# Chapter 14 : Evapouration¶

## Example 14.1 pgno:383¶

In [2]:
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
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

	 example 14.1
approximate values are mentioned in the book
Qevap is  Btu/hr	9610000
Q300 degreeF is  Btu/hr	9600500
Temperature head =  degree F	74
Overall coefficient 	605.5
Surface required is  ft^2	2144.75416788


## Example 14.2 pgno:412¶

In [3]:
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

	 example 14.2
approximate values are mentioned in the book
Total product is lb/hr	10000.0
Total evaporation is lb/hr	40000.0
Total temperature difference is del degree F	119
Average pressure difference is del psi/effect 	8.25
Pressure, psia		 delP, psi 	 Steam or vapor, degree F 	 lambda, Btu/lb		Steam chest, 1st effect 	 26.70 			 .... 		 Ts = 244 		 ls = 949 		Steam chest, 2nd effect 	 18.45 			 8.25 		 t1 = 224 		 l1 = 961 		Steam chest, 3rd effect 	 10.20(20.7 in. Hg) 	 8.25 		 t2 = 194 		 l1 = 981 		Vapor to condenser 		 1.95(26 in. Hg) 	 8.25 		 t2 = 125 		 l1 = 1022
949*Ws + 50000*(100-224) = 961*w1		961*w1 + (50000 - w1)*(224-194) = 981 * w2		981*w2 + (50000-w1-w2)(194-125) = 1022 * w2		w1+w2+w3 = 40000
Solving simultaneously
w1 = 	12400
w2 = 	13300
w3 = 	14300
W1-3 is 	40000
A1 is ft**2 	1510
A2 is ft**2 	1588
A3 is ft**2 	1512
Heat to condenser is Btu/hr	14614600
Water requirement is lb/hr	417560
= gpm 	835


## Example 14.3 pgno:414¶

In [4]:
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"

	 example 14.3

approximate values are mentioned in the book

981*w2 + 50000*(100-125) = 1022*w3
961*w1 + (50000 - w3)*(125-194) = 981 * w2
949*Ws + (50000-w3-w2)(194-224) = 961 * w1
w1+w2+w3 = 40000

Solving simultaneously

w1-3 =
40000
A1 is  ft**2
2010
A2 is  ft**2
2043
A3 is  ft**2
1048
Average surface is  ft**2
1700

With a better distribution temperatures and pressure, Average surface is  %.0f ft**2
5100
Recalculation

A1 is  ft**2
0
A2 is  ft**2
0
A3 is  ft**2
1048
Ts-t3 is  degreeF
119
Pressure, psia		 Steam or vapor, degreeF 	 lambda, Btu/lb
Steam chest, 1st effect 	 26.70 			Ts = 244 		 949
Steam chest, 2nd effect 	 16.0 			 t1 = 216 		 968
Steam chest, 3rd effect 	 16.4 in. Hg) 		 t2 = 175 		 992
Vapor to condenser 		 26 in. Hg 		 t3 = 125 		 l1 = 1022

w1 is
15450
w2 is
13200
w3 is
11350
Ws is
16850
Heat to condenser is  Btu/hr
11599700
Water requirement  lb/hr
331420
= gpm
662
Economy, lb evaporation/lb steam
2
Forward		Backward
Total steam, lb/hr	19100		16850
Cooling water, gpm	840		664
Total surface, ft**2	4800		4500


## Example 14.4 pgno:418¶

In [5]:
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

	example 14.4

approximate values are mentioned in the book

AVERAGE EVAPORATION PER SQUARE FOOT HEATING SURFACE FOR SUGAR EVAPORATORS

Effects		Water evaporated(lb/(hr)*(ft**2))

1		14-16
2		6-8
3		5-6
4		4-5
5		3-4

EVAPORATOR SUMMARY

------------------------------------------------------------------------------------------------------------------------------

Item									Effects
t					----------------------------------------------------------------------------------------------
1A		1B		2		3		4		5

------------------------------------------------------------------------------------------------------------------------------

1.Steam  flow, lb/hr		42600		38000
2.Steam pressure, psi/in.Hg	30		30		15		5		4		14.5

3.Steam temp,degreeF			274		274		250		227		205		181

4.delT,degreeF			23		23		21		20		20		27
5.Liquor temp, degreeF		251		251		229		207		185		164
6.BPR, degreeF			1		1		2		2		4		7
7.Vapor temp, degreeF		250		250		227		205		181		147
8.Vapor pressure, pis/in.Hg	15		15		5		4		14.5		23
9.Lambda, Btu/lb		946		946		960		975		990		1010
10.Liquor in, lb/hr		229000		190200		154000		117100		87800		64000
11.Liqour out, lb/hr		190200		154000		117100		87800		64000		49600
12.Evaporation,lb/hr		38800		36200		36900		29300		23800		14400
13.degreeBrix(out)			15.7		19.4		25.5		34.4		46.5		50.0
14.A,ft**2			3500		3500		5000		5000		5000		3500
15.UD,Btu/(hr)*(ft**2)*(degreeF)	478		425		310		264		219		138
16.UD delT,Btu/(hr)*(ft**2)	11000		9780		6520		5270		4390		3740

Total temperature difference in the evaporator system =  degreeF
127
The sum of all the BPR(from effect 1B to the fifth effect inclusive) =  degreeF
16
Total EFFECTIVE  temperature difference = degreeF
111

SUGAR-JUICE HEATERS

Raw-juice heaters				Clear=juice heaters
-----------------------------------------------------------------------------------------

1.lbh(tp1-tp2)(0.91) =  rj1 Btu/hr 229000 212 184 5834920.0
1.lbh(tj1-tj2)(0.91) = rj2 Btu/hr
229000 243 220 4792970.0
Vapor temp. = 227degreeF	delT=26.6degreeF		Vapor temp. = 250degreeF	delT=15.8degreeF

ud1=.						UD=ud2
231 243
Surface,A=A1 ft**2				Surface,A=A2 ft**2

949.601275917 1248.36432776
2.lbh(tp2-tp3)(0.91) = rj3 Btu/hr 229000 184 144 8244000.0
2.lbh(tj2-tj3)(0.91) = rj4 Btu/hr
229000 220 200 4122000.0
Vapor temp. = 205degreeF	delT=37.6degreeF		Vapor temp. = 227degreeF	delT=14.8degreeF

UD=Ud3						UD=Ud4
230 214
Surface,A=A3 ft**2				Surface,A=A4 ft**2

953.2839963 1301.46501642
2.lbh(tp3-tp4)(0.91) = rj4 Btu/hr 229000 144 82 4122000.0
(Use 2 heaters at 1300 ft**2 each plus 1
heater at 1300 ft**2 as spare)

Vapor temp. = 181degreeF	delT=62.2degreeF
Surface,A=
946.715663757
(Use 3 heaters at 100 ft**2
each plus 1 heater as spare)

HEAT BALANCE

Effect			Btu/hr		Evaporation,l/hr

----------------------------------------------------

1A.Heat in steam........
38388138.0
Heating liquor.......
%.3e	.
1667120.0 36721018.0 38817.1437632
Liqour to 1B
=  lb/hr
190182.856237
1B.Heat in steam........
34979292.7593
Heating liquor........

0.0 34979292.7593 36975.9965744
Liqour to 2d
effect= lb/hr
153206.859662
Lb/hr

(a) Actual evaporation..................................
179400
(b) Equivalent evaporation from vapors of
1st effect used for vaccum pans......................
145500
(c) Equivalent evaporation from 1st effect
vapors used for clarified-juice heaters..............
19700
(d) Equivalent evaporation from 2d effect
vapors used for clarified-and raw-juice heaters......
30600
(e) Equivalent evaporation from 3d effect
vapors used for raw-juice heaters....................
2.71828182846
(f) Equivalent evaporation from 4th effect
vapors used for raw-juice heaters....................
13100
-----

Extrapolated evaporation............................
406200
Estimated steam quantity =  lb/hr
81240
Actual steam required from final heat balance =  lb/hr
80600
Error =  lb/hr
640
Gallons per minute of Water required =  gpm 583


## Example 14.5 pgno:427¶

In [9]:
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

	example 14.5
approximate values are mentioned in the book
Overall temperature difference =  deg F	155
The estimated total BPR =  degF	[10 18 25 31 36 41]
Effective temperature difference = %.0f degF	[145 137 130 124 119 114]
EVAPORATOR SUMMARY		All bodies will consist of 300 2 in. OD, 10 BWG tubes 24 long
------------------------------------------------------------------------------------------------------------------------------
Item									Effects						----------------------------------------------------------------------------------------------						1A		1B		2		3		4		5
------------------------------------------------------------------------------------------------------------------------------
1.Steam  flow, lb/hr		20000		2.Steam pressure, psi/in.Hg	35		14.5		4		7		16.5		22		3.Steam temp,degF			280		249		224		199		174		151		4.delT,degF			21		17		18		19		18		21		5.Liquor temp, degF		259		232		206		180		156		130		6.BPR, degF			10		8		7		6		5		5		7.Vapor temp, degF		259		232		206		180		156		130		8.Vapor pressure, pis/in.Hg	14.5		4		7		6		5		5		9.Lambda, Btu/lb		946		962		978		994		1008		1022		10.Liquor in, lb/hr		73400		88300		101000		113000		72000		72000		11.Liqour out, lb/hr		56200		73400		88300		101100		58300		54700		12.Evaporation,lb/hr		17200		14900		12800		11900		13700		17300		13.Total solids, 		38.9		29.8		24.7		21.6		18.7		20.0		14.A,ft^2			3250		3250		3250		3250		3250		3250		15.UD,Btu/(hr)*(ft^2)*(degF)	262		295		252		251		221		221		16.UD delT,Btu/(hr)*(ft^2)	5510		5000		4530t		4770		3980		4650
Total amount of water evaporated = 	 lb/hr	[17200 32100 44900 56800 70500     0]
Theoretical amount of steam for a six-effect evaporator = 	 lb/hr	[ 2866  5350  7483  9466 11750     0]
Steam used for trail balance = 	 lb/hr	[  3822.22222222   7133.33333333   9977.77777778  12622.22222222
15666.66666667      0.        ]
Estimate of the amount of evaporation in the first effect = 	 lb/hr	[  3295.9    6152.5    8605.45  10885.9   13512.5       0.  ]
Estimated discharge from second effect = 	 lb/hr	[ 59495.9   62352.5   64805.45  67085.9   69712.5   56200.  ]
HEAT BALANCE
Cooling water at 60 degreeF = 	 gpm	710
--------------------------------------------------------
Effect			Btu/hr		Evaporation,l/hr
--------------------------------------------------------
1.a.Heat in steam 		17925600.0
b.Heating liquor 		1625076.0
c.Evaporation						17230.9978858
d.To flash tank	2190591.08245
2239.86818247
e.Flashed vapor=		2239.86818247
f.product 	53929.1339317
2.a.Heat in 1st vapors		16300524.0
b.Heating liqour		1951430.0
c.Evaporation= 14915.8981289
14915.8981289
d.Liquor to 1b=		73384.1018711
total hourly evapouration lb : 90000
economy is lb/lb : 4.5


## Example 14.6 pgno:437¶

In [10]:
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

	example 14.6

approximate values are mentioned in the book

Total temperature difference =  degreeF
159
The effective temperature difference is  degreeF
56

CAUSTIC EVAPORATOR MATERIAL BALANCE

Cell liquour at 120degreeF 		Wash at 80degreeF

---------------------------------------------

8.75 prcnt NaOH = l1
16.6 prcnt NaCl = l2		25 prcnt NaCl = w2
2000 3800 340
74.65 prcnt H20 = l3		75 prcnt H20 = w2
17050 1020
Total cell liquor = lq	Totoal wash = w
22850 1360

-------------------------------------------------------------------------

NaOH		NaCl		H20,Lb	Total,Lb
prcnt	Lb	prcnt	Lb

-------------------------------------------------------------------------

Overall operation:
Cell liquor.......... 8.75	2000 3800 17050 22850
Wash................. ....	....	25.00	340 1020 1360
Total in............. ....	2000 4140 18070 24210
Product.............. 50.00	2000 110 1890 4000
Removed.............. ....	....	....		%.0f	%.0f
4030 16180 20210

CAUSTIC EVAPORATOR SUMMARY

------------------------------------------------------------------------------------

Item					Effects
t					--------------------		Flash Tank
\I		II

------------------------------------------------------------------------------------

1.Steam pressure, psi/in.Hg	30
2.Steam temperature,degreeF		274		169
3.delT,degreeF			28		28
4.Liquor temperature, degreeF	246		141		192
5.BPR, degreeF			77		26		77
6.Vapor temperature, degreeF		169		115		115
7.Lambda, Btu/lb		997		1027		1027
8.Feed, lb/hr			22788		50602		13367
9.Product, lb/hr		13367		40352		12813
10.Evaporation,lb/hr		9421		10250		554
11.Heat flow, Btu/hr		11890000	11020000
12.UD,Btu/((hr)*(ft**2)*(degreeF))	700
13.A,ft**2			683		683
14.Tubes, OD, in. and BWG	1,16		1,16
15.Tube length, ft		7		7
16.No. tubes			432		432
17.Circulating pump. gpm	3200 at 20 ft	3200 at 20ft	167 at 45 ft
18.Apparent efficiency, prcnt	54		64
18.BHP				38		35		8.2
20.Motor,hp			40		40		10.0

------------------------------------------------------------------------------------

G = V(s*62.5*3600) = lb/((hr)*(ft**2))
2700000.0
UC or UD =  Btu/((hr)*(ft**2)*(degreeF))
700

-------------------------------------------------------------------------------------

tx,degreeF	w,lb/hr		delT	UC	A,ft**2	at,flow area	Gcalc		Ucalc
per pass, ft**2

-------------------------------------------------------------------------------------

251	2970000		25.4	700	670	0.87		3420000
252	2480000		25.0	700	680	0.88		2820000
252.5	2290000		24.7	700	685	0.89		2570000		700
253	2120000		24.5	700	695	0.90		2520000

Thee gain per minute is gpm
3200

CAUSTIC EVAPORATION HEAT BALANCE

(Basis = 1ton/hr NaOH)

-------------------------------------------------------------------------------------

EFFECT			Btu/hr		Evaopration, lb/hr

1.a.Heat in steam		hi
b.Heating liquor		hl
c.Resultant heat		rhf
d.Heat of concentrate		hc
e.Heat of vapors		hv	%.0f
9511110.0 1452566.4 8058543.6 300000 7758543.6 7781.8892678

G = V(s*62.5*3600) =  lb/((hr)*(ft**2))
2430000.0
UD =  Btu/((hr)*(ft**2)*(degreeF))
700

-------------------------------------------------------------------------------------

tx,degreeF	w,lb/hr		delT	UC	A,ft**2	at,flow area	Gcalc		Ucalc
per pass, ft**2

-------------------------------------------------------------------------------------

146	2400000		25.4	700	620	0.80		2790000		700
146.5	2160000		25.2	700	683	0.89		2430000



## Example 14.7 pgno:447¶

In [23]:
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

	example 14.7

approximate values are mentioned in the book

M1 =  lb
17900

EVAPORATOR SUMMARY

------------------------------------------------------------------------------------------------------------------------------

Effects					Straight triple effect				Thermocompression
t					----------------------------------------------------------------------------------------------
1		2		3		1		2		3

------------------------------------------------------------------------------------------------------------------------------

Steam  flow, lb/hr		22400						17900
Steam pressure, psi in.Hg	20		9		2		20		9		2
Steam temp,degreeF			258		237		217		258		237		217
tdelT,degreeF			20		18		22		20		18		22
Liquor temp, degreeF			238		219		195		238		219		195
BPR, degreeF				1		2		3		1		2		3
Vapor temp, degreeF			237		217		192		237		215		192
Vapor pressure, pis/in.Hg	9		2		10		9		2		10
Lambda, Btu/lb			954		965		983		954		965		983
Liquor in, lb/hr		100000		79400		56900		109000		70000		52400
Liqour out, lb/hr		79400		56900		33300		70000		52400		33300
Evaporation,lb/hr		20600		22500		23500		30000		17600		19100
degreeBrix(out)													30
Condenser water, gpm				455						365

HEAT BALANCE-STRAIGHT TRIPLE EFFECT
Condenser water = 455 gpm

--------------------------------------------------------

Effect			Btu/hr		Evaporation,l/hr

--------------------------------------------------------

1.a.Heat in steam
20424320.0
b.Heating liquor
736000.0
c.Evaporation		 Evaporation/954
19688320.0 20637.6519916
d.Liquor to 2d =  79362.3480084
total evapouriation 66700