Executive Summary

Given a cogeneration system modified with regeneration, we are tasked to set stream

pressures, temperatures and mass flow rates such that the system generates 3.5 MW of net work as

electricity and 7.7 MW of process heating while the utilization factor is at a minimum of 80%. The

equipment are also limited such that the boiler and the turbine are 85% efficient and the pumps are

70% efficient.

Using an inlet turbine stream of 5.8638 kg/s at 6000 kPa and 500 C, an intermediate stream

of 5.623567 kg/s at 400 kPa and an oulet stream of 0.2402334 kg/s at 10 kPa, we generate the

following values (Table 1 and 2)

Table 1. Pressure and Temperature Values at Each Stream

Parameter 1 2 3 4 5 6 7 8 9 10 11

Temp C 45.83

45.883 143.62 143.62 144.8156

500 188.496 188.496 188.496 45.83 143.62

Pressure (kPa)

10 400 400 400 6000 6000 400 400 400 10 400

Vapor Fraction

0 0 0 0 0 1 1 1 1 0.90867 0

Table 2. Energy Values

Energy Form Value

Wturbine -3550.986 kJ

Wpump,I 0.135179 kJ

Wpump,II 50.85087 kJ

Qboiler 16470.5882 kJ

Qcondenser -522.369744 kJ

With the obtained values, it is possible to meet the required net work, process heat and

utilization factor considering the efficiencies of the equipment while also producing an exhaust stream

with a quality of 90.867%.

Background of the Study

A steam power plant is a power plant whose working fluid is water. Water is heated in a

boiler, turns into steam, spins a steam turbine which in turn produces work, goes to the condenser to

be condensed back into liquid and this liquid is then pumped back to the boiler. This cycle is also

known as the Rankine cycle (Figure 1).

Figure 1. Rankine Cycle1

The steam power plant is used for the production of power and a lot of effort has been made

to increase its efficiency. A way to do this is to make the average temperature of the working fluid as

high as possible during heat addition1. This can be done by regeneration. In this process, steam is

extracted from intermediate points of expansion at the turbine. This steam goes to the feed water

heater and heats the feed water that comes out of the condenser before it is pumped back to the boiler2

(Figure 2). The average temperature of the feed water is increased and in turn, also increase the

thermal efficiency of the cycle.

Figure 2. A Steam Turbine Power Plant with Regeneration1

Some industries however, require heat as an energy input to their devices. This heat energy is

called process heat. The energy from the combustion of fuel is transferred to the steam and this is used

in the process heating (Figure 3).

Figure 3. Process Heating Plant1

Industries which use high amounts of process heat also use high amounts of electricity

(power). Because of this, people thought of using the work potential to produce power while

producing process heat at the same time. This process is an example of cogeneration, which is the

production of more than one useful form of energy from the same energy source. Plants which utilize

cogeneration are called cogeneration plants1.

Either a Rankine or Brayton cycle can be used as a power cycle for the cogeneration plant. In

an ideal steam turbine cogeneration plant, it can be noted that the condenser is missing (Figure 4).

Steam enters the turbine, produces work and is expanded to state 4. The flow rate is adjusted such that

the stream leaves as saturated liquid from the process heater. This liquid is then pumped back to boiler

pressure. The work in the pump is usually small and can be neglected. This implies that all the energy

transferred to the steam in the boiler is transformed into either power in the turbine or heat in the

process heater. Therefore a cogeneration plant is a process heating plant combined with a power plant

that has a thermal efficiency of 100 percent.1

Figure 4. Ideal Cogeneration Plant1

The ideal cogeneration plant is not practical because it cannot adjust to varying loads in terms

of power and process heat1. The power plant to be studied is a cogeneration plant modified with

regeneration (Figure 5). The flow rates can be adjusted according to the required loads of the system.

If the systems requires both power and process heat, part of the stream is extracted from an

intermediate point in the turbine. A part of this steam flows to the process heater to provide process

heat and the rest of the extracted steam goes to the feed water heater. The rest of the steam in the

turbine goes to the condenser and is condensed as feed water. The presence of the condenser accounts

for waste heat and it also lowers the thermal efficiency of the power plant since some of the heat from

the boiler that could have been converted to work or process heat is rejected in the condenser. This is

why the plant was modified to undergo regeneration. As explained earlier part of the steam is

extracted from the turbine to heat the feed water from the condenser. This raises the temperature of

the feed water and causes a higher thermal efficiency. The saturated liquid from the feed water heater

mixes with the saturated liquid from the process heater and are pumped back to boiler pressure. If the

system needs a high amount of process heat, all the steam passes through the process heater (no steam

passes through the condenser and the feed water heater) and no waste heat is produced. If the system

does not need process heat, no steam passes through the process heater and the plant operates like a

steam power plant with regeneration.

Figure 5. A Cogeneration Plant modified with Regeneration

This study will focus on the approximation of boiler pressure and the temperature of the

steam entering the turbine given the required loads in power and process heat while considering the

limits of the devices used in the system. It will also focus on the varying pressures at which steam is

extracted from the turbine and the varying amounts of steam extracted from the turbine.

Proposed Design

As stated earlier, the study focuses on the approximation of boiler pressure and the

temperature of the steam entering the turbine given certain conditions. The conditions specified are as

follows; the power plant must be able to produce 3.5 MW of electric power and 7.7 MW of process

heat, the turbine is to have an efficiency of 85% and the pumps 70%, the working fluid is steam, the

utilization factor (Equation 1) should not go below 80% and the heating value of methane, the natural

gas used to power the system is 35,600 kJ/m3.

ϵ u=|W net|+|Q p|

Q¿

(1)

The group started working with the boiler pressure and the temperature of the steam entering

the turbine. It should be noted that the quality of the steam leaving the turbine exhaust should not be

less than 90%. A certain condition is set and the quality of the steam at the turbine exhaust was

calculated. Once a condition meets the quality required, the group will then determine if the said

conditions are practical and feasible and if this condition is met, the group moves on with the

calculations from the turbine. The pressure at which the steam was extracted from the turbine (State 7)

was determined through trial and error and by consulting similar examples in thermodynamics books.

The pressure was assumed to be around the same pressures as the examples. From this pressure, H7 is

calculated by assuming isentropic expansion. The work in the first section of the turbine can also be

determined from this process. The method used in the determination of P7 was also used in the

determination of P10 and H10. The work in the second section of the turbine however, is calculated

using the change in enthalpy between H10 and H7 and not between H10 and H6. H6 should only be used

in the determination of H10 and not in the calculation of work since the required work is the work

produced after steam has been extracted from the turbine and not from the entrance.

Since all devices, except for the pumps and the turbine are isobaric, the determination of P6,

P7 and P10 is enough to determine the pressures in all of the streams in the power plant. Thus the work

in the pumps can be calculated. The work in the pumps is given by

W Pump=1❑ VΔP=ΔH (2)

The work in pump 1 and 2 can then be used to determine the enthalpy at stream 2 and 5

respectively. At this point the enthalpy at each stream is already known since the other enthalpies can

be taken from the steam tables.

The determination of the mass flow rate in each stream was done next. A utilization factor of

80% was assumed. Using Eq. 1, the value of Qin was determined. From this value, the energy

transferred to the steam in the boiler (Qboiler) was obtained. It should be noted that the value for Qin is

85% of the value of Qboiler. It is also the 85% of the heat generated from fuel combustion that is

transferred to the working fluid. Going back to a previous statement in the instructions, it is stated that

Qin is the energy from fuel combustion that the working fluid needs. If Qin is exactly the energy that the

working fluid needs and only 85% of the Qboiler is transferred to the working fluid, we can express this

relationship through the equation

Q¿=0.85 ×Qboiler (3)

Going back to the process, another equation for Qboiler is given by,

Qboiler=ṁ ( H 6−H 5 ) (4)

Where the total mass flow rate can be determined. The group then works with the equation for Wnet

which is given by the following equations:

Ẇ net=Ẇ turbine+Ẇ Pump 1+Ẇ Pump2 (5)

−3500=ṁW I+xṁW II+ṁW Pump2+xṁ W Pump 1 (6)

the mass fraction (x) can then be calculated. This is used to determine the mass flow rate in the

turbine exhaust and in the pump. The mass flow rate in stream 8 (ṁII) is determined by an enthalpy

balance around the process heater and ṁIII is from the enthalpy balance around the feed water heater.

If the mass flow rates in streams 8, 9 and 10 add up to the calculated ṁ, then the assumed pressures in

stream 7 and 10 are correct. The fraction of steam extracted from the turbine and the fraction of the

extracted steam sent to the feed water heater can then be calculated. From here the T-S Diagram can

be determined and is shown in Figure 6.

Figure 6. T-S Diagram for the Steam Turbine Cogeneration Plant modified with Regeneration.

Since all of the needed enthalpies and mass flow rates are known, the work of the turbine, and

the boiler and condenser duties can be determined using an energy balance around the appropriate

device.

The group then worked on the determination of the rate of fuel requirement. The heating

value of methane was first converted, using its density, to be in terms of kJ/kg. The amount of heat

absorbed by the boiler (Qboiler) is already known and it is in terms of kJ/s. This value is divided by the

heat supplied by the combustion of methane (heating value) to get the rate of fuel requirement of the

power plant.

Lastly, all the calculated values were counter checked with the given conditions to make sure

that each condition was met. These values are presented in Table 3,4, and 5.

Table 3. Calculated Values (Stream 1 – 5)

I 1 2 3 4 5

Temp © 45.83 45.8831549 143.62 143.62 144.815587

Pressure (kPa) 10 400 400 400 6000

Vapor Fraction 0 0 0 0 0

Enthalpy (kJ/kg) 191.832 192.394714 604.67 604.67 613.342

Entropy (kJ/kgK) 0.6493 0.6495 1.7764 1.7764 1.8062328

Mass flow rate (kg/s)

0.24018991 0.24018991 2.41268791 5.86381282 5.86381282

Table 4. Calculated Values (Stream 6 – 11)

I 6 7 8 9 10 11

Temp © 500 188.496 188.496 188.496 45.83 143.62

Pressure (kPa) 6000 400 400 400 10 400

Vapor Fraction 1 1 1 1 0.908671816 0

Enthalpy (kJ/kg) 3422.2 2835.82651 2835.82651 2835.826513 2366.254579 604.67

Entropy (kJ/kgK) 6.8818 7.1176 7.1176 7.1176 7.465974232 1.7764

Mass flow rate (kg/s) 5.86381282 5.62362291 3.45112491 2.172498 0.240189911 3.45112491

Table 5. Other Calculated Values

Extraction Fraction ([mI + mII]/m) 0.9590

(mI/[mI + mII]) 0.3863

Utilization factor, ϵu 0.80

Boiler duty (Qboiler, kW) 16470.588 2

Condenser duty (Qcondenser, kW) −522.369744

Rate of fuel requirement (kg/s) 2.81868427

Assessment of Design

Every power plant’s goal is to produce energy through optimum and economically practical

utilization of machines needed without compromising the safety and health of people living near the

plant or working at the plant itself. This assessment looks into the feasibility of the proposed

Cogeneration System.

Power plants are classified according to the working fluid they used. Steam is a widely

available, and cheap working fluid that also has a high enthalpy of vaporization value which would

then produce a high amount of heat once condensed. These characteristics make steam a practical

choice for a working fluid. As for the fuel used to supply the initial heat into the system, natural gases

are good choices when viewed from an environmental perspective due to their low carbon dioxide

emission.

Besides the working fluid and fuel, the working system itself is just as important a

consideration. A cogeneration system is a system that simultaneously produces more than one useful

form of energy from a single energy source which in this particular case is methane (CH4). As

previously mentioned, a power plant’s main goal is to produce as much energy as it can by using as

few resources as possible. Since a cogeneration system produces much energy by using a single

energy source, it as a desirable system to work with.

The amount of fuel to be combusted is dependent on how much heat the working fluid will

need to reach the high pressure and temperature desired after it goes through the boiler. Since heating

the fuel is an added expenditure, engineers have thought of a way to reduce this cost by designing a

regenerative cycle. The main operation that defines a regenerative cycle is the operation or sets of

operations that heat the liquid that enters the boiler to even higher temperatures compared to the liquid

that would have entered the boiler without those extra sets of operations. Heating the liquid decreases

the amount of fuel to burn because hotter liquids need less heat to reach a higher temperature than

cooler liquids. This reduction of fuel not only reduces the cost of operating the plant but also

decreases the amount of air pollution and greenhouse gases it emits.

The energy from the heat that the working fluid intakes, Q in, is harvested as useful energy in

the forms of net work and process heat,Wnet and Qp respectively. Work is produced from the turbines

through extracting heat from superheated steam which leaves subcooled liquid. In order to increase

the amount of heat extracted, we have to increase the heat of the turbine inlet stream and decrease the

heat in the turbine outlet stream. This is why the turbine inlet stream is set at 6000 kPa and 500 C and

the outlets are at 400 kPa and 10 kPa. Engineers also have to be mindful that the maximum service

pressure and temperature of most power plants is at 10,000 kPa and 600 C. Operating above these

conditions greatly increases corrosion of the equipment. Maintaining a factory at these conditions is

also quite costly due to the amount of heat to supply. Equipment that can withstand these conditions

also require great expenditure.

Most of the equipment in the system work at 400 kPa and temperatures below 200 C which

are conditions well below the maximum limits of operation. Thus, the proposed design is deemed

feasible.

Effect of Steam Extraction Pressure

As the design for the power plant is completed, the effect of varying conditions in the process

became the next focus of the study. The effect of varying the steam extraction pressure was analyzed.

A lower and higher value than the established steam extraction pressure were chosen. The net work,

process heat, boiler and condenser duties and utilization factor were calculated using the chosen

values and compared to the output in Part I. The values obtained are shown in Table 6. It can be

observed that lowering the steam extraction pressure would result to a higher work output, a lower

process heat produced, higher boiler duty, and a higher utilization factor. Increasing the steam

extraction pressure would give the opposite result. It should also be noted that any change in the

steam extraction pressure does not cause any change in the condenser duty. The condenser duty

depends only on the mass flow rate and the enthalpy of the fluid in the inlet and outlet stream. Since

the enthalpy of both the inlet and outlet depend on the same pressure (but are of different states) and

this pressure is maintained despite the changes made in the extraction pressure, the condenser duty

will not be affected.

As for the work output. It was observed that decreasing the steam extraction pressure would

cause an increase in the produced work of the turbine and the required work of the pumps. However,

the produced work in the turbine can account for the increased demand for work in the pumps and still

have a higher value than the work produced by the turbine in Part I. An analysis of the work produced

in each section of the turbine shows that at a lower steam extraction pressure, the work produced by

the turbine in the first section (before extraction) is greater than the obtained value in Part I. The work

in the second section is lower than the original value but this value only constitutes a small part of the

total work of the turbine because of the very small mass flow rate in the second section. So overall,

the decreased steam extraction pressure gave rise to a higher work output.

In the case of the process heat output, a decrease was observed when the steam extraction

pressure was decreased. Since there is no change in the amount of waste it. It is assumed that the rest

of the energy from the boiler was converted to either work output or process heat and since there was

a large increase in the work output, it is expected to have a decrease in the amount of process heat

produced. However, the decrease in the amount of process heat produced is too small compared to the

increase of work output. This can be accounted for by the observed increase in the boiler duty. So

more energy was available for conversion to work or process heat.

The utilization factor was also observed to have a slight increase in value. By analyzing Eq. 1,

it may be expected to have a decrease in ϵu because of the increase in Qin (increase in Qboiler) and the

decrease in process heat produced. However, the increase in Wnet is large enough to account for these

changes and even caused an increase in ϵu.

In increasing the steam extraction pressure, it was observed that the changes were opposite to

those observed earlier. The same explanations apply to these changes so generally, increasing the

steam extraction pressure would just reverse the decrease in steam extraction pressure.

Table 6. Calculated Parameters for Varying P7

300 kPa 475 kPa

Wnet (kW) -3748.4988 -3344.5689

Qp (kW) -7696.637596 -7702.184496

Qboiler (kW) 16723.45762 16312.11395

Qcondenser (kW) -522.36974 -522.36974

ϵu 0.8051 0.7967

Effect Of Extraction Fraction

Increase of the extraction fraction in the intermediate turbine will cause a decrease in the

turbine power output since lesser mass will flow through the second section of the turbine. This will

then cause a decrease in the rate of power generated by the cogeneration plant. Also, the heat supplied

will increase since the mass fraction entering the process heater increases.

In this case, the utilization factor will also increase. Since the enthalpy difference in the

process heater is greater than in the second section of the turbine, hence the increase of the process

heat supply will contribute more to the change of the utilization factor. Furthermore, the heat released

in the condeser will increase, but the heat supplied to the boiler remains the same.

A negative effect was observed when the extraction fraction is decrease.

Table 7. Calculated Parameters for Varying Extraction Fraction

0.93 0.97

Wnet (kW) -3579.8466 -3469.8312

Qp (kW) -7466.9759 -7788.1362

Qboiler (kW) 16470.5882 16470.5882

Qcondenser (kW) -892.5269 -382.5115

ϵu 0.7891 0.8041

Conclusion and Recommendation

The starting point for the design of a cogeneration plant is the selection of the input and

exhaust conditions of the turbine. The lower the exhaust/condensing temperature, the more efficient

the system. But the condensing temperature is restricted by the availabilty of cooling water used in the

condenser, which needs a temperature difference of 10 degrees for effective heat transfer. Also the

selection of the exhaust conditions must satisfy that the moisture content of the turbine will not

exceed 10%.

With determined exhaust conditions and efficiency of the turbine, there's an optimum value of

pressure for the selected temperature of the input stream. The input temperature must be high enough

to generate a higher power rate, however the steam temperature is restricted by the metallurgical

limits of the turbine. Once the stream conditions have been agreed, the preferred mass flow rate of the

working fluid is can be determined using the preliminary conditions. The pressure of the stream

exiting the intermediate turbine is can also be optimized such that the net work of the pumps is

minimum. With the determined value of the mass flow rate and the thermodynamic properties of the

extracted stream, the extraction fraction is can be calculated.

Through this design project, it is found out that decreasing the intermediate pressure will

result to a decrease in the process heat supply (Qp) but an increase in the total rate of power

generation (Wnet) and in the utilization factor (eu). The boiler duty will also increase but the

condenser duty remains the same. Also, by increasing the extraction fraction exiting the intermediate

turbine, the Wnet and condenser duty will decrease, but the Qp will increase along with ϵu.

Setting temperatures and pressures for a power plant are tricky choices to make because of the

trade offs of energy produced and cost to operate as seen in the following recommendations.

If prolonging the service life of the turbines is desired, exhaust quality should be maximized.

This is because water droplets in the turbine cause erosion inside the turbine. However, the only way

to maximize the exhaust quality would be to use an inlet of superheated steam. The difficulty with

handling superheated steam lies in obtaining equipment that can withstand the high pressures and

temperature of such a stream such that it can still operate for a long time.

If much work is desired, we must look at the turbine, pumps and boiler individually. For a

turbine to generate as much heat as possible, we must extract the heat from a superheated steam and

produce a subcooled liquid. To produce an inlet stream of superheated steam, the boiler must intake a

big amount of energy to produce as hot a stream as possible. The problems of having a system work

with such high temperatures and pressures are stated in the previous paragraph.

Another machine used to increase the pressure of a stream is a pump. Pumps take in saturated

liquid at low pressures and excrete subcooled liquids at higher pressures which is a desirable effect

because these high pressure streams go to the boiler. The boiler will then need to intake less energy to

produce a superheated stream. The disadvantage of pumps producing a high pressure stream is that

they tend to consume more work as they produce higher pressure streams. Thus, depending on

whether it is more desired to maximize the work or to minimize the amount of fuel to burn, pumps

play a part in the adjustment to attain these goals.

References:

[1] BIBLIOGRAPHY Cengel, Yunus A and Michael A Boles. Thermodynamics An Engineering

Approach. New York: McGraw-Hill, 2011.

[2] Smith, J.M, H.C Van Ness and M.M Abbott. Introduction to Chemical Engineering

Thermodynamics. New York: McGraw-Hill, 2005.

[3] Clarke Energy. n.d. Retrieved from http://www.clarke-energy.com/natural-gas/commercial-chp/

15 December 2014.

[4] EPA United States Environmental Protection Agency. 4 November 2013. Retrieved from

http://www.siemens.com/innovation/en/publikationen/publications_pof/pof_fall_2007/

materials_for_the_environment/optimizing_turbine_blades.htm 15 December 2014.

[5] Muller, Bernd. Siemens. 2007. Retrieved from http://www.epa.gov/chp/basic/ 15 December 2014.

[6] Weston, Kenneth. Fundamentals of Steam Power. Retrieved from

http://www.personal.utulsa.edu/~kenneth-weston/chapter2.pdf 16 December 2014.

[7] Inkson, M., Misplon, B. Cogeneration Thermodynamics Revisited. Retrieved from

http://www.thermalenergysystems.com/issct07.pdf 16 December 2014.

[8] TS Diagram Retrieved from http://commons.wikimedia.org/wiki/File:T-s_diagram.svg 16

December 2014

[9] Water – Thermal Properties The Engineering ToolBox. Retrieved from

http://www.engineeringtoolbox.com/water-thermal-properties-d_162.html 14 December 2014

APPENDICES

I. CALCULATIONS:

Part I:

H6 = 3422.2 kJ/kg S6 = 6.8818 = 0.85

@Stream 7:

S7=S6=x S7l+ y S7

v

y=6.8818−1.77646.8943−1.7764

=0.998

H 7s=(1−0.998 ) H 7l+(0.998)H 7

v

H 7s=(1−0.998 )604.670+ (0.998 )2737.6

H 7s=2732.3905

W I=( H 7s−H 6 )=−586.3381 kJ /kg

H 7=H 6+(−586.3381)

H 7=2835.8619kJkg

=H 8=H 9 (superheated steam)

@Stream 10:

6.8818=x S10l+ y S10

v

y=6.8818−0.64938.154−0.6493

=0.831

H 10s= (1−0.831 ) H 10l+(0.831)H 10

v

H 10s= (1−0.831 ) 191.832+ (0.831 )2584.8

H 10s=2179.943

W s=( H 10s−H 6 )=−1055.9454

H 10=H 6+(−1055.9454)

H 10=2366.2546 kJ /kg

W II=H 10−H 7

W II=−469.6074 kJ /kg

@Pump 1:

W Pump1=1❑VΔP=ΔH

W Pump1=1

0.7 [1010cm3

kg ( 400−10106 )]

W Pump1=0.5627 kJ /kg

0.562=H 2−H 1

H 2=192.394714 kJ /kg

@Pump 2:

W Pump2=1❑ VΔP=ΔH

W Pump2=1

0.7 [1084cm3

kg (6000−400106 )]

W Pump2=8.672 kJ /kg

8.672=H 5−H 4

H 5=613.342 kJ /kg

ϵ u=|W net|+|Q p|

Q¿=¿Q¿=

0.83.5+7.7

=14 MW

Qboiler=14 MW

0.85=16470.5882 kW =ṁ ( H 6−H 5 )

ṁ=5.8638 kg /s

Ẇ net=Ẇ turbine+Ẇ Pump 1+Ẇ Pump2

−3500=ṁW I+xṁW II+ṁW Pump2+xṁ W Pump 1

x=0.0409689

Mass Flow rates calculations (MATLAB code)

let x=(m-mi-mii)/m

x=(-3500-mdot*(W1+W2pump))/(mdot*(W2+W1pump))

x=0.0410

extraction fraction=1-x=0.9590

m-mi-mii=x*m=0.2402

mii=7700/(H8-H7)

=7700/(561.429-2835.8619)

=3.4510

mi=m-xm-mii

mi=2.1725

mi/(mi+mii)=0.3863

W turbine=ṁW I+xṁW II

W turbine=5.8638 (−586.3381 )+(0.0409689)(5.8638)(−469.6074)

W turbine=−3550.486 kW

Qcondenser=xṁ ( H 1−H 10 )=(0.0409689 ) (5.8638 ) (191.832−2366.2546 )

Qcondenser=−522.369744 kW

HV C H 4=35600

kJ

m3 ρC H 4=0.66

kg

m3

Rate of fuel requirement=Qboiler ρ

HV

Rate of fuel requirement=(16470.5882

kJs )(0.66

kg

m3 )35600

kJm3

Rate of fuel requirement=2.81868427kgs

Part II: ( MATLAB was used in the calculation of the required parameters in this part. Since

the process is generally the same as the calculation process in Part I, instead of hand

calculations, the codes were attached here instead.)

Lower Intermediate Turbine Pressure (300 KPa)

H6=3422.2; S6=6.8818;

S7L=1.6716; S7V=6.9909;

H7L=561.429; H7V=2724.7;

y7=(S6-S7L)/(S7V-S7L);

n=0.85;

H7s=(1-y7)*(H7L)+y7*(H7V);

W1=n*(H7s-H6);

W1=-630.588785

H7=W1+H6;

H7= 2791.611215

S10L=0.6493;

S10V=8.1511;

H10L=191.832;

H10V=2584.8;

y10=(S6-S10L)/(S10V-S10L);

n=0.85;

H10s=(1-y10)*(H10L)+y10*(H10V);

W2a=n*(H10s-H6);

H10=W2a+H6;

W2=H10-H7;

W2=-425.356715

y10actual=(H10-H10L)/(H10V-H10L);

W1pump=((1010*(300-10))/10^6)/.7;

W1pump=0.4184285714

W2pump=((1073*(6000-300))/10^6)/.7;

W2pump=8.737285714

H5=W2pump+604.670;

H5=613.4072857

Qboiler=(H6-H5);

Qboiler=2808.792714

mdot= 5.86380238349326;

Qp=3.4511*(H7L-H7);

Qp=-7696.581842

x= 0.0409689412039485;

WTURBINE=mdot*(W1)+(mdot*x)*W2

WTURBINE=-3799.83307

W1PUMP=(mdot*x)*(W1pump);

W1PUMP=0.1005206753

W2PUMP=(mdot)*(W2pump);

W2PUMP=51.56804533

Qin=mdot*Qboiler*0.85;

Qin=13999.6746

WNET=WTURBINE+W1PUMP+W2PUMP;

WNET=-3748.164534

Qcondenser=mdot*x*(191.832-H10);

Qcondenser=-522.3697

eu=((abs(WNET+Qp)))/(Qin);

eu=.8175

Higher Intermediate Turbine Pressure (475 kPa)

H6=3422.2;

S6=6.8818;

n=0.85;

H7s=2765.0443;

W1=n*(H7s-H6);

W1=-558.582345

H7=W1+H6;

H7=2863.617655

S10L=0.6493;

S10V=8.1511;

H10L=191.832;

H10V=2584.8;

y10=(S6-S10L)/(S10V-S10L);

y10=0.8308006079

n=0.85;

H10s=(1-y10)*(H10L)+y10*(H10V);

H10s=2179.911269

W2a=n*(H10s-H6);

W2a=-1055.945421

H10=W2a+H6;

H10=2366.254579

W2=H10-H7;

W2=-497.3630764

y10actual=(H10-H10L)/(H10V-H10L);

y10actual=0.9086718163

W1pump=((1010*(475-10))/10^6)/.7;

W1pump=.670929

W2pump=((1084*(6000-475))/10^6)/.7;

W2pump=8.55586

H5=W2pump+631.812;

H5=640.36786

Qboiler=(H6-H5);

Qboiler=2781.83214

mdot=5.86380238349326;

Qp=3.4511*(631.812-H7)

Qp=-7702.184496

x= 0.0409689412039485;

WTURBINE=mdot*(W1)+(mdot*x)*W2;

WTURBINE=-3394.8999

W1PUMP=(mdot*x)*(W1pump);

W1PUMP=.1611798065

W2PUMP=(mdot)*(W2pump);

W2PUMP=.2.055406547

Qin=mdot*Qboiler*0.85;

Qin=13865.29686

WNET=WTURBINE+W1PUMP+W2PUMP

WNET=-3344.5689

eu=((abs(WNET+Qp)))/(Qin)

eu=.7967

Qcondenser=mdot*x*(191.832-H10)

Qcondenser=-522.36974

Qboiler=mdot*(H6-H5)

Qboiler=16312.11395

Part III: ( MATLAB was used in the calculation of the required parameters in this part. Since

the process is generally the same as the calculation process in Part I, instead of hand

calculations, the codes were attached here instead.)

Lower Extraction Factor (0.93)

H6=3422.2; S6=6.8818;

S7L=1.7764; S7V=6.8943;

H7L=604.670; H7V=2737.6;

y7=(S6-S7L)/(S7V-S7L);

n=0.85;

H7s=(1-y7)*(H7L)+y7*(H7V);

W1=n*(H7s-H6);

W1=-586.3381

H7=W1+H6;

H7= 2835.8619

S10L=0.6493;

S10V=8.1511;

H10L=191.832;

H10V=2584.8;

y10=(S6-S10L)/(S10V-S10L);

n=0.85;

H10s=(1-y10)*(H10L)+y10*(H10V);

W2a=n*(H10s-H6);

H10=W2a+H6;

W2=H10-H7;

W2=-469.6074

y10actual=(H10-H10L)/(H10V-H10L);

W1pump=((1010*(400-10))/10^6)/.7;

W1pump=0.5627

W2pump=((1084*(6000-400))/10^6)/.7;

W2pump=8.672

H5=W2pump+604.670;

H5=613.342

mdot=5.8638;

x=1-0.93;

m2=0.93*mdot*(0.613679753494816);

Qp=m2*(H11-H8)

Qp= -7466.9115

WTURBINE=mdot*(W1)+(mdot*x)*W2;

Wturbine=-3630.9285

W1PUMP=(mdot*x)*(W1pump);

W1PUMP=0.2309

W2PUMP=(mdot)*(W2pump);

W2PUMP=50.8509

WNET=WTURBINE+W1PUMP+W2PUMP

Wnet=-3579.8466

Qcondenser=mdot*x*(H1-H10);

Qcondenser=-892.5269

Qboiler=mdot*(H6-H5);

Qboiler=16470.5882

eu=((abs(WNET+QP)))/(Qin)

eu=0.7891

Higher Extraction Factor (0.97)

x=1-0.97

WTURBINE=mdot*(W1)+(mdot*x)*W2;

Wturbine=-3520.7811

W1PUMP=(mdot*x)*(W1pump);

W1PUMP= 0.0990

W2PUMP=(mdot)*(W2pump);

W2PUMP= 50.8509

WNET=WTURBINE+W1PUMP+W2PUMP

Wnet=-3469.8312

Qcondenser=mdot*x*(H1-H10);

Qcondenser=-382.5115

Qboiler=mdot*(H6-H5);

Qboiler=16470.5882

eu=((abs(WNET+QP)))/(Qin)

eu=0.8041

extraction fraction=0.93

m-mi-mii=x*m=(1-0.93)= 0.4105

mi=0.3863*(m*0.93)=2.1067

mii=(m*0.93)-2.1067= 3.3467

extraction fraction=0.97

m-mi-mii=x*m=(1-0.97)= 0.1759

mi=0.3863*(m*0.97)=2.1973

mii=(m*0.97)-2.1067= 3.4905

II. STEAM PROPERTIES TABLE FOR PART II

II lower 1 2 3 4 5Temp © 45.83 45.8695252 133.54 133.54 134.713524Pressure (kPa) 10 300 300 300 6000Vapor Fraction 0 0 0 0 0Enthalpy (kJ/kg) 191.832 192.250429 561.429 561.429 570.223286Entropy (kJ/kgK) 0.6493 0.6527772 1.6716 1.6716 5.98154618Mass flow rate (kg/s)

0.24018991 0.24018991 2.41268791 5.86381282 5.86381282

6 7 8 9 10 11500 164.69454

8164.69454

8164.694548 45.83 133.54

6000 300 300 300 10 3001 1 1 1 0.90867181

60

3422.2 2791.61122

2791.61122

2791.611215

2366.254579

561.429

6.8818 7.14875474

7.14875474

7.148754738

7.465974232

1.6716

5.86381282

5.62362291

3.45112491

2.172498 0.240189911

3.45112491

II higher 1 2 3 4 5temp C 45.83 45.8933768 149.92 149.92 151.1177Pressure (kPa) 10 475 475 475 6000Vapor Fraction 0 0 0 0 0Enthalpy (kJ/kg) 191.832 191.899093 631.812 631.812 640.367857Entropy (kJ/kgK) 0.6493 0.65487402 1.8408 1.8408 1.86891439Mass flow rate (kg/s)

0.24018991 0.24018991 2.41268791 5.86381282 5.86381282

6 7 8 9 10 11500 203.34164

3203.34164

3203.341643

245.83 149.92

6000 475 475 475 10 4751 1 1 1 0.90867181

60

3422.2 2863.6177 2863.6177 2863.6177 2366.254579

631.812

6.8818 7.09973665

7.09973665

7.099736646

7.465974232

1.8408

5.86381282

5.62362291

3.45112491

2.172498 0.240189911

3.45112491

III. STEAM PROPERTIES TABLE FOR PART III

0.93 1 2 3 4 5Temp © 45.83 45.8831549 143.62 143.62 144.815587Pressure (kPa) 10 400 400 400 6000Vapor Fraction 0 0 0 0 0Enthalpy (kJ/kg) 191.832 192.394714 604.67 604.67 613.342Entropy (kJ/kgK) 0.6493 0.6495 1.7764 1.7764 6.10338161Mass flow rate (kg/s) 0.4104662 0.4104662 2.5172004 5.86381282 5.86381282

6 7 8 9 10 11

500 188.496 188.496 188.496 45.83 143.626000 400 400 400 10 400

1 1 1 1 0.908671816

0

3422.2 2835.82651

2835.82651

2835.826513

2366.254579

604.67

6.8818 7.1176 7.1176 7.1176 7.465974232

1.7764

5.86381282

5.4533362 3.346602 2.1067342 0.4104662 3.346602

0.97 1 2 3 4 5

Temp © 45.83 45.8831549 143.62 143.62 144.815587

Pressure (kPa) 10 400 400 400 6000

Vapor Fraction 0 0 0 0 0

Enthalpy (kJ/kg) 191.832 192.394714 604.67 604.67 613.342

Entropy (kJ/kgK) 0.6493 0.6495 1.7764 1.7764 6.10338161

Mass flow rate (kg/s) 0.1759141 0.1759141 2.3732605 5.86381282 5.86381282

6 7 8 9 10 11500 188.496 188.496 188.496 45.83 143.62

6000 400 400 400 10 4001 1 1 1 0.90867181

60

3422.2 2835.82651

2835.82651

2835.826513

2366.254579

604.67

6.8818 7.1176 7.1176 7.1176 7.465974232

1.7764

5.86381282

5.687888 3.490542 2.1973464 0.1759141 3.490542