MARINE STEAM POWER PLANT HEAT BALANCE
by
Steven A. Gelardi
An Engineering Project Submitted to the Graduate
Faculty of Rensselaer Polytechnic Institute
in Partial Fulfillment of the
Requirements for the Degree of
MASTER OF ENGINEERING IN MECHANICAL ENGINEERING
Approved:
_________________________________________
Ernesto Gutierrez-Miravete, Engineering Project Adviser
Rensselaer Polytechnic Institute
Hartford, Connecticut
August, 2012
(For Graduation August, 2012)
© Copyright 2012
by
Steven A. Gelardi
All Rights Reserved
ii
CONTENTS
MARINE STEAM POWER PLANT HEAT BALANCE....................................................i
CONTENTS ...................................................................................................................... iii
LIST OF SYMBOLS ........................................................................................................ vii
LIST OF FIGURES ............................................................................................................ ii
LIST OF TABLES ............................................................................................................. iii
ACKNOWLEDGMENT ....................................................................................................iv
ABSTRACT ........................................................................................................................ v
1. INTRODUCTION ......................................................................................................... 1
2. METHODOLOGY ........................................................................................................ 4
2.1
Theory .................................................................................................................. 4
2.1.1
2.2
The First Law of Thermodynamics .......................................................... 4
Problem Description............................................................................................. 5
2.2.1
Preliminary Values ................................................................................... 5
2.2.2
Secondary Values ..................................................................................... 6
3. EQUATIONS/MATRICES ........................................................................................... 7
3.1
3.2
3.3
Boilers .................................................................................................................. 7
3.1.1
Component Detail .................................................................................... 7
3.1.2
Boiler Fuel Oil ......................................................................................... 8
3.1.3
Heat Rate ................................................................................................ 10
3.1.4
Steam Drum ........................................................................................... 11
3.1.5
Boiler Efficiency .................................................................................... 11
Superheater ......................................................................................................... 16
3.2.1
Component Detail .................................................................................. 16
3.2.2
Superheater Outlet Flow ........................................................................ 17
Main Propulsion Turbine ................................................................................... 18
3.3.1
Steam Rate (Non-Extraction) ................................................................. 18
iii
3.4
3.5
3.6
3.7
3.3.2
Available Energy.................................................................................... 19
3.3.3
Efficiency of the State Line.................................................................... 21
3.3.4
Temperature Correction Factor .............................................................. 22
3.3.5
State Line Energy ................................................................................... 22
3.3.6
Exhaust Loss .......................................................................................... 24
3.3.7
Mechanical and External Efficiency ...................................................... 25
3.3.8
Gland Leak Off ...................................................................................... 26
Power ................................................................................................................. 27
3.4.1
Power Equation ...................................................................................... 27
3.4.2
Estimated Total Flow Rate from Superheaters ...................................... 27
3.4.3
Power Matrix .......................................................................................... 29
High Pressure Heater (Economizer)................................................................... 30
3.5.1
Component Detail .................................................................................. 30
3.5.2
Terminal Temperature Difference and Temperature Difference ........... 31
3.5.3
Heat Exchange Pressure, Temperatures and Enthalpies ........................ 31
Direct Contact Heater (Deaerating Feed Tank) ................................................. 34
3.6.1
Component Detail .................................................................................. 34
3.6.2
DC Heater and Auxiliary Exhaust Pressure ........................................... 35
3.6.3
DC Heater Flow Rate to Feed Pump ...................................................... 35
3.6.4
Direct Contact Vent Condenser ............................................................. 36
Low Pressure Heater .......................................................................................... 38
3.7.1
3.8
Component Detail .................................................................................. 38
Feed Water Drain Collection Tank .................................................................... 39
3.8.1
Component Detail .................................................................................. 39
3.8.2
Contaminated Drains .............................................................................. 39
3.8.3
Steam Air Heaters .................................................................................. 40
3.8.4
Make-Up Feedwater ............................................................................... 42
iv
3.9
3.8.5
Steam Atomizers .................................................................................... 43
3.8.6
Soot Blowers .......................................................................................... 43
3.8.7
Distilling Plant ....................................................................................... 44
3.8.8
Distiller Air Ejector Flow....................................................................... 46
3.8.9
Feed Water Drain Collection Tank Flow ............................................... 46
Main Condenser ................................................................................................. 48
3.9.1
Component Detail .................................................................................. 48
3.9.2
Main Condenser Flow ............................................................................ 49
3.10 Auxiliary Condenser .......................................................................................... 50
3.10.1 Component Detail .................................................................................. 50
3.10.2 Auxiliary Condenser Flow ..................................................................... 50
3.11 Desuperheater ..................................................................................................... 51
3.11.1 Component Detail .................................................................................. 51
3.11.2 Desuperheater Steam Flow .................................................................... 51
3.11.3 Desuperheated Steam Mass Flow .......................................................... 52
3.11.4 Desuperheated Temperature and Pressure ............................................. 52
3.12 Domestic / Hotel Loads ...................................................................................... 54
3.12.1 Component Detail .................................................................................. 54
3.12.2 Domestic Water Heating ........................................................................ 54
3.12.3 Galley Heating ....................................................................................... 54
3.12.4 Laundry Heating..................................................................................... 54
3.12.5 Total Domestic Hotel Loads .................................................................. 55
4. COMPONENT CALCULATIONS ............................................................................. 56
4.1
Turbo-Generator ................................................................................................. 56
4.1.1
Component Detail .................................................................................. 56
4.1.2
Rated Kilo-Watt Load ............................................................................ 57
4.1.3
Operating Kilo-Watt Load ..................................................................... 57
v
4.2
4.3
4.4
4.1.4
Generator / Turbine Size ........................................................................ 57
4.1.5
TG Temperatures, Pressures and Enthalpies .......................................... 58
4.1.6
TG Turbine Efficiency ........................................................................... 58
4.1.7
TG Steam Rates and Flows .................................................................... 62
Feed Pump .......................................................................................................... 64
4.2.1
Component Detail .................................................................................. 64
4.2.2
Feed Pump Turbine ................................................................................ 65
4.2.3
Feed Pump Pump ................................................................................... 68
4.2.4
Feed Pump Table Overview ................................................................... 71
Air Ejectors ........................................................................................................ 73
4.3.1
Component Detail .................................................................................. 73
4.3.2
Main Air Ejectors ................................................................................... 73
4.3.3
Auxiliary Air Ejectors ............................................................................ 74
4.3.4
Distillate Air Ejectors ............................................................................. 74
4.3.5
Air Ejector Inter-Condensers ................................................................. 74
4.3.6
Air Ejector After-Condensers ................................................................ 75
Plant Cycle Efficiency........................................................................................ 76
4.4.1
Efficiency Theory................................................................................... 76
4.4.2
Cycle Efficiency ..................................................................................... 76
4.4.3
Plant Cycle Efficiency............................................................................ 76
4.4.4
Carnot Efficiency ................................................................................... 77
5. RESULTS/DISCUSSION ........................................................................................... 78
5.1
MASS BALANCE ............................................................................................. 79
5.2
DISCUSSION .................................................................................................... 82
6. CONCLUSION............................................................................................................ 83
7. REFERENCES ............................................................................................................ 84
vi
LIST OF SYMBOLS
A/F, Air to Fuel Ratio
fb, Correction Factor
Aa, Annulus Area, ft^2
fL, Load Correction Factor
AE, Available Energy, BTU/lb
fp, Initial Pressure Correction Factor
B1, High Pressure Heater’s mass flow,
fs, Superheat Correction Factor
lb/hr
ft, Initial Temperature Correction Factor
B2, Steam Air Heaters flow, lb/hr
G(EST), Estimated Total Flow Rate,
B3, Main Turbine Exhaust flow, lb/hr
BHP)OPER,
Operating
lb/hr
Brake
G, Total Flow Rate, lb/hr
Horsepower, hp
GPD, Gallons per Day, GPD
BHP, Brake Horsepower, hp
Cp(Air),
Heat
Capacity
GPM, Gallons per Minute, GPM
of
Air,
h(B), Mean Enthalpy of B1, B2, and B3,
BTU/lbm*°F
BTU/lb
Cp(FO), Heat Capacity of Fuel Oil,
h(B1), Enthalpy of High Pressure
Heater’s mass flow, BTU/lb
BTU/lbm*°F
E(EST), Estimated Total Energy, lb/hr
h(B2), Enthalpy of Steam Air Heaters
E, Total Energy, lb/hr
flow, BTU/lb
EB, Efficiency of the State Line, %
h(B3),
Eb, TG Turbine Efficiency, %/100%
Main
Turbine
Exhaust flow, BTU/lb
Eff(FP), Efficiency of Feed Pump,
h(COMPONENT),
%/100%
enthalpy
of
component, BTU/lb
Eff(TG TURB))OPER, Turbo Generator
h(CONT DRNS), Contaminated Drains
Turbine at Operated Load, %/100%
Eff(TG
Enthalpy of
TURB))RATED,
enthalpy, BTU/lb
Turbo
h(FP TURB EXH), Feed Pump Turbine
Generator Turbine at Rated Load,
Exhaust, BTU/lb
%/100%
h(MAKEUP
EL, Exhaust Loss, lb/hr
FEED),
enthalpy
of
Makeup Feedwater, BTU/lb
Em, Engine efficiency, %/100%
h(MN LEAK OFF), enthalpy of Total
f(D), Distillate Output Heat Steam
Leak Off Steam, BTU/lb
Factor
h(SAH IN), Steam Air Heaters Inlet
f(t), Temperature Correction Factor
enthalpy, BTU/lb
vii
h(SAH OUT), Steam Air Heaters Outlet
hw, enthalpy of the Wheel Used Energy,
enthalpy, BTU/lb
BTU/lb
h(TG EXH), TG Exhaust Enthalpy,
kW(OPER),
BTU/lb
h(VAP
kilo-Watt
Load, kW
FWDCT),
FWDCT
Vapor
kW(RATED), Rated kilo-Watt Load,
enthalpy, BTU/lb
h(VENT
Operational
COND),
kW
Vent
Condenser
kW(TG EST), kilo-Watt Load of Turbo
enthalpy, BTU/lb
Generator (Estimated), kW
h, enthalpy, BTU/lb
kW(TG), Turbo Generator Load, kW
H, Total Head developed per stage, feet
LHP, Windage Loss, hp
Ha, Heat Added to Combustion Air,
N, Number of Persons
BTU/lb
Ns, Rated Specific Speed, RPM
hc1, enthalpy at point Tc1, BTU/lb
P(AUX
hD3, enthalpy of exiting heating steam,
EXH),
Auxiliary
Exhaust
Pressure, psia
BTU/lb
P(AUX EXH), TG Exhaust Pressure,
hF2, High Pressure Inlet Enthalpy,
psia
BTU/lb
P(COMPONENT),
hF3, enthalpy of the Temperature
pressure
of
component, psig
Outlet, BTU/lb
P(DC HTR), Direct Contact Heater
Hg, Stack Loss, BTU/lb
Pressure, psia
HHV, Higher Heating Value, BTU/lb
P(DESUP),
hi, enthalpy of the State Line End Point,
Desuperheater
Pressure,
psig
BTU/lb
P(DISCH), discharge pressure, psig
HL, Heat Loss, BTU/lb
P(HP HTR SHELL), High Pressure
ho, enthalpy of Initial Pressure and
Heater Shell Pressure, psig
Temperature, BTU/lb
P(LP HTR SHELL), Pressure of the
Ho, Heat Input, BTU/lb
Low Pressure Heater Shell, psia
hp, enthalpy of So and pressure of main
P(STM DRUM), Steam Drum Pressure,
condenser, BTU/lb
psig
HR, Heat Rate, BTU/lb/hp
P(SUCT), suction pressure, psig
Hu, Heat Radiation, BTU/lb
P, Power, hp (SHP) or kW
ii
P’o, TG entering Steam pressure, psia
Q(DOM
Po, Initial Pressure, psig
WATER
HEATING),
Domestic Water Heating flow, lb/hr
Pso, Superheater Outlet Pressure, psig
Q(DOM), Domestic Loads flow, lb/hr
Q(AUX A/E), Auxiliary Air Ejector
Q(DUMP), Dump flow, lb/hr
flow, lb/hr
Q(AUX
AC),
Q(FOH), Fuel Oil Heaters flow, lb/hr
Auxiliary
After-
Q(FWDCT),
Condenser flow, lb/hr
Drain
Collection Tank flow, lb/hr
Q(AUX COND), Auxiliary Condenser
Q(LOST EST), Estimated Lost Energy,
flow, lb/hr
lb/hr
Q(AUX IC), Auxiliary Inter-Condenser
Q(LOST), Lost Energy flow, lb/hr
flow, lb/hr
Q(MAKEUP
Q(AUX LEAK OFF), Auxiliary Leak
FEED),
Makeup
Feedwater, lb/hr
Off flow, lb/hr
Q(MN A/E), Main Air Ejector flow,
Q(BLR EST), Estimated Boiler Flow,
lb/hr
lb/hr
Q(MN AC), Main After-Condenser
Q(COMPONENT), flow of component,
flow, lb/hr
lb/hr
Q(MN COND), Main Condenser flow,
Q(CONT DRNS), Contaminated Drains
lb/hr
flow, lb/hr
Q(DESUP
Feedwater
Q(MN IC), Main Inter-Condenser flow,
EST),
Estimated
lb/hr
Desuperheater flow, lb/hr
Q(MN LEAK OFF), Main Leak Off
Q(DESUP), Desuperheater flow, lb/hr
flow, lb/hr
Q(DISTILL A/E), Distillate Air Ejector
Q(MN LEAK OFF), Total Leak Off
flow, lb/hr
Steam, lbs/hr
Q(DISTILL PLT), Distilling Plant Flow,
Q(SAH), Steam Air Heater flow, lb/hr
lb/hr
Q(SOOT BLOWERS), Soot Blowers
Q(DOM GALLEY), Galley Heating
Flow, lb/hr
flow, lb/hr
Q(STM ATOM), Steam Atomizer Flow,
Q(DOM LAUNDRY), Laundry Heating
lb/hr
flow, lb/hr
Q(TG))OPER, Flow of the Turbo
Generator at Operating Load, lb/hr
iii
Q(TG))RATED, Flow of the Turbo
T(COMPONENT),
Generator at Rated Load, lb/hr
temperature
of
component, °F
Q(TG), Turbo Generator flow, lb/hr
T(DESUP), Desuperheater Temperature,
Q(VAP FWDCT), FWDCT Vapor flow,
°F
lb/hr
T(FO), Fuel Oil Temperature, °F
Q(VENT), Vent Condenser flow, lb/hr
T(H), Highest Point Temperature, °F
Q, flow, lb/hr
T(L), Lowest Point Temperature, °F
R, Feed Pump Recirculation Rate, lb/hr
T(MAKEUP
RL, Astern-Turbine Windage, “Hg
SFC,
Specific
Fuel
FEED),
Makeup
Feedwater Temperature, °F
Consumption,
T(SAH), Steam Air Heater temperature,
lb/hr/hp
°F
SHP, Shaft Horseshp, shp
T(SAT), Saturation Temperature, °F
SLEP, energy of the Desuperheated
T(So), TG Exhaust Temperature, °F
T’o, TG entering Steam temperature, °F
Steam, BTU/lb
So, enthalpy of Initial Pressure and
T1,
Temperature, J/K
Steam
Air
Heaters
intake
temperature, °F
SPC(EST), Estimated Specific Power
T2,
Consumption, kWh/kgal
Steam
Air
Heaters
exhaust
temperature, °F
SPC, Specific Power Consumption,
Tc1,
kWh/kgal
Low
Pressure
Heater
Shell
Temperature plus the Heat Exchange
SR(FP), Steam Rate of the Feed Pump,
Terminal
lb/hr/hp or lb/hr/kW
Temperature
Difference, °F
SR(NE), Steam Rate Non-Extraction,
TD, Temperature Difference, °F
lb/hr/hp or lb/hr/kW
TD3, temperature of exiting heating
SR(TG EST), Steam Rate of Turbo
Steam, °F
Generator (Estimated), lb/hr
TF2,
SR(TG)OPER, Operating Steam Rate of
High
Pressure
Heater
Temperature, °F
the Turbo Generator, lb/hp/kW
TF3, Temperature Outlet, °F
SR(TG)RATED, Rated Steam Rate of
Tg, Uptake Temperature, °F
the Turbo Generator, lb/hp/kW
To, Initial Temperature, °F
iv
Inlet
TTD, Terminal Temperature Difference,
X, Condenser Flow, lb/hr
Δh(SAH), Steam Air Heater difference
°F
TTD1,
Terminal
Temperature
in enthalpies, BTU/lb
ΔH(TOTAL), Change in Head Pump,
Difference of Heat Exchanger, °F
UE(SL), State Line Energy, BTU/lb
BTU/hr
UE(W), Wheel Used Energy, BTU/lb
Δh, change in enthalpy, BTU/lb
v(f), volume per pound, ft^3/lb
ΔP(BP), Change in Boiler Pressure, psig
v(L), Direct Contact Heater Outlet
ΔP(CTRL
volume of Liquid, ft^3/lb
DESUP),
Control
Desuperheater Pressure, psig
ΔP(Superheater), Change in Superheater
v, Final Rated Specific Volume of
Steam, ft^3/lb
Pressure, psig
ΔP, Change in pressure, psig
W(FO), Work Done on the Fuel Oil, ft-
ΔT(FO), Change in Temperature of the
lb
W, Work, J or ft-lb
Fuel Oil, °F
ΔT, Change in Temperature, °F
WHP, Wheel Horsepower, hp (WHP) or
kW
v
LIST OF FIGURES
Figure 1 – Basic Marine Steam Power Plant Cycle ........................................................... 1
Figure 2 - Conservation of Energy (1) ............................................................................... 4
Figure 3 - Fuel Oil Fired Steam Boiler (3) ........................................................................ 8
Figure 4 - Steam Drum Internals (4)................................................................................ 11
Figure 5 - Stack Loss, Hg (5 p. 54).................................................................................. 14
Figure 6 - Superheater Tubes (Exposed) (6).................................................................... 16
Figure 7 - Main Propulsion Turbine (Cross Sectional) (7) .............................................. 18
Figure 8 – Propulsion Turbines Replacement Factors (5 p. 42) ...................................... 20
Figure 9 - Main Propulsion Turbine Basic Efficiency (5 p. 34) ...................................... 21
Figure 10 - Throttle Temperature Correction (5 p. 35) ................................................... 22
Figure 11 - State Line (5 p. 38)........................................................................................ 23
Figure 12 - Typical Economizer (9) ................................................................................ 30
Figure 13 - Economizer Flow (aka "Third Stage Heater") (2 p. 11) ............................... 33
Figure 14 - Typical DC Heater (Deaerating Feed Tank) (10) ......................................... 34
Figure 15 - Direct Contact Heater Flow (2 p. 12)............................................................ 37
Figure 16 - Mean Specific Heat (5 p. 48) ........................................................................ 42
Figure 17 - Extraction Steam for Flash Type Evaporators (5 p. 25) ............................... 45
Figure 18 - Feed Water Drain Collection Tank System Diagram (2 p. 15) ..................... 47
Figure 19 – Single-Pass Condenser (11) ......................................................................... 49
Figure 20 - Turbo Generator (Uninstalled without Piping) (12) ..................................... 56
Figure 21 - Basic Efficiency, Eb (5 p. 60) ....................................................................... 59
Figure 22 - Initial Temperature Correction Factor, ft (5 p. 63) ....................................... 59
Figure 23 - Initial Pressure Correction Factor, fp (5 p. 63) ............................................. 60
Figure 24 - Exhaust Pressure Correction Factor, fb (5 p. 60) .......................................... 61
Figure 25 - Load Correction Factor, fL (5 p. 63)............................................................. 61
Figure 26 - Steam Driven Feed Pump (13) ...................................................................... 64
Figure 27 - Windage Loss, LHP (5 p. 66) ....................................................................... 66
Figure 28 - Feed Pump Rated Efficiency (5 p. 76) .......................................................... 69
Figure 29 - Typical Air Ejector (14) ................................................................................ 73
Figure 30 - Main Air Ejector Flow (5 p. 13) ................................................................... 74
ii
LIST OF TABLES
Table 1 - Design Characteristics (2 p. 1) ........................................................................... 5
Table 2 – Standard Capacity of a Distilling Plant (2 p. 23) ............................................. 44
Table 3 - Desuperheater and its Components’ Flows ...................................................... 51
Table 4 - Rated kW and Standard Size Generator (2 p. 23) ............................................ 57
Table 5 - Optimal Feed Pump Table................................................................................ 72
Table 6- Mass Balance..................................................................................................... 79
Table 7 - Matrix of Equations .......................................................................................... 96
Table 8 - Minverse ........................................................................................................... 96
Table 9 – Main Turbine Extraction Stages ...................................................................... 99
Table 10 - Power Matrix ................................................................................................ 100
Table 11 - Superheater Outlet Flow Matrix ................................................................... 100
Table 12 - High Pressure Heater Matrix ........................................................................ 100
Table 13 - Direct Contact Heater Matrix ....................................................................... 101
Table 14 – Feed Water Drain Collection Tank Matrix .................................................. 101
Table 15 - Main Condenser Matrix ............................................................................... 101
Table 16 - Auxiliary Condenser Flow Matrix ............................................................... 102
Table 17 - Desuperheater Steam Flow Matrix ............................................................... 102
iii
ACKNOWLEDGMENT
I’d first like to acknowledge Torii Schopflin, who introduced me to Rensselaer’s
Master’s Program and who also took the liberty to fill out most of the paperwork to
enroll me without asking. I’d like to also thank my wife, Laurel, who has been a constant
supporter of my education and has kept “gently” pushing me to finish my Master’s
Degree.
iv
ABSTRACT
Ever wondered how a Marine Steam Power Plant works, with all its components
working together in unison? This project will lead you through the world of steam ship’s
engine rooms; giving you step-by-step methods and calculations to perform a full heatbalance of a typical marine steam powered plant.
A power plant with a Shaft Horse Power of 32,500 SHP will generate a certain kiloWatt (kW) and will produce a certain flow to the main condenser. To find this, we will
find various component details, such as the Steam-Water Cycle; what it does, what it
includes, and how to solve for temperatures, pressures, flows, and enthalpies of each
basic component. This project will provide equations, tables, and figures to guide you
through solving the basic components of a steam powered engine room.
In each section you will find defining terms and values that are typical for a steam
ship. My final matrix will show step-by-step the values of each component in the total
heat balance and how the total heat balance itself has an equal heat-in and heat-out.
v
1. INTRODUCTION
Steam Plants have been around for well over one-hundred years, and have paved the way
for modern shipping. Steam plants work by heating water past its boiling point, until it’s
superheated, then driving it through multiple turbines to produce electricity and thrust.
The purpose of this project is to set forth data for a specified marine steam power
plant heat balance. This project is not for all power plants however; it is specifically
applicable to oil-fired steam-turbine marine power plants at maximum continuous
power. Some modifications have been made to suit this plant as was saw fit.
A heat balance is a necessary tool for designing a marine power plant. They can also
be used to determine optimum steam conditions and a cycle design or even to analyze
the performance of a power plant already in service. In order to solve the heat balance
calculation, shaft horsepower, steam conditions and the basic steam-water cycle must be
established. Once these have been solved, various steam, feed and condensate, and
exhaust flow can be determined.
The figure below shows a basic rendition of a typical Marine Steam Power Plant
Cycle.
Figure 1 – Basic Marine Steam Power Plant Cycle
1
The Basic Steam-Water Cycle is as follows:
1. Boiler– A boiler contains the heating element of the system; fire generated by
burning fuel, and applies it to boil the medium, water.
2. Throttle Box – This is the means to control the output of the boiler. In this project
I will be using set values to calculate the heat balance.
3. High Pressure (HP) Turbine – The HP Turbine is where the superheated steam
from the boiler is directed to, in order to generate a spinning action of the blades of the
turbine. The more nozzles open, the greater the flow, the greater the power.
4. Low Pressure (LP) Turbine – The LP Turbine is where the now less-superheated
steam travels through after the HP Turbine.
5. Condenser (Main/Auxiliary) – The Condenser is where the steam is dumped after
the HP and LP Turbine. It interacts with a cooling medium (usually Seawater) and heat
transfer occurs, cooling the steam into water. In this project, a Low Pressure Condenser
will be used (at Vacuum Pressure).
6. Condensate Pump – The means of pumping the condensed steam through the
remainder of the system.
7. Air Ejector (Main/Auxiliary) – Air Ejectors remove the non-condensable gases
from the system and helps to draw and maintain vacuum.
8. Low Pressure Heater (First Stage Heater) – Heats the water for hotel loads,
auxiliary loads, and miscellaneous loads.
9. DC Heater (Direct Contact Heater, or Deaerating Feed Tank) – acts as a holding
tank for the incoming water to the feed pump to pump into the boiler, removes noncondensable gas, and heats the water more.
10. Feed Pump – The Feed Pump serves as the driving force of the water into the
boiler. Its discharge pressure must be greater than that of the boilers in order to have
positive flow.
11. High Pressure Heater (Economizer) – Acts as a reheat to the system so a shock
to the system with the entrance of cold water into the boiler will not occur.
12. Boiler – This is considered the last part of the cycle as well as the beginning of
the cycle.
2
A heat balance contains many variable factors. Some of these include fuel, hotel and
domestic loads, ambient conditions (which I am nullifying this project – atmospheric
pressure is equal to 1 atmosphere), and various other ship components. These variables
will vary from ship to ship and will affect its associated calculated results.
For the purpose of this project, I will be providing you typical variables that will be
explained step-by-step throughout the project.
3
2. METHODOLOGY
2.1 Theory
2.1.1
The First Law of Thermodynamics
The First Law of Thermodynamics states that energy is conserved.
“The first law of thermodynamics is a version of the law of conservation of energy,
specialized for thermodynamical systems. It is usually formulated by stating that the
change in the internal energy of a closed system is equal to the amount of heat supplied
to the system, minus the amount of work performed by the system on its surroundings.
The law of conservation of energy can be stated: The energy of an isolated system is
constant”. (1)
Since energy cannot be created nor destroyed, the Figure below (Figure 2 Conservation of Energy) shows that the Energy ‘IN’ (QH) is equal to the Energy ‘OUT’
(Out being summation of QC and W). The Term QH is the heat energy input into the
system, W is the useful work done by the system, and QC is the heat energy output from
the system (into a low temperature sink).
Figure 2 - Conservation of Energy (1)
𝑄(𝐻) = 𝑄(𝐶) + 𝑊
(2.1)
𝑊 = 𝑄(𝐻) − 𝑄(𝐶)
(2.2)
𝑄(𝐻) − 𝑊 − 𝑄(𝐶) = 0
(2.3)
4
2.2 Problem Description
A design requires that the original design characteristics contain a Shaft Horse Power
(SHP) of 32,500 SHP, a complement of 40 persons, 2 boilers, a throttle temperature of
950 degrees Fahrenheit (F) and pressure of 850 pounds-per-square-inch-gauge (psig).
Condenser pressure is 1.5 Inches of Mercury ("Hg) and Auxiliary Condenser is 2 "Hg.
The required Engine efficiency “Em” is given at 0.96. The objective is to complete the
heat balance. Note: The design must contain a total plant flow “E” for 2 boilers.
2.2.1
Preliminary Values
Given the scope of the work, the given values in the problem description are not a lot to
go on. Some preliminary values must be given before we can proceed. The table below
shows what we already know from the problem statement, plus a few components not
yet unidentified.
The Initial Pressure and Temperature, Shaft Horsepower, and Required Engine
Efficiency were calculated from a specific set of factors already given from MARINE
ENGINEERING DESIGN I REPORT (2 p. 1). The Appendices A.1.5 and A.1.2 give us the
values to solve for the Initial Temperature and Pressure. The Complement (Number of
Persons) “N” was chosen as a typical value of people aboard a cargo ship.
Table 1 - Design Characteristics (2 p. 1)
Item
Pressure, Initial (“Po”)
Value
850
864.696
950
Temperature, Initial (“To”)
Main Condenser Pressure
1.5
0.736765
2
0.982353
Auxiliary Condenser Pressure
Shaft Horse Power (“SHP”)
Units
psig
psia
°F
"Hg
psia
"Hg
psia
32,500 SHP
Required Engine Efficiency (“Em”)
Number of Boilers
0.96
2 Boilers
Number of Persons (“N”)
40 Complement
5
2.2.2
Secondary Values
In order to solve the total heat balance for a marine steam-powered plant, some other
conditions have to be solved for. The subsequent sections in this project will show the
areas that are going to be solved. First we will start with the Main Propulsion Turbine
(Section 3.1), then work our way to the Power Equation (Section 0), then the
Superheater and Desuperheater (Sections 3.2 and 3.11), High Pressure Heater (Section
3.5), Direct Contact Heater (Section 3.6), Low Pressure Heater (Section 3.7), Feedwater
Drain Collection Tank (Section 3.8), Main and Auxiliary Condenser (Sections 3.9 and
3.10), and finally the Domestic Loads (Section 3.12) before getting into the Component
Calculations (Section 4).
6
3. EQUATIONS/MATRICES
Boilers
3.1
3.1.1
Component Detail
The boiler’s main purpose is to provide superheated steam to the Propulsion Turbines
and Turbo Generators for propulsion and electrical power. This plant utilizes two boilers
in parallel. Normally, an operating plant would have one boiler carry most of the load;
however for simplicity in equations we will be assuming equal loads throughout the
entirety of the project.
The Boilers are equipped with fuel oil injectors that supply a concentrated yet
distributed spray into the burners. This fuel is consumed (burned) and its energy is
transferred to the steam in the internal piping in the Boilers. The steam is heated to a
superheated, high pressure state where it is sent to the Main Propulsion Turbines and the
Turbo Generators. It’s equipped with a Superheater system, which increases the output
steam to a state with no moisture, a Desuperheater system which controls the output
temperature of the Superheater system, an Economizer which redirects some lost flow
back into the system to conserve energy, and a steam drum that serves as an expansion
tank to the inlet to the Superheater system.
7
Figure 3 - Fuel Oil Fired Steam Boiler (3)
3.1.2
Boiler Fuel Oil
Bunker “C” (Number 6) is the fuel oil consumed in the boiler. It is a low-grade fuel and
is very robust. It will not light if a match is thrown on an open barrel. Its primary
component is Carbon, but it also contains Hydrogen, Sulphur and trace amounts of
Oxygen and Nitrogen. It is a very good oil to use as it has good burning characteristics
and is cheaper than most other grades of fuel oil. Bunker “C” has a standard higher
heating value of 18,500 BTUs per pound.
3.1.2.1 Specific Fuel Consumption
The Specific Fuel Consumption, SFC is the ratio of the fuel mass flow of an engine to its
output power, in specified units. Specific Fuel Consumption is a widely used measure of
8
atmospheric engine performance. Its units are pound-mass-per-hour-per-horsepower
lbm/hr/hp.
𝑆𝐹𝐶
=
(3.1)
(𝐸 − (𝑄(𝐷𝐸𝑆𝑈𝑃) ∗ ℎ(𝑆𝑂)) + ((𝑄 ∗ ℎ)(𝐷𝐸𝑆𝑈𝑃)) − (𝐸 ∗ (ℎ𝐹3)))
𝐴
((𝐸𝑓𝑓(𝐵𝐿𝑅) ∗ 𝑆𝐻𝑃 ∗ 𝐻𝐻𝑉) + (𝐶𝑝(𝐹𝑂) ∗ 𝛥𝑇(𝐹𝑂)) + ((𝐹 ) ∗ 𝐶𝑝(𝐴𝑖𝑟) ∗ 𝛥𝑇(𝐴𝑖𝑟)))
𝑙𝑏
)
ℎ𝑟
𝑆𝐹𝐶 = 0.204
ℎ𝑝
(
3.1.2.2 Fuel Oil Heaters
To solve for “Q(FOH)”, I need the Work Done on the Fuel Oil “W(FO)”. To get this, we
need the product of the Ships Horsepower “SHP” (32,500 SHP) and the Specific Power
Consumption “SPC(EST)” (0.5).
𝑊(𝐹𝑂) = 𝑆𝐻𝑃 ∗ 𝑆𝑃𝐶(𝐸𝑆𝑇) = 16250 𝑓𝑡 ∗ 𝑙𝑏
(3.3)
Now that we have solved the work, we may multiply in a factor of 0.7 pounds
2
lb
squared per foot-hour(hr∗ft) to get the Flow of the Fuel Oil Heaters.
𝑄(𝐹𝑂𝐻) = 0.07 ∗ 𝑊(𝐹𝑂) = 1137.5
𝑙𝑏
ℎ𝑟
(3.4)
3.1.2.3 Fuel Consumption
Fuel consumption is the rate at which the Fuel Oil is consumed in a Boiler. To calculate
the Flow of Fuel Oil “Q(FO)”, I need to find the amount of fuel I consume.
𝑄(𝐹𝑂) = 𝑆𝐹𝐶 ∗ 𝑆𝐻𝑃 =
(0.203790
ℎ𝑝
𝑙𝑏
)
ℎ𝑟 ∗ (32500) ℎ𝑝
(3.5)
𝑙𝑏
𝑔𝑎𝑙
= 6623.175
= 907.167
ℎ𝑟
ℎ𝑟
Bunker C is a bit lighter than water, as a gallon of it weighs 7.3 pounds. Dividing the
Q(FO) by gallons per pounds gets us 907.167 gallons per hour.
To convert this to barrels-per-hour, we need to divide the gallons per hour by the
amount of gallons in a barrel, which are 42. This yields 2.16 barrels per hour and since
9
(3.2)
there are 24 hours in a day, 51.84 barrels per day are consumed. To summate, this steam
power plant uses around 52 Barrels of Bunker “C” Fuel Oil a day.
𝑔𝑎𝑙
907.167
ℎ𝑟 ) ∗ (24 ℎ𝑟 ) = 51.84 𝑏𝑎𝑟𝑟𝑒𝑙𝑠
𝐵𝑎𝑟𝑟𝑒𝑙𝑠 𝑐𝑜𝑛𝑠𝑢𝑚𝑒𝑑 𝑎 𝑑𝑎𝑦 = (
𝑔𝑎𝑙
𝑑𝑎𝑦
𝑑𝑎𝑦
42
𝑏𝑎𝑟𝑟𝑒𝑙
3.1.3
Heat Rate
The rate of heat added into (IN) the system divided by the power output (OUT) is called
the Heat Rate (HR). This is useful in finding the Ships Heat Rate for the Plant and the
Cycle. Its units are in BTUs-per-pound-per-hp.
𝐻𝑅(𝑊𝐻𝑃)(𝑐𝑦𝑐𝑙𝑒) =
((𝐸 − 𝑄(𝐷𝐸𝑆𝑈𝑃) ∗ ℎ(𝑆𝑂)) + ((𝑄 ∗ ℎ)(𝐷𝐸𝑆𝑈𝑃)) − (𝐸 ∗ (ℎ𝐹3))) (3.6)
𝑊𝐻𝑃
𝑊𝐻𝑃 = 33854.17 ℎ𝑝, (solved in Section 3.4.3)
The Wheel Horsepower is calculated by finding the quotient of “SHP” to “Em”.
𝐵𝑇𝑈
)
𝐻𝑅(𝑊𝐻𝑃)(𝑐𝑦𝑐𝑙𝑒) = 3128.115 𝑙𝑏
ℎ𝑝
(
For the final equation, which is the Heat Rate of the Entire Plant “HR(SHP)(plant)”,
we need the value of the Higher Heating Value “HHV”. The “HHV” is a standard higher
heating value by ‘bomb’ calorimeter corrected for specific heat at constant pressure. Its
value is set at 18,500 BTU per pound. With this value now know, we can solve for the
Plant’s Heat Rate. It’s defined by the equation below
𝐵𝑇𝑈
𝑄(𝐹𝑂) ∗ 𝐻𝐻𝑉 ( 𝑙𝑏 )
𝐻𝑅(𝑆𝐻𝑃)(𝑝𝑙𝑎𝑛𝑡) =
,
𝑆𝐻𝑃
ℎ𝑝
(3.7)
After plugging in the values of 6623.175 lb/hr for “Q(FO)”, 18500 for “HHV” and
32500 for “SHP”, we get a Plant’s Heat Rate “HR(SHP)(plant)” of 9250 BTU/lb/hp.
𝑅(𝑆𝐻𝑃)(𝑝𝑙𝑎𝑛𝑡) =
𝐵𝑡𝑢
𝐵𝑇𝑈
∗ 18500
𝑙𝑏
= 9250 𝑙𝑏
32500ℎ𝑝
ℎ𝑝
6623.175
10
3.1.4
Steam Drum
The Steam Drum is where the incoming feedwater is pumped into. From Section 3.11.4,
the Pressure of the Steam Drum is calculated to be 993 psig and its Temperature 543.59
°F.
.
Figure 4 - Steam Drum Internals (4)
3.1.5
Boiler Efficiency
A steam plant boiler is typically 80- to 95-percent efficient. This range is due to the
different types, classes, and styles of boiler design. The Boiler Efficiency Equation is
given below. Notice how it contains various components, including Desuperheater
values, the air to fuel ratio, temperature values, the Higher Heating Value, Specific Fuel
Consumption, enthalpy values, shaft horsepower, and overall energy. All these factors
determine the efficiency of a boiler - how the fuel is burned, the flows going through it,
how much temperature change is used; the list goes on and on.
The following is the initial equation used to solve the Boiler Efficiency.
𝐸𝑓𝑓(𝐵𝐿𝑅)
=
[ℎ(𝑆𝑂) ∗ (𝐸 − 𝑄(𝐷𝐸𝑆𝑈𝑃)) + {𝑄(𝐷𝐸𝑆𝑈𝑃) ∗ ℎ(𝐷𝐸𝑆𝑈𝑃)} − (𝐸 ∗ ℎ𝐹3)]
𝐴
[𝑆𝐹𝐶 ∗ 𝑆𝐻𝑃 ∗ 𝐶𝑝(𝐹𝑂) ∗ {𝛥𝑇(𝐹𝑂) + (𝐹 ) ∗ 𝐶𝑝(𝐴𝑖𝑟) ∗ 𝛥𝑇(𝐴𝑖𝑟) + 𝐻𝐻𝑉}]
11
(3.8)
It contains many variables that have not been discussed. The change in Temperature
of the Fuel Oil “ΔT(FO)” is 100 °F based on a specified range of 100 °F to 200 °F. The
Heat Capacity of Air “Cp(Air)” is equal to 0.2438 BTU-per-pound-degree-Fahrenheit
and the Heat Capacity of the Fuel Oil ”Cp(FO)” is 0.46 BTU-per-pound-degreeFahrenheit. The Higher Heating Value “HHV” is 18500 BTU per pound and the amount
of Excess Air is 0.05, or 5%.
These are a lot of component values to throw at you at once; therefore we will be
looking at the alternate approach in solving the Boiler Efficiency. It involves fewer
values and a simpler equation:
𝐵𝑜𝑖𝑙𝑒𝑟 𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦, “𝐸𝑓𝑓(𝐵𝐿𝑅)” =
[(𝐻𝑜 + 𝐻𝑎) − 𝐻𝐿]
[𝐻𝑜 − 𝐻𝑎]
(3.9)
This equation will be the equation we will use to solve for the Boiler Efficiency.
Each of these values will be described and solved in the subsequent sections.
3.1.5.1 Heat Input
The Heat Input “Ho” is the higher heating value of the fuel oil burned, corrected for
specific heat at constant pressure, plus the heat in the oil above 100 degrees Fahrenheit.
Its equation is laid out below.
𝐻𝑜 = 𝐻𝐻𝑉 + 𝐶𝑝(𝐹𝑂) ∗ 𝛥𝑇(𝐹𝑂)
(3.10)
With this equation, we only have 3 ‘unknowns’. They are the Higher Heating Value
“HHV”, the Heat Capacity of the Fuel Oil “Cp(FO)”, and the Change in Fuel Oil
Temperature “ΔT(FO)”. Each of these values is known; therefore “Ho” is 18557.5
BTUs-per-pound.
𝐻𝑒𝑎𝑡 𝐼𝑛𝑝𝑢𝑡 “𝐻𝑜”
𝐵𝑇𝑈
𝐵𝑇𝑈
) + (0.46
∗ 100 °𝐹)
𝑙𝑏
°𝐹 ∗ 𝑙𝑏
𝐵𝑇𝑈
= 18557.5
𝑙𝑏
= (18500
3.1.5.2 Heat Added by Air Heater
The Heat Added to the Combustion Air by the Air Heater “Ha” is equal to the product of
the Heat Capacity of the Fuel Oil “Cp(FO)”, the Air to Fuel Ratio “A / F”, and the
12
Difference in the Temperature of Air leaving the Steam Air Heater “T2” and the
Temperature of Air entering the Steam Air Heater “T1”. The equation is listed below for
your convenience.
𝐻𝑎 = 𝐶𝑝(𝐹𝑂) ∗
𝐴
∗ (𝑇2– 𝑇1)
𝐹
(3.11)
The Heat Capacity of the Fuel Oil “Cp(FO)”is equal to 0.46 BTUs-per-pounddegree-Fahrenheit. The Air to Fuel Ratio (A/F) “R” is equal to 15.05 pounds Air to
pounds Fuel. The Temperature of Air leaving the Steam Air Heater “T2” is 200 degrees
Fahrenheit and the Temperature of Air entering the Steam Air Heater “T1” is 75 degrees
Fahrenheit. With all the unknowns solved for “Ha”, we can now solve for the Heat
Added to the Combustion Air by the Air Heater “Ha”. “Ha” is calculated to be 865.375
BTUs-per-pound.
3.1.5.3 Stack Loss
The Stack Loss “Hg” is the amount of energy that is lost to external atmosphere via the
stack. To solve for “Hg”, we need the Temperature of the Uptake (Smoke Stack Gas
Temperature). The Uptake Temperature is designated by the term “Tg”.
To solve for Hg we need to use Figure 5. By finding the point on the 5% Excess
Air Line that meets with the Temperature of the Uptake “Tg”, we can find the
approximate value for the Stack Loss. “Tg”, we will assume is right at the temperature of
boiling, 212 degrees. With that in mind, and lining that up with the excess air curve, we
get a Stack loss of about 7.40 [% * 18546 * BTU / lb]. The units for Stack loss can be
condensed into BTUs per pound. This is done by multiplying the Stack loss initial value
by %-18546/100%. This gets us a Stack Loss of 1372.404 BTUs per pound of Fuel Oil.
13
Figure 5 - Stack Loss, Hg (5 p. 54)
% ∗ 18546 𝐵𝑇𝑈
𝑆𝑡𝑎𝑐𝑘 𝐿𝑜𝑠𝑠 “𝐻𝑔” = 7.40 [(
)∗
]
100%
𝑙𝑏
𝐵𝑇𝑈
= 1372.404
𝑙𝑏(𝐹𝑂)
(3.12)
3.1.5.4 Heat Radiation
The Heat Radiation “Hu” is unaccounted for losses and manufacturer's margin (R-U and
M) therefore it simply is 1.1415 percent of the Heat Input “Ho”. Since “Ho” is 18557.5
BTUs per pound, when we calculate “Hu” we get a value of 262.588625 BTUs per
pound.
𝐻𝑒𝑎𝑡 𝑅𝑎𝑑𝑖𝑎𝑡𝑖𝑜𝑛 “𝐻𝑢” = (1.415% ∗ 𝐻𝑜) = 262.589
14
𝐵𝑇𝑈
𝑙𝑏
(3.13)
3.1.5.5 Heat Loss
The Heat Loss “HL” is equal to the sum of the Heat Radiation “Hu” and the Stack Loss
“Hg”. Since “Hu” is 262.588 BTUs-per-pound and “Hg” is 1372.404 BTUs-per-pound,
we get a Heat Loss “HL” of 1633.323 BTUs-per-pound.
𝐻𝐿 = 𝐻𝑢 + 𝐻𝑔 = 1633.323
𝐵𝑇𝑈
𝑙𝑏
(3.14)
3.1.5.6 Boiler Efficiency Answer
The Boiler Efficiency can now be solved by the formula below.
𝐵𝑜𝑖𝑙𝑒𝑟 𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦, 𝐸𝑓𝑓(𝐵𝐿𝑅) =
[(𝐻𝑜 + 𝐻𝑎) − 𝐻𝐿]
[𝐻𝑜 − 𝐻𝑎]
(3.15)
Once the values from Sections 3.1.5.1 through 3.1.5.5 have been solved and
plugged into the Boiler Efficiency Equation, we get a Boiler with an Efficiency of 91.59
Percent. This efficiency value is rated at maximum continuous power.
𝐸𝑓𝑓(𝐵𝐿𝑅) = 0.9159 = 91.59%
15
3.2 Superheater
3.2.1
Component Detail
The Superheater is a series of tubes inside the Boiler. It’s meant to increase the
temperature of steam past its saturation state and into the superheated state. This
superheated steam is considered ‘dry steam’ as it has no water moisture in it.
Moisture is extremely detrimental to any steam turbine; it can cause excessive
pounding of the turbine, it can cause extreme erosion of the turbine blades, and in some
cases it can rip the turbine apart. A turbine spinning over 8000 revolutions-per-minute, if
shaken loose from the turbine, can easily shred through the casing. There have been
cases of a spinning turbine blade breaking its shell and whirling out into the engine room
and destroying other engine-room components.
Figure 6 - Superheater Tubes (Exposed) shows a section of Superheater Tubes that
resides in typical marine boiler.
Figure 6 - Superheater Tubes (Exposed) (6)
16
3.2.2
Superheater Outlet Flow
This section will continue from Section 3.4.2, which talks about solving for the Total
Flow “G”. Next I need to solve for Superheater Outlet Flow "G," which is the flow of
the superheated steam leaving the boilers' superheaters. What we know so far are the
values “G” and “Q(LOST EST)”.
𝐺 = 𝑄(𝑇𝐺) + 𝑄(𝐷𝐸𝑆𝑈𝑃) + 𝑄(𝐿𝑂𝑆𝑆)
(3.16)
𝑄(𝐿𝑂𝑆𝑇 𝐸𝑆𝑇) = 0.005 ∗ 𝐸(𝐸𝑆𝑇)
(3.17)
Since I do not have “E”, I have to substitute in “E(EST),” which is the Estimated
value for the Energy of the System. “E(EST)” is important to have solved, as it will
provide a good estimation of the marine steam power plant systems energy. However,
since we don't have “E(EST)”, we need to find it using this given equation.
𝐸(𝐸𝑆𝑇) = (𝑆𝑅(𝑁𝐸) ∗ (
𝑆𝐻𝑃
))
𝐸𝑚
𝑆𝐻𝑃
− (0.25 ∗ (𝑆𝑅(𝑁𝐸) ∗ (
)))
𝐸𝑚
+ [0.25 ∗ (𝑆𝑅(𝑁𝐸) ∗ (
= 139648.4
(3.18)
𝑆𝐻𝑃
ℎ𝐵 − ℎ𝑤
))] ∗ [
]
𝐸𝑚
ℎ𝑜 − ℎ𝑤
𝑙𝑏
ℎ𝑟
We have now just solved for our Estimated Energy of the System, “E(EST).” This
value of 139648.4 is for the whole system. With this value we can solve a variety of
other estimated values. For the estimated desuperheated steam flow, “Q(DESUP EST)”.
𝑄(𝐷𝐸𝑆𝑈𝑃 𝐸𝑆𝑇) = 0.04 ∗ 𝐸(𝐸𝑆𝑇)
(3.19)
This last value can now be plugged into the equation to solve for “G(EST)”.
𝐺(𝐸𝑆𝑇) = 𝑄(𝑇𝐺 𝐸𝑆𝑇) + 𝑄(𝐷𝐸𝑆𝑈𝑃 𝐸𝑆𝑇) + 𝑄(𝐿𝑂𝑆𝑆 𝐸𝑆𝑇)
𝐺(𝐸𝑆𝑇) = (12500) ∗ (0.04 ∗ 𝐸(𝐸𝑆𝑇)) + (0.005 ∗ 𝐸(𝐸𝑆𝑇))
= 18910.5
(3.20)
(3.21)
To solve the Flow per one Boiler, the Estimated E “E(EST)” value must be divided
by two, since this system has installed two boilers that will be running simultaneously.
𝑄(𝐵𝐿𝑅 𝐸𝑆𝑇) =
𝐸(𝐸𝑆𝑇)
𝐵𝑇𝑈
= 9455.25
2 𝐵𝑜𝑖𝑙𝑒𝑟𝑠
𝑙𝑏
17
(3.22)
3.3 Main Propulsion Turbine
The Main Propulsion (or High Pressure, “HP”) Turbine is where the superheated steam
from the boiler is directed to and through to generate a spinning action of the blades of
the turbine. This spinning action in turn generates rotation of reduction gears and
pinions, which in turn spins the main shaft and the propeller to generate movement in
water.
The more nozzles open, the greater the flow, the greater the shaft revolutions-perminute, and generally speaking the greater the speed of the ship.
Figure 7 - Main Propulsion Turbine (Cross Sectional) (7)
3.3.1
Steam Rate (Non-Extraction)
Steam turbines differ according to whether or not a portion of the steam flow is extracted
from the turbine’s intermediate portion. Extraction is where a minor portion of the steam
flow leaves the turbine through a different path to partially reheat the water fed back to
the boiler. Extraction can significantly increase the efficiency of the marine steam power
plant. However, this power plant does not call for an Extraction Turbine; therefore this
turbine will be classified as a ‘Straight-Tube Turbine,’ which means Non-Extraction
Turbine.
18
The equation for Steam Rate Non-Extraction “SR(NE)” is defined as
𝑆𝑅(𝑁𝐸) =
2544.4
(ℎ𝑜 − ℎ𝑤)
(3.23)
where the 2544.4 BTUs-per-hour is equal to 1 horsepower, hp.
The SR(NE) should have a value between five and six. In order to solve for
SR(NE), I need the two unknowns, “hw” and “ho”. The first term “ho” is the Enthalpy at
the Initial Pressure and Temperature Point.
ℎ𝑜 = ℎ(𝑃𝑜, 𝑇𝑜) = 1481.998
𝐵𝑇𝑈
𝑙𝑏
(3.24)
The second term “hw” is the Enthalpy of the ‘Wheel Used Energy.’ It is equal to
“hi” (the State Line End Point, from Section 3.3.3) plus “EL” (Exhaust Loss, from
Section 3.3.6).
ℎ𝑤 = ℎ𝑖 + 𝐸𝐿 = 997.911
𝐵𝑇𝑈
𝑙𝑏
(3.25)
This gives us an enthalpy of “hw” to be equal to 997.911 BTUs-per-pound, and an
enthalpy of “ho” to be equal to 1481.998 BTUs-per-pound.
The Steam Rate can now be solved since I have “ho” and “hw”.
𝑆𝑅(𝑁𝐸) =
2544.4
𝑙𝑏
= 5.256
(ℎ𝑜 − ℎ𝑤)
ℎ𝑟
The Steam Rate (Non-Extraction) falls in-between the range of five to six as was
previously noted. The next Section 3.3.2 will show step-by-step how to solve for the
Available Energy of the Main Propulsion Turbine.
3.3.2
Available Energy
Available Energy, AE, is the theoretical total amount of energy (in BTU/lb) that the
turbines see. The difference in enthalpies of points “ho” and “hp” is theoretically what
the Available Energy is equal to.
𝐴𝐸 = ℎ𝑜 − ℎ𝑝
(3.26)
Since “ho” was previously defined (Initial Pressure and Temperature Enthalpy) and
solved in Section 3.3.1, “hp” is the remaining unknown needed solve for AE.
ℎ𝑜 = ℎ(𝑃𝑜, 𝑇𝑜) = 1482
19
𝐵𝑇𝑈
𝑙𝑏
(3.27)
The Entropy, given by the term “So” in this instance, is solved at the Initial Pressure
and Temperature Point.
As described by Wikipedia about Entropy:
“Entropy is a thermodynamic property that can be used to determine the energy
available for useful work in a thermodynamic process, such as in energy conversion
devices, engines, or machines. Such devices can only be driven by convertible energy,
and have a theoretical maximum efficiency when converting energy to work. During this
work, entropy accumulates in the system, which then dissipates in the form of waste
heat.” (8)
𝑆𝑜 = 𝐸𝑛𝑡𝑟𝑜𝑝𝑦 𝑎𝑡 𝑃𝑜𝑖𝑛𝑡𝑠 (𝑃𝑜, 𝑇𝑜) 𝑢𝑠𝑖𝑛𝑔 Figure 8
= 1.651915
𝐽
𝐾
Figure 8 – Propulsion Turbines Replacement Factors (5 p. 42)
20
(3.28)
The Entropy at Point “Po”, “To” is now known; solving for the Enthalpy (“hp”) is
what remains. The term “hp” is the Enthalpy, in BTU/lb, of points “Po”, “To” and of the
Pressure in the Main Condenser “P(MN COND)”.
ℎ𝑝 = ℎ@(𝑆𝑜, 𝑃(𝑀𝑁 𝐶𝑂𝑁𝐷)) = 907.2985
𝐵𝑇𝑈
𝑙𝑏
(3.29)
Now that we have solved for the Enthalpy of the top point (Boiler Temperature and
Pressure) and of the low point (Main Condenser Temperature and Pressure) we can solve
for the difference of the two. The difference we have already identified to be the
Available Energy.
𝐴𝐸 = ℎ𝑜– ℎ𝑝 = 574.6993
3.3.3
𝐵𝑇𝑈
𝑙𝑏
(3.30)
Efficiency of the State Line
The Efficiency of the State Line, “Eb”, is a way to solve for the Basic Efficiency of the
Main Propulsion Turbine Hear for Non-Extraction Operation. It is solved from either
using Figure 9 - Main Propulsion Turbine Basic Efficiency or by using the following
equation.
𝐸𝑏 =
(ℎ𝑜 − ℎ𝑖)
= 0.86494 = 86.494 %
𝐴𝐸
Figure 9 - Main Propulsion Turbine Basic Efficiency (5 p. 34)
21
(3.31)
3.3.4
Temperature Correction Factor
The Temperature Correction Factor “f(t)” is what is used to correct for the Throttle
Temperature (Temperature of the Steam entering the Main Turbine). It’s solved either
using the Polynomial Equation derived in the Appendices A.2.3 - Temperature
Correction Factor or by using Figure 10 - Throttle Temperature Correction.
Figure 10 - Throttle Temperature Correction (5 p. 35)
I used the derived equation in the Appendices to solve for “f(t)” and then doublechecked the answer against the table to ensure the value is correct. The Temperature
Correction Factor was found to be
𝑓(𝑡) = 1.01202
3.3.5
State Line Energy
The State Line Energy “UE(SL)” will determining the actual Energy for the State Line
for the Main Propulsion system. Since we have already solved for the Available Energy
22
“AE”, Basic Efficiency (Eb), and the Temperature Correction Factor (f(t)), the State
Line Energy can be calculated.
𝑈𝐸(𝑆𝐿) = 𝐴𝐸 ∗ 𝐸𝑏 ∗ 𝑓(𝑡) = 497.0778
𝐵𝑇𝑈
𝑙𝑏
(3.32)
State Line End Point “SLEP” is the energy of the Desuperheated Steam. It is solved
using Figure 11 - State Line.
𝑆𝐿𝐸𝑃 = ℎ𝑖 = ℎ𝑜– 𝑈𝐸(𝑆𝐿) = 984.92
Figure 11 - State Line (5 p. 38)
23
𝐵𝑇𝑈
𝑙𝑏
(3.33)
3.3.6
Exhaust Loss
3.3.6.1 Exhaust Loss Theory
The Exhaust Loss, “EL”, is the amount of steam that is ‘lost’ to the turbine system. This
loss, generally speaking, is the steam that isn’t converted to Work through the turbine; it
bypasses the blades one way or another. The Exhaust Loss in this system dumps straight
to the Condenser, whereas with an Extraction Turbine some Exhaust Loss actually is
rerouted to reheat the incoming boiler water.
To start off, an “EL” value between 4,000 and 6,000 pounds-per-hour is the target
range to get what is called an ‘Annulus Area’ “Aa”. The Annulus Area is the average
cross section area of flow. I am going to assume that our Steam Rate, Non-Extraction
“SR(NE)” is going to be 5.5 (for conditions “Po”, “To”, and “P(MN COND)”). After
solving for the two different values (solved in the Appendices), I got the answers. For a
flow of 4000 lb/hr the Exhaust Loss “EL” is equal to 9.81118 lb/hr, and for a flow of
6000 lb/hr, 17.19019 lb/hr. These values provide the general range of where our Exhaust
Loss lies. It is somewhere between approximately 9.8 lb/hr and 17.2 lb/hr. The next
section will solve for the definitive answer.
3.3.6.2 Annulus Area and Solving of Exhaust Loss
Remembering that the Annulus Area “Aa” is the average cross sectional area of flow for
the Main Turbine, we can now solve for it. The new variable below, “X,” is the actual
Condenser Flow.
𝑆𝐻𝑃
( 𝐸𝑚 )
𝐴𝑎 = 𝑆𝑅(𝑁𝐸) ∗
𝑃(𝑀𝑁 𝐶𝑂𝑁𝐷) ∗ 𝑋
(3.34)
“𝐴𝑎” 𝑎𝑡 4,000 “𝐸𝐿” 𝐹𝑙𝑜𝑤 = 31.031 ft²
“𝐴𝑎” 𝑎𝑡 6,000 “𝐸𝐿” 𝐹𝑙𝑜𝑤 = 20.688 ft²
The closest Annulus Area that we can pick will be a value of 25, as it’s the mean of
the two boundaries solved above. The Actual Condenser Flow “X” will come from the
Annulus Area.
𝑋 = 𝑆𝑅(𝑁𝐸) ∗
𝑆𝐻𝑃
( 𝐸𝑚 )
(𝑃(𝑀𝐴𝐼𝑁 𝐶𝑂𝑁𝐷) ∗ 𝐴𝑎)
24
= 4965.019
𝑙𝑏
ℎ𝑟
(3.35)
With this exact Condenser Flow Value, we can use it to solve for the Exhaust Loss
utilizing the curve fit equation (Appendices A.2.5).
𝐸𝐿 = 12.991
3.3.7
𝐵𝑇𝑈
𝑙𝑏
Mechanical and External Efficiency
The Mechanical and External Efficiency “Em” is the ratio of the Shaft-Horsepower to
the Shaft-Horsepower-plus-Losses.
𝑆𝐻𝑃
(𝑆𝐻𝑃 + 𝐿𝑜𝑠𝑠𝑒𝑠)
(3.36)
(100 − (3.5 + 𝑅𝐿))
100
(3.37)
𝐸𝑚 =
𝐸𝑚 =
In order to solve for “Em” I need all of the Losses of the system. The losses are
given by the following equation below.
𝐿𝑜𝑠𝑠𝑒𝑠 =
(𝑆𝐻𝑃 ∗ (3.5 + 𝑅𝐿))
(100 − (3.5 + 𝑅𝐿))
(3.38)
“RL” is the Astern-Turbine Windage. Windage is essentially friction which occurs
when the turbine blades contact the near-stationary steam when at low or no load
conditions. The Windage losses cause heating of the blades in the last few stages of the
turbine. By utilizing the lowest area available exhaust end of the astern casing, windage
losses can be reduced.
𝑅𝐿 = 0.33 ∗ 𝑃(𝑀𝑁 𝐶𝑂𝑁𝐷) = 0.495 "𝐻𝑔
Now that I have solved for “RL,” the Total Loss equation can be solved.
𝐸𝑚 =
𝑆𝐻𝑃
𝑆𝐻𝑃 ∗ (3.5 + 𝑅𝐿)
(𝑆𝐻𝑃 + (
))
100 − (3.5 + 𝑅𝐿)
= 0.96005
This can be verified utilizing the second equation given above.
𝐸𝑚 =
(100 − (3.5 + 𝑅𝐿))
= 0.96005
100
25
(3.39)
3.3.8
Gland Leak Off
Gland Leak Off is just that, steam and air that leaks off from the glands of the turbine.
For a typical cross-compound marine propulsion turbine with single flow low pressure
elements, such as what this system is using, we may make a few assumptions.
The first is that the Total Leak Off Steam “Q(MN LEAK OFF)” is equal to 200
pounds-per-hour multiplied by 0.3 % of the Shaft Horse Power
𝑄(𝑀𝑁 𝐿𝐸𝐴𝐾 𝑂𝐹𝐹) = 200 + (0.003 ∗ 𝑆𝐻𝑃) = 297.5
𝑙𝑏𝑠
ℎ𝑟
(3.40)
and the second is that the Enthalpy of Leak Off Steam “h(MN LEAK OFF)” is
equal to the equation below.
ℎ(𝑀𝑁 𝐿𝐸𝐴𝐾 𝑂𝐹𝐹) = ℎ𝑜 − 0.4 ∗ 𝑈𝐸𝑊
(3.41)
However, to solve for “h(MN LEAK OFF)”, we need to solve for “UE(W).” The
Wheel Used Energy, “UE(W)” is equal to the Initial Enthalpy minus the State Line End
Point (Remaining Energy). This brings us to solve for “UE(W).”
𝑈𝐸(𝑊) = ℎ𝑜 − ℎ𝑤 = 484.087
𝐵𝑇𝑈
𝑙𝑏
(3.42)
Now that we have the Wheel Used Energy, we can solve for the enthalpy of the
Leak-Off Steam.
ℎ(𝑀𝑁 𝐿𝐸𝐴𝐾 𝑂𝐹𝐹) = 1288.363
26
𝐵𝑇𝑈
𝑙𝑏
3.4 Power
3.4.1
Power Equation
Horsepower was originally defined to compare the output of steam engines with the
power of draft horses in continuous operation. The unit is used to measure the output of
piston engines, turbines, electric motors, and various other pieces of machinery. The
power generated is usually defined in terms of kilo-Watts.
𝑊𝐻𝑃 = 𝑃 = (
3.4.2
𝑆𝐻𝑃
) = 33852.4 ℎ𝑝
𝐸𝑚
(3.43)
Estimated Total Flow Rate from Superheaters
Next I want to solve for the Total Flow Rate leaving the Superheaters in the Boilers that
does not reach the Main Turbine, “G.” The equation for “G” is
𝐺 = 𝑄(𝐷𝐸𝑆𝑈𝑃) + 𝑄(𝑇𝐺) + 𝑄(𝐿𝑂𝑆𝑇)
(3.44)
“G” is the Total Flow Rate, as it contains what is used by the Turbo Generator, what
is lost, and what is desuperheated. Since “G” is equal to these three factors, I will be
solving for “Q(DESUP)”, “Q(TG)”, and “Q(LOST)”. The following three equations
which come from MARINE STEAM POWER PLANT HEAT BALANCE PRACTICES (5))
lay out the needed values.
𝑄(𝑇𝐺) = 𝑆𝑅(𝑇𝐺) ∗ 𝑘𝑊(𝑇𝐺)
𝑄(𝐿𝑂𝑆𝑇) = 0.005 ∗ (𝐸) = 824.5402
(3.45)
𝑙𝑏
ℎ𝑟
𝐵𝐻𝑃
𝑄(𝐷𝐸𝑆𝑈𝑃) = 𝑆𝑅(𝐹𝑃) ∗ (
) ∗ 𝐸(𝐸𝑆𝑇) + 3000
𝐸
(3.46)
(3.47)
Where “kW(TG)” is the Turbo Generators kilo-Watt Load and “E(EST)” is the
Energy Estimate. Instead of using “E”, we are going to be using “E(EST)”, which is the
Energy Estimate, instead to find estimated values of “Q(TG)”, “kW(TG)”, and
“Q(DESUP)” (as we don’t have “E” yet).
𝐸(𝐸𝑆𝑇) − 𝐺(𝐸𝑆𝑇) = 𝑆𝑅(𝑁𝐸) ∗ (
𝑆𝐻𝑃
ℎ(𝐵) − ℎ𝑤
(3.48)
) ∗ [1 − 0.25 ∗ (1 −
)]
(ℎ𝑜 − ℎ𝑤)
𝐸𝑚
To solve for “h(B)”, I need the average of “hB1”, “hB2”, and “hB3”. “hB2” is the
Enthalpy of the Steam Air Heaters. “hB1” is equal to the average of “hB2” and the
27
Initial Enthalpy. Finally, “hB3” is equal to “hB2” and the State Line End Point Enthalpy
average.
ℎ𝑜 + ℎ𝐵2
𝐵𝑇𝑈
= 1377.934
2
𝑙𝑏
𝐵𝑇𝑈
ℎ𝐵2 = ℎ(𝑆𝐴𝐻) = 1273.871
𝑙𝑏
ℎ𝐵2 + ℎ𝑖
𝐵𝑇𝑈
ℎ𝐵3 =
= 1129.396
2
𝑙𝑏
ℎ𝐵1 + ℎ𝐵2 + ℎ𝐵3
𝐵𝑇𝑈
ℎ(𝐵) =
= 1260.4
3
𝑙𝑏
ℎ𝐵1 =
(3.49)
(3.50)
(3.51)
(3.52)
Since “E(EST)” is being solved to find a general area value, I need to find its
components in estimated form. “Q(TG EST)” is equal to “SR(TG EST)” multiplied by
“kW(TG EST)”. I need to find “SR(TG EST)” and “kW(TG EST)”. The values listed
below are rough estimates that I am assuming in order to find the Total Flow Rate “G”,
which will be solved later in this project.
𝑆𝑅(𝑇𝐺 𝐸𝑆𝑇) = 10
𝑙𝑏𝑠
ℎ𝑟 ∗ 𝑘𝑊
𝑘𝑊(𝑇𝐺 𝐸𝑆𝑇) = 1250𝑘𝑊
With these two values we may now solve for the Estimated TG Flow:
𝑄(𝑇𝐺 𝐸𝑆𝑇) = 1250𝑘𝑊 ∗ 10
𝑙𝑏
ℎ𝑟 ∗ 𝑘𝑊
𝑙𝑏
= 12500 (~𝑎𝑝𝑝𝑟𝑜𝑥𝑖𝑚𝑎𝑡𝑒𝑙𝑦 0.4 ∗ 𝐸(𝐸𝑆𝑇))
ℎ𝑟
(3.53)
3.4.2.1 Desuperheated Steam Estimate Flow
The last of the things to find for “G” is the Flow of Desuperheated Steam, or in this case,
the Estimated Flow “Q(DESUP EST).”
𝐵𝐻𝑃
𝑄(𝐷𝐸𝑆𝑈𝑃 𝐸𝑆𝑇) = 𝑆𝑅(𝐹𝑃) ∗ (
) ∗ 𝐸(𝐸𝑆𝑇)
𝐸
(3.54)
The Term “DESUP” is Desuperheated steam, or steam that has moisture in it and is
below its superheated state. “Q(DESUP EST)” will not be solved until after the Steam
Rate of the Feed Pump, “BHP / E” and “E(EST)” are found (Section 3.2.2).
28
3.4.3
Power Matrix
Using Table 10 - Power Matrix, Page 100, we can derive the equation needed to solve
the Power.
𝑃
𝐸−𝐺
𝐸– 𝐺 − 𝐵1
= (
)+(
)
𝑎
𝑏
𝐸 − 𝐺 − 𝐵1 − 𝐵2 − 𝑄(𝑀𝑁 𝐿𝐸𝐴𝐾 𝑂𝐹𝐹)
+(
)
𝑐
(3.55)
𝐸 − 𝐺 − 𝐵1 − 𝐵2 − 𝐵3 − 𝑄(𝑀𝑁 𝐿𝐸𝐴𝐾 𝑂𝐹𝐹)
+(
)
𝑑
Since 𝑊𝐻𝑃 =
𝑆𝐻𝑃
𝐸𝑚
, we can solve for “P”:
𝑃 = 𝑊𝐻𝑃 =
𝑆𝐻𝑃
𝐸𝑚
29
= 33852.4 ℎ𝑝
(3.56)
3.5 High Pressure Heater (Economizer)
3.5.1
Component Detail
The High Pressure Heater is also known as the ‘Third Stage Heater’ or the
‘Economizer’. The High Pressure Heater serves as the last reheat before the water is sent
back into the boiler. Usually, the heat stack (comparable to a chimney in a house) is used
to supply the heat for the Economizer; however an Economizer cannot be too efficient. If
the temperature exchange is too great and the stack’s temperature drops below its
condensation point, then the stack will not be able to perform its purpose of carrying out
soot and ash out of the system. An economizer in this way is designed to be inefficient,
so as to allow the stack to perform properly. Figure 12 below shows a Typical
Economizer and its associated components.
In this section I will solve for the High Pressure “HP” Heater’s mass flow, “B1”,
using the First Law of Thermodynamics (Conservation of Energy), with constant volume
equal to the High Pressure Heater “HP HTR”.
Figure 12 - Typical Economizer (9)
30
3.5.2
Terminal Temperature Difference and Temperature Difference
The Terminal Temperature Difference “TTD” (in heaters) is the difference between the
saturation temperature of the condensed steam (heating steam) and the outlet
temperature of the fluid being heated. If the superheat temperature of the heating steam
is two-hundred degrees Fahrenheit or greater than the saturation temperature, a
desuperheating section can be used. In this section, it will be; therefore our “TTD” will
be equal to zero. If counter flow is assured for the drain cooling section of the heater
(which it will), a temperature difference of 10 degrees Fahrenheit will be used between
the feedwater inlet temperature and the drain outlet temperature.
Just to recap what we just learned:
𝑇𝑒𝑟𝑚𝑖𝑛𝑎𝑙 𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝐷𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 “𝑇𝑇𝐷” = 0 °𝐹
𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝐷𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 “𝑇𝐷”
= 10 °𝐹 (𝐶𝑜𝑢𝑛𝑡𝑒𝑟𝑓𝑙𝑜𝑤 𝐻𝑒𝑎𝑡 𝐸𝑥𝑐ℎ𝑎𝑛𝑔𝑒)
(3.57)
(3.58)
Since a Counter-flow Heat Exchanger is being utilized with this Steam Driven
Power Plant, The Temperature Difference “TD” is equal to ten degrees-Fahrenheit.
3.5.3
Heat Exchange Pressure, Temperatures and Enthalpies
For main steam systems, a seven-percent ‘Pressure Loss’ is from the boiler superheater
outlet to the maneuvering throttle valve inlet flange when the turbine handle valves are
utilized. A seven-percent loss is standard for this type of system, as this system is not
considered a ‘Main Steam System,’ a ‘Desuperheated System,’ or a ‘Bleed (Exhaust)
System Less than 30 psia.’ Those values different from our seven percent; therefore they
will not be used. Exhaust steam will in this section of the project be assumed to have no
loss in enthalpy.
𝑃(𝐻𝑃 𝐻𝑇𝑅 𝑆𝐻𝐸𝐿𝐿) = (1.00 − 0.07) ∗ 𝑃(𝑆1)
= 260.6527 𝑝𝑠𝑖𝑔
(3.59)
The temperatures and enthalpies we are looking for in this section are that of the
High Pressure Heater Shell (Outlet) and the Third Stage Heater (“HP Heater”) Inlet. To
find the Temperature Outlet, we need the Temperature of the Saturated Steam at the HP
Heater Shell. That number needs to have the Terminal Temperature Difference
subtracted from it to find “TF3,” which we will use to designate this value.
31
𝑇𝐹3 = (𝑇(𝑆𝐴𝑇)@𝑃(𝐻𝑃 𝐻𝑇𝑅 𝑆𝐻𝐸𝐿𝐿))– 𝑇𝑇𝐷 = 404.67 °𝐹
(3.60)
With this value, we will then find the Enthalpy “hF3” at this point. We get:
ℎ𝐹3 = ℎ(𝑓)@𝑇𝐹3 = 1202.042
𝐵𝑇𝑈
𝑙𝑏
(3.61)
We have now solved for the Temperature (404.67 °F) and Enthalpy (1202.042
BTU/lb) of the High Pressure Heater Outlet. To find the HP Heater Inlet Temperature
and Enthalpy, we need to solve for “TF2,” the Temperature calculated before we add the
TTD to it, and “hF2”, which is the Enthalpy at the Direct Contact Heater. To solve for
the Inlet Enthalpy, “hF2”, we will use the following equation.
ℎ𝐹2 = ℎ𝐹2@𝑃(𝐷𝐶 𝐻𝑇𝑅) + (
𝑣 ∗ 𝛥𝑃 ∗ 144
)
𝐸𝑓𝑓(𝐹𝑃) ∗ 778
(3.62)
The unknowns in this equation are given below and they both are assumptions.
Per the MARINE ENGINEERING DESIGN I REPORT (2 p. 25):
𝛥𝑃 = 70 𝑝𝑠𝑖𝑔
(3.63)
𝐸𝑓𝑓(𝐹𝑃) = 0.7
(3.64)
This leaves us with an Inlet Enthalpy “hF2” around 1170 BTU/lb and an Inlet
Temperature “TF2”, which is the Temperature at Enthalpy “hF2,” around 272 °F.
ℎ𝐹2 = 1171.367
𝐵𝑇𝑈
𝑙𝑏
𝑇𝐹2 = 271.955 °𝐹
Since the Third Stage Heater is a Heat Exchanger, we need to find a couple of
values associated with it. The flow of the incoming heat exchanging liquid (which is
saturated steam from the Main Propulsion Turbine Exhaust) is given from the equation:
ℎ𝐹3 − ℎ𝐹2
𝑙𝑏
𝐵1 = 𝐸(𝐸𝑆𝑇) ∗ (
) = 164108
ℎ𝐵1 − ℎ𝐷3
ℎ𝑟
(3.65)
This flow is considered constant through the Third Stage Heater, however the
Enthalpy will change. The Enthalpy of Heating Steam is given by the value “hB1,”
which has already been solved to be equal to 1377.934 BTUs-per-pound. To solve the
Enthalpy of the Heating Steam Exiting the Heat Exchange “hD3”, it will be solved using
the following equation. Temperature “TD3” is determined by utilizing the Steam Charts
at that Enthalpy.
ℎ𝐷3 = ℎ(𝑓)@((𝑇(𝑆𝐴𝑇)@ℎ𝐹2) + 𝑇𝐷) = 1176.884
32
𝐵𝑇𝑈
𝑙𝑏
(3.66)
𝑇𝐷3 = 𝑇(𝑆𝐴𝑇)@ℎ𝐷3 = 282.955 °𝐹
(3.67)
Figure 13 - Economizer Flow (aka "Third Stage Heater") shows the flows we have
identified and labels the flow path for both the Main Steam Flow and the Heat
Exchanger Fluid Flow.
Figure 13 - Economizer Flow (aka "Third Stage Heater") (2 p. 11)
33
3.6 Direct Contact Heater (Deaerating Feed Tank)
3.6.1
Component Detail
The Direct Contact (DC) Heater (aka Deaerating Feed Tank) acts as a holding tank for
the incoming water to the feed pump to pump into the boiler. It provides a positive head
pressure for the Feed Pump, as well as removes non-condensable gas, and preheats the
water before it’s pumped into the boiler.
The figure below, (Figure 14), show a cross section of a DC Heater with a water and
steam flow path. Notice how it acts as a retention tank for water, as well as allowing
non-condensable gases to be vented out the top.
Figure 14 - Typical DC Heater (Deaerating Feed Tank) (10)
34
3.6.2
DC Heater and Auxiliary Exhaust Pressure
The Pressure of the Direct Contact Heater is a very small fraction of the Initial Pressure
(usually 1/10th the pressure), as one of its functions is meant to provide a positive head
pressure on the Feed Pump so that no air has the opportunity to be entrapped in the
water.
“𝑃𝑜”, 𝑖𝑛 𝑝𝑠𝑖𝑎
𝑃(𝐷𝐶 𝐻𝑇𝑅) = 0.1 ∗ (
)
2
(3.68)
This value should fall within a range of forty to fifty psia. After substituting in the
Initial Pressure, “Po” (in psia), we get:
864.696
𝑃(𝐷𝐶 𝐻𝑇𝑅) = 0.1 ∗ (
) = 43.235 𝑝𝑠𝑖𝑎
2
This value falls within the range we set; therefore our DC Heater Pressure is an
acceptable value. To solve for the Auxiliary Exhaust Pressure “P(AUX EXH)”, which is
the pressure of the Steam Exhaust of the Feed Pump, it’s found by adding five psia to
that of the DC Heater.
𝑃(𝐴𝑈𝑋 𝐸𝑋𝐻) = 𝑃(𝐷𝐶 𝐻𝑇𝑅) + 5 = 48.235 𝑝𝑠𝑖𝑎
(3.69)
3.6.2.1 Brake Horsepower
The Brake Horsepower “BHP” is equal to the specific volume at the DC “v(DC HTR)”
multiplied by the Change in Pressure Overall (highest to lowest point) “ΔP,” multiplied
by 144, divided by 778, and finally divided by the Efficiency of the Feed Pump
“Eff(FP)”. The specific volume is solved at the Pressure of 43.2348, we get a specific
volume equal to 0.0171872 ft^3/lb. The change in pressure “ΔP” we are going to be
assuming is approximately 1000 psi, so this will be our value. The Efficiency of the Feed
Pump will be assumed to be 70%, or 0.7. This leaves us with the equation for “BHP”
with all factors solved.
𝐵𝐻𝑃 =
3.6.3
𝑣(𝐷𝐶 𝐻𝑇𝑅) ∗ 𝛥𝑃 ∗ 144
= 1706.42 ℎ𝑝
778 ∗ 𝐸𝑓𝑓(𝐹𝑃)
(3.70)
DC Heater Flow Rate to Feed Pump
The mass flow of the Direct Contact Heater to the Feed Pump is solved by multiplying
the Steam Rate Non-Extraction with the Brake Horsepower divided by the Estimated
35
Enthalpy and then multiplied by the Estimated Enthalpy. The value of E(EST) divided
by E(EST) cancels out to equal one. Therefore:
𝐵𝐻𝑃
𝑄(𝐹𝑃) = 𝑆𝑅(𝑁𝐸) ∗ (
) ∗ 𝐸(𝐸𝑆𝑇) = 𝑆𝑅(𝑁𝐸) ∗ 𝐵𝐻𝑃
𝐸(𝐸𝑆𝑇)
𝑙𝑏
= 4354.290
ℎ𝑟
(3.71)
3.6.3.1 Feed Pump Recirculation
The Feed Pump Recirculation Rate “R” will not be used for this system, so it will have a
value of zero.
𝑅 = 0.00
3.6.4
𝑙𝑏
ℎ𝑟
Direct Contact Vent Condenser
The Low Pressure Heater (also known as the First Stage Heater, which will be
explained more in Section 3.7) plays an important role to the Direct Contact Heater. Its
exit flows directly into what is called the Direct Contact Vent Condenser. This action
creates an effect similar to the Air Ejectors, where a change in pressure (from high to
low) causes the non-condensable gases to come out of solution and be vented out the DC
Heater Vent. This action not only provides this feature, but it also helps maintain a
cooling action on the DC Heater, a more consistent pressure, and allows for storage. If
this feed line were to stop allowing flow into the DC Heater, there would be a time-lapse
before the Feed Pump would stop receiving flow. That’s why the DC Heater acts as a
storage tank. It will allow some time to recuperate whatever losses to the DC Heater
before a loss of feed into the boilers occurs.
36
Figure 15 - Direct Contact Heater Flow (2 p. 12)
The flow through the vent condenser on the Direct Connect Heater is also needed to
be found. A typical flow set by the manufacturer is one-hundred pounds (mass) per hour,
therefore:
𝑄(𝑉𝐸𝑁𝑇 𝐶𝑂𝑁𝐷) = 100
𝑙𝑏
ℎ𝑟
The enthalpy at the Vent Condenser “VENT COND” is going to be denoted by the
term “h(VENT COND)”. It’s equal to the enthalpy of saturated steam at the Pressure of
the DC Heater “P(DC HTR)”.
ℎ(𝑉𝐸𝑁𝑇 𝐶𝑂𝑁𝐷) = ℎ(𝑔)@𝑃(𝐷𝐶 𝐻𝑇𝑅) = 1171.367
𝐵𝑇𝑈
𝑙𝑏
(3.72)
Therefore h(VENT COND) is equal to approximately 1170 BTUs-per-pound.
37
3.7 Low Pressure Heater
3.7.1
Component Detail
The Low Pressure Heater is also known as the First-Stage Heater, and generally it
operates at ambient pressure. It’s usually the first stage of heaters (hence the nickname
“First-Stage Heater”) that preheats the water before being pumped back into the steam
drum in the boiler. The Gland Exhaust Condenser supplies the flow to the Low Pressure
Heater, after a Heat Exchange from vapor from the Feed Water Drain Collection Tank
(“FWDCT”). In the next section, we will discuss the FWDCT’s purpose and equations
associated with it.
Since I chose a Counter-Flow Heat Exchanger for my project (as it has the best heat
exchange properties per single pass), I will be using the Counter-Flow values for this
“TTD1” and each Temperature Difference:
𝑇𝑇𝐷1 = 10 °𝐹 (𝐶𝑜𝑢𝑛𝑡𝑒𝑟 − 𝐹𝑙𝑜𝑤)
(3.73)
Now I need to solve for hc1, which is found by using the steam tables to find the
enthalpy at “Tc1,” the Temperature at the Pressure of the Low Pressure Heater Shell
“P(LP HTR SHELL).” The Pressure of the Low Pressure (LP) Heater Shell is equal to
90 % (10% Loss) of the value “B3”, Exhaust Flow.
𝑃(𝐿𝑃 𝐻𝑇𝑅 𝑆𝐻𝐸𝐿𝐿) = (1.0– 0.1) ∗ 𝑃(𝐵3) = 9.931 𝑝𝑠𝑖𝑎
(3.74)
The Temperature “Tc1” is equal to the Temperature of the Low Pressure Heater
Shell plus the Terminal Temperature Difference of the Heat Exchange “TTD1”. The
equation below shows each value.
𝑇𝑐1 = (𝑇(𝑆𝐴𝑇)@𝑃(𝐿𝑃 𝐻𝑇𝑅 𝑆𝐻𝐸𝐿𝐿)) + 𝑇𝑇𝐷1
(3.75)
𝑇𝑐1 = 192.83 °𝐹 + 10 °𝐹 = 202.83 °𝐹
Finally, to solve for the Enthalpy, we just use the Steam Tables to find the Enthalpy
at Saturated Temperature point “Tc1”.
ℎ𝑐1 = ℎ(𝑔)@𝑇𝑐1 = 1147.783
38
𝐵𝑇𝑈
𝑙𝑏
(3.76)
3.8 Feed Water Drain Collection Tank
3.8.1
Component Detail
The Feed Water Drain Collection Tank “FWDCT” serves as a dump for the Distiller Air
Ejector(s), Make-Up Feed Water, and the Vapor generated in it after it passes through
the Gland Exhaust Condenser (however it will not be used in this system). The vapor
flow exiting the FWDCT is equal to zero in this system; therefore the enthalpy of the
vapor will be the enthalpy at 212 °F, which is the boiling point of water at atmospheric
pressure. The flow of this vapor is zero and that and the enthalpy are solved from the
Appendices (Section A.2.9).
𝑄(𝑉𝐴𝑃 𝐹𝑊𝐷𝐶𝑇) = 0.00
𝑙𝑏
ℎ𝑟
ℎ(𝑉𝐴𝑃 𝐹𝑊𝐷𝐶𝑇) = ℎ(𝐹)@212 °𝐹 = 180.180
(3.77)
𝐵𝑇𝑈
𝑙𝑏
(3.78)
To solve for the overall flow through the Feed Water Drain Collection Tank, a
variety of values need to be calculated prior to solving it. Given by the Equation below,
the flow of each of the following components needs to be found: Contaminated Drains,
Steam Air Heaters, Make-Up Feed Water, Auxiliary After-Condenser, Main AfterCondenser, Distiller Air Ejector, Main Leak Off, Auxiliary Leak Off, and finally Ventflow.
𝑄(𝐹𝑊𝐷𝐶𝑇) = 𝑄(𝐶𝑂𝑁𝑇 𝐷𝑅𝑁) + 𝑄(𝑆𝐴𝐻)
+ 𝑄(𝑀𝐴𝐾𝐸𝑈𝑃 𝐹𝐸𝐸𝐷) + 𝑄(𝐴𝑈𝑋 𝐴𝐶)
+ 𝑄(𝑀𝑁 𝐴𝐶) + 𝑄(𝐷𝐼𝑆𝑇𝐼𝐿𝐿 𝐴/𝐸)
(3.79)
+ 𝑄(𝑀𝑁 𝐿𝐸𝐴𝐾 𝑂𝐹𝐹) + 𝑄(𝐴𝑈𝑋 𝐿𝐸𝐴𝐾 𝑂𝐹𝐹)
+ 𝑄(𝑉𝐸𝑁𝑇)
3.8.2
Contaminated Drains
3.8.2.1 Component Detail
The Contaminated Drains “CONT DRNS” purpose is to collect the Exhaust from the
Cargo Heating and the Fuel Oil Heating and to direct it into the Feed Water Drain
Collection Tank.
39
3.8.2.2 Flow and Enthalpy
The first thing we on our list to solve is the Contaminated Drains flow, “Q(CONT
DRNS)”. To solve for this value, we need the flows of the Fuel Oil Heaters “Q(FOH)”
and the flow of the Domestic Loads “Q(DOM)”. Both of these values are solved later in
this project, as their Sections haven’t come up yet. The Flow of the Fuel Oil Heaters is
solved in Section 3.1.2.2 and the Flow of the Domestic Loads is solved in Section 3.12.
With the Flow of Fuel Oil Heaters and the Domestic Loads solved in their respective
section, we have the necessary values for the Contaminated Drain Flow. The equation
below is a reminder of what we have learned at the beginning of this section.
𝑄(𝐶𝑂𝑁𝑇 𝐷𝑅𝑁𝑆) = 𝑄(𝐹𝑂𝐻) + 𝑄(𝐷𝑂𝑀)
(3.80)
Substituting the values we have gathered into the above equation, we get:
𝑄(𝐶𝑂𝑁𝑇 𝐷𝑅𝑁𝑆) = 1137.5
𝑙𝑏
𝑙𝑏
𝑙𝑏
+ 100
= 1237.5
ℎ𝑟
ℎ𝑟
ℎ𝑟
This flow has an associated enthalpy with it “h(CONT DRNS)”. It’s equal to the
enthalpy at 200 degrees Fahrenheit; a value chosen because of the water/vapor mixture
state of the steam.
ℎ(𝐶𝑂𝑁𝑇 𝐷𝑅𝑁𝑆) = ℎ(𝑓)@200 °𝐹 = 168.099
𝐵𝑇𝑈
𝑙𝑏
(3.81)
The Flow of the Contaminated Drains will be plugged into the equation for the
“FWDCT” Flow, but not until we finish solving the remainder of the flows. The next
flow we will discuss is the Steam Air Heater “SAH” Flow.
3.8.3
Steam Air Heaters
3.8.3.1 Component Detail
The Steam Air Heaters are piping lines that come off the boiler and are installed to
utilize the thermal energy from the boiler to heat ‘low-temperature flue gas’ (air) before
being dumped into the “FWDCT” and effectively increases the thermal efficiency of the
system. They recover the heat from the boiler flue gas (the gases that leave the stack)
and redirect some of that heat and energy back into the system.
40
3.8.3.2 Temperature and Flow
The Temperature of the Steam Air Heaters “T(SAH)” is usually chosen to be ~25
degrees Fahrenheit greater than the Temperature of the Fuel Oil “T(FO)”. Since this
plant’s “T(FO)” is 200 degrees Fahrenheit, our “T(SAH)” is equal to 225 degrees
Fahrenheit.
𝑇(𝑆𝐴𝐻) = 225 °𝐹
The enthalpy of the Steam Air Heaters Inlet “h(SAH IN)” is equal to the Enthalpy at
“B2”, which is the exhaust steam from the medium between the High and Low Pressure
Turbines.
ℎ(𝑆𝐴𝐻 𝐼𝑁) = ℎ(𝐵2) = 1273.871
𝐵𝑇𝑈
𝑙𝑏
(3.82)
The enthalpy of the Steam Air Heaters Outlet “h(SAH OUT)” is equal to the
enthalpy found at the Temperature of the Steam Air Heaters, 225 degrees. This enthalpy
yields us around 193 BTU/lb.
ℎ(𝑆𝐴𝐻 𝑂𝑈𝑇) = ℎ(𝑓)@225 °𝐹 = 193.297
𝐵𝑇𝑈
𝑙𝑏
(3.83)
To find the difference in the Enthalpies “Δh(SAH)”, the Outlet Enthalpy needs to be
subtracted from the Inlet Enthalpy.
𝛥ℎ(𝑆𝐴𝐻) = ℎ(𝑆𝐴𝐻 𝐼𝑁) − ℎ(𝑆𝐴𝐻 𝑂𝑈𝑇) = 1080.57
𝐵𝑇𝑈
𝑙𝑏
(3.84)
This system has saved around 1100 BTUs-per-pound of energy that would have
otherwise been lost out the stack of the ship. That’s a pretty hefty amount; it will help
with retaining energy and increasing overall thermal and energy efficiency.
The last thing to solve is the flow of the Steam Air Heaters “Q(SAH)”. To do so, the
Air to Fuel Ratio “A / F” is needed, along with the Specific Fuel Consumption “SFC”,
the Work “W”, and finally the Heat Capacity of Air “Cp(Air)”. The Heat Capacity of Air
is the only value that still has to be solved, as the others have been solved prior to this
section.
The Heat Capacity of Air, or Mean Specific Heat, is how much energy air can hold
at a given temperature and humidity. Figure 16 - Mean Specific Heat, below, shows the
graph of mean specific heat as a function of final air temperature.
41
Figure 16 - Mean Specific Heat (5 p. 48)
𝐶𝑝(𝐴𝑖𝑟)@(200 𝑑𝑒𝑔𝑟𝑒𝑒𝑠) = 0.244
(3.85)
Now that we have the Heat Capacity of Air we can solve for the Flow of the Steam
Air Heaters Drains.
𝐴
𝐶𝑝(𝐴𝑖𝑟)
𝑄(𝑆𝐴𝐻 𝐷𝑅𝑁𝑆) = ( ) ∗ 𝑆𝐹𝐶 ∗ 𝑊 ∗ (
)
𝐹
2
𝑙𝑏
= 5576.231
ℎ𝑟
3.8.4
(3.86)
Make-Up Feedwater
3.8.4.1 Component Detail
Make-Up Feedwater is a means of providing Feedwater back into the system. Normally,
there are losses in the system and with an operating system, water levels in the
condensers will eventually run dry and the boiler will be starved for feedwater. To
counter this, makeup feed is added back into the Condensers. Feedwater is usually made
up of system ‘pure’ water, or water that meets the requirements of system use (usually
free of chromates and salt.) Makeup water isn’t preheated before entering the condenser,
so it will have ambient temperature. In our equations, we will use the temperature value
of 75 degrees Fahrenheit for “T(MAKEUP FEED)”.
42
3.8.4.2 Temperature, Enthalpy, and Flow
As discussed in Section 3.8.4.1, the temperature of the Make-Up Feedwater
“T(MAKEUP FEED)” will be 75 degrees.
𝑇(𝑀𝐴𝐾𝐸𝑈𝑃 𝐹𝐸𝐸𝐷) = 75 °𝐹
The Enthalpy at this point “h(MAKEUP FEED)” is 43.07439 BTUs-per-pound. To
solve for the flow of the makeup feed, the sum of the Steam Atomizer Flow “Q(STM
ATOM”, Soot Blowers Flow “Q(SOOT BLOWERS)”, and Lost Flow “Q(LOST)” is
needed. The flows for the Steam Atomizers and Soot Blowers are solved in Sections
3.8.5 and 3.8.6, respectfully. Once we solve these two values, we can determine the
Make-Up Feedwater flow to be around 1925 pounds-per-hour.
𝑄(𝑀𝐴𝐾𝐸𝑈𝑃 𝐹𝐸𝐸𝐷)
= 𝑄(𝑆𝑇𝑀 𝐴𝑇𝑂𝑀) + 𝑄(𝑆𝑂𝑂𝑇 𝐵𝐿𝑂𝑊𝐸𝑅𝑆)
+ 𝑄(𝐿𝑂𝑆𝑇) = 1923.623
(3.87)
𝑙𝑏
ℎ𝑟
The next section will discuss the Distilling Plant “DISTILL PLT”, which is the next
thing needed for the total flow of the Feedwater Drain Collection Tank “FWDCT”.
3.8.5
Steam Atomizers
The flow of the Steam Atomizers is the remainder of the flow from the High Pressure
Turbine “HP TURB” that has not gone to the Fuel Oil Heaters.
The flow from the HP Turbine is 1529.6 pounds-per-hour. If you subtract the flow
to the Fuel Oil Heaters (1137.5 pounds-per-hour) we get our answer of 392.1 poundsper-hour flow to the Steam Atomizers.
3.8.6
Soot Blowers
The soot blowers’ function is to clear the boilers from settled dirt, silt, and unwanted
particles. Soot blowing steam is drawn from the Desuperheated Steam and is forced
through the system, to help break away sediment and foreign particles that, over time,
would be detrimental to piping due to corrosion. Its flow has been solved in the
Appendices (Section A.1.8). Its value is 706.9844 lb/hr for the system (two Boilers) and
353.4922 lb/hr for one Boiler.
43
3.8.7
Distilling Plant
3.8.7.1 Component Detail
The Distilling Plant “DISTILL PLT” is the amount of water used per day for the ship
and crew. It’s necessary to have a 24-hour estimate of the average load on the distilling
plant so the steam consumption rate can be determined in our Heat Balance. Solving the
estimated service load will be performed in the next subsection.
3.8.7.2 Estimated Service Load
The equation for finding the Gallons per Day “GPD” is by adding the Makeup Feed
“MAKEUP FEED” water amount used in a day with the Potable Water Allowance per
person “PWA”. I will assume a typical person uses 80 Gallons per Day. I will also be
assuming 8.33 Pounds are in one Gallon.
24 𝐻𝑜𝑢𝑟𝑠
( 1 𝐷𝑎𝑦 )
𝑙𝑏𝑠
𝐺𝑃𝐷 = (𝑀𝐴𝐾𝐸𝑈𝑃 𝐹𝐸𝐸𝐷,
)∗(
)
8.33 𝑃𝑜𝑢𝑛𝑑𝑠
ℎ𝑟
(
)
𝐺𝑎𝑙𝑙𝑜𝑛
(
)
80 𝐺𝑎𝑙𝑙𝑜𝑛𝑠
+ ((
) ∗ 𝑁)
1 𝐷𝑎𝑦
= 8740 𝐺𝑎𝑙𝑙𝑜𝑛𝑠 𝑝𝑒𝑟 𝐷𝑎𝑦, (
(3.88)
𝑔𝑎𝑙
)
𝑑𝑎𝑦
This plant has been determined to use over 8500 Gallons of water per day. This is its
Service Load. The rated Capacity of the Evaporators must be 130% that of the Service
Load. Therefore, 11400 Gallons-per-Day minimum is the Rated Load for the
Evaporators. Rounding that number to the nearest on Table 2 – Standard Capacity of a
Distilling Plant, we get a value of our Distilling Plant Capacity of 12000 Gallons per
Day.
Table 2 – Standard Capacity of a Distilling Plant (2 p. 23)
Standard Capacity of a Distilling Plant (GPD Range)
8000
10000
12000
15000
20000
25000
Special
USE
44
3.8.7.3 Distiller Plant Flow
The Distiller Air Ejector Flow “Q(DISTILL PLT)” needed for this is solved by the
following Equation.
𝑄(𝐷𝐼𝑆𝑇𝐼𝐿𝐿 𝑃𝐿𝑇) =
(8.33 ∗ 𝐺𝑃𝐷 ∗ 𝑓(𝐷))
(24 ∗ ℎ𝐵1)
(3.89)
The Distillate Output Heat Steam “f(D)” is set by Figure 17 - Extraction Steam for
Flash Type Evaporators to be 780 BTU-per-pound Distillate. “hB1” is the Enthalpy of
the heating Steam, which is 1377.9 BTUs-per-pound
Figure 17 - Extraction Steam for Flash Type Evaporators (5 p. 25)
Now that we have each and every value, we can now solve the Distilling Plant
Extraction Steam (Heating Steam) Flow required. The flow, “Q(DISTILL PLT)” is
calculated to be 1807.55 pounds-per-hour.
(8.33
(𝐷𝐼𝑆𝑇𝐼𝐿𝐿 𝑃𝐿𝑇) =
𝑔𝑎𝑙
𝑔𝑎𝑙
𝐵𝑇𝑈
∗ 12000
∗ 780
)
𝑙𝑏
𝑙𝑏
𝑑𝑎𝑦
𝑙𝑏
= 1807.55
ℎ𝑟
𝐵𝑇𝑈
ℎ𝑟
(24
∗ 1377.9
)
𝑑𝑎𝑦
𝑙𝑏
45
3.8.8
Distiller Air Ejector Flow
To solve the Distiller Air Ejector “DIST A/E” Flow, “f(D)” needs to be altered
from 780 to 60 in the equation used for the Distilling Plant.
60
𝑙𝑏
𝑄(𝐷𝐼𝑆𝑇 𝐴/𝐸) = 8.33 ∗ 𝐺𝑃𝐷 ∗ (
) = 139.042
24 ∗ ℎ𝑖
ℎ𝑟
(3.90)
Using the equation above, we get a value of 139.042 pounds-per-hour for the Air
Ejector Flow of the Distiller.
3.8.9
Feed Water Drain Collection Tank Flow
This section will wrap up the “FWDCT” Section and will solve for the flow through the
Feed Water Drain Collection Tank. The remaining values we have already solved are
listed below:
𝑄(𝐿𝑂𝑆𝑇) = 824.540
𝑙𝑏
ℎ𝑟
𝑙𝑏
ℎ𝑟
𝑙𝑏
𝑄(𝐴𝑈𝑋 𝐿𝐸𝐴𝐾 𝑂𝐹𝐹) = 50
ℎ𝑟
𝑙𝑏
𝑄(𝑀𝑁 𝐴𝐶) = 196
ℎ𝑟
𝑙𝑏
𝑄(𝐴𝑈𝑋 𝐴𝐶) = 60
ℎ𝑟
𝑙𝑏
𝑄(𝑉𝐸𝑁𝑇) = 100
ℎ𝑟
𝑄(𝑀𝑁 𝐿𝐸𝐴𝐾 𝑂𝐹𝐹) = 297.5
With the remaining values we can finally solve the “FWDCT” Flow Equation with
the equation given in Section 3.8.1. We calculate our Feed Water Drain Collection
Tank Flow to be equal to around 11,000 pounds-per-hour.
𝑄(𝐹𝑊𝐷𝐶𝑇) = 11135.70
𝑙𝑏
ℎ𝑟
The figure on the next page (Figure 18 - Feed Water Drain Collection Tank System
Diagram) shows the generalized flow paths in and out of the Feed Water Drain
Collection Tank.
46
Figure 18 - Feed Water Drain Collection Tank System Diagram (2 p. 15)
47
3.9 Main Condenser
3.9.1
Component Detail
The Main Condenser “MN COND” is where the High and Low Pressure Turbines
saturated steam is dumped to. It is a crucial part of the Steam Water Cycle. Without this
component, a working steam power plant is unattainable. A heat exchange occurs between
the exhaust steam dumped and a cooling medium (usually seawater) - this cools the steam
into water. When this occurs, a vacuum is created. Typically, the more condensing that
occurs, the greater the vacuum is created (Higher inches of Mercury Vacuum) and in turn,
a better overall heat balance.
The Main Condenser is where makeup feedwater is pumped into, and it usually
operates with a condensed-steam-turned-water level between one-third and two-thirds of
the height. It consists of numerous tubes where the seawater flows through so there is no
cross-contamination of the two fluids.
In this project I am using a counter-flow, single-pass Condenser. Figure 19 – SinglePass Condenser shows the cross-sectional flow path of a typical single-pass condenser.
Notice how there are many sets of tubes. This is to allow for a large surface area and yet
still maintain sufficient flow through the system. Keeping these tubes clean and free of
debris is vital to the operation of a Condenser. If these tubes were to get clogged, heat
transfer would be reduced, which would hinder the cooling effects of it, which would
decrease the vacuum and the overall capabilities of the ships engine room. This is why
condensers are one of the components that have to be cleaned meticulously.
The figure on the next page shows the cross sectional configuration of a Single-Pass
Condenser with its inlets and outlets identified. Typically, a cross flow single pass
condenser is the most efficient for a single pass as it allows for the greatest heat transfer.
48
Figure 19 – Single-Pass Condenser (11)
3.9.2
Main Condenser Flow
The flow is given by the following equation. It is missing a variety of values that will be
discussed and solved in later sections in this project. However, the Main Condenser Flow
will be solved now to allow for an answer in this section. The values needed will be
calculated in later sections.
𝑄(𝑀𝑁 𝐶𝑂𝑁𝐷)
= [𝐸 − 𝐺 − 𝐵1 − 𝐵2 − 𝑄(𝑀𝑁 𝐿𝐸𝐴𝐾 𝑂𝐹𝐹) + 𝑄(𝑀𝑁 𝐼𝐶)
𝑙𝑏
− 𝑄(𝐷𝐼𝑆𝑇𝐼𝐿𝐿 𝑃𝐿𝑇)] = 168715.93
ℎ𝑟
49
(3.91)
3.10 Auxiliary Condenser
3.10.1 Component Detail
The Auxiliary Condenser’s purpose is to condense the steam driven to the Turbo
Generators. It acts in the same way as the Main Condenser, with a heat exchange
occurring and a vacuum drawn. The only difference is the size of the Auxiliary
Condensers when compared to that of the Main Condensers. They are smaller and
therefore receive and condense less steam.
3.10.2 Auxiliary Condenser Flow
The Auxiliary Condenser Flow “Q(AUX COND)” is made up of flow from all the
Auxiliary Steam Dumps. It includes the Turbo Generators, Make-Up Feedwater, the
Auxiliary Inter-Condensers, the Distilling Plant and Dump Flow. Its output is the
Auxiliary Leak Off.
𝑄(𝐴𝑈𝑋 𝐶𝑂𝑁𝐷)
= 𝑄(𝑇𝐺) + 𝑄(𝑀𝐴𝐾𝐸𝑈𝑃 𝐹𝐸𝐸𝐷)
+ 𝑄(𝐴𝑈𝑋 𝐼𝐶) + 𝑄(𝐷𝑈𝑀𝑃)
(3.92)
− 𝑄(𝐴𝑈𝑋 𝐿𝐸𝐴𝐾 𝑂𝐹𝐹) + 𝑄(𝐷𝐼𝑆𝑇𝐼𝐿𝐿 𝑃𝐿𝑇)
Each one of these components of the Auxiliary Condenser Flow Equation has been
previously solved in other sections; therefore the total is equal to around 10,500 pounds
per hour.
𝑄(𝐴𝑈𝑋 𝐶𝑂𝑁𝐷) = 10619.91
50
𝑙𝑏
ℎ𝑟
3.11 Desuperheater
3.11.1 Component Detail
The Desuperheater’s purpose is to control the temperature of a portion of the boiler
superheated steam output. In this plant, we will be utilizing an internal Steam Drum, as it
is more efficient (due to less radiation losses).
3.11.2 Desuperheater Steam Flow
The Desuperheated Steam that exits the Desuperheater is exactly that - desuperheated
steam. That means it contains moisture and isn’t superheated. The purpose of this is to
control how much flow is directed back into the boilers’ Superheater system so that the
Superheater Outlet Temperature can be controlled.
If the Desuperheater was not implemented, the Superheater would keep increasing the
system temperature until component failure would occur due to melting. This regulating
system helps to maintain overall temperature in the boiler. The Desuperheater Steam Flow
“Q(DESUP)” consists of eight parts, which are listed in the equation below and in Table 3
and if not already solved, will be solved in the following subsections.
𝑄(𝐷𝐸𝑆𝑈𝑃) = 𝑄(𝑆𝑂𝑂𝑇 𝐵𝐿𝑂𝑊𝐸𝑅𝑆) + 𝑄(𝑆𝑇𝑀 𝐴𝑇𝑂𝑀)
+ 𝑄(𝐷𝑂𝑀) + 𝑆𝑅(𝐹𝑃𝑇)
+ [𝑄(𝑀𝑁 𝐴/𝐸), 𝑄(𝐴𝑈𝑋 𝐴/𝐸), 𝑄(𝐷𝐼𝑆𝑇 𝐴/𝐸)]
+ 𝑄(𝐹𝑂𝐻) + 𝑄(𝑆ℎ𝑖𝑝𝑠 𝐻𝑒𝑎𝑡𝑖𝑛𝑔)
+ 𝑄(𝑂𝑡ℎ𝑒𝑟𝑠)
Table 3 - Desuperheater and its Components’ Flows
1. “Q(SOOT BLOWERS)” (Section 3.8.6)
2. “Q(STM ATOM)” (Section 3.8.5)
3. “Q(DOM)” (Section 3.12.5)
4. “SR(FP TURB)” (Section 4.2.2)
5. Σ(“Q(MN A/E)”,“Q(AUX A/E)”,“Q(DIST A/E)”)
(Sections 4.3.2, 4.3.3, 4.3.4)
6. “Q(FOH)” (Section 3.1.2.2)
7. Ships Heating (Assume 0.00 lb/hr flow)
8. Others (Assume 0.00 lb/hr flow)
Total Desuperheater Flow, “Q(DESUP)”
51
+
+
+
706.984
392.09827
100
6.3248
+
+
+
+
729.31544
1137.5
0
0
=
3072.22 lb/hr
(3.93)
3.11.3 Desuperheated Steam Mass Flow
To solve for “Q(DESUP)”, we simply add up the eight components that make it up.
“Q(DESUP)” is 3072.22 pounds per hour (as shown in Table 3).
𝑄(𝐷𝐸𝑆𝑈𝑃) 𝑓𝑜𝑟 𝑇𝑤𝑜 − 𝐵𝑜𝑖𝑙𝑒𝑟𝑠 = 3077.22
𝑄(𝐷𝐸𝑆𝑈𝑃)𝑝𝑒𝑟 𝐵𝑜𝑖𝑙𝑒𝑟 =
𝑙𝑏
ℎ𝑟
𝑄(𝐷𝐸𝑆𝑈𝑃) 𝑓𝑜𝑟 𝑇𝑤𝑜−𝐵𝑜𝑖𝑙𝑒𝑟𝑠
(3.94)
2
𝑙𝑏
=1538.61 ℎ𝑟 𝑓𝑜𝑟 𝑂𝑛𝑒 𝐵𝑜𝑖𝑙𝑒𝑟
This means that the Flow for the cycle of the desuperheated steam is equal to around
1540 pounds per hour for one boiler.
3.11.4 Desuperheated Temperature and Pressure
As we have learned, the temperature and pressure of desuperheated steam is less than that
of the superheated steam (superheater output.) Again, this is to control the outlet
temperature as to have a controlled system. This section will begin by calculating the
Pressure of the Desuperheated Steam “P(DESUP),” then will end with calculating the
Temperature “T(DESUP)”.
To calculate the Desuperheated Steam Pressure, the values for “Pso” (886.3134 psia)
and the Pressure Drop (Change) in Control Desuperheater Pressure “ΔP(CTRL DESUP)”
are needed. “ΔP(CTRL DESUP)” comes from Table 5 (Section 4.2.3.1), already solved
for this type of plant.
𝛥𝑃(𝐶𝑇𝑅𝐿 𝐷𝐸𝑆𝑈𝑃) = 30 𝑝𝑠𝑖
Given “To” and “Po” of 950 °F and 864.696 psia, respectively, we are going to be
solving for the temperature and Pressure of the desuperheated system. I need to solve for
the Superheater Outlet Steam Pressure first, which is 102.5 percent of “Po”.
𝑃𝑠𝑜 = 1.025 ∗ (𝑃𝑜) = 886.313 𝑝𝑠𝑖𝑎 = 871.6 𝑝𝑠𝑖𝑔
(3.95)
The Superheater Outlet Pressure is around 872 pounds per square inch gauge. This
pressure, when it has the Control Desuperheater Pressure subtracted from it, will solve for
the Pressure of the Desuperheater.
𝑃(𝐷𝐸𝑆𝑈𝑃) = 𝑃𝑠𝑜 − 𝛥𝑃(𝐷𝐸𝑆𝑈𝑃) = 841.6 𝑝𝑠𝑖𝑔
(3.96)
The Pressure of the Desuperheater has been found to be around 842 psig. Now all that
is left is to find the Temperature (and Enthalpy) of the Desuperheater.
52
The Temperature of the Desuperheater is found at the Saturation Temperature of the
Steam Drum Pressure. The Pressure of the Steam Drum is the summation of “P(DESUP)”
and the pressure drop at maximum Boiler Rate. There are a few steps before reaching the
solving the Pressure of the Steam Drum “P(STM DRUM)”.
First, using the Estimated Overall Energy “E(EST)” of 220,000 pounds per hour, the
Estimated Rate of Flow for each of the two Boilers “Q(BOILER)EST” is needed.
“Q(BOILER)EST” is simply half the estimated overall flow. Two Boilers share the load
evenly in our assumptions; therefore the flow is half of 220,000 pounds per hour, or
110,000 pounds per hour.
At 110,000 pounds per hour, the Change in Pressure at the Superheater
“ΔP(Superheater)” is equal to the Superheater pressure drop at maximum continuous
power plus the pressure change through the orifice. At 110,000 pounds and a Temperature
of 950 degrees, we get a “ΔP(Superheater)” of 69 psig. This includes the Orifice Pressure
of 7 psig. With this value, we need to convert it into the Pressure Drop at the Maximum
Boiler Rate. To do so, we simply take the Superheater Change in Pressure of 69 psig and
multiply it by the maximum steaming rate percentage (1.15 or 115%).
1.15 2
69 ∗ (
) = 91 𝑝𝑠𝑖𝑔
1.00
(3.97)
This Pressure Drop (91 psig), when added with the Superheater Outlet Pressure (872
psig) and the Pressure Drop due to Steam Temperature Control Desuperheater (30 psig),
gives us our Steam Drum Pressure “P(STM DRUM)” at Maximum Flow of 993 psig.
This value of 993 psig is used to measure the Temperature of the Desuperheater
“T(DESUP)” - by finding the Saturation Temperature “T(SAT)” at the Pressure of the
Steam Drum, we will find our answer. It comes out to about 544 degrees, which is
substantially lower than that of the Superheater’s Temperature or even the Boilers’.
𝑇(𝐷𝐸𝑆𝑈𝑃) = 𝑇(𝑆𝐴𝑇)@𝑃(𝑆𝑇𝑀 𝐷𝑅𝑈𝑀) = 543.59 °𝐹
(3.98)
When we use our steam tables at the Saturation Temperature of the Pressure at the
Steam Drum, we get our enthalpy for Desuperheater Steam. It calculates out that
“h(DESUP)” is equal to 1484.205 BTUs per pound.
ℎ(𝐷𝐸𝑆𝑈𝑃) = ℎ(𝑇@𝑃(𝑆𝑇𝐸𝐴𝑀 𝐷𝑅𝑈𝑀)) = 1484.205
53
𝐵𝑇𝑈
𝑙𝑏
(3.99)
3.12 Domestic / Hotel Loads
3.12.1 Component Detail
In this plant, the Domestic Flow Load is the Sum of the Domestic Water Heating, Galley,
and Laundry Loads for a crew of 40 persons “N”. The following paragraphs will walk you
through how to solve for each load and then will give you the total sum of all Domestic
Loads.
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑃𝑒𝑟𝑠𝑜𝑛𝑠, 𝑁 = 40
3.12.2 Domestic Water Heating
Domestic Water Heating “Q(DOM WATER HEATING)” is a necessity of each person
aboard this ship, for heat for showering and sink. It is rated for 0.9 pounds steam per hour
per person. Therefore the total for the ship is equal to 0.9 pounds-per-hour of steam
consumption time the Complement of forty persons.
𝑄(𝐷𝑂𝑀 𝑊𝐴𝑇𝐸𝑅 𝐻𝐸𝐴𝑇𝐼𝑁𝐺) = 0.9
𝑙𝑏
𝑙𝑏
∗ 𝑁 = 36
ℎ𝑟
ℎ𝑟
(3.100)
3.12.3 Galley Heating
Galley Heating “Q(DOM GALLEY)” is the steam consumption used to heat the water
used in the Galley (Ship’s Kitchen). It is rated that for every person, a half-a-pound of
steam consumption will be used.
𝑄(𝐷𝑂𝑀 𝐺𝐴𝐿𝐿𝐸𝑌) = 0.5
𝑙𝑏
𝑙𝑏
∗ 𝑁 = 20
ℎ𝑟
ℎ𝑟
(3.101)
3.12.4 Laundry Heating
Laundry Heating “Q(DOM LAUNDRY)” is the steam consumption used to heat the water
used for laundry services. It is rated at one-and-a-half pounds per person.
𝑄(𝐷𝑂𝑀 𝐿𝐴𝑈𝑁𝐷𝑅𝑌) = 0.1
54
𝑙𝑏
𝑙𝑏
∗𝑁 =4
ℎ𝑟
ℎ𝑟
(3.102)
3.12.5 Total Domestic Hotel Loads
Now that we have found each component of the Total Domestic Load, we can solve for
the sum. The Total Domestic Hotel Loads “Q(DOM)” is equal to the sum of the steam
consumption rates multiplied by the Complement of 40 persons.
𝑄(𝐷𝑂𝑀) = (0.9 + 0.5 + 0.1)
𝑙𝑏
𝑙𝑏
𝑙𝑏
∗ 𝑁 = 1.5 ∗ 𝑁 = 60
ℎ𝑟
ℎ𝑟
ℎ𝑟
(3.103)
We arrive at 60 pounds per hour, however since “Q(DOM)” is less than a rate of 100
pounds per hour, we are going be rounding up to 100 pounds per hour for the 40-person
crew of the ship as the Domestic Loads Total Rating.
𝑄(𝐷𝑂𝑀) = 100
55
𝑙𝑏
ℎ𝑟
4. COMPONENT CALCULATIONS
4.1 Turbo-Generator
These next sections will walk you step-by-step through the Turbo Generator (TG)
calculations. Rated conditions are used to select the generator and turbine ‘size’. Power for
the Turbo-Generator will be in kilo-Watts, as this is an electrical generator.
4.1.1
Component Detail
The purpose of the Turbo Generator “TG” is to provide electrical power to the plant and
the ship. Steam enters through a steam strainer and passes through a throttle valve and
controlling valves to the first-stage nozzle plate. Steam expands in successive stages from
Initial Pressure “Po” to Condenser Pressure “P(MN COND)”.
The generator in this project that is connected to the turbine generates the power via
3-phase, 3-wire system with a continuous kilo-Watt (kW) rating. Each generator is
enclosed and is cooled by circulated air. The turbine can spin at speeds around 10,000
RPM and each Generator spins at around 1000 RPM.
Figure 20 - Turbo Generator (Uninstalled without Piping) (12)
56
4.1.2
Rated Kilo-Watt Load
This section will solve for the Rated kilo-Watt (kW) Load of the Turbo Generator. Some
things need to be discussed prior to, however. First, this ship will not be fitted with a
“scoop.” A scoop is an alternate means of pumping seawater into the system. When the
ship moves forward, it drags seawater up into ship via drag forces and creates a positive
pressure, allowing it to feed into the system to cool the main and auxiliary condensers.
After utilizing the equations in Appendices A.2.16, we get a kilo-Watt Rating of
1218.177 kW. Using the Table below (Table 4), we need to round off the calculated kW
rating with the closest match on the Table. Doing so, we get a Rated kW of 1250 kW. This
value is highlighted green on the Table for your convenience.
Table 4 - Rated kW and Standard Size Generator (2 p. 23)
Rated kW and Standard Size Generator
Most Common ………………………………………………Least Common
350
4.1.3
500
750
1000
1250
1500
2000
Special
Operating Kilo-Watt Load
Using Appendices A.2.17, we solve our Operating kilo-Watt Load for our TG’s. This
value is calculated to be 941.921 kW. Using Table 4 to round to the nearest size, we get a
closest match of 1000 kW. This value is highlighted purple on the Table for your
convenience.
4.1.4
Generator / Turbine Size
A Generator's Rated Size is need so that [kW(OPER) / (kW(RATED)] is greater than twothirds but less than three-fourths, when possible. To start with the 1000 kW Rating, we
will divide the 941.921 kW Rating and divide it by the 1000 kW. We get a decimal of
0.941921. This choice will not work, as its value is greater than the higher band of threefourths.
Next, we will try the 1250 kW Rating. Dividing 941.921 kW by the 1250 kW Rating
we get a decimal value of 0.75354 which is just a hair above our upper band. However,
when rounded down to 0.75 (75 %), it falls into band. Using 1250 kW instead of 1000 kW
57
will give the more optimal choice; therefore 1250 KW is the Operating Kilo-Watt Load.
The Percent Rated Load is equal to 75% Rated kW Load.
The Turbine Generator has a kW Rating for 1250 kW and will operate at 942 kW.
4.1.5
TG Temperatures, Pressures and Enthalpies
The temperature of the entering steam “T’o” is 950 degrees, which is the boilers’ output.
The pressure of the entering steam “P’o” is 864.696 psia, but in this case we will round to
865 psia. The enthalpy of the steam leaving the boiler is 1484.102 BTU per pound, and
when it reaches the boiler, it’s calculated to be 1481.988 BTU per pound. A little energy
loss is expected along this path. As this steam passes through the Turbines of the Turbo
Generator, its energy is converted into work and then into electrical power to supply the
electrical demands and loads of the ship.
The steam temperature after it passes through the Turbo Generators “T(So)” is 955
degrees, which is a five degree increase. The pressure of the exiting steam “P(AUX
EXH)” 48.2348 psia (round to 50 psia). This means that there is an 800 psia drop in
pressure. The enthalpy of the Turbo Generator Exhaust “h(TG EXH)” is calculated out to
be 921.6939 BTU per pound. This goes to show that the work done on the system is
primarily due to the pressure of the steam.
4.1.6
TG Turbine Efficiency
To solve the TG Turbine Efficiency, five values are needed to be calculated. The values
are the Basic Efficiency at Generator Terminals “Eb”, Initial Temperature Correction
Factor “ft”, Initial Pressure Correction Factor “fp”, Correction Factor “fb”, and the Load
Correction factor “fL”.
4.1.6.1 Basic Efficiency
The Basic Efficiency at Generator Terminals “Eb” is solved by using Figure 21 - Basic
Efficiency, Eb at the Turbo Generator Rated Capacity of 0.75 (75%).
58
Figure 21 - Basic Efficiency, Eb (5 p. 60)
𝐸𝑏 = 0.649
4.1.6.2 Initial Temperature Correction Factor
The Initial Temperature Correction Factor “ft” is solved by using the Steam Temperature
of 950 degrees Fahrenheit.
Figure 22 - Initial Temperature Correction Factor, ft (5 p. 63)
𝑓𝑡 = 1.027
59
4.1.6.3 Initial Pressure Correction Factor
The Initial Pressure Correction Factor “fp” is solved by Figure 23 - Initial Pressure
Correction Factor, fp at Pressure “Po” (850 psig).
Figure 23 - Initial Pressure Correction Factor, fp (5 p. 63)
𝑓𝑝 = 1.011
4.1.6.4 Correction Factor
The Correction Factor “fb” is calculated by utilizing Figure 24 - Exhaust Pressure
Correction Factor, fb, where it states that if you use the 2 “Hg line, it is always 1.00.
Therefore this project will be using a Correction Factor “fb” of 1.00, or 100% (No
Change).
60
Figure 24 - Exhaust Pressure Correction Factor, fb (5 p. 60)
4.1.6.5 Load Correction Factor
The Load Correction Factor “fL” is solved by using Figure 25 - Load Correction Factor,
fL at the point of 0.7535368 Rated Load.
Figure 25 - Load Correction Factor, fL (5 p. 63)
𝑓𝐿 = 0.965
61
4.1.6.6 Solving for the Efficiency of the TG Turbine
With all the factors of Sections 4.1.6.1 through 4.1.6.5 solved, we may now solve for the
Efficiency of the Turbo Generator Turbine at Operated Load “Eff(TG TURB))OPER” and
at Rated Load “Eff(TG TURB))RATED”. The product of the five values solved will give
the Efficiency of the Turbo Generator Turbine at Operation. To get the Rated Efficiency,
“fL” has to be substituted with 1.00, which would be 100% capacity. The equations and
answers are given below.
𝐸𝑓𝑓(𝑇𝐺 𝑇𝑈𝑅𝐵))𝑂𝑃𝐸𝑅 = 𝐸𝑏 ∗ 𝑓𝑡 ∗ 𝑓𝑝 ∗ 𝑓𝑏 ∗ 𝑓𝐿
= 0.650173 = 65.02%
𝐸𝑓𝑓(𝑇𝐺 𝑇𝑈𝑅𝐵))𝑅𝐴𝑇𝐸𝐷 = 𝐸𝑏 ∗ 𝑓𝑡 ∗ 𝑓𝑝 ∗ 𝑓𝑏 ∗ (1.00)
= 0.673754 = 67.38%
4.1.7
(4.1)
(4.2)
TG Steam Rates and Flows
The Turbo Generator “TG” Theoretical Steam Rate “TSR” is solved with the following
information included: “ho’” is equal to the enthalpy at points “T’o”, “P’o” (1481.9877
BTU/lb) and “hp’” is equal to the enthalpy at points “P(AUX EXH)”, “So” (921.694
BTU/lb). The Theoretical Steam Rate “TSR” is solved to be approximately 6 lb/hr/kW
𝑙𝑏
( )
3412.1
ℎ𝑟
𝑇𝑆𝑅 =
= 6.090
(ℎ𝑜′ − ℎ𝑝′ )
𝑘𝑊
(4.3)
Now that we have the “TSR”, we can solve for the Operating Steam Rate of the
Turbo Generator “SR(TG)OPER”. It’s equal to the “TSR” divided by the product of the
variables in Section 4.1.6.1 through 4.1.6.5.
𝑙𝑏
( )
𝑇𝑆𝑅
ℎ𝑟
𝑆𝑅(𝑇𝐺)𝑂𝑃𝐸𝑅 =
= 9.366
𝐸𝑏 ∗ 𝑓𝑡 ∗ 𝑓𝑏 ∗ 𝑓𝑝 ∗ 𝑓𝐿
𝑘𝑊
(4.4)
Our Steam Rate while Operating has been calculated to approximately 9.4 pounds per
hr per kW. All that remains are the Flows of Rated and Operating Loads.
To solve the Flow of the Turbo Generator at Rated Load “Q(TG))RATED”, we need
the product of the Rated Steam Rate of the Turbo Generator and Rated kW Load. To
solve for the Rated Steam Rate of the Turbo Generator, we simply take “SR(TG)OPER”
62
and multiply it by “fL”. When this is done, a Rated “TG” Steam Rate of 9.038503
lb/hr/kW is calculated, which is slightly less than the Operating Steam Rate.
Now that both needed values are found, “Q(TG))RATED) can be solved. It is equal
to 8513.56 pounds per hour. The operating Turbo Generator Flow is equal to the
Operating Steam Rate of the TG multiplied with the Operating kW Load of the TG. When
done, “Q(TG)OPER” is equal to 8822.337 pounds per hour.
𝑄(𝑇𝐺)𝑂𝑃𝐸𝑅 = 𝑆𝑅(𝑇𝐺 𝑂𝑃𝐸𝑅) ∗ 𝑘𝑊(𝑇𝐺 𝑂𝑃𝐸𝑅)
= 8822.337
63
𝑙𝑏
ℎ𝑟
(4.5)
4.2 Feed Pump
4.2.1
Component Detail
The Feed Pump supplies water from the DC Heater into the Steam Drum of the Boilers via
a steam feed turbine, which drives the pump. The Feed Pump has to overcome the pressure
of the boiler, as pressure is driven from high to low; therefore it must supply water at a
pressure greater than that of the boilers (850 psig).
The Feed Pump Turbine calculations are done utilizing a component called a constant
Discharge Pressure Governor. The control of a turbine with a governor is needed, as
turbines have to be started up slowly to prevent damage and for speed control. An overspeed trip is installed to prevent the turbine from accelerating into dangerous speeds which
could cause damage. A ‘Governor’ is a device that senses the speed of the turbine and will
automatically adjust the speed (RPM) to produce a constant load.
Figure 26 - Steam Driven Feed Pump (13)
64
4.2.2
Feed Pump Turbine
4.2.2.1 Feed Pump Turbine Speed and Wheel Diameter
The Feed Pump Turbine works just like an impulse turbine; there are multiple stages of
wheels that convert pressure into useable energy. The Feed Pump Turbine supplies the
Feed Pump’s Pump with power to enable the Feed Pump to supply Feed Water into the
Boiler.
The Feed Pump Turbine’s speed is determined by selecting a typical value between
8,000 RPM and 9,000 RPM. The average of the boundaries will be used as its speed.
𝑅𝑒𝑣𝑜𝑙𝑢𝑡𝑖𝑜𝑛𝑠 − 𝑝𝑒𝑟 − 𝑀𝑖𝑛𝑢𝑡𝑒 =
8000 + 9000
2
(4.6)
= 8500 𝑅𝑃𝑀
Determining the Feed Pump Turbine’s Wheel Diameter is also done with a preselected
set of values. The typical mean values of wheel diameters are twelve, sixteen, twenty, or
twenty-five inches. Twelve inches would be too small a size for this Turbine, so a sixteeninch mean wheel diameter will be used. Therefore:
𝑊ℎ𝑒𝑒𝑙 𝐷𝑖𝑎𝑚𝑒𝑡𝑒𝑟 = 16 𝑖𝑛𝑐ℎ𝑒𝑠
(4.7)
4.2.2.2 Feed Pump Turbine Values Needed to Solve Efficiency
Now that I have determined my Feed Pump Turbine’s speed and Wheel Diameter, I need
to determine its Windage Loss (LHP), measured in horsepower. The LHP is determined
using Figure 27 - Windage Loss, LHP (see below).
65
Figure 27 - Windage Loss, LHP (5 p. 66)
The Windage Loss comes out to about 12.5 loss-horsepower as determined by Figure
27. This Loss is considered ‘average,’ and is a factor in solving the Efficiency of the Feed
Pump Turbine.
The Total Head developed per stage is ‘H,’ which is measured in feet, is another
component needed to determine the Efficiency of the Turbine. As determined in Section
4.1.6.6, the Net developed Pump Pressure is approximately 1027 psig. For the sake of
easier calculations, 1000 psig will be used.
Since the change in Pressure (∆P) is proportional (α) to the change in Head Pump
“∆H”, then “∆H(TOTAL)” is equal to 1000 BTUs-per-hour-difference. However, since
there are two stages to the Feed Pump, that number needs to be halved. This leaves us with
a “∆H” equal to 500 BTUs-per-hour-difference.
66
Now that I have these values, I can solve for the Basic Efficiency “Eb” as determined
by this equation:
𝐸𝑏 = (𝑅𝑃𝑀) ∗
𝑊ℎ𝑒𝑒𝑙 𝐷𝑖𝑎𝑚𝑒𝑡𝑒𝑟
√∆𝐻
= 0.55
(4.8)
There are two variables left to solve before being able to calculate the Turbine
Efficiency. They are “fs”, which is the Superheat Correction Factor and “v”, which is the
Final Rated Specific Volume of steam.
𝑓𝑠 =
𝑃(𝑇ℎ𝑟𝑜𝑡𝑡𝑙𝑒)
= 0.913
𝑃(𝐸𝑥ℎ𝑎𝑢𝑠𝑡)
(4.9)
𝑣 = 𝑣(𝑔)@(𝐹𝑃 𝑇𝑈𝑅𝐵 𝐸𝑋𝐻 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒)
= 𝑣(𝑓)@(𝐷𝐶 𝐻𝐸𝐴𝑇𝐸𝑅) = 0.017
(𝑓𝑡)3
𝑙𝑏
(4.10)
The BHP is the Brake Horsepower at rated load, measured in brake horsepower.
𝐵𝐻𝑃)𝑂𝑃𝐸𝑅 =
𝐸(𝑂𝑃𝐸𝑅)
144 ∗ 𝑣 ∗ 𝛥𝑃 ∗ (
) ∗ 550 ∗ 𝐸𝑓𝑓(𝑃𝑢𝑚𝑝)𝑂𝑃𝐸𝑅
3600
(4.11)
The “E” calculated earlier of 375449.2 pounds per hour and an assumed Efficiency of
the Pump of 0.7 yields a “BHP” of 294.511 Brake Horsepower.
4.2.2.2.1 Feed Pump Turbine Efficiency
The Efficiency of the Feed Pump Turbine “Eff FP TURB)” is determined by the following
equation:
𝐵𝐻𝑃
𝐸𝑓𝑓(𝐹𝑃 𝑇𝑈𝑅𝐵) = 𝐸𝑏 ∗ 𝑓𝑠 ∗ (
)
13.7
𝐵𝐻𝑃 + 𝐿𝐻𝑃 ( 𝑣 )
(4.12)
Using the values solved from Section 4.2.2.2, we arrive at our answer. The Efficiency
of the Feed Pump Turbine “Eff(FP TURB)” is equal to 0.502137, or around 50 %. This
efficiency is considered ‘normal’ as most turbines are generally 80 percent or less
efficient. Given the losses of the stages and the Feed Pump itself, a 50% Efficient Feed
Pump Turbine is considered a good calculated value.
67
4.2.2.3 Feed Pump Turbine Energy Calculations
The Steam Rate of the Operating Feed Pump “SR(FP)OPER” is equal to the equation
below. The term “ho”” is equal to “h(DESUP)” and the term “hw”” is equal to “h(AUX
EXH)” (Feed Pump Exhaust).
𝑆𝑅(𝐹𝑃)𝑂𝑃𝐸𝑅 =
2544.4
𝑙𝑏
= 6.323
ℎ𝑜-hw
ℎ𝑟
(
)
𝐸𝑓𝑓(𝐹𝑃)
(4.13)
With this, we can solve for the Flow of the FP Turbine (while at Operation) “Q(FP
TURB)OPER”. It’s equal to the product of Steam Rate and the Brake Horsepower. Once
the product has been done, we get our “Q(FP TURB)OPER” calculated out to be 1862.483
pounds per hour.
4.2.2.4 Feed Pump Turbine Exhaust
The Pressure of the Turbine Exhaust “P(FP EXH)” is equal to the sum of the Direct
Contact Heater Pressure “P(DC HTR)” and the change in pressure of the change in Boiler
Pressure “ΔP(BP)”. The change in Boiler Pressure is usually a designated value between
three and five psi. In this project, a value of five will be chosen for “ΔP(BP)”.
Using the Appendices Section A.2.19, we are able to solve for the Feed Pump Turbine
Exhaust Enthalpy “h(FP TURB EXH)”. We get a value of 999.35, or around 1000 BTUs
per pound. The Pressure of the Exhaust is around 40 psia and the Temperature is around
900 degrees Fahrenheit.
In the next section, the Feed Pump’s Pump, which is the driving force of the feed
water into the boilers steam drum will be discussed.
4.2.3
Feed Pump Pump
The Feed-Pump Pump is the driving force to feed the boiler, and its motion is driven by
the Feed Pump Turbine. The next two sections are going to solve in a step-by-step manner
of how to solve for the Feed Pump’s Pump values.
68
4.2.3.1 Feed Pump Pump Efficiency
To solve the Feed Pump Pump’s Efficiency, the Rated Specific Speed “Ns” needs to be
solved and used on the Graph in Figure 28 - Feed Pump Rated Efficiency to solve for the
Expected Average Efficiency.
Figure 28 - Feed Pump Rated Efficiency (5 p. 76)
The Equation for “Ns” is shown in Figure 28 - Feed Pump Rated Efficiency, but I
have rewritten it below for your convenience.
𝑁𝑠 = 𝑅𝑃𝑀 ∗
√𝐺𝑃𝑀(𝑅𝑎𝑡𝑒𝑑)
3
𝐻4
(4.14)
The RPM has been determined from Section 4.2.2.1 to be 8,500 RPM. The Gallonsper-Minute (Rated) flow is determined by the following equation.
𝐺𝑃𝑀|(𝑟𝑎𝑡𝑒𝑑) = 𝐸(𝐸𝑠𝑡) ∗ 1.25 ∗ (
𝑙𝑏
ℎ𝑟
𝑓𝑡 3
𝑔𝑎𝑙
)∗(
) ∗ 𝑣(𝑓) ∗ ( ) ∗ (7.48 3 )
ℎ𝑟
60𝑚𝑖𝑛
𝑙𝑏
𝑓𝑡
After reworking this Equation, we get the GPM|(rated) equal to “E|(rated)(est)”
multiplied by (1/60min) multiplied by the v(f) of water and finally multiplied by 7.48
Gallons. The v(f) of water will be equal to 0.17187. With this, we get a Rated GPM of
69
205.5854 gallons-per-minute. With this, we only need to find the last unknown, “H”. That
unknown “H” is the ‘Total Head Developed per Stage’ and is measured in feet.
The ‘Total Head Developed per Stage “H” (measured in feet) is the total pressure
increase per both stages in the Feed Pump. This Feed Pump is double staged, so the
number of stages is equal to two. When solving “H”, it will be solved overall and then per
each stage. To start off, it’s necessary to find the difference in pressure of the Discharge
and Suction of the Pump. 1000 pounds of differential pressure will be assumed. With this,
we need to multiply these values by 144 (constant) and then by “v(L)” (volume (of Liquid)
at DC Heater Outlet Pressure). This will get the answer to Total Head Developed. The
equation below will show what you have just learned.
144 ∗ 𝑣
𝑝𝑒𝑟
𝐻 (𝑠𝑡𝑎𝑔𝑒) = (𝑃(𝐷𝐼𝑆𝐶𝐻) − 𝑃(𝑆𝑈𝐶𝑇)) ∗
(# 𝑠𝑡𝑎𝑔𝑒𝑠)
(4.15)
= 1270.889 𝑓𝑡
“H” can now be solved for two stages. It calculates out to 2541.7773 Total Head. To
solve for the Head Developed per stage, that number needs to be divided by two. This
leaves us with an “H” per one stage equal to 1270.889 feet.
With “H” solved, we may now solve for the Rated Specific Speed “Ns”. Since
Specific speed “Ns” is used to compare different pumps at different conditions, it is found
to be a dimensionless number that identifies the geometric and hydraulic similarity of
pumps.
𝑁𝑠 = 𝑅𝑃𝑀 ∗
√𝐺𝑃𝑀(𝑅𝑎𝑡𝑒𝑑)
3
𝐻4
𝑁𝑠 = (8500 𝑅𝑃𝑀) ∗
√205.5854 𝐺𝑃𝑀
(4.16)
3
(1270.88865 𝑓𝑡)4
𝑁𝑠 = 572.6 𝑅𝑎𝑡𝑒𝑑 𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑆𝑝𝑒𝑒𝑑
A value of around 570 is calculated as the Rated Specific Speed. Using the curve fit
that is shown in Figure 28 - Feed Pump Rated Efficiency at the 200 GPM mark at points
“Ns” and “Eff|(rated)” we get a Feed Pump Pump Efficiency “Eff(FPP)|rated” of 0.554
(around 55%).
70
4.2.4
Feed Pump Table Overview
Table 5 on the next page shows the pressure required to pump water into the boiler as well
as the process of each of the main steps from the Superheater Outlet Pressure all the way
through.
71
Table 5 - Optimal Feed Pump Table
Constant Discharge Pressure Governing
850 psig, (950 °F)
Superheater Outlet Pressure
872
psig
62
psig
7
psig
69
psig
91
psig
30
psig
993
psig
16
psig
Feed stop and check valve loss
7
psig
Feed regulator pressure drop at max flow
40
psig
Loss for High Pressure Feed Heater(s)
5
psig
Feed line pressure loss
5
psig
10
psig
Total = Pump Discharge Pressure “FP(DISCH)”
1076
psig
Deaerating Feed Heater Pressure
29
psig
21
psig
50
psig
1
psig
49
psig
1027
psig
Superheater pressure drop at maximum continuous power
(Including saturated pipe to Superheater)
Orifice in piping to Superheater
+
Pressure drop at maximum Boiler Rate

69 * (1.15 / 1.00)²
Pressure drop due to steam temperature control
Desuperheater (assuming constant loss above full power)
Steam Drum Pressure at max flow
Economizer pressure drop including piping to drum
12 * (1.15 / 1.00)²
Static head, pump discharge to Boiler Drum

+
Static Head, DFT to Pump Suction
+
Summation
Less suction line pressure loss
-
Net Suction Pressure
Net developed pump pressure,
(Feed Pump Discharge Pressure - Net Suction Pressure) =
𝐸𝑓𝑓(𝑇𝑈𝑅𝐵) = 𝐸𝑏 ∗ 𝑓𝑠 ∗
𝐵𝐻𝑃|𝑟𝑎𝑡𝑒𝑑
13.7
𝐵𝐻𝑃 + 𝐿𝐻𝑃 ∗ ( 𝑣 )
72
4.3 Air Ejectors
4.3.1
Component Detail
An Air Ejector (A/E) is made up of two condensers, each at a different pressure (vacuum).
Its purpose is to remove non-condensable gases from the system (which can cause severe
damage to the main or auxiliary condensers) and to help draw and maintain the required
vacuum pressure on the Main and Auxiliary Condenser(s).
Figure 29 - Typical Air Ejector (14)
4.3.2
Main Air Ejectors
Using Figure 30 (Below), we will find the Main Air Ejector’s Flow (in pounds-per-hour).
As stated in Note (3) in the figure, this is based on 4.5 pounds of steam per pound of air
vapor mixture rounded to the nearest 10 pounds.
The Main Air Ejector sees about 170,000 pounds per hour maximum condensed
steam (from the Boilers) therefore it falls in the <100,001 to 250,000> pounds per hour
Category. This yields a flow for the Main Air Ejector “Q(MN A/E)” of 490 pounds per
hour.
73
Figure 30 - Main Air Ejector Flow (5 p. 13)
4.3.3
Auxiliary Air Ejectors
Using Figure 30 (Above), we will find the Auxiliary Air Ejector’s Flow (in pounds per
hour). As stated in Note (3) in the figure, this value is also based on 4.5 pounds of steam
per pound of air vapor mixture rounded to the nearest 10 pounds.
The Auxiliary Air Ejector sees about 10,000 pounds per hour maximum condensed
steam (from the Turbo Generators) therefore it falls in the <Up to 12,500> Category. This
yields a flow for the Auxiliary Air Ejector “Q(AUX A/E)” of 100 pounds per hour.
4.3.4
Distillate Air Ejectors
From Section 0, we solve for the Distillate Air Ejector Flow “Q(DIST A/E)”. As a quick
refresher, it’s equal to the product of 8.33 (conversion factor), the Gallons per Day rate,
Distillate Output Steam “f(D)” divided by 24 hours (in this project 60/24 hours), and
finally “hi”. We get a flow of the Distillate Air Ejectors to be equal to around 140 pounds
per hour.
𝑄(𝐷𝐼𝑆𝑇 𝐴/𝐸) = (8.33) ∗ 𝐺𝑃𝐷 ∗ (
4.3.5
60
𝑙𝑏
) ∗ ℎ𝑖 = 139.042
24
ℎ𝑟
(4.17)
Air Ejector Inter-Condensers
The Inter-Condenser is the first of two condensers in a typical Air Ejector, separated from
the After-Condenser by a loop seal, to allow for different pressures in each. Its purpose is
to draw steam into it for condensing so as to draw and maintain a vacuum in the Main or
74
Auxiliary Condenser(s) throughout operation. The Inter-Condenser’s flow two-fifths of
the total flow of its respecting condenser.
𝑄(𝑀𝑁 𝑜𝑟 𝐴𝑈𝑋 𝐼𝐶 𝐷𝑅𝑁)
2
= ( ) ∗ 𝑄(𝑀𝑁 𝑜𝑟 𝐴𝑈𝑋 𝐴/𝐸 𝐶𝑂𝑁𝐷)
5
By using this Equation we can solve for both the Main Condenser (MN IC DRN) and
Auxiliary Condenser (AUX IC DRN). Solving the Inter-Condenser Calculations we get a
value of 196 pounds per hour for the Main Inter-Condenser Drain and 40 pounds per hour
for the Auxiliary Inter-Condenser Drain.
2
𝑙𝑏
𝑄(𝑀𝑁 𝐼𝐶 𝐷𝑅𝑁) = ( ) ∗ 𝑄(𝑀𝑁 𝐴/𝐸) = 196
5
ℎ𝑟
2
𝑙𝑏
𝑄(𝐴𝑈𝑋 𝐼𝐶 𝐷𝑅𝑁) = ( ) ∗ 𝑄(𝐴𝑈𝑋 𝐴/𝐸) = 40
5
ℎ𝑟
4.3.6
(4.18)
(4.19)
Air Ejector After-Condensers
The Air Ejector After-Condenser is the second stage condenser, as the steam’s flow path
enters the After-Condenser after the Inter-Condenser. Its purpose is meant to draw an even
higher vacuum (negative pressure) in the Main or Auxiliary Condenser(s). The AfterCondenser’s flow is three-fifths the total flow of its respecting condenser.
𝑄(𝑀𝑁 𝑜𝑟 𝐴𝑈𝑋 𝐴𝐶 𝐷𝑅𝑁)
3
= ( ) ∗ 𝑄(𝑀𝑁 𝑜𝑟 𝐴𝑈𝑋 𝐴/𝐸 𝐶𝑂𝑁𝐷)
5
So by using this Equation, we can solve for both the Main Condenser (MN AC DRN)
and Auxiliary Condenser (AUX AC DRN). Solving the After-Condenser Calculations we
get a value of 294 pounds per hour for the Main After-Condenser Drain and 60 pounds per
hour for the Auxiliary After-Condenser Drain.
3
𝑙𝑏
𝑄(𝑀𝑁 𝐴𝐶 𝐷𝑅𝑁) = ( ) ∗ 𝑄(𝑀𝑁 𝐴/𝐸) = 294
5
ℎ𝑟
3
𝑙𝑏
𝑄(𝐴𝑈𝑋 𝐴𝐶 𝐷𝑅𝑁) = ( ) ∗ 𝑄(𝐴𝑈𝑋 𝐴/𝐸) = 60
5
ℎ𝑟
75
(4.20)
(4.21)
4.4
Plant Cycle Efficiency
4.4.1
Efficiency Theory
The Efficiency of the Plant is equal to the Power Output divided by the Rate of Heat
Added. This is true in all Plants, not just this one. The correction factor for units
throughout these Sections is 2544.4 BTU/hp-hr. Given that these sections are all plug-andchug equations with values that have already been previously solved in other sections,
there will no explanations of terms here.
4.4.2
Cycle Efficiency
The Cycle Efficiency is given in the Equation below and is the Efficiency of the Steam
Water Cycle, without most losses.
𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 𝑜𝑓 𝐶𝑦𝑐𝑙𝑒
=
𝑊𝐻𝑃 ∗ 2544.4
(4.22)
[𝐸 − (𝑄(𝐷𝐸𝑆𝑈𝑃) ∗ ℎ(𝑆𝑂)) + ((𝑄 ∗ ℎ)(𝐷𝐸𝑆𝑈𝑃)) − 𝐸(ℎ𝐹3)]
𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 𝑜𝑓 𝐶𝑦𝑐𝑙𝑒 =
2544.4
𝐻𝑅
𝐶𝑦𝑐𝑙𝑒
(4.23)
= 0.81339 = 81.34%
A Cycle Efficiency of 81.34% has been found for this plant.
4.4.3
Plant Cycle Efficiency
The Plant Cycle Efficiency is the Efficiency of the Plant, all in all.
𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 𝑜𝑓 𝐶𝑦𝑐𝑙𝑒 =
2544.4
= 0.27507 = 27.51%
𝐻𝑅
𝐶𝑦𝑐𝑙𝑒
The calculated overall efficiency of 27.51% has been found for this plant. This value
is substantially lower than that of the Cycle Efficiency, as losses are accounted for in this
equation. It is considered ‘normal’ for a plant to only receive 20 to 40 percent efficiency.
No marine steam power plant is perfect, and many today are old plant systems that have
been around for many years and have efficiencies even less than which was solved for in
this project.
76
4.4.4
Carnot Efficiency
As defined by Wikipedia:
“The Carnot cycle is a theoretical thermodynamic cycle proposed by Nicolas
Léonard Sadi Carnot in 1824 and expanded by Benoit Paul Émile Clapeyron in the 1830s
and 1840s. It can be shown that it is the most efficient cycle for converting a given amount
of thermal energy into work, or conversely, creating a temperature difference by doing a
given amount of work.” (15)
Simply by using the High Temperature “T(H)” and Low Temperature “T(L)” of the
System, in Rankine, we can determine the Carnot Efficiency. The Temperature extremes
are listed below, both in degrees Fahrenheit and Rankine.
𝑇(𝐿) = 91.68 °𝐹 = 551.68 𝑅𝑎𝑛𝑘𝑖𝑛𝑒
𝑇(𝐻) = 955 °𝐹 = 1415 𝑅𝑎𝑛𝑘𝑖𝑛𝑒
With these values, we can determine the Efficiency of the Carnot Cycle. It’s equal to
the difference of 1.00 and the quotient of T(L) over T(H). With these values, we get a
Carnot Efficiency of 61 %. This represents the best efficiency this system can produce.
𝐶𝑎𝑟𝑛𝑜𝑡 𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 = 1 −
𝑇(𝐿)
𝑇(𝐻)
(4.24)
= 0.61011 = 61.01%
Ways to increase system efficiency is by lagging piping to insulate, using a colder
coolant (i.e. sailing in the arctic, as seawater temperatures will be lower), or by using
machinery that run at higher efficiencies. Given the fact that this is Marine Steam Power
Plant, a lot of these components are based off of older technology and with today’s newage technology, there are plants out there with higher efficiency and better performance.
The next Section is my Heat Balance Spreadsheet, where I lay out the values in a
clear manner to show you my ending values and overall error.
77
5. RESULTS/DISCUSSION
The following Table displays my Mass Balance for this project. It starts with the Feed
Pump outlet mass flow and works its way through the Superheater and Desuperheater
mass flows, then proceeds through the Main Steam, Turbo-Generator, Turbines,
Exhaust, Condensers, Condensate System, Feedwater Drain Collection Tank, and Direct
Contact Heater system.
78
Table 6- Mass Balance
5.1 MASS BALANCE
Sub
Total
DC Heater Outlet
(to Feed Pump)
Running
Total
375449
E
375449
375449
Super Heater Outlet
E
375449
3094
Desuperheater Outlet
Q(DESUP)
3094
372355
Main Steam
Superheater Outlet
- Desuperheater Outlet
375449
-3094
8822
Flow to TG
Q(TG)
8822
362708
Flow to Main Turbine
Main Steam
- Q(TG)
- Q(LOST)
372355
-8822
-825
174058
Flow to LP Turbine
Flow to Main Turbine
- B1
- B2
- Q(MN LEAK OFF)
362708
-164108
-24244
-298
-16903
Exhaust from LP Turbine
Flow to LP Turbine
- B3
174058
-190961
8772
Exhaust from TG
Flow to TG
- Q(AUX LEAK OFF)
79
8822
-50
Sub
Total
MASS BALANCE
Running
Total
172446
Main Condensate
Exhaust from LP Turbine
+ B3
- Q(DISTILL PLT)
+ Q(MN IC DRN)
-16903
190961
-1808
196
10620
Auxiliary Condensate
Exhaust from TG
+ Q(DISTILL PLT)
+ Q(AUX IC DRN)
8772
1808
40
183066
Total Condensate
Main Condensate
+ Auxiliary Condensate
172446
10620
1238
Flow to CONT DRNS
Q(FOH)
+ Q(DOMESTIC)
1138
100
9580
Flow to FWDCT
Q(MN AC DRN)
+ Q(AUX AC DRN)
+ Q(VENT)
+ Q(MN LEAK OFF)
+ Q(AUX LEAK OFF)
+ Q(CONT DRNS)
+ Q(MAKEUP FEED)
+ Q(DISTILLER A/E)
+ Q(SAH)
196
60
100
298
50
1238
1923
139
5576
379776
Flow to DC Heater
Total Condensate
+ B1
+ B2
- Q(SAH)
+ Q(FP)
+ Flow to FWDCT
80
183066
164108
24244
-5576
4354
9580
Sub
Total
MASS BALANCE
379676
Flow from DC Heater
(to Feed Pump)
Flow to DC Heater
- Q(VENT)
379776
-100
1.12585
Error
Flow
Flow
Running
Total
Calculated
Actual
375449
379676
81
5.2 DISCUSSION
My calculated “E” was calculated to be 375,449 pounds-per-hour. The actual energy put
out by this plant came out to just over 379,000 pounds-per-hour, which per 2 boilers
ends up being 187,500 pounds-per-hour. The boiler values are on the high end of the
range for flow; however the calculations end up with a 1.12585 % error, so that is a good
indication that these calculations were done correctly.
𝑂𝑣𝑒𝑟𝑎𝑙𝑙 𝐸𝑟𝑟𝑜𝑟 =
(𝐴𝑐𝑡𝑢𝑎𝑙 𝑀𝑎𝑠𝑠 𝐹𝑙𝑜𝑤) − (𝐶𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒𝑑 𝑀𝑎𝑠𝑠 𝐹𝑙𝑜𝑤)
= 1.126%
𝐶𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒𝑑 𝑀𝑎𝑠𝑠 𝐹𝑙𝑜𝑤
I validated my initial results (Section EQUATIONS/MATRICES) by comparing
them against the Appendices Section A.1 - Initial Values. My value for “E(EST)” was a
little low, but given the fact that most of my estimate values were low, it worked out. I
feel that if I had more time, I would have been able to iron out every last detail; maybe
even just to use higher decimal values to get an even more accurate number. Some of my
values were rounded to 3 or 4 decimal places, even though they sometimes had 8 or
more. If all these round offs were not done, and the exact value was kept, maybe the
final Error would be even less.
All-in-all, given the scope of the project and the fact that I had to calculate many
various components for a complete Steam Power Plant, I think any small changes
wouldn’t have affected the overall result. I put a lot of time and effort into this project,
and since my Error percentage was only 1.12%. Given the scope of the project and
rounding values to four decimal places, having an overall error of about 1% is
considered negligible.
I want to thank you for taking the time to read this, as many hours was put into this.
I hope this project enabled you to learn more about Steam Powered Marine Power
Plants.
82
6. CONCLUSION
In a quick summation, this plant was calculated to have a total plant flow of 379,676
pounds per hour, or 189,838 pounds per hour per Boiler. I calculated the Turbo
Generators to have an Operational Load of 942 kilo-Watts.
Some of the major calculated efficiencies are the Boiler with a calculated efficiency
of 91.59%, the Cycle Efficiency with a calculated efficiency of 81.34%, the Plant Cycle
with a calculated efficiency of 27.51%, and the Carnot Cycle with a calculated efficiency
of 61%.
This project is consistent with the First Law of Thermodynamics – which is the
Energy Conservation Theory. As the Law states:
“The first law of thermodynamics is a version of the law of conservation of energy,
specialized for thermodynamical systems. It is usually formulated by stating that the
change in the internal energy of a closed system is equal to the amount of heat supplied
to the system, minus the amount of work performed by the system on its surroundings.
The law of conservation of energy can be stated: The energy of an isolated system is
constant.” (1)
My project supports this, as my ‘Energy In’ is equal to my ‘Energy Out.’
83
7. REFERENCES
1. Various. First Law of Thermodynamics. Wikipedia. [Online] December 11, 2011.
[Cited: Decemeber 12, 2011.]
http://en.wikipedia.org/wiki/First_law_of_thermodynamics.
2. Heath, K., Gelardi, S. and Nurnberger, J. Marine Engineering Design I. Bronx :
SUNY Maritime Collehe, 2007. MED Project.
3. Various. Main Propulsion Plant DD445 and 692 Classes and Converted Types,
Operation Manual - Sectional V. Historic Naval Ships Association. [Online] 2007.
[Cited: December 8, 2011.] http://www.hnsa.org/doc/destroyer/steam/sec05.htm.
4. —. INFORMATION SHEET. Federation of American Scientists. [Online] [Cited:
December 6, 2011.] http://www.fas.org/man/dod-101/navy/docs/swos/eng/62B203I.html.
5. Holm, J.T. Marine Steam Power Plant Heat Balance Practices. New York City : The
Society of Naval Architects and Marine Engineers, 2006. Technical & Research
Bulletin.
6. Group, EEC. Boiler Tubes. EEC GROUP. [Online] February 12, 2010. [Cited:
December 12, 2011.]
http://www.eecgroup.com.vn/home/products.asp?iCat=531&iChannel=58&nChannel=Pr
oducts.
7. Randall. steam turbine. [Online] 2011. [Cited: November 22, 2011.]
http://www.steam-boilers.org/boiler/steam-turbine.html.
8. Various. Entropy. Wikipedia. [Online] December 2, 2011. [Cited: December 8, 2011.]
http://en.wikipedia.org/wiki/Entropy.
9. Farthing, David C. Improving Boiler Room Efficiencies. Federal Corp. [Online]
Federal Corporation, September 2000. [Cited: Decemeber 1, 2011.]
http://www.federalcorp.com/tech/econpapr.htm.
10. Rowen, Alan L. Marine Machinery. Access Science. [Online] McGraw-Hill, 2008.
[Cited: December 1, 2011.] http://accessscience.com/content/Marine-machinery/406400.
11. Various. Shell and tube heat exchanger. Wikipedia. [Online] October 6, 2011.
[Cited: December 13, 2011.]
http://en.wikipedia.org/wiki/Shell_and_tube_heat_exchanger.
84
12. Anish. Starting Procedure for Turbo Generator on Ship. Marine In Sight. [Online]
March 6, 2011. [Cited: Decemember 1, 2011.]
http://www.marineinsight.com/tech/proceduresmaintenance/starting-procedure-forturbine-generator-on-ship/.
13. Fyre, Milt. 5MW Steam Turbine Generator Sets, 60Hz for sale! Utility Warehouse.
[Online] 2011. [Cited: December 3, 2011.]
http://www.utilitywarehouse.com/info2/5mwstg/5mwstg11.htm.
14. Sebald, Joseph F. Steam Jet Ejector. Access Science. [Online] McGraw-Hill, 2011.
[Cited: Decemeber 12, 2011.] http://accessscience.com/search.aspx?rootID=797318.
15. Various. Carnot Cycle. Wikipedia. [Online] October 19, 2011. [Cited: December 13,
2011.] http://en.wikipedia.org/wiki/Carnot_cycle.
16. lex. Navy showers and the Steam Cycle. Neptunus Lex. [Online] January 8, 2005.
[Cited: December 1, 2011.] http://www.neptunuslex.com/wpcontent/uploads/2006/08/second43.gif.
85
A.1
A.
Initial Values
APPENDICES
A.1.1 Main Propulsion Turbine Gear Basic Efficiency:
y = a + bx + cx^2
a=
0.76647319
b=
0.005653846
c=
-9.91E-05
x=
3.25E+01 = SHP
A.1.2 Throttle Temperature Correction 850 psig
f(t) = a + bx + cx^2
a=
0.75786182
b=
0.000433256
c=
-1.76E-07
x=
9.50E+02 = To
A.1.3 Exhaust Loss
y = a + bx + cx^2 + dx^3 + ex^4 + fx^5 + gx^6
a=
73.6139
b=
-0.0702697
c=
3.06E-05
d=
-7.17E-09
e=
9.82E-13
f=
-7.29E-17
g=
2.26E-21
x=
5.10E+03 = X
A.1.4 Basic Efficiency (TG)
Eb = a + bx + cx^2
x=
0.75
a=
5.20E-05
86
b=
1.1051853
c=
-0.45642039
Eb =
1
A.1.5 Initial Temperature
f(t) = a + bx + cx^2
a=
0.636609
b=
6.96E-01
c=
-3.32E-01
x=
950 = To
A.1.6 TG Initial Pressure Factor 750 KW
y = a + bx + cx^2 + dx^3 + ex^4 + fx^5
a=
1.0105385
b=
7.33E-05
c=
-3.68E-07
d=
3.55E-13
e=
-1.50E-13
f=
2.26E-17
x=
8.50E+00
A.1.7 Load Correction Factor 1000 KW
y = a + bx + cx^2 + dx^3 + ex^4 + fx^5 + gx^6
a=
-0.005019998
b=
0.069283177
c=
-0.002340548
d=
4.40E-05
e=
-4.63E-07
f=
2.57E-09
g=
-5.88E-12
x=
71.44209979
87
A.1.8 Soot Blower
y = a + bx
a=
204.54239
b=
1.7523506
x=
85
A.1.9 Pressure Drop across Superheater
y = a + bx
a=
32.445652
b=
0.95186335
x=
85
A.1.10 Temperature
y = a + bx + cx^2
a=
43.13285
b=
0.041315
c=
-1.36E-05
x=
816.8
A.1.11 Auxiliary Turbine Efficiency
y = a + bx + cx^2 + dx^3
a=
0.91592775
b=
-0.000537166
c=
-1.19E-06
d=
9.12E-10
x=
9130.365193
A.1.12 Superheat Correction Factor for Value 40
y = a + bx + cx^2 + dx^3 + ex^4 + fx^5
a=
0.9027
b=
0.000571862
88
c=
-1.40E-06
d=
1.30E-09
e=
9.18E-13
f=
-1.85E-15
x=
1.88E+01
A.1.13 Load Factor
fL= a + bx + cx^2
a=
3.381818
b=
2.331667
c=
-0.014318
x=
87.83487
A.1.14 Heat Capacity of Air
Cp(Air) = a + bx + cx^2
a=
0.243084
b=
7.56E-07
c=
1.52E-08
x=
200
A.1.15 Efficiency of Single Stage Auxiliary Turbines
y = a + bx + cx^2 +dx^3
a=
-1.17E+00
b=
9.69E-01
c=
-2.04E-01
d=
6.26E-02
x=
118728.1489
A.1.16 Low Pressure Heater
y = a + bx + cx^2
a=
8.700205
b=
-
4.948946
89
c=
0.901032
x=
8.5
A.2
Values
A.2.1 Steam Rate (Non-Extraction)
In order to calculate the Steam Rate Non-Extraction, SR(NE), I need the following
unknowns:
Need: a, b, c, d, ho, h(B1), h(B2), h(B3), hi, hw
a = 2544.4 / (ho - h(B1))
b = 2544.4 / (h(B1) - h(B2))
c = 2544.4 / (h(B2) - h(B3))
d = 2544.4 / (h(B3) - hw)
In order to solve for a, b, c and d, I need to find ho, h(B1), h(B2), h(B3), and hw
SR(NE DESIGN) = 2544.4 / (ho - hw)
SR(NE DESIGN) = 5.256086
Now, I can solve for a, b, c, and d.
a=
24.45049
b=
24.45049
c=
17.61129
d=
19.35137
A.2.2 Efficiency of the State Line
A second check can be done against to verify that the value is accurate and plausible:
Weibull Model:
y=a-b*exp(-c*x^d)
Coefficient Data
a=
0.887102
b=
0.921078
c=
1.448821
d=
0.24045
SHP/1000, x = 32.5
Eb = 0.854662339
This second check validates and proves that the value I calculated for the Efficiency
of the State Line is correct.
A.2.3 Temperature Correction Factor
91
Using the 4th Polynomial Fit Equation:
y = a + b*x + c*x^2 + d*x^3 + e*x^4
substituting ‘f(t)’ for ‘y’ and ‘To’ for ‘x’:
f(t) = a + b*(To) + c*(To)^2 + d*(To)^3 + e*(To)^4
Coefficient Data:
e=
a=
0.95766
b=
-0.000588
c=
0.00000174
d=
-1.58E-09
4.82E-13
To =
950
f(t) = 1.01202
A.2.4 State Line Equations
Si = S(COND) = 1.792699
h = h1-((S1-s)/(S1-Si))*(h1-hi))
s = S1-((h1-h)/(h1-hi))*(S1-Si))
hB1 = (ho+hB2)/2 = 1377.934
hB2 = (hB2+hi)/2 = 1129.396
A.2.5 Annulus Area
Using the 4th Degree Polynomial Fit:
y = a + b*x + c*x^2 + d*x^3 + e*x^4
Coefficient Data:
a = -51.86644
b = 0.051456
c = -0.0000171
d = 2.6E-09
e = -1.42E-13
For a Flow of 4000:
X = 4000
EL, y = 9.81118
For a Flow of 6000:
X = 6000
EL, y = 17.19019
92
A.2.6 Turbine Horsepower
P = ((E - G) / a) + ((E – G - B1) / b) + ((E - G - B1 - B2 - Q(MN LEAK OFF)) / c) + ((E
- G - B1 - B2 - B3 - Q(MN LEAK OFF)) / d)
High-Pressure Turbine Wheel Horsepower:
WHP HP = ((E - G ) / a) + ((E - G - B1) / b)
Low-Pressure Turbine Wheel Horsepower:
WHP LP = ((E - G - B1 - B2 - Q(MN LEAK OFF)) / c) + ((E - G - B1 - B2 - B3 - Q(MN
LEAK OFF))
/ d)
A.2.7 E(EST) and G(EST)
Assume:
hB = (hB1+hB2+hB3)/3
hB =
1259.426
Q(TG(EST)) = SR(TG)*KW(TG)  #(10)(1000)
Q(DESUP EST) = SR(FP)*(BHP/E)*E(EST) + 3000
Q(DESUP EST) = 48212.7 BTU/lb ????
When I substitute all the variables I know, the following equation is what's left:
20*(0.013) = 0.04 * E(EST)
From the equation for G(EST),
G(EST) = (0.005 + 0.04) * E(EST) + 13000
Therefore:
G(EST) = 19284.18
A.2.8 Low Pressure Heater Theory
(B3-Q(DIST))*hB2 + (Q*h)(MN LEAK OFF) + (Q*h)(AUX LEAK OFF) +
(Q*h')(MN COND) + (Q*h')(AUX COND)+(Q(MN A/E) + Q(AUX A/E) +
Q(DIST A/E))(h(DESUP) + (Q*h)(STM AIR HTR DRN) +
(Q*h)(CONTMND DRN) + (Q*h)(MAKEUP FEED) = (B3-Q(DIST)*hD1 + [Q(MN
IC) + Q(AUX IC)]*h(IC DRN) + (Q*h)(FWDCT) + [(Q(MN COND) + Q(AUX
COND)]*hc1
First Law of Thermodynamics
[Q(MN LO) + Q(VAP FWDCT) + Q(VENT COND)]*h(f)@200°F + [Q(MN AE) +
Q(AUX AE) + Q(DIST AE)]*h(f)@200°F + (Q(CONT DRNS)*h(COND DRN)) +
93
((Q(SAH DRNS)*h(SAH DRNS)) + (Q*h)(MAKEUP FEED) = Q(VAP
FWDCT)*h(g)@212°F + (Q*h)(FWDCT)
A.2.9 Feed Water Drain Collection Tank Vapor Flow
Q(VAP FWDCT) ≠
therefore
0
h(FWDCT) = h(f)@212 °F
Need to solve for Q(VAP FWDCT)
If Q(VAP FWDCT) >
0
OK
If Q(VAP FWDCT) <
0
ERROR
(Impossible)
therefore Q(VAP FWDCT) = 0
Q(VAP FWDCT) = 0
therefore h(FWDCT) = h(f)@212°F
If h(FWDCT) >
0
OK
If h(FWDCT) <
0
ERROR
(Impossible)
Q(VAP FWDCT) = 180.1802
W = 0.5 * SHP
W = 16250
A.2.10 Mass Flow of Air
m(Air) = m(FO) * (A / F)
= 16250 * 15.05 = 244562.5
A.2.11 Evaporator Flow
This Section is going to show the calculations if there would be only one Evaporator
installed. This system has two Evaporators installed, as shown in Section 2.7.1.5.1
If there is one evaporator installed and operating:
GPH(RATED) = 1.3 * GPH(OPER) * 1.5
(Round it to the nearest largest size)
1.3 * 11500 * 1.5
=?=
22425
I need to round to the nearest Range. Doing so, I have chosen a Gallons Per Day
Rate of GPD(RATED) = 25000
94
A.2.12 Desuperheater
Constants = Q(DIST) + Q(AUX IC) + Q(TG) - Q(AUX LO)
Q(AUX IC) = 40
Q(AUX LO) = 50
Q(TG) =
8822.337
Constants =
1166.898
Q(DESUP) = Q(SOOT BLOWERS) + Q(STM ATOM) + Q(DOM) + Q(FP) + Q(MN
A/E) + Q(AUX A/E) + Q(DIST A/E) + Q(FOH) + Q(OTHERS)
Q(TOTAL) = E(EST)
Q per boiler is equal to Q(Boiler), which is equal to E(EST) divided by the number of
boilers (in this case 2)
E=
0.0264044
Constants =
3087.9401
A.2.13 Soot Blowers Verification Check
Q(SOOT BLOWERS) = # Blowers * (Q(SOOT BLOWERS) / Boilers)
100,000 < Q(SOOT BLOWERS) < 300,000
Q(SOOT BLOWERS) = 110,000
Q = 120 + (0.00165) * (Q(SOOT BLOWERS))
Q = 392.098
For a standard burner, four-thousand pounds per hour of fuel is considered
"maximum."
0.5 = SFC/Est[lb/hr/SHP]
0.5 * 32500 = 16250 lb/hr per 2 boilers
16250 / 2 Boilers
8125 lb/hr per 1 boiler
4062.5 lb/hr per burner (2 Installed per Boiler, 4 Total)
This 4000 lb/hr is considered to be the maximum flow, and isn’t reached under
normal operating conditions.
[SR * E * (5.5) * 325000 = E(Est)]
8125 / 4000 = 2.03125(which is approximately 2)
95
A.2.14 Matrix of Equations
Table 7 - Matrix of Equations
Equation
E
Q(MN
Q(AUX
COND)
COND)
G
B1
B2
B3
-0.1494
-0.1085
-0.0517
0
Q(DESUP)
Constants
0
0
33886.4
3.1.2
0.1902
0.1903
3.2.2
0
1
0
0
0
0
0
-1
9646.88
3.3.2
87.87.875
0
-201.05
0
0
0
0
0
0
3.4.2
-1171.3
0
1176.88
1273.87
0
1147.78
1147.78
0
5414076
3.5.2
0
0
0
0
1105.76
-1088
-1078.6
0
1605936
3.6.2
1
-1
-1
-1
0
-1
0
0
3250.14
3.7.2
0
0
0
0
0
0
1
0
1168.98
3.8.2
0.0264
0
0
0
0
0
0
1
3087.94
A.2.15 Minverse
A Minverse returns the inverse matrix (above) for the matrix stored in an array.
Table 8 - Minverse
9.18039
-0.2424
4.01273
-1.4178
6.71847
6.82787
0
-0.2424
6.09424
0.83909
2.66378
8.16187
-5.4812
-5.5705
0
-0.1609
-0.0389 0.0042
0.00103 -0.0001
-0.0067 0.00183
0.00175 0.00728
0.00092 -0.4734
0.00093 -0.0048
0
0
0.0001 -0.0001
0.000429 4.34762 431651 6.09424 33886.4
-1.13E-05 -0.1148 0.1149 0.83909 9646.88
0.000188 1.90034 -1.9021 2.66378
0
-6.63E-05 8.43162
-8.431 8.16187 5414076
0.001218 -6.7595 6.74802 -5.4812 1605936
3.19E-04 -6.8695 5.86654 -5.5705 3250.14
0
0
1
0
11669
-1.13E-05 -0.1148 0.1149 0.83909 3087.94
A.2.16 Rated kW Load
KW Rated = ((A + B + C) * SHP) + (1.6 * N) + (9 * √(N)) + 80 + “Other”
Other" may include the refrigerant system or special electrical demands.
A = 0.017 (Given)
B (Option 1) = 0.0042 with scoop
B (Option 2) = 0.007 with no scoop  Chose
C = 0.0004 * (Rated / Normal)^3 "H2O Draft Loss
where Draft Loss is the pressure needed to push air through the Boiler
Draft Loss = 12 "H2O
96
Rated/Normal = 1.15
C = 0.0073
N = 40
kW Rated = 1218.177 kW
A.2.17 Operating kW Load
KW(OPER) = (A + B + C)*SHP + 1.6*N + 9√(N) + 80 + other
A=
0.011 (Given)
B (Option 1) =
0
B (Option 2) =
0.007 with no scoop
C=
0.0004 * (Normal / Normal)^3 "H2O Draft Loss
with scoop
where Draft Loss is the pressure needed to push air through the Boiler
Draft Loss = 12 "H2O
and
C=
0.0048
N=
40
KW Operating = 941.921
** 1000 KW is closest match **
A.2.18 Turbo Generator Temperatures, Pressures, and Enthalpies
T’o = 950°F
T(So) = T’o + 5°F = 955 °F
h(SO) = 1484.1015\
ho’ = h(T’o,P’o)
ho' = 1481.9877
hi = ho’ - Eff(TG TURB OPER) * (ho’ – hp’)
hi = 1117.7
hp’ = h(P(AUX EXH), So) = 921.6939
A.2.19 Feed Pump Exhaust Enthalpy, Pressure, Temperature
ΔP ~ ΔP α H
α H = Pump Head
97
Δh(ACTUAL) = Δh(FP TURB)OPER
h(FP TURB EXH) = ho" - Δh(ACTUAL) = ho"(ACTUAL) - (Eff(FP)oper * (h(PUMP) - hp") = hw
Δh(ACTUAL) = UEW (Wheel Used Energy)
UEW = 484.087
Δh = 484.87
h(FP TURB EXH) = ho” - Δh
ho" = h(DESUP)
ho" = 1484.205
h(FP TURB EXH) = 999.35
hw = 997.911
P"o = P(DESUP) / 0.975 = 44 45 (Rounded to the nearest 5 psig)
P"1 = 0.9 * P"o = 40.5
To" = T(P"o, ho") = 903.2053
So" = 1.974451
S1" = S(P"o, h(DESUP)) = 1.986037
hp" = h(FP EXH), S1”) = 1456.517
A.2.20 First Law of Thermodynamics
The First Law of thermodynamics, which is the Conservation of Work Law states that
whatever energy enters a system will leave the system as denoted by this equation:
Ʃ(Q * h)(IN) = Ʃ(Q * h)(OUT)
W(Main or Aux A/E) = 0
Q(Main or Aux A/E) = 0
A.2.21 Drain Cooler Total Condensate Enthalpy.
h(MN or AUX A/E) = h'(MN or AUX COND) + [Q(MN or AUX A/E)] *
[h(DESUP) - (2/5)*h(IC) - (3/5)*h(AC)]
where
h'(MN or AUX COND) = h(MN or AUX COND) + W(PUMP)
h(MN or AUX COND) = h(f) @ P(MN or AUX COND)
W(PUMP) = v*ΔP*144/(36000*550*Eff(PUMP)(1.0))
W(PUMP) = 0 (approx.)
98
therefore
h'(MN COND) =
59.7354
h'(AUX COND) =
69.134517
h(MIX), which is the Enthalpy of the Total Condensate into the Drain Cooler (1st
Stage Combo) is needed for the check of TC1.
h(MIX) =
[Q(MN COND)*h(MN A/E) + Q(AUX COND)*h(AUX A/E)] /
[Q(MN COND) + Q(AUX COND)]
h(MIX) = 398316.7719
A.2.22 Condenser Flow Derivations
The total heat flow through the Main or Auxiliary Condenser(s) is equal to the sum of
both stages of the Air Ejectors as derived by the following equation.
Q(MN or AUX COND)*h'(MN or AUX COND) +
Q(MN or AUX COND)*h(DESUP) =
(2/5)*Q(MN or AUX AUX)*h(IC) + (3/5)*Q(MN or AUX A/E)*h(AC) +
Q(MN or AUX COND)*h(MN or AUX A/E)
A.2.23 Inter- and After-Condenser Enthalpies
IC = (2/5)*Q(MN or AUX COND)
h(IC) = h(f)@125°F = 92.993847
AC = (3/5)*Q(MN or AUX COND)
h(AC) = h(f)@200°F = 168.09914
A.2.24 Main Turbine Extraction Stages
Table 9 – Main Turbine Extraction Stages
P
TURB IN 864.696
h(SL)
S
v
1482 1.65192 0.92807
TOP PT
778.226 944.917
1482 1.66304 1.03174
B1
280.272 715.751 1377.93 1.69018 2.42592
1377.13 -0.08001
B2
86.4696 483.608 1273.87 1.71733 6.63272
1272.43
B3
11.0344
1128.72 -0.67852
COND
0.73677 91.6838
197.85
1129.4 1.75501 34.4641
984.92
1.7927 395.165
99
x
h*
Δh
T
950
1.44467
A.3
Appendices
A.3.1 Power Matrix
Table 10 - Power Matrix
+
+
+
+
+
+
+
1/a+1/b+1/c+1/d
(-1)(1/a+1/b+1/c+1/d)
(-1)(1/b+1/c+1/d)
(-1)(1/c+1/d)
(-1)(1/d)
0
0
0
SHP/Em + Q(MN LEAK OFF)*(1/c+1/d)
E
G
B1
B2
B3
Q(MN COND)
Q(AUX COND)
Q(DESUP
Constants
A.3.2 Superheater Outlet Flow Matrix
Table 11 - Superheater Outlet Flow Matrix
0
1
0
0
0
0
0
-1
+
+
+
+
+
+
+
E
G
B1
B2
B3
Q(MN COND)
Q(AUX COND)
Q(DESUP)
0.005*E(EST) + SR(TG)*KW|operating Constants
A.3.3 High Pressure Heater Matrix
Table 12 - High Pressure Heater Matrix
+
+
+
+
+
+
+
hF3-h'F2
0
hD3-hB1
0
0
0
0
0
0
100
E
G
B1
B2
B3
Q(MN COND)
Q(AUX COND)
Q(DESUP)
Constants
A.3.4 Direct Contact Heater Matrix
Table 13 - Direct Contact Heater Matrix
SR(FP)*(BHP/E)-hF2
0
+
hD3
+
hB2
+
0
+
hc1
+
hc1
+
0
+
Q(SAH)*hB2 + Q(VENT COND)*h(VENT COND) Q(FWDCT)*h(FWDCT)-R*h'F2-Q(LIVE STEAM
MAKEUP)*hDS
E
G
B1
B2
B3
Q(MN COND)
Q(AUX COND)
Q(DESUP)
Constants
A.3.5 Feed Water Drain Collection Tank Matrix
Table 14 – Feed Water Drain Collection Tank Matrix
0
0
+
0
+
0
+
hB2-hD1
+
h'(MN COND) - hc1
+
h'(AUX COND) - hc1
+
0
+
Q(DIST)*(hB2-hD1) - (Q*h)(SAH DRN) - (Q*h)(CNTMD
DRN) - [(Q*h)(VENT COND) + (Q*h)(MN LO) + (Q*h)(AUX
LO)] - (Q*h)(MAKEUP FEED) - Q(MN A/E) + Q(AUX A/E) +
Q(DISTILL A/E)|h(DESUP)
E
G
B1
B2
B3
Q(MN COND)
Q(AUX COND)
Q(DESUP)
Constants
A.3.6 Main Condenser Matrix
Table 15 - Main Condenser Matrix
1
-1
-1
-1
0
-1
+
+
+
+
+
101
E
G
B1
B2
B3
Q(MN COND)
+
+
0 Q(AUX COND)
0 Q(DESUP)
Q(MN LEAK OFF) - Q(MN IC) + Q(DIST) Constants
A.3.7 Auxiliary Condenser Flow Matrix
Table 16 - Auxiliary Condenser Flow Matrix
0
0
+
0
+
0
+
0
+
0
+
1
+
0
+
Q(DIST) + Q(AUX IC) + Q(TG) - Q(AUX LO)
E
G
B1
B2
B3
Q(MN COND)
Q(AUX COND)
Q(DESUP)
Constants
A.3.8 Desuperheater Steam Flow Matrix
Table 17 - Desuperheater Steam Flow Matrix
SR(FP)*(BHP/E)
0
+
0
+
0
+
0
+
0
+
0
+
1
+
Q(SOOT BLOWERS) + Q(STM ATOM) + Q(DOM)
+ Q(MN A/E) + Q(AUX A/E) + Q(DIST A/E) +
Q(FOH)
102
E
G
B1
B2
B3
Q(MN COND)
Q(AUX COND)
Q(DESUP)
Constants