# RPI_Project_Hotel2

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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
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

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)
Rated conditions are used to select the generator and turbine "size"
see T&R Bull.
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
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 =
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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