print"\t example 14.1 \t"
print"\t approximate values are mentioned in the book \t"
t1 = 300; # degreeF
t2 = 226; #degree F
bs = 700; # Btu/((hr)(ft^2)(ddegree F))
#Heat Balance
Qv = 10000 * 961; # Btu/hr
print"\tQevap is Btu/hr\t",Qv
Q3 = 10550 * 910; #Btu/hr
print"\tQ300 degreeF is Btu/hr\t",Q3
delT = t1-t2; # degree F
print"\tTemperature head = degree F\t",delT
Ud = bs * 0.865;
print"\tOverall coefficient \t",Ud
A = Qv*10/(Ud * delT); #ft^2
print"\tSurface required is ft^2\t",A#Wrong calculation in book
#end
print"\t example 14.2 \t"
print"\t approximate values are mentioned in the book \t"
wf = 50000; # lb/hr
sf = wf * 0.10; # lb/hr
tp = sf/0.50; # lb/hr
print"\tTotal product is lb/hr\t",tp
te = wf - tp;
print"\tTotal evaporation is lb/hr\t",te
cf = 1.0;
tF = 100; # degree F
T1 = 244; # degree F
T2 = 125; # degree F
U1=600; # Btu/((hr)*(ft**2)*(degree F))
U2=250; # Btu/((hr)*(ft**2)*(degree F))
U3=125; # Btu/((hr)*(ft**2)*(degree F))
T = T1-T2;
print"\tTotal temperature difference is del degree F\t",T
df = (26.70- 1.95)/3; # psi/effect
print"\tAverage pressure difference is del psi/effect \t",df
print"\t\t\t\t\t\tPressure, psia\t\t delP, psi \t Steam or vapor, degree F \t lambda, Btu/lb\t\tSteam chest, 1st effect \t 26.70 \t\t\t .... \t\t Ts = 244 \t\t ls = 949 \t\tSteam chest, 2nd effect \t 18.45 \t\t\t 8.25 \t\t t1 = 224 \t\t l1 = 961 \t\tSteam chest, 3rd effect \t 10.20(20.7 in. Hg) \t 8.25 \t\t t2 = 194 \t\t l1 = 981 \t\tVapor to condenser \t\t 1.95(26 in. Hg) \t 8.25 \t\t t2 = 125 \t\t l1 = 1022 \t"
print"\t949*Ws + 50000*(100-224) = 961*w1\t\t961*w1 + (50000 - w1)*(224-194) = 981 * w2\t\t981*w2 + (50000-w1-w2)(194-125) = 1022 * w2\t\tw1+w2+w3 = 40000\t"
print"\tSolving simultaneously\t"
w1=12400;
print"\tw1 = \t",w1
w2=13300;
print"\tw2 = \t",w2
w3=14300;
print"\tw3 = \t",w3
Wt = w1+w2+w3;
print"\tW1-3 is \t",Wt
Ws = 19100;
lms = 949;
lm1 = 961;
lm2 = 981;
lm3 = 1022;
Ts = 244;
t1 = 224;
t2 = 194;
t3 = 125;
A1 = (Ws * lms)/(U1*(Ts-t1)); #ft**2
print"\tA1 is ft**2 \t",A1
A2 = (w1*lm1)/(U2*(t1-t2)); #ft**2
print"\tA2 is ft**2 \t",A2
A3 = (w2 * lm2)/(U3*(t2-t3)); #ft**2
print"\tA3 is ft**2 \t",A3
hc = w3 * lm3; # Btu/hr, WRONG CALCULATION IN TEXT BOOK
print"\tHeat to condenser is Btu/hr\t",hc
wr = hc/(120-85); #lb/hr
print"\tWater requirement is lb/hr\t",wr
wr1 = wr/500;
print"\t= gpm \t",wr1
#end
print"\t example 14.3 \n"
print"\t approximate values are mentioned in the book \n"
#Same conditions as example 14.2
U1 = 400; #Btu/((hr)*(ft**2)*(degreeF))
U2 = 250; #Btu/((hr)*(ft**2)*(degreeF))
U3 = 175; #Btu/((hr)*(ft**2)*(degreeF))
w1 = 50000; # lb/hr From example 14.2
wt = 40000; # lb/hr From example 14.2
cf = 1; # From example 14.2
print"\t981*w2 + 50000*(100-125) = 1022*w3\n\t961*w1 + (50000 - w3)*(125-194) = 981 * w2\n\t949*Ws + (50000-w3-w2)(194-224) = 961 * w1\n\tw1+w2+w3 = 40000\n"
print"\tSolving simultaneously\n"
w1 = 15950;
w2 = 12900;
w3 = 11150;
lms = 949;
lm1 = 961;
lm2 = 981;
lm3 = 1022;
wt = w1+w2+w3;
print"\tw1-3 = \n",wt
Ws = 16950;
A1 = (Ws*lms)/(U1*20); #ft**2
print"\tA1 is ft**2\n",A1
A2 = (w1*lm1)/(U2*30); #ft**2
print"\tA2 is ft**2\n",A2
A3 = (w2*lm2)/(U3*69); #ft**2
print"\tA3 is ft**2\n",A3
Avs = (A1 + A2 + A3)/3; #ft**2
print"\tAverage surface is ft**2\n",Avs
Av1 = 3 * Avs; #ft**2
print"\n\tWith a better distribution temperatures and pressure, Average surface is %.0f ft**2\n",Av1
print"\tRecalculation\n"
Av2 = 1500; #ft**2, assume
dT1 = 28; #degreeF
A4 = (20/dT1)*A1; #ft**2
print"\tA1 is ft**2\n",A4
dT2 = 41; #degreeF
A5 = (30/dT2)*A2; #ft**2
print"\tA2 is ft**2\n",A5
dT3 = 50; #degreeF
A6 = (69/50)*A3; #ft**2
print"\tA3 is ft**2\n",A6
del1 = 119; #degreeF
print"\tTs-t3 is degreeF\n",del1
print"\t\t\t\t\tPressure, psia\t\t Steam or vapor, degreeF \t lambda, Btu/lb\n\tSteam chest, 1st effect \t 26.70 \t\t\tTs = 244 \t\t 949 \n\tSteam chest, 2nd effect \t 16.0 \t\t\t t1 = 216 \t\t 968 \n\tSteam chest, 3rd effect \t 16.4 in. Hg) \t\t t2 = 175 \t\t 992 \n\tVapor to condenser \t\t 26 in. Hg \t\t t3 = 125 \t\t l1 = 1022 \n"
w1 = 15450; #Solving again for
print"\tw1 is \n",w1
w2 = 13200;
print"\tw2 is \n",w2
w3 = 11350;
print"\tw3 is \n",w3
Ws = 16850;
print"\tWs is \n",Ws
Hc = w3 * 1022;
print"\tHeat to condenser is Btu/hr\n",Hc
wr = Hc/(120-85); #lb/hr
print"\tWater requirement lb/hr\n",wr
wr1 = wr/500;
print"\t\t\t= gpm\n",wr1
ec = wt/Ws;
print"\tEconomy, lb evaporation/lb steam \n",ec
#comparision of forward and backward feed
print"\t\t\t\tForward\t\tBackward\n\tTotal steam, lb/hr\t19100\t\t16850\n\tCooling water, gpm\t840\t\t664\n\tTotal surface, ft**2\t4800\t\t4500"
print"\texample 14.4 \n"
print"\tapproximate values are mentioned in the book \n"
from math import e
#Assumed that 37500 lb/hr of 15 psig vapor is bled from the first effect for use in thevaccum pans
print"\n\tAVERAGE EVAPORATION PER SQUARE FOOT HEATING SURFACE FOR SUGAR EVAPORATORS\n"
print"\tEffects\t\tWater evaporated(lb/(hr)*(ft**2))\n"
print"\t1\t\t14-16\n\t2\t\t6-8\n\t3\t\t5-6\n\t4\t\t4-5\n\t5\t\t3-4\n"
print"\n\tEVAPORATOR SUMMARY\n"
print"\t------------------------------------------------------------------------------------------------------------------------------\n"
print"\tItem\t\t\t\t\t\t\t\t\tEffects\nt\t\t\t\t\t----------------------------------------------------------------------------------------------\n\t\t\t\t\t1A\t\t1B\t\t2\t\t3\t\t4\t\t5\n"
print"\t------------------------------------------------------------------------------------------------------------------------------\n"
print"\t1.Steam flow, lb/hr\t\t42600\t\t38000\n\t2.Steam pressure, psi/in.Hg\t30\t\t30\t\t15\t\t5\t\t4\t\t14.5\n"
print"\t3.Steam temp,degreeF\t\t\t274\t\t274\t\t250\t\t227\t\t205\t\t181\n"
print"\t4.delT,degreeF\t\t\t23\t\t23\t\t21\t\t20\t\t20\t\t27\n\t5.Liquor temp, degreeF\t\t251\t\t251\t\t229\t\t207\t\t185\t\t164\n\t6.BPR, degreeF\t\t\t1\t\t1\t\t2\t\t2\t\t4\t\t7\n\t7.Vapor temp, degreeF\t\t250\t\t250\t\t227\t\t205\t\t181\t\t147\n\t8.Vapor pressure, pis/in.Hg\t15\t\t15\t\t5\t\t4\t\t14.5\t\t23\n\t9.Lambda, Btu/lb\t\t946\t\t946\t\t960\t\t975\t\t990\t\t1010\n\t10.Liquor in, lb/hr\t\t229000\t\t190200\t\t154000\t\t117100\t\t87800\t\t64000\n\t11.Liqour out, lb/hr\t\t190200\t\t154000\t\t117100\t\t87800\t\t64000\t\t49600\n\t12.Evaporation,lb/hr\t\t38800\t\t36200\t\t36900\t\t29300\t\t23800\t\t14400\n\t13.degreeBrix(out)\t\t\t15.7\t\t19.4\t\t25.5\t\t34.4\t\t46.5\t\t50.0\n\t14.A,ft**2\t\t\t3500\t\t3500\t\t5000\t\t5000\t\t5000\t\t3500\n\t15.UD,Btu/(hr)*(ft**2)*(degreeF)\t478\t\t425\t\t310\t\t264\t\t219\t\t138\n\t16.UD delT,Btu/(hr)*(ft**2)\t11000\t\t9780\t\t6520\t\t5270\t\t4390\t\t3740\n"#BPR values from fig 14.34a
#Saturate vapor pressure above the liquour determined from Table 7
#Saturated steam pressure in the following effect determined from Table 7
t1 = 274; #degreeF
t2 = 147; #degreeF
t = t1-t2; #degreeF
print"\tTotal temperature difference in the evaporator system = degreeF\n",t
bpr1 = 1; #degreeF
bpr2 = 2; #degreeF
bpr3 = 2; #degreeF
bpr4 = 4; #degreeF
bpr5 = 7; #degreeF
bpr = bpr1 + bpr2 + bpr3 + bpr4 + bpr5; #degreeF
print"\tThe sum of all the BPR(from effect 1B to the fifth effect inclusive) = degreeF\n",bpr
tf = t-bpr; #degreeF
print"\tTotal EFFECTIVE temperature difference = degreeF\n",tf
lbh = 229000; #lb/hr
tp1=212; #degreeF
tp2=184; #degreeF
tp3=144; #degreeF
tp4=82; #degreeF
tj1=243; #degreeF
tj2=220; #degreeF
tj3=200; #degreeF
Ud1=231;
Ud2=243;
Ud3=230;
Ud4=214;
Ud5=217;
print"\n\t\t\t\tSUGAR-JUICE HEATERS\n"
print"\tRaw-juice heaters\t\t\t\tClear=juice heaters\n\t-----------------------------------------------------------------------------------------\n"
rj1=lbh*(tp1-tp2)*(0.91); #Btu/hr
print"\t1.lbh(tp1-tp2)(0.91) = rj1 Btu/hr",lbh,tp1,tp2,rj1
rj2=lbh*(tj1-tj2)*(0.91); #Btu/hr
print"\t1.lbh(tj1-tj2)(0.91) = rj2 Btu/hr\n",lbh,tj1,tj2,rj2
print"\tVapor temp. = 227degreeF\tdelT=26.6degreeF\t\tVapor temp. = 250degreeF\tdelT=15.8degreeF\n"
print"\tud1=.\t\t\t\t\t\tUD=ud2\n",Ud1,Ud2
A1=rj1/(26.6*Ud1);#ft**2
A2=rj2/(15.8*Ud2);#ft**2
print"\tSurface,A=A1 ft**2\t\t\t\tSurface,A=A2 ft**2\n\n",A1,A2
rj3=lbh*(tp2-tp3)*(0.90);#Btu/hr
print"\t2.lbh(tp2-tp3)(0.91) = rj3 Btu/hr",lbh,tp2,tp3,rj3
rj4=lbh*(tj2-tj3)*(0.90);#Btu/hr
print"\t2.lbh(tj2-tj3)(0.91) = rj4 Btu/hr\n",lbh,tj2,tj3,rj4
print"\tVapor temp. = 205degreeF\tdelT=37.6degreeF\t\tVapor temp. = 227degreeF\tdelT=14.8degreeF\n"
print"\tUD=Ud3\t\t\t\t\t\tUD=Ud4\n",Ud3,Ud4
A3=rj3/(37.6*Ud3);#ft**2
A4=rj4/(14.8*Ud4);#ft**2
print"\tSurface,A=A3 ft**2\t\t\t\tSurface,A=A4 ft**2\n\n",A3,A4
rj5=lbh*(tp3-tp4)*(0.90);#Btu/hr
print"\t2.lbh(tp3-tp4)(0.91) = rj4 Btu/hr",lbh,tp3,tp4,rj4
print"\t(Use 2 heaters at 1300 ft**2 each plus 1\n\t\t\t\t\t\t\theater at 1300 ft**2 as spare)\n"
A5=rj5/(62.2*Ud5);#ft**2
print"\tVapor temp. = 181degreeF\tdelT=62.2degreeF\n\tSurface,A=\n",A5
print"\t(Use 3 heaters at 100 ft**2\n\teach plus 1 heater as spare)\n\n"
v1=42600;#lb/hr
tt1=251;#degreeF
print"\t\t\t\tHEAT BALANCE\n"
print"\tEffect\t\t\tBtu/hr\t\tEvaporation,l/hr\n"
print"\t----------------------------------------------------\n"
hia=v1*929*0.97;#Btu/hr
print"\t1A.Heat in steam........\n",hia
hla=lbh*(tt1-tj1)*0.91;#Btu/hr
hh=hia-hla;#Btu/hr
lb1=946;#Btu/lb
dif=hh/lb1;#lb/hr
print"\t Heating liquor.......\n\t\t\t\t%.3e\t.\n",hla,hh,dif
ltob=lbh-dif;#lb/hr
print"\t Liqour to 1B\n\t = lb/hr\n",ltob
hia1=dif*929*0.97;#Btu/hr
print"\t1B.Heat in steam........\n",hia1
hla1=ltob*(tt1-tt1)*0.91;#Btu/hr
hh1=hia1;#Btu/hr
dif1=hh1/lb1;#lb/hr
print"\t Heating liquor........\n\t\t\t\t\n",hla1,hh1,dif1
dif2=ltob-dif1;#lb/hr
print"\t Liqour to 2d \n\t effect= lb/hr\n",dif2
#Similarily the values in the table are calculated
print"\t\t\t\t\t\t\t\tLb/hr\n"
aa=179400;#lb/hr
bb=145500;#lb/hr
cc=19700;#lb/hr
dd=30600;#lb/hr
ee=17900;#lb/hr
ff=13100;#lb/hr
tto=aa+bb+cc+dd+ee+ff;#lb/hr
print"\t(a) Actual evaporation..................................\n",aa
print"\t(b) Equivalent evaporation from vapors of \n\t 1st effect used for vaccum pans......................\n",bb
print"\t(c) Equivalent evaporation from 1st effect \n\t vapors used for clarified-juice heaters..............\n",cc
print"\t(d) Equivalent evaporation from 2d effect \n\t vapors used for clarified-and raw-juice heaters......\n",dd
print"\t(e) Equivalent evaporation from 3d effect \n\t vapors used for raw-juice heaters....................\n",e
print"\t(f) Equivalent evaporation from 4th effect \n\t vapors used for raw-juice heaters....................\n",ff
print"\t -----\n"
print"\t Extrapolated evaporation............................\n",tto
esq=tto/5;#lb/hr
print"\t\tEstimated steam quantity = lb/hr\n",esq
aesq=80600;#lb/hr
err = esq-aesq;#lb/hr
print"\t\tActual steam required from final heat balance = lb/hr\n",aesq
print"\t\t\t\t\t\t\tError = lb/hr\n",err
ta=15;
Q=14575000; #Btu/hr Total hourly evaporation
Gpm=Q/(500*(t2-tp4-ta));#From equation 14.4
print"\tGallons per minute of Water required = gpm",Gpm
print"\texample 14.5\t"
print"\tapproximate values are mentioned in the book \t"
st1=280; #degF
vt6=125; #degF
odT=st1-vt6; #degF
print"\tOverall temperature difference = deg F\t",odT #corresponding to 35 psig and 26 in. Hg
import numpy
bpr=numpy.array([10, 8, 7, 6, 5, 5])
#bpr(1)=10; #degF
#bpr(2)=8; #degF
#bpr(3)=7; #degF
#bpr(4)=6; #degF
#bpr(5)=5; #degF
#bpr(6)=5; #degF
i=1;
tbpr=numpy.array([0, 0, 0, 0, 0, 0]);
tbpr[0]=bpr[0];
while (i<6):
tbpr[i]=tbpr[i-1]+bpr[i];
i=i+1;
print"\tThe estimated total BPR = degF\t",tbpr #from fig. 14.36a
edT=odT-tbpr;
print"\tEffective temperature difference = %.0f degF\t",edT
print"\t\t\t\t\tEVAPORATOR SUMMARY\t\tAll bodies will consist of 300 2 in. OD, 10 BWG tubes 24 long\t"
print"\t------------------------------------------------------------------------------------------------------------------------------\t"
print"\tItem\t\t\t\t\t\t\t\t\tEffects\t\t\t\t\t\t----------------------------------------------------------------------------------------------\t\t\t\t\t\t1A\t\t1B\t\t2\t\t3\t\t4\t\t5\t"
print"\t------------------------------------------------------------------------------------------------------------------------------\t"
print"\t1.Steam flow, lb/hr\t\t20000\t\t2.Steam pressure, psi/in.Hg\t35\t\t14.5\t\t4\t\t7\t\t16.5\t\t22\t\t3.Steam temp,degF\t\t\t280\t\t249\t\t224\t\t199\t\t174\t\t151\t\t4.delT,degF\t\t\t21\t\t17\t\t18\t\t19\t\t18\t\t21\t\t5.Liquor temp, degF\t\t259\t\t232\t\t206\t\t180\t\t156\t\t130\t\t6.BPR, degF\t\t\t10\t\t8\t\t7\t\t6\t\t5\t\t5\t\t7.Vapor temp, degF\t\t259\t\t232\t\t206\t\t180\t\t156\t\t130\t\t8.Vapor pressure, pis/in.Hg\t14.5\t\t4\t\t7\t\t6\t\t5\t\t5\t\t9.Lambda, Btu/lb\t\t946\t\t962\t\t978\t\t994\t\t1008\t\t1022\t\t10.Liquor in, lb/hr\t\t73400\t\t88300\t\t101000\t\t113000\t\t72000\t\t72000\t\t11.Liqour out, lb/hr\t\t56200\t\t73400\t\t88300\t\t101100\t\t58300\t\t54700\t\t12.Evaporation,lb/hr\t\t17200\t\t14900\t\t12800\t\t11900\t\t13700\t\t17300\t\t13.Total solids, \t\t38.9\t\t29.8\t\t24.7\t\t21.6\t\t18.7\t\t20.0\t\t14.A,ft^2\t\t\t3250\t\t3250\t\t3250\t\t3250\t\t3250\t\t3250\t\t15.UD,Btu/(hr)*(ft^2)*(degF)\t262\t\t295\t\t252\t\t251\t\t221\t\t221\t\t16.UD delT,Btu/(hr)*(ft^2)\t5510\t\t5000\t\t4530t\t\t4770\t\t3980\t\t4650\t"#BPR values from fig 14.36a
#Specific-heat data are given in Fig. 14.36b
ev=numpy.array([17200, 14900, 12800, 11900, 13700, 17300]);
#ev(1)=17200; #lb/hr
#ev(2)=14900; #lb/hr
#ev(3)=12800; #lb/hr
#ev(4)=11900; #lb/hr
#ev(5)=13700; #lb/hr
#ev(6)=17300; #lb/hr
i=1;
tev =numpy.array([0, 0, 0, 0, 0, 0])
tev[0]=ev[0];
while (i<5):
tev[i]= tev[i-1]+ev[i];
i=i+1;
print"\t\tTotal amount of water evaporated = \t lb/hr\t",tev
ttev=tev/6;#lb/hr
print"\tTheoretical amount of steam for a six-effect evaporator = \t lb/hr\t",ttev
tev2=tev/(6*0.75); #lb/hr . order of 75 percent of theoretical
print"\tSteam used for trail balance = \t lb/hr\t",tev2
lq=(tev/6);
lq=lq+(lq*0.15);
print"\tEstimate of the amount of evaporation in the first effect = \t lb/hr\t",lq
lout6=54000;#lb/hr
lq2=lout6+lq+2200;#lb/hr
print"\tEstimated discharge from second effect = \t lb/hr\t",lq2
print"\t\t\t\t\tHEAT BALANCE\t"
cw = 17750000/(500*(125-15-60)); #gpm, values from table 14.6
print"\t\tCooling water at 60 degreeF = \t gpm\t",cw
print"\t--------------------------------------------------------\t"
print"\tEffect\t\t\tBtu/hr\t\tEvaporation,l/hr\t"
print"\t--------------------------------------------------------\t"
sf=20000;#lb/hr
lqi=73400;#lb/hr
the=90000;
lqi2=88300
lt1=259;#degreeF
lt2=232;#degreeF
lt3=206;#degreeF
ecn=90000./20000.;
ev=17200;#lb/hr
his=sf*924*0.97;#Btu/hr
print"\t1.a.Heat in steam \t\t",his
hl=lqi*(lt1-lt2)*0.82;#Btu/hr
print"\t b.Heating liquor \t\t",hl
hh=his-hl;
ev1=(hh)/946;#lb/hr
print"\t c.Evaporation\t\t\t\t\t\t",ev1
dif=lqi-ev1;
tft=(dif)*(lt1-209)*0.78;
print"\t d.To flash tank\t",tft
ev2=tft/978;#lb/hr
print"\t\t\t\t",ev2
print"\t e.Flashed vapor=\t\t",ev2
p=dif-ev2;
print"\t f.product \t",p
print"\t\t2.a.Heat in 1st vapors\t\t",hh
hl2=lqi2*(lt2-lt3)*0.85;
print"\t b.Heating liqour\t\t",hl2
ev3=(hh-hl2)/962;
print"\t c.Evaporation=",ev3
print"\t\t\t\t\t",ev3
lto1=lqi2-ev3;
print"\t d.Liquor to 1b=\t\t",lto1
print"total hourly evapouration lb :",the
print"economy is lb/lb :",ecn
#end
print"\texample 14.6\n"
print"\tapproximate values are mentioned in the book \n"
st1=274; #degreeF
vt6=115; #degreeF
odT=st1-vt6; #degreeF
print"\tTotal temperature difference = degreeF\n",odT #corresponding to 35 psig
eb1=77;#degreeF, From fig.14.38
eb2=26;#degreeF, From fig.14.38
etd=odT-(eb1+eb2);#degreeF
print"\tThe effective temperature difference is degreeF\n",etd
print"\n\t\t\tCAUSTIC EVAPORATOR MATERIAL BALANCE\n"
#Basis: 1 ton/hr NaOH
print"\tCell liquour at 120degreeF \t\tWash at 80degreeF\n"
print"\t---------------------------------------------\n"
l1=2000;#Lb
l2=3800;#Lb
l3=17050;#Lb
lq=l1+l2+l3;#Lb
w1=340;#Lb
w2=1020;#Lb
w=w1+w2;#Lb
print"\t8.75 prcnt NaOH = l1\n\t16.6 prcnt NaCl = l2\t\t25 prcnt NaCl = w2\n",l1,l2,w1
print"\t74.65 prcnt H20 = l3\t\t75 prcnt H20 = w2\n",l3,w2
print"\tTotal cell liquor = lq\tTotoal wash = w\n",lq,w
print"\n\t-------------------------------------------------------------------------\n"
print"\t\t\t\tNaOH\t\tNaCl\t\tH20,Lb\tTotal,Lb\n\t\t\t\tprcnt\tLb\tprcnt\tLb\n"
print"\t-------------------------------------------------------------------------\n"
print"\tOverall operation:\n\t Cell liquor.......... 8.75\t",l1,l2,l3,lq
print"\t Wash................. ....\t....\t25.00\t",w1,w2,w
wl1=l2+w1;#Lb
wl2=l3+w2;#Lb
wlt=lq+w;
print"\t Total in............. ....\t",l1,wl1,wl2,wlt
prn=110;#Lb
prh=1890;#Lb
prt=4000;#Lb
print"\t Product.............. 50.00\t",l1,prn,prh,prt
r1=wl1-prn;#Lb
r2=wl2-prh;#Lb
r3=wlt-prt;#Lb
gain=3200;#gpm
print"\t Removed.............. ....\t....\t....\t\t%.0f\t%.0f\n",r1,r2,r3
#Rest of the table is calculated similarily
print"\n\t\t\t\t\tCAUSTIC EVAPORATOR SUMMARY\n"
print"\t------------------------------------------------------------------------------------\n"
print"\tItem\t\t\t\t\tEffects\nt\t\t\t\t\t--------------------\t\tFlash Tank\n\t\t\t\t\t\I\t\tII\n"
print"\t------------------------------------------------------------------------------------\n"
print"\t1.Steam pressure, psi/in.Hg\t30\n\t2.Steam temperature,degreeF\t\t274\t\t169\n\t3.delT,degreeF\t\t\t28\t\t28\n\t4.Liquor temperature, degreeF\t246\t\t141\t\t192\n\t5.BPR, degreeF\t\t\t77\t\t26\t\t77\n\t6.Vapor temperature, degreeF\t\t169\t\t115\t\t115\n\t7.Lambda, Btu/lb\t\t997\t\t1027\t\t1027\n\t8.Feed, lb/hr\t\t\t22788\t\t50602\t\t13367\n\t9.Product, lb/hr\t\t13367\t\t40352\t\t12813\n\t10.Evaporation,lb/hr\t\t9421\t\t10250\t\t554\n\t11.Heat flow, Btu/hr\t\t11890000\t11020000\n\t12.UD,Btu/((hr)*(ft**2)*(degreeF))\t700\n\t13.A,ft**2\t\t\t683\t\t683\n\t14.Tubes, OD, in. and BWG\t1,16\t\t1,16\n\t15.Tube length, ft\t\t7\t\t7\n\t16.No. tubes\t\t\t432\t\t432\n\t17.Circulating pump. gpm\t3200 at 20 ft\t3200 at 20ft\t167 at 45 ft\n\t18.Apparent efficiency, prcnt\t54\t\t64\n\t18.BHP\t\t\t\t38\t\t35\t\t8.2\n\t20.Motor,hp\t\t\t40\t\t40\t\t10.0\n"
print"\t------------------------------------------------------------------------------------\n"
V=8;
s=1.5;
G=V*s*62.5*3600;#lb/((hr)*(ft**2))
print"\tG = V(s*62.5*3600) = lb/((hr)*(ft**2))\n",G
UD=700;#Btu/((hr)*(ft**2)*(degreeF))
#Combining with a steam film coefficient of approximately 1500
print"\tUC or UD = Btu/((hr)*(ft**2)*(degreeF))\n",UD
print"\n\t-------------------------------------------------------------------------------------"
print"\n\ttx,degreeF\tw,lb/hr\t\tdelT\tUC\tA,ft**2\tat,flow area\tGcalc\t\tUcalc\n\t\t\t\t\t\t\tper pass, ft**2\n"
print"\t-------------------------------------------------------------------------------------\n"
print"\t251\t2970000\t\t25.4\t700\t670\t0.87\t\t3420000\n\t252\t2480000\t\t25.0\t700\t680\t0.88\t\t2820000\n\t252.5\t2290000\t\t24.7\t700\t685\t0.89\t\t2570000\t\t700\n\t253\t2120000\t\t24.5\t700\t695\t0.90\t\t2520000\n"
print"\tThee gain per minute is gpm\n",gain
print"\n\t\t\t\tCAUSTIC EVAPORATION HEAT BALANCE\n"
print"\t\t\t\t(Basis = 1ton/hr NaOH)\n"
print"\t-------------------------------------------------------------------------------------\n"
print"\t\tEFFECT\t\t\tBtu/hr\t\tEvaopration, lb/hr\n"
hi=10500*930*0.974;#Btu/hr
hl=18230*(246-150)*0.83;#Btu/hr
rh=hi-hl;#Btu/hr
hc=300000;#Btu/hr
hv=rh-hc;#Btu/hr
evv=hv/997;#lb/hr
print"\t1.a.Heat in steam\t\thi\n\t b.Heating liquor\t\thl\n\t c.Resultant heat\t\trhf\n\t d.Heat of concentrate\t\thc\n\t e.Heat of vapors\t\thv\t%.0f\n",hi,hl,rh,hc,hv,evv
s1=1.35;
G1=V*s1*62.5*3600;#lb/((hr)*(ft**2))
print"\n\tG = V(s*62.5*3600) = lb/((hr)*(ft**2))\n",G1
UD1=700;#Btu/((hr)*(ft**2)*(degreeF))
#Using thermal characteristics for this solution
print"\tUD = Btu/((hr)*(ft**2)*(degreeF))\n",UD1
#As for effect I:
print"\n\t-------------------------------------------------------------------------------------"
print"\n\ttx,degreeF\tw,lb/hr\t\tdelT\tUC\tA,ft**2\tat,flow area\tGcalc\t\tUcalc\n\t\t\t\t\t\t\tper pass, ft**2\n"
print"\t-------------------------------------------------------------------------------------\n"
print"\t146\t2400000\t\t25.4\t700\t620\t0.80\t\t2790000\t\t700\n\t146.5\t2160000\t\t25.2\t700\t683\t0.89\t\t2430000\n"
#end
print"\texample 14.7\n"
print"\tapproximate values are mentioned in the book \n"
M2=14300;#From fig.14.43 and heat balance above
M1=32200-14300;#From fig.14.43 and heat balance above
print"\tM1 = lb\n",M1
print"\n\t\t\t\tEVAPORATOR SUMMARY\n"
print"\t------------------------------------------------------------------------------------------------------------------------------\n"
print"\tEffects\t\t\t\t\tStraight triple effect\t\t\t\tThermocompression\nt\t\t\t\t\t----------------------------------------------------------------------------------------------\n\t\t\t\t\t1\t\t2\t\t3\t\t1\t\t2\t\t3\n"
print"\t------------------------------------------------------------------------------------------------------------------------------\n"
print"\tSteam flow, lb/hr\t\t22400\t\t\t\t\t\t17900\n\tSteam pressure, psi in.Hg\t20\t\t9\t\t2\t\t20\t\t9\t\t2\n\tSteam temp,degreeF\t\t\t258\t\t237\t\t217\t\t258\t\t237\t\t217\n\ttdelT,degreeF\t\t\t20\t\t18\t\t22\t\t20\t\t18\t\t22\n\tLiquor temp, degreeF\t\t\t238\t\t219\t\t195\t\t238\t\t219\t\t195\n\tBPR, degreeF\t\t\t\t1\t\t2\t\t3\t\t1\t\t2\t\t3\n\tVapor temp, degreeF\t\t\t237\t\t217\t\t192\t\t237\t\t215\t\t192\n\tVapor pressure, pis/in.Hg\t9\t\t2\t\t10\t\t9\t\t2\t\t10\n\tLambda, Btu/lb\t\t\t954\t\t965\t\t983\t\t954\t\t965\t\t983\n\tLiquor in, lb/hr\t\t100000\t\t79400\t\t56900\t\t109000\t\t70000\t\t52400\n\tLiqour out, lb/hr\t\t79400\t\t56900\t\t33300\t\t70000\t\t52400\t\t33300\n\tEvaporation,lb/hr\t\t20600\t\t22500\t\t23500\t\t30000\t\t17600\t\t19100\n\tdegreeBrix(out)\t\t\t\t\t\t\t\t\t\t\t\t\t30\n\tCondenser water, gpm\t\t\t\t455\t\t\t\t\t\t365\n"
print"\n\t\t\t\tHEAT BALANCE-STRAIGHT TRIPLE EFFECT\n\t\t\t\tCondenser water = 455 gpm\n"
print"\t--------------------------------------------------------\n"
print"\tEffect\t\t\tBtu/hr\t\tEvaporation,l/hr\n"
print"\t--------------------------------------------------------\n"
sf=22400;#lb/hr
lc=100000;#lb/hr
t1=238;#degreeF
t2=230;#degreeF
te=30000+17600+19100;
his=sf*940*0.97;#Btu/hr
hlq=lc*(t1-t2)*0.92;#Btu/hr
hd=his-hlq;#Btu/hr
eva=(hd)/954;#lb/hr
l2d=lc-eva;
print"\t1.a.Heat in steam\t\n\t" ,his
print"b.Heating liquor\t\n\t",hlq
print"c.Evaporation\t\t Evaporation/954\t\n\t",hd,eva
print"d.Liquor to 2d = ",l2d
print"total evapouriation",te
#end