 * * 24) * ones(1,2),'linestyle','--','color','k'); %% Cycle Three clear all; % RANKINE PORTION OF THE CYCLE % Pressures P8 = 20; % BAR P9 =.14; % BAR P10 = P9; % BAR P11 = P8; % BAR % Temperatures T8 = 370; % deg C % Enthalpy and Entropy for State 8 as a % function of Pressure and Temperature H8 = XSteam('h_pT',P8,T8); S8 = XSteam('s_pT',P8,T8); % Ideal Entropy for State 9 S9_ideal = S8; % Finding ideal enthalpy at State 9 H9_ideal = XSteam('h_ps',P9,S9_ideal); % Finding H9 rankine_turbine_efficiency = 0.85; H9 = H8 - rankine_turbine_efficiency * (H8 - H9_ideal); T9 = XSteam('T_ph',P9,H9); % deg C % Finding enthalpy and entropy for State 10 H10 = XSteam('hL_p',P10); S10 = XSteam('sL_p',P10); T10 = XSteam('T_ph',P10,H10); % deg C % Finding ideal entropy for State 11 S11_ideal = S10; % Finding ideal enthalpy for State 11 H11_ideal = XSteam('h_ps',P11,S11_ideal); Page 28 of 32

29 Matlab Code % Finding actual enthalpy for State 11 rankine_pump_efficiency = 0.65; H11 = H10 + (H11_ideal - H10) / rankine_pump_efficiency; % Finding the temperature at State 11 T11 = XSteam('T_ph',P11,H11); % C % BRAYTON PORTION OF CYCLE % Temperatures T1 = 298; % K T3 = 1273; % K T5 = T3; % K T7 = T ;% K % Known Enthalpies and Standard Entropies S1 = IdealAir(T1,'T','so'); H1 = IdealAir(T1,'T','h'); S3 = IdealAir(T3,'T','so'); H3 = IdealAir(T3,'T','h'); S5 = IdealAir(T5,'T','so'); H5 = IdealAir(T5,'T','h'); S7 = IdealAir(T7,'T','so'); H7 = IdealAir(T7,'T','h'); step =.1; startpointc = 1; endpointc = 30; i=1; for compressor_pressure_ratio = startpointc :step: endpointc % Finding Standard Entropies of Ideal States R = ; S2_ideal = S1 + R * log(compressor_pressure_ratio); % Finding ideal enthalpies h = IdealAir(S2_ideal,'so','h'); H2_ideal = h(1); % Finding actual enthalpies brayton_compressor_efficiency = 0.85; brayton_turbine_efficiency = 0.87; H2 = H1 - (H1 - H2_ideal) / brayton_compressor_efficiency; T2 = IdealAir(H2,'h','T'); H4 = H1 + H3 - H2; T4 = IdealAir(H4,'h','T'); S4 = IdealAir(H4,'h','so'); H4_ideal = H3 - (H3- H4)/brayton_turbine_efficiency; S4_ideal = IdealAir(H4_ideal,'h','so'); Brayton_T1_Ratio = exp( -1 * (S4_ideal-S3)/R); Brayton_T2_Ratio = compressor_pressure_ratio / Brayton_T1_Ratio; S6_ideal = S5 + R * log(1 / Brayton_T2_Ratio); H6_ideal = IdealAir(S6_ideal,'so','h'); H6 = H5 - brayton_turbine_efficiency * (H5 - H6_ideal); T6 = IdealAir(H6,'h','T'); % Create Temperature Matrix T = [T1, T2, T3, T4, T5, T6, T7, T8, T9, T10, T11]; % Determining Mass Flow Rates desired_output = 80 * 10^3; % KW waste_heat_generator_efficiency = 1 ; constant = H1 - H2 + H3 - H4 + H5 - H6 + waste_heat_generator_efficiency * ((H6-H7)/(H8-H11))* (H8-H9+H10-H11); m_brayton = desired_output / constant; m_rankine = ( (H6 - H7)/(H8-H11) ) * m_brayton * waste_heat_generator_efficiency; Page 29 of 32

30 Matlab Code % Populating the pressure ratio matrices cpr_matrix(i) = compressor_pressure_ratio; % Cost Equation % Initial Costs % Rankine Turbine c = 40; Wt3 = m_rankine * abs(h8 - H9); % KW out of T3 ic_rankine_turbine = c * Wt3; % Rankine Pump d = 150; Wp = m_rankine * abs(h10 - H11); % KW into pump ic_rankine_pump = d * Wp; % Waste Heat Steam Generator Air Flow Cost ic_waste_heat_airflow = 5 * (m_rankine * 3600 * ); % Condenser Air Flow Cost ic_rankine_condenser_airflow = 6 * (m_rankine * 3600 * ); Rankine_Initial_Costs = ic_rankine_turbine + ic_rankine_pump + ic_waste_heat_airflow + ic_rankine_condenser_airflow; Rankine_Power = Wt3 - Wp; % Brayton Compressor a = 50; Wc = m_brayton * abs((h1 - H2) ); % KW into comp ic_brayton_compressors = a * Wc; % Brayton Turbines b = 45; Wt1 = m_brayton * abs(h3 - H4); % KW out of T1 Wt2 = m_brayton * abs(h5 - H6); % KW out of T2 ic_brayton_turbines = b * (Wt1 + Wt2); % Brayton Air Flow Costs ic_brayton_airflow = 2 * ( 2 * (m_brayton * 3600 * )); Brayton_Initial_Costs = ic_brayton_compressors + ic_brayton_turbines + ic_brayton_airflow; Brayton_Power = Wt1 + Wt2 - Wc; initial_costs = Rankine_Initial_Costs + Brayton_Initial_Costs; % Maintenance Costs for one year Maintenance = (0.10) * initial_costs; % Fuel Costs for one year Qin1 = m_brayton * abs(h2 - H3); %KW into C1 Qin2 = m_brayton * abs(h4 - H5); %KW into C2 Qin = Qin1 + Qin2; Fuel_Cost = Qin * 1000 * * 24 * 3600 * (1/ ) * (4 / ); cycle_efficiency(i) = desired_output / Qin * 100; brayton_efficiency(i) = (Wt1 + Wt2 - Wc) / Qin * 100; heat_transferred = m_rankine * abs(h11 - H8); rankine_efficiency(i) = (Wt3 - Wp)/ heat_transferred * 100; % Total Cost for 10 years Total_Cost(i) = initial_costs + 10 * (Maintenance + Fuel_Cost); carnot = (1 - min(t)/max(t)) * 100; Page 30 of 32

31 Matlab Code if Total_Cost(i) == min(total_cost) ideal_m_brayton = m_brayton; ideal_m_rankine = m_rankine; Ideal_Compressor_Ratio = compressor_pressure_ratio; Ideal_Turbine_Ratio = Brayton_T2_Ratio; Ideal_Cycle_Efficiency = cycle_efficiency(i); Ideal_Brayton_Efficiency = brayton_efficiency(i); Ideal_Rankine_Efficiency = rankine_efficiency(i); Ideal_Carnot_Efficiency = carnot; Minimized_Cost = Total_Cost(i); end if i == 2 % Plotting Price Vs. Compressor Pressure Ratio figs(5)=figure('windowstyle','docked','name', 'Cycle 3 - CPR versus Cost per KW','NumberTitle','off'); h9 = plot(cpr_matrix(2),total_cost(2)/desired_output, 'LineWidth',2); hold on; ax(2) = gca; xlabel('compressor Pressure Ratio','FontSize',20) ylabel('total Cost ($/KW)','FontSize',20) title('compressor Pressure Ratio Versus Cost','FontSize',22) set(gca,'fontsize',15) axis([ 0 endpointc 0 max(total_cost)/desired_output]) h10 = stem(ideal_compressor_ratio,minimized_cost/ desired_output, 'color','r','linewidth',2); h11 = line( [ 0 endpointc], Minimized_Cost/desired_output * ones(1,2),'linestyle','--','color','k'); h12 = plot(cpr_matrix(2),total_cost(2)/desired_output, '*','LineWidth',2,'MarkerFaceColor','y', 'MarkerEdgeColor','r'); tilefigs(figs) elseif i>2 set(h9,'xdata',cpr_matrix,'ydata',total_cost/ desired_output) set(h10,'xdata',ideal_compressor_ratio,'ydata', Minimized_Cost/desired_output) set(h11,'ydata',minimized_cost/desired_output * ones(1,2)) set(h12,'xdata',cpr_matrix(end),'ydata', Total_Cost(end)/desired_output) end pause(.001) i = i + 1; end % Converting the money into a readable format dollars = commas(floor(minimized_cost)); cents = Minimized_Cost - floor(minimized_cost); cents = ceil(cents * 100); money = ['$',dollars,'.',num2str(cents)]; % Converting the price per KW into a readable format perkw = Minimized_Cost / desired_output; dollars = commas(floor(perkw)); cents = perkw - floor(perkw); cents = ceil(cents * 100); perkw = ['$',dollars,'.',num2str(cents)] Page 31 of 32

32 Matlab Code % Finding price per kilowatt hour kwhr = desired_output * 10 * * 24; priceperkwhr = Minimized_Cost / kwhr ; % Display the results disp(' ') cprintf('_black',' Results for Cycle 3 \n') cprintf('black',[' Ideal Compressor Pressure Ratio: ', num2str(ideal_compressor_ratio),'\n']) cprintf('black',[' Ideal Turbine Pressure Ratio: ', num2str(ideal_turbine_ratio),'\n\n']) cprintf('black',[' Ideal Cycle Efficiency: ', num2str(ideal_cycle_efficiency),'\n']) cprintf('black',[' Ideal Brayton Efficiency: ', num2str(ideal_brayton_efficiency),'\n']) cprintf('black',[' Ideal Rankine Efficiency: ', num2str(ideal_rankine_efficiency),'\n']) cprintf('black',[' Carnot Efficiency: ', num2str(ideal_carnot_efficiency),'\n\n']) cprintf('black',[' Ideal Brayton Mass Flow Rate: ', num2str(ideal_m_brayton),' (kg/s)\n']) cprintf('black',[' Ideal Rankine Mass Flow Rate: ', num2str(ideal_m_rankine),' (kg/s)\n\n']) cprintf('black',[' Optimized Cost: ',money,'\n']) cprintf('black',[' Optimized Cost Per KW HR: $', num2str(priceperkwhr),'\n']) cprintf('black',[' Optimized Cost per KW: ',perkw,'\n\n']) cprintf('black',[' Net Power From Brayton Cycle: ', num2str(brayton_power,'%.2f'),' KW (', num2str(brayton_power/desired_output * 100,'%.2f'),'%%) \n']) cprintf('black',[' Net Power From Rankine Cycle: ', num2str(rankine_power,'%.2f'),' KW (', num2str(rankine_power/desired_output * 100,'%.2f'),'%%) \n']) disp(' ') % Plotting Price Vs. Compressor Pressure Ratio figs(6)=figure('windowstyle','docked','name', 'Cycle 3 - CPR versus Cost per KW- HR','NumberTitle','off'); h5 = plot(cpr_matrix,total_cost/(desired_output * 10 * * 24),'LineWidth',2); hold on; ax(2) = gca; xlabel('compressor Pressure Ratio','FontSize',20) ylabel('price per KW-HR ($)','FontSize',20) title('compressor Pressure Ratio Versus Cost','FontSize',22) set(gca,'fontsize',15) axis([ 0 endpointc 0.05]) h6 = stem(ideal_compressor_ratio,minimized_cost/ (desired_output * 10 * * 24),'color','r','LineWidth',2); h7 = line( [ 0 endpointc], Minimized_Cost/(desired_output * 10 * * 24) * ones(1,2),'linestyle', '--','Color','k'); Page 32 of 32