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					Marine Steam Power Plant Heat Balance
Steven Gelardi
RPI Project                                     X - STEAM NOT INSTALLED
Fall 2011

Introduction
A heat balance is all the calculations required for an engine room. These calculations include the
mass flows and enthalpy's throughout the plant and at certain times other variables such as
temperature and pressure are necessary. The idea is to complete the heat balance with a mass
balance. I will design as if we will not lose anything (heat, water or steam). Essentially, what goes in
will theoretically come out. I will also will include a flow diagram to help illustrate.

Statement
A design requires that the original design characteristics contain a Ships Horse Power (SHP) of
32,500 SHP, a complement of 40 persons, 2 boilers, a throttle temperature of 950 degrees Fahrenheit
and pressure of 850 psig. Condenser pressure is 1.5 Inches of Mercury ("Hg) and Auxilary Condenser
is 2"Hg.
The required Engine efficiency (Em) is given at 0.96. The objective is to complete the heat balance
and flow diagram. Note: The design must contain a total plant flow (E) for 2 boilers.


     Design Characteristics
                                     Units
Po                            850 psig
                          864.696 psia
To                            950 °F

Main Condenser                 1.5   "Hg
                         0.736765    psia
Auxiliary Condenser              2   "Hg
                         0.982353    psia

Shaft Horse Power           32500 SHP

Engine Effic., Em             0.96

Number of Boilers                2 Boilers

No. of Persons, N               40 Complement
Main Turbine (HP)

I need to calculate the Steam Rate (SR) Non-Extraction

Need:      a, b, c, d, ho, h(B1), h(B2), h(B3), hi, hw

           Units       lbs/hr/WHP

           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)

I need SR(NE), which is the Steam Rate(NE)
           SR(NE)=     2544.4/(ho-hw)                    This value should be between 5 and 6




                                           INSERT PICTURE




                                           INSERT PICTURE
Available Energy

           AE =       ho-hp                    574.6993

           ho =       h(Po,To)                 1481.998

           hp =       h(So,P(MN COND))         907.2985

           So =       So(Po,To)                1.651915

Top Point (P1)                                                             see T&R Bull.
           P1 =       .9*(Po)                  778.2264                    pg       39
                                                                           sect     3.3.5
                                                                           fig      8

Efficiency of the State Line (Eb)                                          see T&R Bull.
Eb =       (ho-hi)/(AE)           =            0.854662                    pg       34
                                                                           fig      5
       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 =        0.854662 OK


Temperature Correction Factor (f(t))

Need the 4th Polynomial Fit Equation:          y=a+b*x+c*x^2+d*x^3+e*x^4


                      Coefficient Data:
                      a=             0.95766
                      b=            -0.00059
                      c=            1.74E-06
                      d=           -1.58E-09
                      e=            4.82E-13
                      To =               950

                      f(t) =          1.01202 OK                           see T&R Bull.
                                                                           pg       35
                                                                           fig      6
State Line Energy (UE(SL))
           UE(SL) = AE*Eb*ft
                        497.0778

           State Line End Point (SLEP)
           SLEP = hi
           SLEP = hi       =      ho-UE(SL)
                                      984.92 OK

           Si =        S(COND)      1.792699

EXTRACTION STAGES
             P              T         hSL          S           v       x            h*      delta(h)
Turb Inlet 864.696           950    1481.998   1.651915    0.928074
Top Pt      778.2264    944.9168    1481.998   1.663038    1.031744
B1          280.2717    715.7506    1377.934   1.690182    2.425921              1377.134 -0.08001
B2           86.4696    483.6083    1273.871   1.717327    6.632715              1272.426 1.444668
B3           11.0344    197.8497    1129.396   1.755013     34.4641              1128.717 -0.67852
COND        0.736765    91.68379      984.92   1.792699    395.1653

State Line Equations :
           h=h1-((S1-s)/(S1-Si))*(h1-hi))
           s=S1-((h1-h)/(h1-hi))*(S1-Si))

           hB1 =       (ho+hB2)/2              hB2 =      h(SAH)
                        1377.934                           1273.871

           hB3 =       (hB2+hi)/2
                        1129.396


Need Exhaust Loss (EL)

First, I will pick an EL between 4000 and 6000 to get what is called             see T&R Bull.
an "annulus area"                                                                pg       37
                                                                                 sect     3.3.4.3


Annulus Area, (Aa)
I am going to assume that our SR(NE) is going to be (for conditions
Po, To, and P(MN COND):                                                    5.5

           Using the 4th Degree Polynomial Fit:           y = a+bx+cx^2+dx^3+ex^4
                      Coefficient Data:
                      a=            -51.8664
                      b=           0.051456
                      c=           -1.71E-05
                      d=            2.60E-09
                      e=           -1.42E-13
            For a Flow of 4000      X=                  4000
                                    EL =             9.81118

            For a Flow of 6000      X=               6000
                                    EL =        17.19.019

            Now I can solve for Aa (Annulus Area)
                       Aa = SR(NE)*(SHP/Em)/(Cond Press in "Hg*X)

                    Aa @ 4000 =      31.03137
                    Aa @ 6000 =      20.68758

            The closest Aa that we can pick will be:               25 = Aa

            The Condenser Flow, (X) will come from the Aa we got:
                        X = SR(NE)*(SHP/Em)/(Cond Press in "Hg*Aa)

                        X=           4965.019

            I can calculate the El with our solved X from our curve fit equation
                        El =          12.99132

            Next I need the Mechanical and External Efficiency (Em)                see T&R Bull.
                       Em =SHP/(SHP+Losses)                                        pg       31
                                                                                   sect     3.3.2
So in order to find Em, I need all of the Losses:

                        Losses are found by this calculation:
                                   Losses = SHP(3.5+RL)/100-(3.5+RL)

                                    Where RL is the Astern Turbine Windage
                                    RL = .33*(P(MN COND) in "Hg)                   see T&R Bull.
                                                                                   pg       33
                                    RL =               0.495 OK                    sect     3.3.3

                                    Em = (100-(3.5+RL))/(100)
                                    Em =         0.96005 %             OK

            Now I will solve for hw                                                see T&R Bull.
                        hw = hi+EL                                                 pg       43
                        hw =        997.9113                                       sect     3.3.8

            SR(NE) Design can be solved since I now have hw

                        SR(NE DESIGN) = 2544.4/(ho-hw)

                        SR(NE DESIGN) =             5.256086
Now, the next thing I need to solve for is a, b, c, and d.

            a=           24.45049
            b=           24.45049
            c=           17.61129
            d=           19.35137

I want to establish the unknowns in the matrix to be used to find E

Power Equations (1M)
         Power, P = WHP
                    WHP = Wheel Horse Power
Horsepower was originally defined to compare the output of steam engines      see Wikipedia
with the power of draft horses in continuous operation. The unit was widely   Horsepower
                                                                              Definition of WHP
adopted to measure the output of piston engines, turbines, electric motors,
and other machinery. The definition of the unit varied between geographical
regions. Most countries now use the SI unit Watt for measurement of power.


            WHP = (SHP/Em)
                     WHP =            33852.4 Kw             OK

(1M) = (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
            WHP HP = (E-G)/a+(E-G-B1)/b
Low Pressure Turbine
            WHP LP = (E-G-B1-B2-Q(MN LEAK OFF))/c+(E-G-B1-B2-B3-Q(MN LEAK OFF))/d

I need to solve for enthalpy of the Main Leak Off, h(MN LEAK OFF)             see T&R Bull.
            h(MN LEAK OFF) = ho-(.4)*UEW                                      pg       43
                        where UEW = ho-hw                                     sect     3.3.8
            Wheel Used Energy, UEW =
                     UEW = ho-hw                 484.0865

            h(MN LEAK OFF) =         1288.363


I can solve for the Heat Flow of the Main Leak Off, Q(MN LEAK OFF)
            Q(MN LEAK OFF) = 200+0.003(SHP)

            Q(MN LEAK OFF) =             297.5 lbs/hr        OK
I want to solve for G, which is the Total Flow Rate leaving the Superheater in
the Boilers that doesn't reach the Main Turbine


           G = Q(DESUP) + Q(TG) + Q(LOST)
DESUP is Desuperheated steam, or steam that has moisture in it and is below its superheated state

G is the Total Flow Rate, as it contains what is used by the TG, what is lost, and what is desuperheated

           G = Q(DESUP) + Q(TG) + Q(LOST)
                    Q(TG) = SR(TG)*(Kw)
                    Q(LOST) = 0.005*(E)                                          see T&R Bull.
                    Q(DESUP) = SR(FP)*(BHP/E)*E(EST)+3000                        pg       11
                                                                                 sect     3.2.6

Instead of using E, I am going to be using E(EST), which is the Estimate of the Energy

           E(EST)-G(EST) = SR(NE)*(SHP/Em)[1-0.25(1-(h(B)-hw)/(ho-hw)]

                       h(B) =         1260.4

                       E(EST)-G(EST) =         160948.7

                       E(EST) = G(EST) - [E(EST)-G(EST)]

                       Q(TG EST) = SR(TG)*(Kw(TG))
                            SR(TG EST) =         10 lbs/hr
                            Kw(TG EST) =       1250 Kw

                                Q(TG EST) =       12500 OK



           Q(DESUP EST) = SR(FP)*(BHP/E)*(E(EST))
EQUATION 1 and 1M
Solving for "E"
In this first set of Equations, I am going to be using the Power Equation to solve for "E"


           Q(MN LEAK OFF) =             297.5 lbs/hr                              see T&R Bull.
                                                                                  pg       43
           h(MN LEAK OFF) =         1288.363                                      sect     3.3.8

      <1> P = WHP = SHP/Em          33854.17

           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

    <1M>                            1/a+1/b+1/c+1/d        E
                            + (-1)(1/a+1/b+1/c+1/d)        G
                            +      (-1)(1/b+1/c+1/d)       B1
                            +           (-1)(1/c+1/d)      B2
                            +                (-1)(1/d)     B3
                            +                        0     Q(MN COND)
                            +                        0     Q(AUX COND)
                            +                        0     Q(DESUP
                SHP/Em + Q(MN LEAK OFF)*(1/c+1/d)          Constants
Equations 2, 2.1 and 2M
Next I need to solve for "G", which is the Superheater Outlet Flow (The flow of
the superheated steam leaving the boilers' superheaters.)
Solving for "G", Superheater Outlet Flow
      <2> G = Q(TG) + Q(DESUP) + Q(LOSS)

           Q(LOST) = 0.005*E              =      824.5402                            see T&R Bull.
Since I do not have E, I have to substitute in E(EST); and since I don't have        pg       11
E(EST), I need to find it.                                                           sect     3.2.6


    <2.1> E(EST) =
    E(EST)-G(EST) = SR(NE)*(SHP/Em)-0.25*(SR(NE)*(SHP/Em))+[0.25(SR(NE)*(SHP/Em))]*[(hb-hw)/(ho-hw)]

           E(EST) =       139648.4 OK

           Assume:
                          hB = (hB1+hB2+hB3)/3
                          hB =       1259.426

                          E(EST) = G(EST)+SR(NE)*(SHP/Em)[1-0.25{(hB-hw)/(ho-hw)}]

           Q(TG(EST)) = SR(TG)*KW(TG)                       #(10)(1000)

           Q(DESUP) = SR(FP) * (BHP/E) * E(EST) + 3000
                            (Feed Pump)


When I substitute all the variables I know, the following equation is what's left:
                          20*(0.013) = 0.04 * E(EST)

From <2> G(EST) = (0.005+0.04) * E(EST) + 13000             Therefore: G(EST) =      78963.22

Now that I have G(EST), I can plug and chug utilizing <2.1> to solve for E(EST)
           E(EST) = (0.005+0.4)*E(EST)+13000+(SHP/Em)[1-0.25{1-((hB-hw)/(ho/hw))}]
                  = 13000 + SR(NE)*(SHP/Em)[1-0.25{1-((hB-hw)/(ho/hw))}]/(0.955)

           Q(BLR)
                    EST    = E(EST)              82454.02

    <2M>                                              0     E
                              +                       1     G
                              +                       0     B1
                              +                       0     B2
                              +                       0     B3
                              +                       0     Q(MN COND)
                              +                       0     Q(AUX COND)
                              +                     -1      Q(DESUP)
                    0.005*E(EST) + SR(TG)*KW| operating     Constants
Equations 3 and 3M
Solving for the High Pressure (HP) Heater, "B1"

In Equations 3 and 3M, I will be solving for B1 using the 1st Law of
Thermodynamics with constant volume equal to the High Pressure Heater (HP
HTR)
      <3> B1 = E(hF3-h'F2)/(hB1-hD3)
          B1 =         164908                                                 see T&R Bull.
                                                                              pg       7
           P(HP HTR SHELL) = 093*P(S1)                                        sect     3.2.3
                                  260.6527 psi          OK

           Terminal Temperature Difference, TTD =
                    =         0 degrees Fahrenheit (°F), when T(B1)>T(SAT)@P(HP HTR SHELL)+200
            otherwise
                    =       10 degrees Fahrenheit; High Pressure and Low Pressure Surface
                    =       25 degrees Fahrenheit; Steam Air Heaters

           Temperature Difference, TD =                                       see T&R Bull.
                  =          10 °F         OK                                 pg       8
                                Counterflow Heat Exchange (Utilizing)         sect     3.2.4.2
                     otherwise
                  =          25 °F         BAD
                                 Parallel Heat Exchange (Not Utilizing)

               TD =           10 °F

           hF3 = h(f)@TF3 = T(SAT)@P(HP HTR SHELL) - TTD3
                      hF3 =     1202.042
           TF3 = T(SAT)@P(HP HTR SHELL)
                      TF3 =     404.6738 °F

                      TF2 =       271.9545 °F
                      hF2 =       1171.367

          hD3 = h(f)@T(SAT)@h'F2 + TD
          TD3 = T(SAT)@h'F2+        10
                     TD3 =    282.2893
                     hD3 =    1176.884

           h'F2 = hF2@P(DC HTR) + (v*ΔP*144)/(Eff(FP)*778)
           h'F2 =     1171.547
Since the Pressure of the Deaerating Feed Tank, also known as the Direct Contact Heater (DC HTR),
is a fraction of the Pressure of the Boilers (usually 0.5), I am going to be using the equation:


       P(DC HTR) = 0.1*Po
       P(DC HTR) =    850          +        14.696        *           0.1         /           2
                   Po, in Psig Converted to Psia then multiplied by 0.1 and divided by 2 Boilers
       P(DC HTR) =   43.2348 psia         OK         Range of <40-45> psia
<3M>          hF3-h'F2   E
       +             0   G
       +      hD3-hB1    B1
       +             0   B2
       +             0   B3
       +             0   Q(MN COND)
       +             0   Q(AUX COND)
       +             0   Q(DESUP)
                     0   Constants




           INSERT PICTURE
Equations 4 and 4M
Solving for the DC Heater, "B2"




                                         INSERT PICTURE




      <4> Q(FP) = SR(NE)*(BHP/E)*E
                      Q(FP) = 4354.29

           P(DC HTR) = 0.5 * P(B2)
                  P(DC HTR) = 43.2348
           P(AUX EXH) = P(DC HTR) + 5
                P(AUX HTR) =       48.2348

            BHP/E = (v*ΔP*144)/(778*Eff(FP))*(E/E)
                    BHP/E =     0.004545

           hD3 = h(f)@TD3
                         hD3 = 1176.884
           TD3 = T'F2 + TD3
                         TD3 = 554.5758 °F

Since I chose a Counter-Flow Heat Exchanger for my project (as it has the best heat exchange per
once pass), I will be using the Counter-Flow values for each Temperature Difference

           TTD3 =             10 °F          (Counter-Flow) (*Chose*)         see T&R Bull.
                              25 °F          (Parallel-Flow)                  pg       8
                                                                              sect     3.2.4.2

           T'F2 = T(SAT)@h'F2
           hF2 = h(f)@P(DC HTR)
               hF2 = 1171.363

           h'F2 = hF2 + (v*ΔP*144)/(Eff(FP)*778)
                          where         v = 0.017187
               h'F2 = 1171.3635
Now I need to solve for hc1, which is found by the steam tables from Tc1 (Temperature
set using Pressure in the Low Pressure Heater)


           Tc1 = T(SAT)@P(LP HTR SHELL) + TTD1
              and
           P(LP HTR SHELL) = 0.9*P(B3)

                       P(LP HTR SHELL) = 9.930963 psia
                                     and
                                   Tc1 = 202.8328 °F

           hc1 =      1147.7834


           TTD1 =             10 °F          (Counter-Flow) (*Chose*)          see T&R Bull.
                              25 °F          (Parallel-Flow)                   pg       8
                                                                               sect     3.2.4.1

           Feed Pump Recirculation, R
                 R=         0


Next I need to add in the value for the flow through the vent condenser on the Direct Connect Heater.
               Q(VENT COND) =              100

Therefore the enthalpy at the Vent Condenser is going to be denoted by the value h(VENT COND)
           h(VENT COND) = hg @ P(DC HTR)
              h(VENT COND) = 1171.367

    <4M>                       SR(FP)*(BHP/E)-hF2       E
                             +                  0       G
                             +                hD3       B1
                             +                hB2       B2
                             +                  0       B3
                             +                hc1       Q(MN COND)
                             +                hc1       Q(AUX COND)
                             +                  0       Q(DESUP)
              Q(SAH)*hB2 + Q(VENT COND)*h(VENT          Constants
              COND) - Q(FWDCT)*h(FWDCT)-R*h'F2-
                      Q(LIVE STEAM MAKEUP)*hDS


                               Constants =    5414076
Equations 5 and 5M
Solving for the Low Pressure Heater, "B3"

      <5> (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)) + ((Q(SAH DRNS)*h(SAH DRNS)) + (Q*h)(MAKEUP FEED) =
           Q(VAP FWDCT)*h(g)@212°F + (Q*h)(FWDCT)

Option A     Q(VAP FWDCT) ≠                0

                      therefore h(FWDCT) = h(f)@212°F

                      Solve for   Q(VAP FWDCT)

                                     If Q(VAP FWDCT) >            0 OK
                                     If Q(VAP FWDCT) <            0 ERROR   (Impossible)

Option B     Q(VAP FWDCT) =                0

                      therefore h(FWDCT) = h(f)@212°F

                                     If Q(VAP FWDCT) >            0 OK
                                     If Q(VAP FWDCT) <            0 ERROR   (Impossible)

                         Q(VAP FWDCT) =        180.1802

            h(MAKEUP FEED) = h(f)@75°F
                      h(MAKEUP FEED) =                    43.07439          see T&R Bull.
             h(CONT DRNS) = h(f)@200°F                                      pg       5
           (I set the Temperature)                                          sect     3.1.1
                           h(CONT DRNS) =                 168.0991

           h(SAH DRN) = h @ T(SAH DRN) = T(SAT) @ P(SAH) > T(Fuel Oil) [approx. 200-225°F]
                                               25°F Above air temperature

      Q(FWDCT) = Q(CONT DRN) + Q(SAH) + Q(MAKEUP FEED) + Q(AUX AC) + Q(MN AC) +
                 Q(DISTILLER A/E) + Q(MN LEAK OFF) + Q(AUX LEAK OFF) + Q(VENT)

           W = 0.5*SHP
                           W=          16250

              Q(DOMESTIC) =              100

           Q(CONT DRNS) = Q(FOH) + Q(DOMESTIC)
             Q(CONT DRNS) = 105.3259
   M(air) = MFO*A/F
              M(air) =     244562.5

   Δh(SAH) = h(SAH IN) - h(SAH OUT)
             h(SAH OUT) = h(f)@225°F
                  h(SAH OUT) = 193.2969
             h(SAH IN) = h(B2)
                     h(SAH IN) = 1273.871

        ΔT =          100 °F
                                      Q(SAH) =   5576.231

               Q(SOOT BLOWERS) [per boiler] = 353.4922
                Q(SOOT BLOWERS) = Q(SOOT BLOWERS[per boiler] * (2)
                       Q(SOOT BLOWERS) = 706.9844

                               Q(STM ATOM) =  392.0983
                                    Q(LOST) = 824.5402

   Q(MAKEUP FEED) = Q(STM ATOM) + Q(SOOT BLOWERS) + Q(LOST)
          Q(MAKE UP FEED) = 1923.623

                           Q(MN LEAK OFF) =         297.5             see T&R Bull.
                          Q(AUX LEAK OFF) =           50              pg       65
                                 Q(MN AC) =          196              sect     3.5.1.3
                                Q(AUX AC) =           60
                                  Q(VENT) =          100

   GPD| operating = 2.88*(MAKEUP FEED) + 35N + 45N
                    GPD|operating = 8740.034 Gallons Per Day (Operating)

                Q(DISTILLER AE) = 8.33*GPD*F(d)
                   where     F(d) =      780 for this 2-stage flash
                                                          see T&R Bull.
                therefore Q(DISTILLER AE) = 1717.1685     pg        25
                                                          sect      3.2.17.1
Q(FWDCT) = Q(CONT DRN) + Q(SAH) + Q(MAKEUP FEED) + Q(AUX AC) + Q(MN AC) +
           Q(DISTILLER A/E) + Q(MN LEAK OFF) + Q(AUX LEAK OFF) + Q(VENT)
        Q(FWDCT) = 10025.85




                                 INSERT PICTURE
<5M>                                          0    E
                        +                     0    G
                        +                     0    B1
                        +                     0    B2
                        +             hB2-hD1      B3
                        + h'(MN COND) - hc1        Q(MN COND)
                        + h'(AUX COND) - hc1       Q(AUX COND)
                        +                     0    Q(DESUP)
          Q(DIST)*(hB2-hD1) - (Q*h)(SAH DRN) -     Constants
       (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(DIST A/E)| h(DESUP)


                           Q(DIST) =    2756.64
                               hB2 =   1273.871
                               hD1 =   168.1063
                      Q(SAH DRN) =     5576.231
                      h(SAH DRN) =     193.2969
                   Q(CNTMD DRN) =      105.3259
                   h(CNTMD DRN) =      168.1063
                   Q(VENT COND) =           100
                   h(VENT COND) =      1171.367
                         Q(MN LO) =        297.5
                         h(MN LO) =    1288.363
                        Q(AUX LO) =          50
                        h(AUX LO) =          50
                           Q(MUF) =    1923.623
                           h(MUF) =    43.07439
                        Q(MN A/E) =         490
                       Q(AUX A/E) =         100
                       Q(DIST A/E) =   161.3574
                         h(DESUP) =    1484.205
                               TD1 =        200
                                B3 =   1105.765
                     Q(MN COND) =      -1088.05
                    Q(AUX COND) =      -1078.65


                     CONSTANTS =       1605936
Equations 6 and 6M
Solving for FWDCT, "Q(MN COND)"


     <6> Q(MN COND) = E - G - B1 - B2 - B3 - Q(MN LO) + Q(MUF) + Q(MN IC) + (B3 - Q(DIST))

                 Q(MN LO) =      297.5
                 Q(MN IC) =       196
                  Q(DIST) =    2756.64

   <6M>                                        1    E
                          +                   -1    G
                          +                   -1    B1
                          +                   -1    B2
                          +                    0    B3
                          +                   -1    Q(MN COND)
                          +                    0    Q(AUX COND)
                          +                    0    Q(DESUP)
                   Q(MN LO) - Q(MN IC) + Q(DIST)    Constants


                           Constants =    3250.14
Equations 7 and 7M
Solving for QTG, "Q(AUX COND)"

      <7> Q(AUX COND) = Q(TG) + Q(MUF) + Q(AUX IC) + Q(DUMP) - Q(AUX LO) + Q(DIST)

    <7M>                                         0     E
                             +                   0     G
                             +                   0     B1
                             +                   0     B2
                             +                   0     B3
                             +                   0     Q(MN COND)
                             +                   1     Q(AUX COND)
                             +                   0     Q(DESUP)
                      Q(DIST) + Q(AUX IC) + Q(TG) -    Constants
                                        Q(AUX LO)

       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) + Ships Heating + Others

Q(DESUP) is broken down into Eight (8) Parts:
          1. Soot Blowers, Q(SOOT BLOWERS)
          2. Steam Atomization, Q(STM ATOM)
          3. Domestic Services, Q(DOM)
          4. Feed Pump Turbine Steam Consumption (Function of "E"), SR(FPT)
          5. Steam to Air Ejectors (Motivating Steam), {Q(MN A/E), Q(AUX A/E), Q(DIST A/E)}
          6. Steam to Fuel Oil Heaters, Q(FOH)
          7. Ships Heating, Assume 0
          8. Others

           1. Q(SOOT BLOWERS)                                     706.9844
           2. Q(STM ATOM)                                         392.0983
           3. Q(DOM)                                                   100
           4. SR(FPT)                                             5.810133
           5. SUM[Q(MN A/E), Q(AUX A/E), Q(DIST A/E)]             751.3574
           6. Q(FOH)                                                 1137.5
           7. Ships Heating                                               0
           8. Others                                     +                0
                                                Q(DESUP) =         3093.75

           Constants = Q(DIST) + Q(AUX IC) + Q(TG) - Q(AUX LO)

                                Q(DIST) = 2756.64
                             Q(AUX IC) =       40
                             Q(AUX LO) =       50
                                 Q(TG) = 8822.337



                             Constants =    1166.898
Equations 8 and 8M
Solving for "Q(DESUP)"

      <8> Q(DESUP) = Q(SOOT BLOWERS) + Q(STM ATOM) +


    <8M>                          SR(FP)*(BHP/E)            E
                             +                  0           G
                             +                  0           B1
                             +                  0           B2
                             +                  0           B3
                             +                  0           Q(MN COND)
                             +                  0           Q(AUX COND)
                             +                  1           Q(DESUP)
               Q(SOOT BLOWERS) + Q(STM ATOM) +              Constants
                  Q(DOM) + Q(MN A/E) + Q(AUX A/E)
                            +Q(DIST A/E) + Q(FOH)

           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=      0.026404
                     Constants =     3087.94

           Q(SOOT BLOWERS) = # Blowers*(Q(SOOT BLOWERS)/Boilers)

              100000        <      Q(SOOT BLOWERS)              <         300000

                     Q(SOOT BLOWERS) =            110000

           Q = 120 + (0.00165)*(Q(SOOT BLOWERS))
                            Q=      301.5

           (110,000)/(1000) =               11

                                            11 = E(EST)/2(1000)

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

           [SR*E(5.5)*325000 = Est]
                8125      /              4000       =        2.03125 which is approximately 2
A value or 2 or 3 operating is estimated for this cycle
           <0.5 approx~ 0.45>SFC (4 Installed though)

           Q(STM ATOM) = #Boilers * (2 or 3) burners * 300
                       Q(STM ATOM) =             1200
Domestic and Hotel Loads

           Complement <20-40>               N=            40

           Domestic Water Heating              Q = 0.9*N                          36
           Galley                              Q = 0.5*N                          20
           Laundry                           + Q = 0.1*N              +            4
           Total Hotel Loads              Q(DOM) = 1.5*N                          60

However, since Q(DOM) is less than 100, we are going be using 100 as the Domestic Loads
                                Q(DOM) =          100

           Steam to the Air Ejectors
        A. Main, Q(MN A/E) =            490 lb/hr
        B. Aux, Q(AUX A/E) =            100 lb/hr                                      see T&R Bull.
        C. Dist, Q(DIST A/E) =                                                         pg       25
                      Q(DIST A/E) = (8.33)*(GPD)*(60/24)*hi                            sect     3.2.17.1
                           =       161.3574 lb/hr

           Maximum Steam Condensed estimated SR(NE) ) WHP = Q(NE)
              or
                    SR(NE))SHP = Q(NE)

           SHP/Em = WHP

           RF = (hB1-ho)/(ho-hw)

         GPD(OPER) = GPD(MAKEUP FEED) + 45*N + 35*N
   where         GPD(MAKEUP FEED) = Q(MAKEUP FEED)*[24*vf @ 75°F * 7.48]
   where                  Q(MAKEUP FEED) = Q(LOST) + Q(STM ATOM) + Q(SOOT BLWRS)
              GPD(OPER) =    11500

                            Standard Capacity of a Distilling Plant Range
              8000        10000     12000     15000        20000      25000             Special

If there are two evaporators installed and one evaporator is running:
          GPD rate = 1.3*GPD(OPER)
(Round it to the nearest largest size)
               1.3          *        11500          =?=          14950 OK
                                                                 15000 TRUE
                                 GPD Rate =          15000

           Since the GPD Rate is         15000 , I have to select two of this units, as there are 2 Boilers

If there is one evaporator installed and operating:
           GPD(RATED) = 1.3 * GPD(OPER) * 1.5
(Round it to the nearest largest size)
               1.3          *        11500            *         1.5         =?=           22425 OK
                                                                                          25000 TRUE
                             GPD(RATED) =            25000
           Q(DIST) =          8.33 * GPD(OPER) * FD /             24 * hi

           where       FD =          1.3*600 =         780
                              and
                       hi = hB3
                              and since        "hi" = h(DESUP)
                                             h(DESUP) = 1484.205

                       therefore                       hi = 1484.205

Two-Stage Flash
                       Q(DIST) =          2756.6405

      Q(DIST A/E) =           8.33      * GPD(OPER) * (FD/10) /             24   * hi

                   Q(DIST A/E) =     161.3574


Q(FUEL OIL HEATERS)
The next thing I am looking for is Q(FOH) which is the Heat Flow of the Fuel Oil Heaters
           Q(FOH) = ?

Since I know that
           SFC)est * SHP = W(FO)
           C(FO) [T(FO) * {(selected 200°F)-(75°F)}]
                       C(FO) =          0.46
                         SHP =        32500
                          SPC =            0.5

           Q(FOH) * Δh steam (1000 BTU/lb)

           Q(FOH) =        0.057 * W(FO)
Settler Heat Lost approximately 18%
           Q(FOH) =        0.007 * W(FO)
                                                   =         1137.5

           T(SAH DRN) = T(AIR BRNR) + TTD(SAH)
                T(SAH DRN) = 225

          h(SAH DRN) = h(f) @ T(SAH DRN)
                   h(f) @ T(SAH DRN) = T(SAT)                                    see T&R Bull.
therefore      h(SAH DRN) = 193.2969                                             pg       48
                                                                                 fig      13
           Q(SAH) * (hB2 - h(SAH DRN)) =                                         see T&R Bull.
                      (A/F*SHP*SHP)*[Cp Air]*{T(AIR BRNR) - 100°F}               pg       48
                                                                                 fig      13
           where       Cp Air =      .2438 @ 200°F Terminal Temperature
                                      0.24638
Equation        E          G          B1          B2         B3       Q(MN COND) Q(AUX COND) Q(DESUP)   CONSTANTS

  <1M>      0.190256 -0.19026 -0.14936 -0.10846 -0.05168        0        0                            0 33886.43
  <2M>             0        1        0        0        0        0        0                           -1 9646.877
  <3M>     87.87.875        0 -201.05         0        0        0        0                            0        0
  <4M>      -1171.34        0 1176.884 1273.871        0 1147.783 1147.783                            0 5414076
  <5M>             0        0        0        0 1105.765 -1088.05 -1078.65                            0 1605936
  <6M>             1       -1       -1       -1        0       -1        0                            0 3250.14
  <7M>             0        0        0        0        0        0        1                            0 1168.98
  <8M>      0.026404        0        0        0        0        0        0                            1 3087.94

Minverse    9.180389 6.094239 -0.03891 0.004195 0.000429 4.347618 -431651 6.094239 33886.43
              -0.2424 0.839086 0.00103 -0.00011 -1.13E-05   -0.1148 0.114903 0.839086 9646.877
            4.012731 2.663781 -0.00668 0.001833 0.000188 1.900336     -1.9021 2.663781         0
            -1.41781 8.16187 0.001749 0.007283 -6.63E-05 8.431616 -8.43099 8.16187 5414076
             6.71847 -5.48125 0.000917    -0.4734 0.001218 -6.75947 6.748021 -5.48125 1605936
            6.827868    -5.5705 0.000932 -0.00481 3.19E-04 -6.86954 5.866538    -5.5705 3250.14
                    0         0        0        0        0        0         1         0 11668.98
              -0.2424 -0.16091 0.000103 -0.00011 -1.13E-05  -0.1148 0.114903 0.839086 3087.94

                         E=        375449.2   lb/hr
                         G=        2821.324   lb/hr
                        B1 =       164108.2   lb/hr
                        B2 =       24243.65   lb/hr
                        B3 =       190960.7   lb/hr
                Q(MN COND) =       181025.9   lb/hr
               Q(AUX COND) =       11668.98   lb/hr
                  Q(DESUP) =       -6825.55   lb/hr
             Q(DESUP REAL) =       6825.553   lb/hr

Since I got a negative value for Q(DESUP), I took the absolute value of the number. Having a negative 'pounds
per hour' value is technically correct for this; as it is taking heat from the system, however to make it a real
number, I put it as a positive integer
Turbo-Generator, (TG)
Q(TG) and Electric Load
Rated conditions are used to select the generator and turbine "size"
                                                                                  see T&R Bull.
KW Load                                                                           pg       21-24
           KW Rated = (A + B + C)*SHP + 1.6*N + 9√(N) + 80 + other                fig      3.2.16
"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
I opted to have no "SCOOP," which is an alternate means of pumping seawater into the system. A
scoop is basically just that; a giant scoop that when the ship moves forward, it scoops seawater and
creates a positive pressure, allowing it to feed into the system.
           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
                and          Rated/Normal =         1.15                          see T&R Bull.
                 C=        0.0073                                                 pg       21-24
                 N=            40                                                 fig      3.2.16

                                KW Rated = 1218.177 KW         OK
                                         **    1250 KW is closest match           **

                                Rated KW and Standard Size Generator
                     Most Common ………………………………………………Least Common

              3560        500         750        1000       1250         1500          2000   Special

KW OPERATING
       KW(OPER) = (A + B + C)*SHP + 1.6*N + 9√(N) + 80 + other

                 A=       0.011 (Given)
       B (Option 1) =         0 with scoop
       B (Option 2) =     0.007 with no scoop
           C = 0.0004*(Normal/Normal)^3 "H2O Draft Loss
where Draft Loss is the pressure needed to push air through the Boiler
                     Draft Loss =         12 "H2O
                 C=        0.0048
     and         N=            40

                           KW Operating =       941.921 KW
                                                   1000 KW is closest match

I have to pick a Generator's Rated Size so that [KW(RATED) / (KW(OPER)] is greater than 2/3 but less
than 3/4, when possible

     (KW Operating)/(KW Choice of 1000) = 941.921      /                   1000
                                        = 0.941921 CHECK
This choice cannot work is it is way greater than 3/4. I will show how using 1250 KW instead of 1000
KW gives the more optimal choice

     (KW Operating)/(KW Choice of 1250) = 941.921     /                     1250
                                        = 0.753537 OK
This choice is better as it is just about 3/4, which is the top band of my choice. 1250 KW is the
Operating Kilo-Watt


Turbine for Generator
            %KW =       75.35 %
This KW Rated at 1250KW Operating at 942 KW

        Q(TG) ) RATED = [SR(TG)*KW] ) RATED

           SR(TG) = 2544.4/(ho-hw)                 in [lb/hr/HP]
                  = 3412.1/(ho-hw)                 in [lb/hr/KW]

           Pso = Po/0.975                                                        see T&R Bull.
                   = 886.8677                                                    pg          7
                                                                                 sect        -
                P'o = 0.975(Pso) =       864.696         865    psia   (Rounded to nearest 5 psi)
                Po =                                     865    psia
                T'o =                                    950    °F
               Tso = To + 5°F                            955    °F
                To =                                     950    °F
             h(SO) =                                1484.102

                ho' = h(To',P'o)
                            ho' = 1481.988
                 hi = ho' - Eff(TG TURB OPER)*(ho'-hp')
                             hi =    1117.7

           Eff(TG TURB) = (ho'-hi)/(ho'-hp)

           Eff(TG TURB))OPER         =             Eb ft fp fb fL       0.650173       65.02 %
           Eff(TG TURB))RATED        =             Eb ft fp fb (1)      0.673754       67.38 %

           hp' = h(P(AUX EXH), So)
                          hp' = 921.6939
EB = fig 19 @ <1200-1800> RPM Generator
    KWr/1000 =       0.75
   = 0.648817
                                                      T&R Bull.
                                                      pg        60
                                                      sect      19



            INSERT PICTURE




 ft = fig 21 @ P'o, T'o
           P'o =        850 psia          T'o =   950 °F
    =      1.027
                                                      T&R Bull.
                                                      pg        63
                                                      sect      21



            INSERT PICTURE




fp = fig 20 @ KW, Po
          KW =       1250 KW              Po =    850 psia
   = 1.011135
                                                      T&R Bull.
                                                      pg        62
                                                      sect      20



            INSERT PICTURE
                 fb = # @ 2"Hg @ Po
Since it is always 1.00 on 2"Hg, I will use 1.00 as the number, #
                              #=              1                 Po =         850 psia
                      =          1
                                                                                   T&R Bull.
                                                                                   pg        60
                                                                                   sect      19



                              INSERT PICTURE




                 fL = fig 21 @ % Rated
                           KW =      1250 KW                           % Rated =   0.753537 %/100%
                    =      0.965
                                                                                   T&R Bull.
                                                                                   pg        63
                                                                                   sect      21



                              INSERT PICTURE




                       ho' - hw' =    3412.1

           Theoretical SR = 3412.1/(ho'-hp')
                          =       6.089731
           SR(TG)OPER = TSR'/(Eb ft fb fp fL)                                      see T&R Bull.
                          =       9.366324 OK                                      pg       65
           SR(TG OPER) = SR(RATED)/fL                                              sect     3.5.1.3
                          =       9.038503 OK
           Q(TG)OPER = SR(TG OPER)*KW(TG OPER)
                          =       8822.337
           h(TG EXH) = hw' = hi'
                          =          1117.7
           Q(AUX LO) =                   50
           h(AUX LO) = ho' - 0.3*(3412.1/SR(L))
           where ho - hi' = 3412.1 and L = "OPER"
                           =      1372.699
Feed Pump Turbine Calculations of Steam Consumption Q(FP), and Exhaust Enthalpy, h(FP EXH)
  NOTE: Q(FP) ) OPER = SR(NE) ) OPER *(BHP/Em) OPER *E( OPER )

           h"(FP EXH) = ho" - Eff(FP TURB)*(ho"-hw")
           h(FP EXH) = hw"

           SR(FP)OPER = 2544.4/{(ho"-hw")/(Eff(FP))}
                        =      5.810133

           Bhp)OPER = 144*v*ΔP*(E(OPER))/(3600*550*Eff(Pump))OPER
                        =      294.4731

           Eff(FP TURB))RATED = EB fs BHP)RATED/(BHP(RATED)) + LHP(13.7/v)
                       where LHP = Windage Loss                     see T&R Bull.
                                                                    pg       6
                                                                    fig      23 Left
           Wheel Diameter =      <12,16,25>                         see T&R Bull.
                           =            16 OK                       pg       6
           Δh = ho" - hp"                                           fig      23 Top
           P(FP EXH) = P(DC HTR) + ΔP(BP)
                       ΔP(BP) = <3-5>
                           =             5 OK
                       ΔP ~ ΔP α H
                          α H = Pump Head
           ΔP max per Stage is ~50

           ho" = ?
           Need P"o and To" @ FP Throttle
           ho, To@ Main Turbine, ho', To'@ TG, ho", To"@ FP

Given To, Po
           Pso = Po + ΔP
                   = 886.3134
           ΔP(MN STEAM) = 0.025 * Po                                               see T&R Bull.
           Pso = Po/0.975                                                          pg       7
           Tso = To + 5°F                                                          sect     3.2.3
                   =      955                                                      see T&R Bull.
           P(DESUP) = Pso + ΔP(DESUP)                                              pg       56
                   = 886.3132                                                      sect     3.4.6
           T(DESUP) = Tso + ΔT(DESUP)                                              fig      17
                   = 955.1502

                ΔP =           70
                ΔT =         77.5

           Max Desup Flow =
                  A.    20000 lb/hr            BAD
                  B.    40000 lb/hr            OK
Maximum Desuperheated Flow is determined by a set value, in this case
either 20,000 lb/hr or 40,000 lb/hr. I chose 40,000 lb/hr as it works easier for
these Boilers and allows for a greater flow.
            Q(DESUP))NORM = Ʃ Q(AUX) + Q(FP)
            Q(FP) = SR(FP)*(BHP)
                      <0.04-0.05>E(EST)

            Q(DESUP) per Boiler = Q(DESUP))*(2 Boilers)/(2)                see T&R Bull.
              Q(DESUP) Cycle = 1546.875 lb/hr for 1 Boiler                 pg       56

            Pso + (ΔP(SH) + "plus") = P(STM DRUM)                          see T&R Bull.
                        "plus" = ΔP(CTRL DESUP)                            pg       72
                         ΔP(CTRL DESUP) at 950°F and 850 psig              sect     3.5.4
                                           =         30                    fig      25
                        "plus" =          30 at 950°F and 850 psig         see T&R Bull.
                 ΔP(ORIFICE) =             7                               pg       72
                                                                           fig      25
                P(STM DRUM) =     991.3134

            ΔP(SH) =          68 @ 110,000 lb/hr = Q(BOILER EST) &
                                 950°F Superheater Outlet Temperature
       If            E(EST) =       220
    Then       Q(BOILER)EST = 220/2
                            =       110

Next, h(DESUP) needs to be solved for. H(DESUP) can be found using T(DESUP) or P(DESUP)
            h(DESUP)
                       T(DESUP) = T(SAT) @ P(STM DRUM)
                       T(SAT) @ P(STM DRUM) =      543.5944 °F
                               = 543.5944 °F
                       P(DRUM) = Pso - ΔP(DESUP)

                    h(DESUP) =    1484.205
Feed Pump Turbine and Feep Pump Pump                                         see T&R Bull.
                                                                             pg       66
           Δh = h(DESUP) - h"
           SR(FP))oper = 2544.4/(Δh(ACTUAL)oper*η
           Δh(ACTUAL) = Δh(FP TURB)oper

           P(FP EXH) = P(DC HTR) + ΔPP(BP)
              where ΔP(BP) = <3-5>
                 P(FP EXH) =    48.2348

Assuming        η(Eff)FP TURB)oper =               0.5

           Q(FP)oper = SR(FP)oper (BHP/E)E)oper

      BHP/E)oper = [144*v*ΔP - (E/E)]/[(550*3600)*(η(PUMP)oper)]

Assuming Eff(FP PUMP))oper =              0.7                                see T&R Bull.
           ΔP(oper) = ΔP(max)     for a constant Discharge Pressure Governor pg       73
A Governor is a device that senses downstream flow/pressure, and             sect     3.5.4
automatically adjusts to maintain constant speed on a generator to produce
consistency
           v = v(f) @ P(DC HTR)
                   v = 0.017187

           h(FPT EXH) = ho" - Δh(ACTUAL) = ho"(ACTUAL) - (Efficiency of Pump)oper*(h(PUMP) - hp")

Assuming Shortcut
           Eff Turb =        0.5
             Eff Fp =        0.7
                ΔP =       1000

           SR(FP) = 2544.4/(ho" - hp")*Eff(TURB))oper*(h(pump) - hp")
           SR(FP) = 64.32591

           ho" = h(DESUP)
                ho" = 1484.205

           P"o = P(DESUP)/0.975         Rounded to the nearest 5 psi
               P"o =          44
                    =         45
           To" = T(P"o,ho")
               To" = 903.2053
           P"1 = 0.9*P"
               P"1 =        40.5
           hp" = h(FP EXH),S1)
                hp" = 1456.517
           S1" = S(P"o,h(DESUP))
               S1" = 1.986037
               So" = 1.974451
           P(FP EXH) = P(DC HTR) + ΔP(BP)
Required(New)
Applied `st Law
            CV = Main or Aux A/E
            Ʃ(Q*h)(IN) = Ʃ(Q*h)(OUT)
            W(CV) = 0
            Q(CV) = 0

           IC = (2/5)*Q(MN or AUX COND)
           h(IC) = h(f)@125°F                 92.99385

           AC = (3/5)*Q(MN or AUX COND)
           h(AC) = h(f)@200°F           168.0991

           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)

Find       Q(MN IC DRN)
                    Q(MN IC DRN) = (2/5)*Q(MN A/E)                             see T&R Bull.
                            =       196                                        pg       13
                                                                               sect     3.2.7.1
Find       Q(MN AC DRN)
                    Q(MN AX DRN) = (3/5)*Q(MN A/E)                             see T&R Bull.
                           =       294                                         pg       13
                                                                               sect     3.2.7.1
Find       Q(AUX IC DRN)
                    Q(AUX IC DRN) = (2/5)*Q(AUX A/E)                           see T&R Bull.
                            =        40                                        pg       13
                                                                               sect     3.2.7.1
Find       Q(AUX AC DRN)
                    Q(AUX AC DRN) = (3/5)*Q(AUX A/E)                           see T&R Bull.
                           =        60                                         pg       13
                                                                               sect     3.2.7.1

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)]
                                                                    [h(MIX) Part 0]
h'(MN or AUX COND) = h(MN or AUX COND) + W(PUMP)
where     h(MN or AUX COND) = h(f)@P(MN or AUX COND)
and       W(PUMP) = v*ΔP*144/(36000*550*Eff(PUMP)-->(1.0))
           W(PUMP) = 0 (approx.)
           therefore     h'(MN COND) =         59.7354
                        h'(AUX COND) =        69.13452
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) Part 1]      [h(MIX) Part 2]       [h(MIX) Part 3]
     Q(MN COND) =        -1088.05        Q(MN A/E) =          490          h(DESUP) = 1484.205
    Q(AUX COND) =        -1078.65       Q(AUX A/E) =          100
     h(MIX) Part 0 =    1346.148       h(MIX) Part 1 =   -7.2E+08
     h(MIX) Part 2 =     -1.5E+08      h(MIX) Part 3 =    -2166.7            h(MIX) = 398316.8
HP Heater

P(HP Shell) = 0.9*P(B1)
T(F3) = T(SAT) - TTD3

Next thing to find is the efficiency of the Boiler. By finding it, I will be able to calculate the fuel
consumption (Q(FO) and the Specific Fuel Consumption (SFC)
Eff(BLR) =
[h(SO)*(E - Q(DESUP)) + {Q(DESUP)*h(DESUP)} - E*hF3]/[SFC*SHP*Cp(FO)*{ΔT(FO)+(A/F)*Cp(Air)*ΔT(Air)+HHV}]

where       Q(FO) = SFC*SHP
               A/F =     15.05 (ration)
            ΔT(FO) =      100 °F                 <100°F - 200°F >                      see T&R Bull.
                                                                                       pg       47
             Cp(Air) =      0.2438                                                     sect     3.4.2.2
               HHV =        18500

         Excess Air =             5%                                                   see T&R Bull.
           Cp(FO) =            0.46                                                    pg       49
                                                                                       sect     3.4.3.3




                                             INSERT PICTURE
Boiler Efficiency with Steam Air Heater                                           see T&R Bull.
                                                                                  pg       46
             Eff = [(Ho+Ha)-HL]/[Ho-Ha]                                           sect     3.4
where        HL = Hg+Hu

Ho is the Heat Input; 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
    Ho = HHV + Cp(FO)*ΔT(FO)
             where:      HHV is the Heater Heating Value
                           HHV =        18500
                            18500 + Cp(FO)*ΔT(FO)
         =     18557.5

Ha is the heat added to the combustion air by the air heater
    Ha = Cp*R*(T2-T1)                                                             see T&R Bull.
             where:      Cp is the mean specific heat of air at T(FINAL)          pg       48
                                Cp =      0.46 BTU/lb/°F                          fig      13

                         R is the Air to Fuel Ratio                              see T&R Bull.
                                R=         15.05 lb(Air)/lb(FUEL)                pg       51
                                                                                 sect     3.4.3.5
                         T2 is the Temperature of Air leaving the Steam Air Heater        see T&R Bull.
                                T2 =      200 °F                                          pg       53
                                                                                          sect     3.4.5.1
                         T1 is the Temperature of Air entering the Steam Air Heater
                               T1 =        75 °F

         =     865.375

Hg is the Stack Loss                                                              see T&R Bull.
    Hg = 5.15 + (0.021475 + 0.000187*(% Excess Air))*Tg - 100                     pg       53
         Sack loss, % * 18546/100                   BTU/lb(FO)                    sect     3.4.4.1
       =      7.391 %

Hu is the radiation, which is unaccounted for losses and manufacturer's margin (R-U and M)
    Hu = 1.415% * Ho                                                              see T&R Bull.
where:                        Ho =      18557.5                                   pg       46
         =   262.5886                                                             sect     3.4.1
                                                                                  pg       52
HL is the Heat Loss                                                               fig      15
    HL = Hu + Hg                                                                  see T&R Bull.
       = 269.9796                                                                 pg       46
                                                                                  sect     3.4.1

         Eff Boiler =       0.9861

                              98.61 %
Specific Fuel Consumption

SFC =
[(E - Q(DESUP)*h(SO) + (Q*h)(DESUP) - E*(hF3)]/[(Eff(BLR)*SHP*HHV + (Cp(FO)*ΔT(FO)) + (A/F)*Cp(Air)*ΔT(Air)]

where:       Q(FO) = (SFC)*(SHP)                              =            16250 lb/hr
                 W=        16250                              =             8.125 tons/day
                                                              =                6.7 barrels/day

                SFC =   0.199309

The rate heat added into (IN) 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 BTU/lb/HP
     HR WHP (cycle) = [(E - Q(DESUP)*h(SO)) + ((Q*h)(DESUP)) - E(hF3)]/WHP

where:                   WHP = SHP/Em
                             = 33854.17

    HRWHP(cycle) =      3128.115

      HR SHP (plant) = (Q(FO)*HHV)/SHP
                             9250

                                                Units Refresher:
                                            1 hp =         2544.4 BTU/hr
                                            1 Kw =         3412.1 BTU/hr
                                            1 hp =           0.745 Kw

Efficiency = Power Out / Rate of Heat Added
            Efficiency of Cycle = WHP(2544.4)/[(E - Q(DESUP)*h(SO)) + ((Q*h)(DESUP)) - E(hF3)]

         *     2544.4 is the correction factor for units             *


              Efficiency/Cycle =      2544.4 / HR/Cycle =                0.813397 81.33973 %

               Efficiency/Plant =     2544.4 / HrR/Plant =                0.27507      27.51 %

             Efficiency Carnot =        1 - (TL/TH)
                                         °F to R Conversion
                            TL =     91.6838 °F + 460    551.684 R
                            TH =         955 °F + 460       1415 R
                                                         =        0.610118             61.01 %
Optimal Feed Pump Calcs
          Eff(TURB) = (Eb)*(fs)*(BHP/(BHP + LHP(13.7/v)))                           see T&R Bull.
                                                                                    pg       67
where:     BHP is the Break Horse Power at Rated Load                               sect     3.5.3.1
           LHP is the Windage Loss, in Horse Power                                  fig      23
           fs is the Superheat Correction Factor at P1
           v is the final Rated Specific Volume, in cubic-feet per pound of steam
                        v = v(g)@FPT EXH Press

                    Eff(TURB) =    0.502137

For Rated: [BHP/E]|(rated) = [144*ΔP*v*(E/E)]/[3600*500)*(Eff(FP)|(rated))]

           ΔP = ΔP(rated) for a constant discharge pressure governor             see T&R Bull.
                                                                                 pg       73
                                                                                 sect     3.5.4.2
           Constant Discharge Pressure Governing                       850PSIG / 950°F
           Superheater Outlet Pressure                                      872 psig
           Superheater pressure drop at maximum
           continuous power (including saturated pipe
           to superheater)                                                     62 psig
           Orifice in piping to superheater                                     7 psig
                                                                               69 psig
           Pressure drop at maximum Boiler Rate*
                    69 * (1.15/1.00)^2 --->                                    91 psig
           Pressure drop due to steam temperature
           control desuperheat (assuming constant loss
           above full power)                                       +           30 psig
           Steam Drum Pressure at max flow                                    993 psig
           Economizer pressure drop including piping
           to drum
                    12 * (1.15/1.00)^2 --->                                    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)                          5   psig
           Feed line pressure loss                                              5   psig
           Static head, pump dischard to Boiler Drum                           10   psig
           Total = Pump Discharge Pressure,                        +
           FP(DISCH)                                                        1076 psig
                                                                        1090.696 psia

           Deaerating Feed Heater Pressure                                     29 psig
           Static Head, DFT to Pump Suction                                    21 psig
           Summation                                                           50 psig

           Less suction line pressure loss                         -            1 psig
           Net Suction Pressure                                                49 psig
           Net developed pump pressure =
           Feed Pump Disch Press - Net Suct Press                          1027 psig
Eff(TURB) = (Eb)*(fs)*(BHP|(rated)/(BHP + LHP(13.7/v)))

SR(FP) = 2544.4/(ho"-hp")(Eff(FP))T

            ho" = h(DESUP)
                =       1484.205
            P1 = 0.9*Po"
                =            40.5
            S1" = S(P1", h(DESUP))
                =       1.986037
            hp" = h(P(FP EXH), S1")
                =       1456.517

  BHP|(rated)(est) = [144*ΔP*v]/[3600*500)*(Eff(FPP))]*E(rated)(est)
          E|(rated)(est) =      1.15*E(est)

            *ΔP = Δpmax

            v = v(f) @ P(DC HTR)
                 =       0.017187

Eff|(rated)FPP =          121875.1 @ GPM (expected average)              see T&R Bull.
                                   + RPM*√(GPM)/(H^(3/4))                pg       76
                                                                         pg       26
  RPM = < 8000 - 9000 >
                8500 Average
GPM|(rated) = E(est)*1.25 (lb/hr)(hr/60min) *v(f) (ft^3/lb) *7.48Gal /ft^3
               =      (E|(rated)(est))*(1/60)*(v(f)water)*7.48
               =       205.5854

            H/Stage = (P(DISCH) - P(SUCT))*144*v/(# stages)

            *ΔP =                  500 max per stage

                         if ΔP ~            1000 then uses             2 stages

            H = per Stage
                             Ns =          571
                        Use curve fit at Ns, Eff|(rated) @ average Eff to solve
                                = 144*v*ΔP*v(f)
                   H2 = 2541.777
                    H = 1270.889




                                   INSERT PICTURE
           LHP =         8500                                                 see T&R Bull.
                                                                              pg       66
                                                                              fig      23
        Wheel Diameter =           12     inches (common value)

        fs = P(THROTTLE)/P(EXH)
                = 0.912976

        Eb(FP) = RPM*(Wheel Diameter)/√(ΔH)
                =    0.55

           SR(FP)|(rated) = 2544.4/(ho"-hp")*(Eff(FPP)|(rated))
                 Constant =     2544.4
                        ho" = 1484.205
                        hp" = 1456.517
           Eff(FPP)|(rated) =       0.7
                            = 131.2792

            SR(FP)|(oper) = SR(FP)/fL
            SR(FP)|(rated) = 131.2792
                        fL =    0.965
                           = 136.0406                                         see T&R Bull.
                                                                              pg       66
                                                                              fig      23 Bott-R

Bhp/Bhp|(rated) = (ΔP/ΔP(rated))*(E/E(rated))*(Eff|(rated)/Eff(oper))         see T&R Bull.
                          1         (1/1.15)           (1/1.15)               pg       76
                = 0.756144                                                    fig      26

         Bhp/E = 144*v(f)*ΔP/(3600*550*Eff(FPP)|(oper))

                   Eff(FPP)|(oper) =      Eff(rated)*Eff(oper) / Eff(rated)

          Q(FP) = (SR(FP)|(oper))*(BHP/E(oper))*E(unknown)

           hw|FP = h(DESUP) - (ho"-hp")*Eff(FPT)|(oper)
                                        Sub    Running
Mass Balance                           Total      Total
DC Heater Outlet (to Feed Pump)                375449.2
          E                         375449.2

Super Heater Outlet                            375449.2
          E                         375449.2

Desuperheater Outlet                            6825.55
         Q(DESUP)                    6825.55

Main Steam                                     368623.7
          Superheater Outlet        375449.2
           - Desuperheater Outlet    6825.55

Flow to TG                                      8822.34
             Q(TG)                   8822.34

Flow to Main Turbine                           358976.8
           Main Steam               368623.7
           - Q(TG)                   8822.34
           - Q(LOST)                  824.54

Flow to LP Turbine                             170327.5
           Flow to Main Turbine     358976.8
            - B1                    164108.2
            - B2                    24243.65
            - Q(MN LEAK OFF)           297.5

Exhaust from LP Turbine                        -20633.2
           Flow to LP Turbine       170327.5
           - B3                     190960.7

Exhaust from TG                                 8772.34
           Flow to TG                8822.34
           - Q(AUX LEAK OFF)              50

Main Condensate                                167766.8
         Exhaust from LP Turbine    -20633.2
          + B3                      190960.7
          - Q(DISTILLER)             2756.64
          + Q(MN IC DRN)                 196

Auxiliary Condensate                           11568.98
            Exhaust from TG          8772.34
            + Q(DISTILLER)           2756.64
            + Q(AUX IC DRN)               40

Total Condensate                               179335.8
          Main Condensate           167766.8
           + Auxiliary Condensate   11568.98

Flow to CDT                                      1237.5
          Q(FOH)                      1137.5
           + Q(DOMESTIC)                 100
                                       Sub    Running
Mass Balance                          Total      Total
Flow to FWDCT                                 11158.02
          Q(MN AC DRN)                  196
          + Q(AUX AC DRN)                60
          + Q(VENT)                     100
          + Q(MN LEAK OFF)            297.5
          + Q(AUX LEAK OFF)              50
          + Flow to CDT              1237.5
          + Q(MAKEUP FEED)          1923.62
          + Q(DISTILLER A/E)        1717.17
          + Q(SAH)                  5576.23

Flow to DC Heater                             377623.7
           Total Condensate        179335.8
           + B1                    164108.2
           + B2                    24243.65
           - Q(SAH)                 5576.23
           + Q(FP)                  4354.29
           + Flow to FWDCT         11158.02

Flow from DC Heater to Feed Pump              377523.7
           Flow to DC Heater       377623.7
           - Q(VENT)                    100

Error, %                                      0.552535
           Calculated              375449.2
           Actual                  377523.7

				
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