The City College of New York (CUNY)

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ME 430
THERMAL SYSTEM DESIGN

Section

Design Project #1: - Steam Turbine Optimization

Prepared For: - Professor Rishi Raj

Prepared By: - Ali Khokhar

Due Date: - 3/23/2012

Abstract

Table of Contents

Nomencluture

Introduction

RANKINE CYCLE

[pic]

Fig 1.0: - Ideal Rankine Cycle

In the Ideal Rankine Cycle the objective is to find the value of the exhaust pressure (P4) that would give a steam quality of 87%. The picture and the T-S diagram of an ideal Rankine cycle are shown in ‘Figure 1.0’ above. From the T-S diagram it can be seen that the ideal Rankine cycle consists of the following processes

• 1-2 Isentropic compression in a pump. • 2-3 Constant pressure heat addition in a boiler. • 3-4 Isentropic expansion in a turbine. • 4-1 Constant pressure heat rejection in a condenser.
Analyzing the cycle it can be seen that the T-s diagram shows four point that are actually four states. It can be seen that point ‘1’ and ‘4’ are on the same pressure line and hence pressure at point ‘1’ is equal to the pressure at point ‘4’ (P1 = P4). It can also be seen that point ‘2’ and ‘3’ are on the same pressure line and hence the pressures at these two points are equal (P2 = P3). The turbine pressure and temperature are given (P3 = 1000 psi and T3 = 1000 (F), knowing the turbine inlet pressure and temperature the analysis can be performed for the ideal Rankine cycle.

Ideal Rankine Cycle Analysis

State 1: - First the pressure at this point is selected ‘P1’, looking at this point it can be seen that it is on the saturated liquid line. Knowing this fact the state at this point can be fixed by finding two intensive properties (specific volume & specific enthalpy) from the saturated water table.
For P1 {v1 = vf@P1 {h1 = hf@P1

State 2: -The pressure at this point is the same as the turbine inlet pressure (constant pressure line). The enthalpy is obtained by the relation
[pic]
Since the pump is isentropic
[pic]
The relation becomes
[pic]
The difference in pressures is
[pic]
The enthalpy in now given by
[pic]
State 3: - At this point the turbine inlet pressure ‘P3’ and temperature ‘T3’ are specified. Knowing these two intensive properties the enthalpy and entropy could be calculated form the superheated steam table.
For P3 {h3 = h T3 {s3 = s
State 4: - The entropy at point ‘4’ is the same as the entropy at point ‘3’(S4 = S3) and the pressure at this point is the same as point ‘1’ (P4 =P2). It can be seen that this point is in the saturated water-vapor mixture region; so in order to find the enthalpy at this point the quality needs to be determined. First the following properties are obtained from the saturated water table.
For P4 {hf S4 {hfg {Sf {Sfg
Now we use the equation
[pic]
This equation can be solved to obtain ‘x4’ and ‘h4’ can be obtained by using the equation
[pic]
Efficiency: - Now that the enthalpies at each point are known the efficiency of the cycle can be calculated. The efficiency is given by
[pic]
[pic]
[pic]
Where ‘Qin’ is the heat supplied to the boiler and ‘Qout’ is the heat rejected in the condenser.

Ideal Rankine Cycle Calculations

CASE 1: - P1 = 1 psi (Fixed To Do The Calculations) P3 = 1550 psi (Remains Fixed For Other Values of ‘P1’) T3 = 1000 (F (Remains Fixed For Other Values of ‘P1’)

State 1: - [pic] [pic] [pic]
State 2: - [pic] [pic] [pic] [pic]
State 3: - [pic] [pic] [pic] [pic]
State 4: - [pic] [pic] [pic] [pic] [pic] [pic] [pic] [pic] [pic] [pic] [pic] [pic]
Efficiency: - [pic] [pic] [pic]

It can be seen that the efficiency for an inlet pump pressure ‘P1’ of 1 psi is quiet high 42.50%. However the steam quality is below 87%, so in order to obtain the steam quality of 87% different values of ‘P1’ were used and the results obtained for the steam quality ‘x4’ and cycle efficiency ‘(’ were tabulated. From the results two graphs were constructed ‘Efficiency vs. Exhaust Pressure’ & ‘Steam Quality vs. Exhaust Pressure’. The graphs were used to determine the exhaust pressure ‘P4’ at which the steam quality is 87%.
| |Exhaust Pressure ‘P4’ |Steam Quality ‘x4’ |Efficiency ‘(’ |
|Case1 |1 |0.7932 |0.4188 |
|Case 2 |5 |0.8463 |0.3784 |
|Case 3 |10 |0.8730 |0.3522 |
|Case 4 |15 |0.8903 |0.3377 |

Table 1.0: - Ideal Rankine Cycle Results

[pic]
Figure 2.0: - Steam Quality vs. Exhaust Pressure

[pic]
Figure 3.0: - Efficiency vs. Exhaust Pressure

The equation corresponding to both the graphs is shown, in order to determine exactly the exhaust pressure at which the steam quality is 87% the equation shown in the graph of ‘Steam Quality vs. Exhaust Pressure’ was solved for ‘x’ by using MATLAB. The pressure at which the steam quality is 87% came out to be 9.4382 psi. This pressure was now plugged in the equation shown on the graph of ‘ ‘Efficiency vs. Exhaust Pressure’ and the efficiency at this pressure came out to be 0.3551 or 35.51%.

REHEAT CYCLE

[pic][pic]
Figure 4.0: - Ideal Reheat Cycle

The reheat cycle is used to obtain a higher cycle efficiency than the ideal Rankine cycle. In the reheat cycle once the steam expands in the high-pressure turbine it is reheated in the boiler and used in the low-pressure turbine. The objective of this cycle is to find the reheat pressure ‘P4’ that would maximize the cycle efficiency. The second objective is to keep ‘P4’ obtained from the first part fixed and vary the exhaust pressure ‘P6’ so that the overall steam quality of 87% is obtained. From the ‘T-s’ it can be seen that the reheat cycle consists of the following processes

• 1-2 Isentropic compression in a pump. • 2-3 Constant pressure heat addition in a boiler. • 3-4 Isentropic expansion in a high-pressure turbine. • 4-5 Constant pressure heat addition in a boiler. • 5-6 Isentropic expansion in a low-pressure turbine and constant pressure heat rejection in a condenser.

Analyzing the cycle it can be seen that the ‘T-s’ diagram shows 6 points. Point ‘1’ and ‘6’ are on the same pressure line thus these two point have the same pressure (P1= P6). Points ‘2’ and ‘3’ are on the same pressure line hence the pressures at these two points are the same (P2 = P3). Points ‘4’ and ‘5’ are on the same pressure line and they have the same pressures (P4 = P5). The temperature at point ‘3’ and ‘5’ is the same (T3 = T5). The entropy at point ‘3’ is equal to the entropy at point ‘4’ (S3 = S4) and the entropy at point ‘5’ is equal to the entropy at point ‘6’. The turbine inlet pressure and temperature are fixed ‘P3 = 1550 psi, T3 = 1000(F). In the first part ‘P1’ is selected from the ideal Rankine cycle and ‘P4’ is varied according to the relation ‘P4 = (0.1( 0.5)*P3’. This means that different values of ‘P4’ are obtained by multiplying ‘P3’ with numbers in the range for 0.1 to 0.5. With these parameters known the reheat cycle can be analyzed.

Reheat Cycle Analysis Part A

State 1: - The pressure ‘P1’ is fixed from the ideal Rankine cycle, since point one is on the saturated liquid line tow intensive properties specific volume and specific enthalpy are obtained for the saturated liquid table.
For P1 = 11.64 psi {v1 = vf@P1 {h1 = hf@P1
State 2: -The pressure at this point is the same as the turbine inlet pressure (constant pressure line). The enthalpy is obtained by the relation
[pic]
Since the pump is isentropic
[pic]
The relation becomes
[pic]
The difference in pressures is
[pic]
The enthalpy in now given by
[pic]
State 3: - At this point the turbine inlet pressure ‘P3’ and temperature ‘T3’ are specified. Knowing these two intensive properties the enthalpy and entropy could be calculated form the superheated steam table.
For P3 {h3 = h T3 {s3 = s
State 4: - The pressure at this point is obtained by multiplying the turbine inlet pressure ‘P3’ by values ranging from 0.1(0.4. And from the ‘T-s’ diagram it can be seen that the entropy at this point is the same as the entropy at point ‘3’. Since this point is in the superheated region, knowing the pressure and the entropy the enthalpy can be obtained from the superheated steam tables and using linear interpolation.
For P4 {h4 = h S4
State 5: - This point is also in the superheated region and the pressure at this point is equal to the pressure at point ‘4’ because of the constant pressure line (P4 = P5). The temperature at this point is also equal to the turbine inlet pressure (T3 = T5). Knowing these two extensive the enthalpy and entropy can be obtained from the superheated steam tables.
For P5 {h5 = h T5 {s5 = s
State 6: - This point is in the saturated liquid-vapor mixture region, the pressure at this point is the same as point ‘1’ (P1 =P6) and the entropy at this point is the same as point ‘5’ (S5 = S6). So in order to determine the enthalpy at this point the quality need to be determined.
For P6 {hf S6 {hfg {Sf {Sfg
Now we use the equation
[pic]
This equation can be solved to obtain ‘x4’ and ‘h4’ can be obtained by using the equation
[pic]
Efficiency: - Now that the enthalpies at each point are known the efficiency of the cycle can be calculated. The efficiency is given by
[pic]
[pic]
[pic]

Reheat Cycle Calculations Part A

Case 1 P1 = 9.4382 psi (Fixed from ideal Rankine cycle, it gave a quality of 87%)

P3 = 1550 psi (Given and is fixed) T3 = 1000 (F P4 = 160 psi (Obtained from multiplying 0.1 by 1550 and rounding it off to an appropriate value that is available in the steam table.)

State 1: - [pic] [pic] [pic]
State 2: - [pic] [pic] [pic] [pic]
State 3: - [pic] [pic] [pic] [pic]
State 4: - [pic] [pic] [pic]
State 5: - [pic] [pic] [pic] [pic]
State 6: - [pic] [pic] [pic] [pic] [pic] [pic] [pic] [pic] [pic] [pic] [pic] [pic]
Efficiency: - [pic] [pic] [pic]

The analysis is repeated for different values of reheat pressure ‘P4’ and the graph of ‘Efficiency vs Reheat Pressure’ was constructed. The ‘Table 2.0’ below shows the results of the calculations.

| | Exhaust Pressure ‘P6’ |Reheat Pressure ‘P4’ |Efficiency ‘(’ |Quality ‘X6’ |
|Case 1 |9.4382 |160 |0.3699 |1.061 |
|Case 2 |9.4382 |300 |0.3714 |1.003 |
|Case 3 |9.4382 |450 |0.3755 |0.9718 |
|Case 4 |9.4382 |600 |0.3743 |0.9494 |
|Case 5 |9.4382 |800 |0.3718 |0.9264 |
|Case 6 |9.4382 |1000 |0.3685 |0.9081 |
|Case 7 |9.4382 |1500 |0.3599 |0.8732 |

Table 2.0: - Reheat Cycle Results For ‘Part A’

From the results above it can be seen that as the reheat pressure ‘P4’ increases the efficiency first increases and than decreases. The quality keeps on decreasing as the exhaust pressure increases. The objective of ‘Part A’ of the reheat cycle is to plot the graph of ‘Efficiency vs Reheat Pressure’ and determine what value of reheat pressure will give the highest efficiency. The ‘Figure 5.0’ below shows the graph of ‘Efficiency vs Reheat Pressure’.

[pic]Figure 5.0: - The Graph Of ‘Efficiency vs Reheat Pressure’

From the figure above it can be seen that the maximum efficiency is attained around the pressure of 450 psi. So in doing calculation for ‘Part B’ the reheat pressure ‘P4’ was fixed to be 450 psi.

Reheat Cycle Analysis Part B

State 1: - First the pressure ‘P1’ at this point is selected between the values of 100 psi ( 1 psi, looking at this point it can be seen that it is on the saturated liquid line. Knowing this fact the state at this point can be fixed by finding two intensive properties (specific volume & specific enthalpy) from the saturated water table.
For P1 {v1 = vf@P1 {h1 = hf@P1

State 2: -The pressure at this point is the same as the turbine inlet pressure (constant pressure line). The enthalpy is obtained by the relation
[pic]
Since the pump is isentropic
[pic]
The relation becomes
[pic]
The difference in pressures is
[pic]
The enthalpy in now given by
[pic]
State 3: - At this point the turbine inlet pressure ‘P3’ and temperature ‘T3’ are specified. Knowing these two intensive properties the enthalpy and entropy could be calculated form the superheated steam table.
For P3 {h3 = h T3 {s3 = s
State 4: - The pressure at this point is fixed from ‘Part A’ of the reheat cycle (P4 = 600 psi). And from the T-s diagram it can be seen that the entropy at this point is the same as the entropy at point ‘3’ (S3 = S4). Knowing these two extensive properties the enthalpy at this point can be determined form the superheated steam table by using linear interpolation.
For P4 {h4 = h S4
State 5: - This point is also in the superheated region and the pressure at this point is equal to the pressure at point ‘4’ because of the constant pressure line (P4 = P5). The temperature at this point is also equal to the turbine inlet pressure (T3 = T5). Knowing these two extensive the enthalpy and entropy can be obtained from the superheated steam tables.
For P5 {h5 = h T5 {s5 = s
State 6: - This point is in the saturated liquid-vapor mixture region, the pressure at this point is the same as point ‘1’ (P1 =P6) and the entropy at this point is the same as point ‘5’ (S5 = S6). So in order to determine the enthalpy at this point the quality need to be determined.
For P6 {hf S6 {hfg {Sf {Sfg
Now we use the equation
[pic]
This equation can be solved to obtain ‘x4’ and ‘h4’ can be obtained by using the equation
[pic]
Efficiency: - Now that the enthalpies at each point are known the efficiency of the cycle can be calculated. The efficiency is given by
[pic]
[pic]
[pic]

Reheat Cycle Calculations Part B

Case 1 P1 = 100 psi (Varied from 100 psi( 1 psi)

P3 = 1550 psi (Given and is fixed) T3 = 1000 (F (Given and is fixed) P4 = 600 psi (Fixed for all calculations)

State 1: - [pic] [pic] [pic]
State 2: - [pic] [pic] [pic] [pic]
State 3: - [pic] [pic] [pic] [pic]
State 4: - [pic] [pic] [pic]
State 5: - [pic] [pic] [pic] [pic]
State 6: - [pic] [pic] [pic] [pic] [pic] [pic] [pic]
We know that the lower the exhaust pressure the higher the cycle efficiency, but as the exhaust pressure decreases the steam quality also decreases. In subsequent reheat cycle analysis steam quality would be calculated for different exhaust pressures and a graph would be tabulated of ‘Steam Quality vs Exhaust Pressure’. From this graph the exhaust pressure at which steam quality is exactly 87% would be calculated and the efficiency of the cycle would than be calculated at this exhaust pressure.
| |P1 & P6 |P4 |X |
|Case 1 |100 |450 |1.1290 |
|Case 2 |80 |450 |1.1102 |
|Case 3 |50 |450 |1.0730 |
|Case 4 |10 |450 |0.9750 |
|Case 5 |5 |450 |0.9416 |
|Case 6 |1 |450 |0.8769 |
|Case 7 |0.50 |450 |0.8523 |

Table 3.0: - Reheat Cycle Results For ‘Part B’

The ‘Figure 6.0’ below shows the graph of the ‘Steam Quality vs Reheat Pressure’. Using linear interpolation it was determined that the exhaust pressure of 0.8598 psi gave the steam quality of 87.%.

[pic]
Figure 6.0: - The Graph Of ‘Steam Quality vs Reheat Pressure’

Now with the exhaust pressure of 0.8598 psi the efficiency of the cycle is calculated [pic] [pic] [pic] [pic] [pic]
It can be seen that in the reheat cycle the reheat pressure of 450 psi and an exhaust pressure of 0.8598psi gives an efficiency of 44.63 % which is quiet high as compared to the efficiency of the ideal Rankline cycle.

Regenerative Cycle

Regenerative cycles are used to increase the efficiency of the reheat cycle by using feed water heaters (FWD). In the regenerative cycles some steam is taken out of the turbine and used in the FWH. So in order to determine the efficiency of the cycle the enthalpies at each points and the mass flow rates in the FWH’s needs to be determined. The regenerative cycle is analyzed by first using one FWH, than using two FWH’s and finally using four FWH’s.

Regenerative Cycle with One FWH

[pic]
Figure 7.0: - Regenerative Cycle1 with One FWH

The ‘T-s’ diagram of the cycle is shown in ‘Figure 7’ above, from the diagram it can be seen that the cycle consists of ‘8’ points. In this cycle one FWH is connected to the high- pressure turbine.

Regenerative Cycle1 Analysis

State 1: - The pressure ‘P1’ is the exhaust pressure and is fixed from the reheat cycle, since point one is on the saturated liquid line two intensive properties specific volume and specific enthalpy are obtained for the saturated liquid table.
For P1 = 1.85 {v1 = vf@P1 {h1 = hf@P1
State 2: -The pressure at this point is the same as reheat pressure ‘P4’ obtained from the reheat cycle (constant pressure line). The enthalpy is obtained by the relation
[pic]
Since the pump is isentropic
[pic]
The relation becomes
[pic]
The difference in pressures is
[pic]
The enthalpy in now given by
[pic]
State 3: - The pressure at this point is the same as the pressure at point ‘2’ (P2 = P3). Using saturated liquid table the specific volume and enthalpy at this point can be calculated.

For P3 {v3 = vf@P3 {h3 = hf@P3

State 4: - The pressure at this point is the same as the turbine inlet pressure the enthalpy at this point can be determined as follows
[pic]
Since the pump is isentropic
[pic]
The relation becomes
[pic]
The difference in pressures is
[pic]
The enthalpy in now given by
[pic]
State 5: - At this point the turbine inlet pressure and temperature are known, and from the superheated steam tables the enthalpy and the entropy is obtained.
For P5 {h5 = h T5 {s5 = s
State 6: - This pressure at this point is the reheat pressure obtained form the reheat cycle, and the entropy at this point is the same as the entropy at point ‘5’
For P6 {h6 S6
State 7: - The pressure at this point is the same as point ‘6’ and the temperature at this point is the same as turbine inlet pressure at point ‘3’. Knowing these two properties the enthalpy and entropy at this point can be determined from the superheated steam table.
For P7 {h7 T7 {s7
State 8: - This point is in the saturated water-vapor mixture region, so in order to determine the enthalpy the quality needs to be determined. The pressure at this point is known from the reheat cycle and the entropy at this point is the same as the entropy at point ‘7’.
For P8 {hf S8 {hfg {Sf {Sfg
Now we use the equation
[pic]
This equation can be solved to obtain ‘x8’ and ‘h8’ can be obtained by using the equation
[pic]

Efficiency: - Now that the enthalpies at each point are known the efficiency of the cycle can be calculated. The efficiency is given by
[pic]
[pic]

where ‘m’ is the mass flow rate and is calculated as follows
By mass conservation equation

[pic]
By energy conservation equation
[pic]

Having two equations and two unknowns, we solve this system of linear equations and solve for [pic] which is the mass fraction we are interested in:

[pic]

Regenerative Cycle1 Calculations

State 1: - [pic] [pic] [pic]
State 2: - [pic] [pic] [pic] [pic]
State 3: - [pic] [pic] [pic]
State 4: - [pic] [pic] [pic] [pic]
State 5: - [pic] [pic] [pic] [pic]
State 6: - [pic] [pic] [pic]
State 7: - [pic] [pic] [pic] [pic]
State 8: - [pic] [pic] [pic] [pic] [pic] [pic] [pic] [pic] [pic] [pic] [pic] [pic]
Mass Rate: - [pic]

[pic]

[pic]
Efficiency of the cycle: -

[pic]

[pic]
[pic]

Regenerative Cycle with Two FWH’s
[pic][pic]
Figure 7.0: - Regenerative Cycle2 with Two FWH’s

Regenerative Cycle2 Analysis

State 1: - The pressure ‘P1’ is the exhaust pressure and is fixed from the reheat cycle, since point one is on the saturated liquid line two intensive properties specific volume and specific enthalpy are obtained for the saturated liquid table.
For P1 = 1.85 {v1 = vf@P1 {h1 = hf@P1
State 2: -The pressure at this point is the same as the inlet pressure of low-pressure turbine. The inlet pressure of the low-pressure turbine is selected to be 300 psi. The enthalpy is obtained by the relation
[pic]
Since the pump is isentropic
[pic]
The relation becomes
[pic]
The difference in pressures is
[pic]
The enthalpy in now given by
[pic]
State 3: - The pressure at this point is the same as the pressure at point ‘2’ (P2 = P3). Using saturated liquid table the specific volume and enthalpy at this point can be calculated.

For P3 {v3 = vf@P3 {h3 = hf@P3

State 4: - The pressure at this point is the same as the reheat pressure obtained from the reheat cycle. The enthalpy at this point can be determined as follows.
[pic]
Since the pump is isentropic
[pic]
The relation becomes
[pic]
The difference in pressures is
[pic]
The enthalpy in now given by
[pic]
State 5: - At this point the turbine inlet pressure and temperature are known, and from the superheated steam tables the enthalpy and the entropy is obtained.
For P5 {v5 = vf@P5 {h5 = hf@P5
State 6: - This pressure at this point is the same as the reheat pressure obtained form the reheat cycle. The enthalpy is obtained as follows.
[pic]
Since the pump is isentropic
[pic]
The relation becomes
[pic]
The difference in pressures is
[pic]
The enthalpy in now given by
[pic]

State 7: - The pressure and temperature at this point are for the high-pressure turbine and are given. The enthalpy can be determined form the superheated steam tables.
For P7 {h7 T7 {s7
State 8: - The pressure at this point is the reheat pressure form the reheat cycle and the entropy at this point is the same as the entropy at point ‘7’.
For P8 {h8 S8
State 9: - The pressure at this point is the same as the pressure at point ‘8’ and the temperature at this point is the same as the inlet temperature of the high-pressure turbine. Knowing these two properties the enthalpy and entropy can be determined.
For P9 {h9 T9 {s9
State 10: - This point is in the saturated water-vapor mixture region, so in order to determine the enthalpy the quality needs to be determined. The pressure at this point is known from the reheat cycle and the entropy at this point is the same as the entropy at point ‘9’.
For P10 {hf S10 {hfg {Sf {Sfg
Now we use the equation
[pic]
This equation can be solved to obtain ‘x8’ and ‘h8’ can be obtained by using the equation
[pic]

Efficiency: - Now that the enthalpies at each point are known the efficiency of the cycle can be calculated. The efficiency is given by
[pic]
[pic]

By mass conservation equation

[pic]
By energy conservation equation

[pic]

Having two equations and two unknowns, we solve this system of linear equations and solve for [pic] which is the mass fraction we are interested in:

[pic]

Subsequently, the mass fraction [pic] is calculated as follows:

By mass conservation equation

[pic]
By energy conservation equation

[pic]

Having two equations and two unknowns; and knowing the value[pic], we solve this system of linear equations and solve for [pic] which is the other mass fraction we are interested in:

[pic]

Regenerative Cycle2 Calculations

State 1: - [pic] [pic] [pic]
State 2: - [pic] [pic] [pic] [pic]
State 3: - [pic] [pic] [pic]
State 4: - [pic] [pic] [pic] [pic]
State 5: - [pic] [pic] [pic]
State 6: - [pic] [pic] [pic] [pic]
State 7: - [pic] [pic] [pic] [pic]
State 8: - [pic] [pic]

[pic]
State 9: - [pic] [pic] [pic] [pic]

State 10: - [pic] [pic] [pic]

State 11: - [pic] [pic] [pic] [pic] [pic] [pic] [pic] [pic] [pic] [pic] [pic] [pic]

Mass Rate [[pic]]

[pic] [pic]

[pic]
Mass Rate [[pic]] [pic]

[pic] [pic]
Efficiency

[pic]

[pic]

[pic]

Regenerative Cycle with Four FWH’s

[pic]

[pic]
Figure 8.0: - Regenerative Cycle3 with Four FWH’s

In this cycle there are four FWH’s, ant they are connected to the high-pressure turbine, intermediate-pressure turbine and low-pressure turbine. The analysis of this cycle is similar to the previous cycles and for every FWH an additional pump needs to be added. The pressure lines of the FWH’s are 600 psi, 300 psi, 100 psi & 50 psi. Following a similar technique employed in the analysis of the previous regenerative cycles the results of this cycle are shown below. From the ‘T-s’ diagram it can be seen that the cycle consists of 17 points, so in order to calculate the efficiency the enthalpies at all ‘17’ points need to be calculated along with the mass flow rate of each feed water heater.

Regenerative Cycle3 Calculations

State 1: - [pic] [pic] [pic]
State 2: - [pic] [pic] [pic] [pic]
State 3: - [pic] [pic] [pic] [pic]
State 4: - [pic] [pic] [pic] [pic]
State 5: - [pic] [pic] [pic] [pic]
State 6: - [pic] [pic] [pic] [pic] [pic]
State 7: - [pic] [pic] [pic] [pic]
State 8: - [pic] [pic] [pic] [pic] [pic]
State 9: - [pic] [pic] [pic] [pic]
State 10: - [pic] [pic] [pic] [pic] [pic]
State 11: - [pic] [pic] [pic] [pic]
State 12: - [pic] [pic] [pic]
State 13: - [pic] [pic] [pic] [pic]
State 14: - [pic] [pic] [pic]
State 15: - [pic] [pic] [pic]
State 16: - [pic] [pic] [pic]
State 17: - [pic] [pic] [pic]
State 18: - [pic] [pic] [pic] [pic] [pic] [pic] [pic] [pic] [pic] [pic] [pic] [pic]
By energy conservation equation

[pic]

[pic]

[pic]

[pic]

[pic]

[pic]

[pic]

[pic]

[pic]

[pic]

[pic]

[pic]

[pic]

[pic] [pic]

[pic]

Now with these enthalpies the mass flow rate was determined and efficiency of the cycle was calculated. The efficiency of this cycle came out to be ‘(=48.62%’

| |Efficiency |# of FWH |
|Case 1 |46.64 |1 |
|Case 2 |47.68 |2 |
|Case 3 |48.62 |4 |

Table 4.0: - Regenerative Cycle

[pic]
Figure 9.0: - Efficiency vs Number of Feed Water Heaters

Cost Effectiveness

Let us find the cost effectiveness of the steam turbine cycle with 4 feed water heaters.

Revenue = (ηth2 – ηth1)*Power*POE*Y

Where

ηth1 = 0. 4664 (Efficiency of the cycle with one FWH)

ηth2 = 0. 4768 (Efficiency of the cycle with Two FWH)

ηth3 = 0. 4862 (Efficiency of the cycle with Four FWH’s)

Power = 160,000 kwt

POE = $0.20/ kwt

Y = 8760 hours

1) Revenue = (0.4664)*(0.20)(160000)(8760)
Revenue = $130,741,248

2) Revenue = (0. 4768)*(0.20)(160000)(8760)
Revenue = $133,656,576

3) Revenue = (0. 4862)*(0.20)(160000)(8760)
Revenue = $136,291,584

Conclusion

This projects enabled the students to work on a project that they are more likely to encounter in the work place. This project enabled the students to understand what cycles are really used for and by varying certain parameters and adding certain devices the overall efficiency can be dramatically increased. Through tedious time consuming calculations the pressures at various states were determined that would increase the efficiency of the cycle. The cost effectiveness of the regenerative cycle was also determined and it was determined that the revenue generated was large enough and the use of four FWH’s in a power plant is justified.

REFERENCES

Yunus A. Cengel, Michel A. Boles. Thermodynamics: an Engineering Approach. Fourth Edition. 2002. Mc Graw Hill

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