MEMORANDUMTo: Talmage ClementsFrom: Team 7: Eric Canonge; Alexander Lanka; Andrew Neill; Andrew TamayoDate: November 27, 2007Subj: Feedwater Heater Replacement

The purpose of this memo is to inform you that we have completed our analysis of the feedwater heater and have come to a conclusion about how to proceed from here. We have determined that the best course of action is to install a new feedwater heater in a configuration with ¾” OD piping and 3 passes. This was determined to have the lowest initial cost, as well as the lowest net present value cost.

Please see the attached report for details on how we arrived at our conclusion about the feedwater heater replacement project.

College of Engineering

Feedwater Heater Replacement

Team 7Eric Canonge - Engineer

Alexander Lanka - EngineerAndrew Neill - Engineer

Andrew Tamayo - Engineer

MAE 412, Section 002, Mech & Aero Eng Dept, NCSU

11/27/2007

Table of Contents

Nomenclature....................................................................................................................................2Executive Summary..........................................................................................................................3Introduction.......................................................................................................................................4Current Operation.............................................................................................................................5Objectives.........................................................................................................................................5Solution.............................................................................................................................................6Summary of Results..........................................................................................................................8Recommendation..............................................................................................................................9Appendices......................................................................................................................................10 Appendix A - Heat Exchanger Design.................................................................................10 Appendix B - Condensate System Design............................................................................18 Appendix C – Cost-Optimization and Selection of Tubing Option......................................25 Appendix D – Bibliography..................................................................................................30

Nomenclature

BWG – Birmingham Wire Gauge. The wall thickness of the individual pipes. NPSH – Net Positive Suction Head. The amount of pressure needed to overcome the liquid

vapor pressure inside the piping, thereby avoiding cavitation in the pump. OD – Outer diameter of the individual pipes in the feedwater heater O & M - Operating and maintenance costs Passes – the number of feedwater heater lengths the water travels Salvage Value – the raw material value of the heat exchanger at the end of its life TTD – Terminal temperature difference. The difference between the shell side temperature

and the feedwater outlet temperature.

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Executive Summary

Our assignment was to determine the best option regarding the broken, and consequently bypassed, feedwater heater. We considered four replacement options, as well as doing nothing in this report.

The four replacement options were different configurations for the feedwater heater and are as follows:

1. ½” OD piping, 2 passes2. ½” OD piping, 3 passes3. ¾” OD piping, 2 passes4. ¾” OD piping, 3 passes

After considering these four options as well as the possibility of doing nothing, we have determined the best option is option 4. This option requires the new feedwater heater to contain 939 tubes per pass. The tubes should be made of stainless steel, have an outer diameter of ¾”, and have a 22 BWG wall thickness. Each tube pass is 40 feet long, therefore the total length of piping inside the feedwater heater is 112,680 feet and will result in an inner shell diameter of 53 inches. The estimated cost of the tubing for this option is $78,876 and would cost $56,340 to install. The cost of the shell and other internals is $79,500. The total cost is then $214,716. The net present worth cost for this option including taxes, insurance, O & M and salvage is $271,442. The total cost advantage over the next best option of ¾” OD piping with 2 passes is $25,532.

The capacity of the generator without the heat exchanger is 733.5675 MW, and is 735 MW with the heat exchanger. It was found that the cost of not replacing the feedwater heater would result in a cost of 8.39 million dollars. It should be noted that a cost analysis of option 2 was not necessary because the currently installed pumps could not handle the pressure drop caused by this particular configuration. Once this was determined, it was eliminated from our analysis.

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Introduction

The reason a replacement for the feedwater heater must be considered is that the current feedwater heater has developed so many tube leaks that is has been taken out of service and is being bypassed completely. Because of this, the plant is not as effective due the reduction of cycle performance because of the loss of feedwater heating prior to entering the boiler. Our team was asked to provide an analysis of four replacement options and to determine the best course of action regarding the feedwater heater.

For the four options we have considered, each of these combines different pipe wall thicknesses as well as different numbers of tube passes. The other option we have considered is leaving the feedwater heater as it is, should we find that it is the most economical option.

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Current Operation

The current setup provided many constraining factors that we had to consider when seeking the best solution to the current problem. The first factor considered was our four piping options for the replacement feedwater heater. Option one was a heater with ½” OD piping and 2 passes. Option two was a heater with ½” OD piping and 3 passes. Option three was a heater with ¾” OD piping and 2 passes. Option four was a heater with ¾” OD piping and 3 passes. All of these options would use stainless steel piping with 22 BWG wall thickness. All these options also were limited by a space requirement of 40 feet in length. We were also given incoming flow and outgoing flow temperature and pressure parameters that helped us to calculate the required length of piping for adequate heat transfer as well as being given the extraction steam and shell drain flow for each of the four options. These parameters and the feedwater layout can be seen in Figure A1.1. A diagram of the piping layout for the flow from the condenser well to the feedwater heater that was given to us can be seen in Figure B2.2. Using this layout and the Byron Jackson pump curves seen in Figure B2.1, we were able to perform calculations to determine information about the condensate system design, the head delivered by the pumps, and the pressure losses occurring throughout the piping system. Finally, we performed the economic analysis using the following set of cost factors.

Cost of tubing: 0.875 304 SS - $0.70/ft 0.500 304 SS - $0.50/ft

Installation Cost: $20/tubeShell & internals except tubing: $1500/inch of diameterInterest: 5%/yearInflation: 3%/yearTaxes: 1% of initial cost/year, flatInsurance: 0.08%/yr of initial installed cost, escalated for inflationOperation & Maintenance cost: 1% of initial cost/year, escalated for inflationSalvage Value: 5% of initial cost, escalated to the tubing lifetimeTubing life: 20 yearsImprovement in cycle generation and heat rate with TTD: 0.004% of capacity/TTD degreeUnit maximum capacity penalty factor, 60 $/kW-yr escalated with inflation from Year 1Unit efficiency penalty factor, 50 $/MWh escalated with inflation from Year 1Operating time: equivalent full-load for 70% of the year (8760 hr/std year)

Objectives

For this project, we have been asked to provide the following information:

The number of tubes and their characteristics (material, diameter, length, wall thickness) The resulting shell diameter of the feedwater heater The design flow on the tube side and inlet and outlet temperatures The present value total cost analysis of the four options The cost of not replacing the feedwater heater

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Solution

Phase 1 – Feedwater Heater Design (For detailed calculations, please see Appendix A.)

1) The first step was the calculation of total thermal resistance area in the tubing for each option. This was done using the Log Mean Temperature Difference method on the feedwater heater.

2) Then we calculated the internal pipe resistance. By finding the Reynolds and Nusselt numbers for the flow in each option, the internal resistance could be calculated.

3) The heat transfer areas per tube were then found for both the inside and outside of each pipe configuration.

4) The verification of thermal resistance area per tube with determined number of tubes was then completed. This was done using a spreadsheet and iteration on the number of tubes until the previously calculated total thermal resistance area matched the one found by iteration.

5) The next step was the calculation of the shell diameter. Trigonometry was used to determine the number of triangles that could fit in a circle. This was done because the most efficient packing method for the tubes is to stagger them so that triangles are formed between any three tubes.

6) The extraction steam and shell drain flow were then found. This was done by performing an energy balance on the feedwater heater using the given parameters.

7) Finally, the feedwater outlet temperature under winter conditions was calculated. Given the new conditions, new properties were determined. Also, a new value for thermal resistance was calculated. From this, the NTU method was used to determine the maximum heat transfer rate. From that, the new outlet temperature was calculated.

Phase 2 – Condensate System Design(For detailed calculations, please see Appendix B)

1) The first step was the calculation of the condensate pump discharge pressure. This was done by reading the pump chart in Figure B2.1 at the design volume flow rate. The volume flow rate had to be halved because the flow coming from the condenser hotwell into the pumps was along a parallel path with identical pumps and piping.

2) The next step was the calculation of the outlet feedwater pressure. By referring to the piping schematic in Figure B2.2, the pressure at the outlet was found by first finding the pressure drop from the pump outlet to the feedwater inlet and then by finding the pressure drop from the feedwater inlet to the feedwater outlet. The piping system was divided into 4 sections as seen in the figure. By finding the pressure drop in each section, the outlet feedwater pressure was determined.

3) The calculation of the depth of submersion to not violate NPSH was done by reading the NPSH requirement off of the pump curve in Figure B2.1 and then solving for the minimum required pump depth using the NPSH formula.

4) The next step was the calculation of the maximum flow rate without tripping the generating unit. This was done by simplifying Bernoulli’s equation due to the constant elevation and constant volume flow in the section of piping after the feedwater heater. Because all the piping between the feedwater outlet and the boiler feedpump inlet is unknown and the

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pressure drop is known, a flow loss constant K was found that “described” the unknown piping. After finding a value for K, the maximum flow rate was found by simply rearranging the equation to solve for V. It was found that the maximum flow rate for the ½”, 3 pass case was low enough to trip the low suction alarm, and therefore was eliminated as a possible replacement option.

5) The final step was to calculate the diameter of a hole in the partition in the feedwater heater such that half of the flow bypassed the tubes and therefore was not heated. As can be seen in Figure B2.3, half of the flow went through the pipes, while the other half went through the hole. Knowing that the pressure drop through the pipes had to be equal to the pressure drop across the hole, the volume flow rate through the hole could be found. Based on this, the area and consequently, the diameter, could also be found.

Phase 3 – Cost-Optimization and Selection of Tubing Option(For detailed calculations, please see Appendix C)

1) The first step is to determine the cost of the materials involved in building each of the feedwater heater configurations. The number of tubes in each of the configurations, as well as the physical dimensions of the feedwater heater configurations are used from Phases I and II of this report. The material costs are a function of these dimensions and these are considered the initial costs for installing each of the configurations.

2) Once the feedwater heater has been installed, it is subject to taxation, which has been assumed to be a constant rate of 1% of the initial cost and is paid on a yearly basis at the end of the year. The assumption that the tax is constant is due to the fact that taxes are based on many different factors including depreciation, inflation, and changes in the tax rates. The actual cost as reported is based on the present value method, which takes into account the fact that money can be invested now and carry a higher value at the time it will actually be spent. The Present Worth of a Uniform Series method of calculation can be used in this case.

3) The next step is to determine the cost of insurance for the feedwater heater. The insurance covers any unforeseen damage to the feedwater heater and the cost is a factor of the initial cost. The cost of replacement for the feedwater heater increases with time due to inflation. As a result, the insurance costs take inflation into account. The reported costs also take the present value method into account. The Present Worth of a Geometric Series method can be used to calculate the reported costs involved with insuring the feedwater heater over the life expectancy of the apparatus.

4) The feedwater heater also requires regular maintenance and operational procedures that add to the cost of operating the heater. These costs are determined to be approximately equal to 1% of the initial cost of the heater. As the parts increase in cost due to inflation, the cost to maintain the heater also increases as a result of inflation. The present value method was used to determine the reported costs involved in operating and maintaining the feedwater heater. The Present Worth of a Geometric Series method can be used to take inflation and the present value method into account when calculating the costs of maintaining and operating the feedwater heater over the life of the apparatus.

5) The salvage value is the final factor affecting the costs and benefits of the material aspect of the feedwater heater. At the end of the tubing life for the feedwater heater, the materials can be sold for 5% of the initial cost of the feedwater heater, taking inflation into account. The present value method is also used to determine how this return would compare to a dollar amount invested today at current interest rates over the tube life expectancy.

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6) The base case for this analysis is the process of bypassing the feedwater heater. This results in a much higher Terminal Temperature Difference and a resulting decrease in plant efficiency and maximum capacity. The design capacity for the plant is known from when the previous feedwater heater was in service. The reduction of capacity of the plant as a factor of the TTD is determined based on the design capacity. This reduction in capacity and efficiency results in higher operation costs.

7) The Maximum Capacity penalty factor is a function of the loss of overall capacity involved with bypassing the feedwater heater. This will result in the need for additional capacity to be built into the system in order to meet system demand and an additional cost for when considering bypassing the feedwater heater. This plays in as a benefit when considering heater replacement. Inflation and present worth are considered when determing the effects of the reduction in capacity over the life of the tubes.

8) The Efficiency Penalty Factor takes into account the fact that more fuel is needed to create the same rise in temperature in the boiler if the feedwater is not heated by the feedwater heater first. This is a cost associated with not replacing the feedwater heater and is therefore a benefit when considering heater replacement. Inflation and present worth are considered when determining the effects of increased fuel expenses over the life of the feedwater heater.

9) Once all of these factors have been calculated, the present worth costs and benefits of each option are compared and the option with the lowest cost and highest benefit is chosen.

Summary of Results

After analyzing the four feedwater configuration options, the number of tubes per pass, total length of tubing and inside shell diameter were found for each option. The tubing in each option is made of stainless steel with a wall thickness of 22 BWG. The results are shown below in Table 1.

Table 1: Feedwater Heater Design Parameters  Configuration 1 Configuration 2 Configuration 3 Configuration 4  1/2'' pipe, 2 Passes 1/2'' pipe, 3 Passes 3/4'' pipe, 2 passes 3/4'' pipe, 3 passesNumber of Tubes per Pass 2139 1335 1572 939Length of Tubing Required (ft) 171,120 160,200 125,760 112,680Inside Shell Diameter (in) 43 42 56 53

Design flow on the tube side of the heat exchanger: 4,000,000

Feedwater inlet temperature: 110°FFeedwater outlet temperature: 168.82°F

Table 2: Configuration Costs  Configuration 1 Configuration 3 Configuration 4  1/2'' pipe, 2 Passes 3/4'' pipe, 2 passes 3/4'' pipe, 3 passesInitial Cost $235,620 $234,912 $214,716Net Present Value Cost $297,740 $296,845 $271,324

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From the data in Table 2, it can be seen that configuration 4 has the lowest initial cost and the lowest net present value cost. It should be noted that configuration 2 is not included in the cost analysis because the pumps could not handle the pressure drop created by this configuration.

The cost of not replacing the feedwater heater due to loss of efficiency and maximum capacity is $8.39 million.

Recommendation

For the four configurations considered, it was determined that the ¾” OD piping, 3 pass configuration has the lowest net present value cost in addition to the lowest initial cost. Also, this configuration has a cost advantage of $25,521 over the next least expensive option, configuration 3. For these reasons, we recommend installing a feedwater heater with a ¾”, 3 pass configuration. Over the 20 year life of this feedwater heater, the savings would be approximately $8.1 million over the alternative of not replacing the feedwater heater.

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Appendices

Appendix A - Heat Exchanger Design

Section I: Heat Exchanger Fluid Properties

Extraction steam pressure at shell inlet:

Table A1.1: Steam properties at heat exchanger shell inletInlet Outlet

Property Units Value Property Units ValueTemp F 173.82 Temp F 173.82Pressure psia 6.58 Pressure Psia 6.58Enthalpy BTU/lbm 1141 Enthalpy BTU/lbm 141.85Quality 1 Quality 0

Terminal temperature difference and feedwater outlet temperature:

Table A1.2: Temperatures of feedwater in heat exchangerInlet Outlet

Property Units Value Property Units ValueTemp F 120 Temp F 168.82Pressure psia 300 Pressure psia  

Mean feedwater temperature:

Table A1.3: Feedwater Properties at mean inlet and outlet temperatureProperties at Tmean=144.41oF

Property Units ValueDyn. Visc. (lbm/in∙s) 2.53E-05Them. Cond. (Btu/in∙ft∙R) 8.77E-06Specific Heat (Btu/lbm∙R) 1.00E+00Prandlt 2.85

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Figure A1.1: Heat Exchanger Configuration

The information about the inlet drain flow is included in Figure A1.1.

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Section II: Sample Calculations for ½” Pipe with 2 Passes

Calculation of Total Thermal Resistance Area:

The heat transfer to the feedwater is calculated using the equation:

The log mean temperature difference method is used to find UA:

The correction factor F = 1 because of the phase change.

Solving for UA,

Calculation of Internal Pipe Resistance:

The Reynolds number can be found using the equation:

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The Nusselt number can be found from the Reynolds and Prandlt numbers:

The coefficient of thermal resistance due to convection inside the pipe is a function of the Nusselt number.

Calculation of Heat Transfer Areas per Tube:

Verification of Thermal Resistance Area per Tube with Determined Number of Tubes:

Knowing the necessary UA value from above, the number of tubes required was found by iteration. The resistance method serves to verify the accuracy of the previous calculation.

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Verification of Total Thermal Resistance Area with Determined Number of Tubes:

By multiplying the thermal resistance area per tube by the number of tubes, the total thermal resistance area was confirmed.

Calculation of Shell Diameter:

The shell diameter for the feedwater heater is a function of the number of tubes as well as the method in which they are packed. The most effective method for packing the most tubes into the least amount of space is by staggering the tubes so that their centers are connected by an equilateral triangle, as seen below in Figure A2.1:

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Figure A2.1: Arrangement of Tubes in the ShellWhile the tubes could be packed so that they are touching, this would impede the flow of the hot fluid, steam in this case. For this problem, the ligament, or distance between the tubes, is set at ¼ of the diameter. From this, it can be concluded that the distance between the centers of the tubes

is of the outer Diameter. This distance is also the side length of the equilateral triangle. The

figure also shows that 1/6th of each of the three tubes is contained within the area of the triangle. From this, each triangle accounts for ½ of a tube. Thus, for the final shell area, there will be twice as many equilateral triangles as tubes.

The area of each triangle now needs to be determined. The triangle can be seen below in Figure A2.2:

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Figure A2.2: Triangle Between TubesThe Pythagorean Theorem is done on the left half of the triangle:

The unknown height of the triangle is isolated:

The area of the triangle is now calculated as:

As was mentioned earlier, there are two triangles for every tube. Thus, the total inside cross sectional area of the shell of the heat exchanger is as follows:

The inside diameter of the shell would then be found using the area of a circle, which would be found as follows:

For example, when analyzing the ½” 2-Pass set up, the following values are found:

Calculation of Extraction Steam and Shell Drain Flow:

By performing and energy balance on the feedwater heater, the outlet drain mass flow was calculated.

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The outlet drain mass flow is equivalent to the inlet drain mass flow combined with the turbine stage mass flow.

Rearranging the energy balance,

The mass flow from the turbine stage was calculated to be:

The outlet drain mass flow was calculated to be:

Calculation of Feedwater Outlet Temperature under Winter Conditions:

For a feedwater flow of at 110°F entering the tubes, a guess of was

made to find the water properties.

Table A2.1: Feedwater Properties at mean inlet and outlet temperatureProperties at Tmean=140oF

Property Units ValueDyn. Visc. (lbf*s/ft2) 1.004E-05Specific Heat (Btu/lbm∙R) 0.9983Prandlt 2.99

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Using similar calculation as before, the Reynolds and Nusselt numbers are found to be 91821

and 327 respectively. The internal thermal resistance was found to be 2138 . Substituting

this new resistance value into the earlier equation, U is found to be .

Using the NTU method,

where,

Using the equation,

q is found by,

It is known that

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Appendix B - Condensate System Design

Section I: Fluid and Material Properties

Table B1.1: Water Properties at 120oFProperty Units ValueViscosity lbf.s/ft2 1.17E-05Density slug/ft3 1.918

Pv Psia 1.695

Table B1.2: Water Properties at 145oFProperty Units ValueViscosity lbf.s/ft2 9.39E-06Density slug/ft3 1.905

Table B1.3: Water properties at 4 in. HgProperty Units ValueViscosity lbf.s/ft2 5.97E-06Density lb/ft3 61.632

Table B1.4: Roughness numbersMaterial Roughness

Commercial Steel 0.00015 ftBrass 0.000007 ft

Design mass flow rate:

Design volume flow rate:

Section II: Sample Calculations

Calculation of Condensate Pump Discharge Pressure:

The condensate discharge pressure was found by reading the pump chart at the design volume flow rate. The volume flow rate had to be halved since the flow coming from the condenser hotwell into the pumps is along a parallel path with identical pumps and piping.

Design volume flow rate per pump: Qpump=Qtot/2=8081.44 GPM/2=4040.72 GPM

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Figure B2.1: Condensate Pump Curve with marked design flow and corresponding pump head

Calculation of Outlet Feedwater Pressure:

Referring to the piping schematic, the outlet pressure at the outlet can be found by first finding the pressure drop from the pump outlet to the feedwater inlet and then by finding the pressure drop from the feedwater inlet to the feedwater outlet:

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Figure B2.2: Piping schematic between condenser and feedwater heater

There are two main sections for which to calculate a drop in pressure; the section from the condenser outlet to the feedwater heater inlet (point A to C) and then the pressure drop across the feedwater heater (point C to D). The pressure drop from A to C can be subdivided into the following sections of piping:

A-P: The condenser outlet to the pump inlet, of which there are two parallel sections necessitating the design flow to be halved per section, and containing 24’’ pipe 20 ft long and containing minor losses due to a sharp entrance

P-B: The pump outlet to the tee joint that joins the two parallel flows, which has pipe of 16’’ in diameter and a total length of 14 ft containing minor losses due to one gate valve and a 90o ell.

AP

B

C D`

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B-C: From the tee joint to the feedwater inlet; the flow from the parallel pumps combines and the flow is equal to the design flow. The piping contains 140 ft of 20’’ pipe and contains minor losses due to one gate valve, three 90o ells, branch flow out of a tee joint, and a sharp exit.

The formula to find pressure loss can be written as:

Isolating the pressure terms yields:

Since pa is assumed to be 0.0 psia, the equation for the inlet feedwater pressure, pc can be reduced to:

Where hp is the discharge pressure of the pump, Zc-Za is the elevation change, the velocity term is the change in velocity from the condenser outlet to the feedwater inlet, and hA-C is the head loss due to major frictional losses and minor losses. This term can also be written as:

Where for any of the individual head loss terms

A detailed numerical solution for pc solving for individual head loss terms can be found in the attached Excel spreadsheet. The final value for the feedwater heater inlet pressure was found to be 530.01 psia.

Once the inlet feedwater heater pressure is known, the pressure drop across the feedwater heater needs to be found to find the feedwater heater outlet pressure. Since there is no elevation change across the feedwater heater and no extra flow is added into the tube side of the heat exchanger, the equation for the pressure drop across the heat exchanger can be written as:

Where the subscripts c and d refer to the piping schematic diagram. The head loss can be solved similarly to the method above, where frictional losses occur from sudden expansions and contractions as fluid moves from the tube to the plenum in each pass of the heat exchanger. This K value is 1.5, yielding a total minor loss K factor of 3 for 2-pass cases and 4.5 for 3-pass cases.

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The numerical values for each configuration are shown in the attached spreadsheet; a sample calculation for the ½’’ 2-pass configuration is shown below:

The

outlet pressure, pd, is the inlet pressure minus this pressure drop:

Calculation of the depth of submersion to not violate NPSH:

Reading from the pump curve, the NPSH requirement at the design flow per pump of 4040.72 GPM is found to be 11 ft. The NPSH formula is given by:

Solving for Zi, the minimum depth of submersion, yields:

Pa is given as 4 in. Hg, which converts to Pv is given by water properties and hfi was found as the head loss from the condenser hotwell to the condensate pump; this was solved for when finding the pressure drop across the piping system to the feedwater heater. Substituting these values yields

Calculation for max flow rate without tripping generating unit:

The pressure drop from the feedwater outlet to the boiler pump inlet can be given by the equation:

Since there is assumed to be no change in elevation or pipe diameter and the volume flow is assumed to be constant throughout the pipe, the equation can be reduced to:

, where can be written as

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Since all the piping between the feedwater outlet and the boiler feedpump inlet is unknown and the pressure drop is known, a flow loss constant K can be found that “describes” this unknown piping. This is found with the equation:

WhereP1 = 510.74 psia (feedwater outlet pressure for ¾’’ 3 pass case)p2 = 260 psig = 274.7psia

Assuming that the piping exiting the feedwater heat exchanger is 20’’ pipe, the design velocity coming out of the heat exchanger can be given by the equation:

V = Q/A =

Substituting values for K equation yields:

With this K value, the maximum flow rate can be found changing the velocity flow rate. The effect on the pressure drop to the feedwater heat exchanger outlet can then be determined and with the new pressure drop to the outlet the K value found above can be used to determine the pressure drop to the boiler feedpump suction. This process can be done easily using the pump curve to determine new discharge pressures and an excel spreadsheet to find pressure drops through the line. Iterating using this process, the maximum flow is found to be approximately 4,100,000 lbm/hr

Calculation for partition hole diameter:

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The situation where there is a hole in the partition can be modeled as a parallel flow to the flow in the tubes. Because of this, half of the flow goes through the pipes, and the other half goes through the hole as shown in Figure AI.3.

Figure B2.3: Flow through Feedwater Heater with Hole in Partition

The flow rate along each path is . Knowing that the pressure drop across the hole must

be equal to the pressure drop in the fluid that travels through the tubes, they can be set equal to each other as such:

The hole can be approximated as a very short tube, therefore having a length of zero. Also the factor can be removed from both sides of the equation yielding:

For the case with the ¾” piping with 3 passes, the fluid velocity, friction factor, tube length and diameter, and losses are all known. Also, the hole can be considered to have a sharp entrance and a sharp exit, thereby causing to have a value of 1.5. Substituting the known values reveals and solving for the velocity of the fluid in the hole reveals:

Dividing the flow rate by this velocity gives the area of the hole.

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The diameter of the hole can easily be found from the area.

Appendix C – Cost-Optimization and Selection of Tubing Option

The cost analysis for the heat exchanger is divided into two fundamental categories; finding the cost of replacing the heat exchanger, and finding the cost of doing nothing to replace the heat exchanger.

Within the cost of the heat exchanger, there are different initial costs depending upon the configurations under consideration for the heat exchangers. Of the four initial possible configurations, one was determined to be a non-starter since the pressure drop across it was too great. Of the remaining configurations, the parameters are listed in the table below:

Table C.1: Heat exchanger physical parametersConfiguration 1 Configuration 2 Configuration 31/2'' pipe, 2 Passes

3/4'' pipe, 2 passes

3/4'' pipe, 3 passes

Number of Tubes 4278 3144 2817Length of Tubing Required (ft) 171120 125760 112680Inside Shell Diameter (in) 43 56 53

The expenses associated with these heat exchangers as well as other financial information are listed in the tables below:

Table C.2: Material and installation costs for heat exchanger3/4'' 304SS $0.70  dollars/ft1/2'' 304SS $0.50  dollars/ftInstallation cost $20  dollars/tubeShells and internals except tubing $1,500  inch of diameter

Table C.3: Rates applicable to replacement heat exchanger

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Interest 5%Inflation 3%Taxes 1% initial cost/year, flatInsurance 0.08% initial cost/year, escalated for inflationO&M cost 1% initial cost/year, escalated for inflation

Table C.4: Relevant financial informationSalvage value 5% initial cost, escalated to tubing lifetimeTubing life 20 yearsImprovement in cycle generation 0.004% of capacity/TTD degreeUnit maximum capacity penalty factor 60 dollars/kW-yr escalated with inflation from year 1Unit effecienty penalty factor 50 dollars/MWh escalated with inflation from Year 1Operating time: 70% equivalent full-load (8760 hr/std year)

By applying the cost of materials and installation to the different heat exchanger configurations, the results for the initial cost of these configurations could be obtained. These costs are shown in the table below:

Table C.5: Initial heat exchanger costsConfiguration 1 Configuration 2 Configuration 3

1/2'' pipe, 2 Passes 3/4'' pipe, 2 passes 3/4'' pipe, 3 passesCost of Tubing material $85,560.00  $88,032.00  $78,876.00 Cost of Tubing Installation $85,560  $62,880  $56,340 Cost of shell and internals $64,500  $84,000  $79,500 Total heat x initial cost $235,620.00  $234,912.00  $214,716.00 

With the initial cost of the heat exchanger calculated, the tax, insurance, and operation and maintenance rates could be applied to project the expense that the heat exchanger would occur in present day dollars. The salvage value of the heat exchanger was also calculated based on the initial cost of each configuration.

Each rate was applied uniquely. The expense of the heat exchanger due to taxes was a flat one percent of the initial cost, which had to be paid yearly starting at the end of the first year. The insurance expense was 0.08 percent of the initial cost, and increased each year due to the three percent inflation specified. The insurance was paid at the beginning of each year. The operation and maintenance cost was one percent of the initial cost, escalated with inflation and paid annually starting at the end of the first year. These maintenance costs for each year were summed and the present worth of these costs for each year was taken at five percent interest assuming annual compounding. These present worth values were summed to obtain the total project expense of the heat exchanger due to taxes, insurance, and operation and maintenance for the 20 year life of the heat exchanger. This was done for each configuration. An example for the ½’’ 2 pass heat exchanger configuration is shown below:

Table C.6: Maintenance costs of ½’’ 2 pass heat exchanger over life of exchanger

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Year Tax Expense Insurance expense O&M Cost Total Maintenance Costs Present Worth of Maintenance0 $0.00 $188.50 $0.00 $188.50 $188.501 $2,356.20 $194.15 $2,356.20 $4,906.55 $4,672.912 $2,356.20 $199.98 $2,426.89 $4,983.06 $4,519.783 $2,356.20 $205.97 $2,499.69 $5,061.87 $4,372.634 $2,356.20 $212.15 $2,574.68 $5,143.04 $4,231.195 $2,356.20 $218.52 $2,651.92 $5,226.64 $4,095.216 $2,356.20 $225.07 $2,731.48 $5,312.76 $3,964.467 $2,356.20 $231.83 $2,813.43 $5,401.45 $3,838.718 $2,356.20 $238.78 $2,897.83 $5,492.81 $3,717.759 $2,356.20 $245.94 $2,984.76 $5,586.91 $3,601.37

10 $2,356.20 $253.32 $3,074.31 $5,683.83 $3,489.3811 $2,356.20 $260.92 $3,166.54 $5,783.66 $3,381.5912 $2,356.20 $268.75 $3,261.53 $5,886.48 $3,277.8113 $2,356.20 $276.81 $3,359.38 $5,992.39 $3,177.8914 $2,356.20 $285.12 $3,460.16 $6,101.48 $3,081.6615 $2,356.20 $293.67 $3,563.96 $6,213.83 $2,988.9616 $2,356.20 $302.48 $3,670.88 $6,329.56 $2,899.6517 $2,356.20 $311.56 $3,781.01 $6,448.76 $2,813.5718 $2,356.20 $320.90 $3,894.44 $6,571.54 $2,730.6119 $2,356.20 $330.53 $4,011.27 $6,698.00 $2,650.6320 $2,356.20 $0.00 $4,131.61 $6,487.81 $2,445.19

This analysis resulted in a present value cost of $70,139 for the projected expenses of the heat exchanger.

The salvage value was determined in a similar way, except that the salvage value helped to reduce the overall cost of the heat exchanger. The salvage value was taken as 5% of the original cost of the heat exchanger, escalated due to inflation up to the lifetime of the tubing. The method for calculating this was done exactly as for the maintenance costs in Excel, with the present worth of the salvage value at year 20 representing the present worth expected salvage value, which would reduce the overall initial cost of the heat exchanger. For the ½’’ 2 pass heat exchanger, this salvage value was found to be $8,019.

With the maintenance costs over the life of the heat exchanger and the salvage value determined, a projection for the total present value cost of the heat exchanger could be obtained by adding the maintenance costs to the initial cost and subtracting the salvage value. The resulting final cost for each configuration is tabulated below:

Table C.7: Initial heat exchanger costsConfiguration 1 Configuration 2 Configuration 3

1/2'' pipe, 2 Passes 3/4'' pipe, 2 passes 3/4'' pipe, 3 passesNet present value cost $297,740.07 $296,845.41 $271,324.83

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In order for the economic analysis to be complete, the cost of not replacing the heat exchanger also had to be determined. This was done by determining the plant output lost and applying penalties based on this loss of output.

The plant output was assumed to be 735 MW with the heat exchanger function. The relationship between the heat exchanger performance and the plant output improvement was given as 0.004% increase in capacity for each reduction in degrees Fahrenheit of the terminal temperature difference. With the heat exchanger in place, the terminal temperature difference is 5 degrees Fahrenheit by design. With the removal of the heat exchanger, the terminal temperature difference is equal to the saturation temperature of steam leaving the shell subtracted by the temperature of the water entering the heat exchanger, or 173.82oF-120oF = 53.82oF. Subtracting this value from the TTD with the heat exchanger, it is found that the feedwater heat exchanger offers a 48.82oF reduction in the terminal temperature difference.

The plant capacity prior to the introduction of the heat exchanger can be then be calculated by the formula:

MW

Solving for the plant capacity yields:

MW

This is a reduction in plant capacity of 1.43 MW

There are two penalties that result from this loss in capacity. The first is the unit maximum capacity penalty factor, which results from the need to build additional units to meet the demand that the lessened capacity plant is unable to fulfill. This is given as $60/kW-year, escalated with inflation. This value is calculated per year by multiplying the factor with the reduction in plant capacity (taking care to ensure that the units are in agreement).

The second penalty is the unit efficiency penalty factor which results from the extra fuel required to make up for the loss in heating provided by the heat exchanger. This is given as $50/MW-hr, and like the maximum capacity penalty is escalated with inflation and calculated on a yearly basis. For the units to be in agreement it was necessary to find the amount of time in hours which the plant is run per year; this was given as 70% of the hours in a standard year equivalent full load. Multiplying these factors yielded the cost of the efficiency penalty.

By adding up these two penalty factors and taking the present worth on a yearly basis, the total cost of not replacing the heat exchanger can be determined. A table presenting this calculation is shown below:

Table C.7: Costs of not replacing feedwater heat exchanger

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YearUnit Maximum Capacity Penalty Cost

Unit Efficiency Penalty Cost Combined Penalty Cost Present Worth of Cost

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0 $0.00 $0.00 $0.00 $0.001 $85,950.64 $439,207.75 $525,158.38 $500,150.842 $88,529.15 $452,383.98 $540,913.14 $490,624.163 $91,185.03 $465,955.50 $557,140.53 $481,278.944 $93,920.58 $479,934.16 $573,854.74 $472,111.725 $96,738.20 $494,332.19 $591,070.39 $463,119.116 $99,640.34 $509,162.16 $608,802.50 $454,297.807 $102,629.55 $524,437.02 $627,066.57 $445,644.518 $105,708.44 $540,170.13 $645,878.57 $437,156.049 $108,879.69 $556,375.23 $665,254.93 $428,829.26

10 $112,146.08 $573,066.49 $685,212.58 $420,661.0811 $115,510.47 $590,258.49 $705,768.95 $412,648.4912 $118,975.78 $607,966.24 $726,942.02 $404,788.5213 $122,545.05 $626,205.23 $748,750.28 $397,078.2614 $126,221.41 $644,991.39 $771,212.79 $389,514.8715 $130,008.05 $664,341.13 $794,349.17 $382,095.5316 $133,908.29 $684,271.36 $818,179.65 $374,817.5217 $137,925.54 $704,799.50 $842,725.04 $367,678.1418 $142,063.30 $725,943.49 $868,006.79 $360,674.7519 $146,325.20 $747,721.79 $894,046.99 $353,804.7520 $150,714.96 $770,153.44

Total cost of not replacing feedwater heater $8,384,039.92

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Appendix D – Bibliography

Clements, Talmage. Engineering Economic Topics. 2007

Electronic Steam Tables “XSteam_Excel_v2.4_US.xls”.

Incropera, Frank P., and David P. Dewitt. Fundamentals of Heat and Mass Transfer, 6th Edition. Hoboken, NJ: John Wiley & Sons, Inc., 2007.

White, Frank M. Fluid Mechanics. 5th ed. McGraw-Hill Professional, 2003.

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