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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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