# Chapter 9 : Evaporation and Evaporators¶

## Example 9.1 Page No : 391¶

In :
# Variables
ro = 1020.       			# kg/m**3, density of feed
sf = 4.1        			#kj/kg C,specific heat of the feed
sp = 3.9        			#kj/kg C,specific heat of the product
ci = 5.       			#initial concentration
cw = 100.-ci    			#conc. of  water
cf = 40.        			#final conc.
rate = 100.      			#m**3/day, rate of conc. of aq. solution
ft = 25.         			# C, feed temp.

#calculation and results
#materiel balance
Wf = rate*ro    			#Kg. feed entering
Ms = ro*ci      			#Kg mass of solute
Mw = ro*cw      			#kg,mass of water
fc = cw/ci      			#kg,feed concentration
pc = (100-cf)/cf 			# kg,product concentration
wlwp = Ms*pc      			#Kg, water leaving with the product
Ws = Mw-wlwp     			#kg,water evaporated
Wp = wlwp+Ms     			# kg, product
#energy balance
rt = 0.            			#C reference temp.
ef = sf*(ft-rt)   			#kj/kg,enthlpy of the feed
#case i
Tp = 100.          			#temp. of the product (because the solute has a 'high molecular wt' the boiling pt elevation is neglected)
ip = sp*(Tp-rt)   			#kj/kg, enthalpy of the product
iv = 2680.         			#kj/kg, enthalpy of the vapour generated at 100 C and 1 atm pr. from the steam table
#refer to fig. 9.23
#from energy balance eq. (Wf*if+qs = Wv*iv+Wp*ip)
qs = Ws*iv+Wp*ip-Wf-ef  			#Wv = Ws
print "The rate at which heat must be supplied at 1 atm pressure is %1.3e kj/ day"%(qs)

#case ii
#650 mm Hg vaccum = 110 mmHg pressure
bp = 53.5         			#C, boiling point of water
ip2 = sp*(bp-rt)  			#kj/kg, enthalpy of the product
es = 2604.         			#kj/kg, enthalpy of the saturated steam (from steam table)
#from energy balnce eq.
qs2 = Wp*ip+Ws*es-Wf-ef
print "The rate at which heat must be supplied at a pressure of 600 mm Hg is %1.3e kj/day "%(qs2)

# note : rounding off error.

The rate at which heat must be supplied at 1 atm pressure is 2.441e+08 kj/ day
The rate at which heat must be supplied at a pressure of 600 mm Hg is 2.373e+08 kj/day


## Example 9.2 Page No : 393¶

In :
import math

# Variables
ci = 10.           			#%,initial concentration
cf = 40.           			#%, final conc
Wf = 2000.        			#kg/h, feed rate
ft = 30.           			#C feed temp.
rp = 0.33         			#kg/cm**2, reduced pressure
bt1 = 75.           			#C,boiling point temp.
sst = 115.          			#C, saturated steam temp.
l = 1.5             			# m,height of calandria
sh = 0.946        			#kcal/kg C, specific heat of liquir
lh = 556.5        			#kcal/kg latent heat of steam
bt2 = 345.         			#K, boiling point of water
h = 2150.          			#kcal/h m**2 C, overall heat transfer coefficient
si = 2000.*(ci/100)  			#kg/h, solids in
wi = 1800.          			#kg/h,wate in

# Calculations
Wp = si/(cf/100)   			#kg/h, product out
Wv = Wf-Wp         			#evaporation rate
ef = sh*(ft-bt1)
ip = 0
lamda_s = 529.5   			#kcal/kg, lamda_s = is-il
bpe = (273+bt1)-345  			#boiling point elevation.
#from eergy balance eq.
Ws = (Wp*ip+Wv*lh-Wf*ef)/lamda_s
q = Ws*lamda_s       			#kcal/h,rate of heat transfer
A = q/(h*(sst-bt1))   			# m**2
di = 0.0221            			#m,inside diameter
At = math.pi*l*di        			#m**2, area of a math.single tube
N = A/At             			#no. of tubes

# Results
print "The steam required is %.0f kg/h"%(Ws)
print "No. of tube are %d"%(N)

The steam required is 1737 kg/h
No. of tube are 102


## Example 9.3 Page No : 393¶

In :
# Variables
Wf = 2000.        			#kg/h, feed rate
ci = 8.          			#% initial conc.
cf = 40.          			#% final conc.
ft = 30.         			#C, feed temp.
vp = 660.        			#mm Hg, vaccum pressure
ssp = 8.         			# bar absolute, saturated steam pr.

#calculation
sr = Wf*(ci/100)  			#kg/h, solid rate
Wp = sr/(cf/100)   			#kg/h,concentrated product rate
ap = 760-vp        			#mm Hg, absolute pressure in the evaporator
bt = 325.          			#K,boiling temp. of water
l_s = 2380.        			#kj/kg, latent heat
R = 8.303         			#gas consmath.tant
w = 40.             			#g,mass of solute
M = 18.           			#g,molecular wt of solvent
W = 60.           			#g,mass of the solvent
m = 2000.         			#g,molecular wt of solute
dtb = (R*bt**2*w*M)/(l_s*W*m)     			#C, boiling point elevation
bp = bt+dtb      			#k,boiling point of 40% solution
dt = 70.        			#C, from given data flux becomes maximum at a temp. drop  = 70 C
st = bp+dt     			#K,saturation temp. of steam in the steam chest
Sp = 2.15      			# bar, from steam table, saturation lr. of steam at this temp.

sh = 4.2       			#kj/kg C, specific heat of product
rt = 0.         			#C reference teml.
ef = sh*(ft-rt)  			# kj/kg, enthalpy of the feed
ip = sh*(54-rt)  			#kj/kg, enthalpy of the product
iv = 2607.        			#kj/kg, enthalpy of vapour produced
#from eq 9.6
Wv = 1600.       			#enthalpy of evaporation
q = Wp*ip+Wv*iv-Wf*ef   			#kj/h, heat transfe rate required
hvp = 2188.      			#kj/kg, heat of vaporization of saturated steam at 397 K
rs = q/hvp      			#kg/h, rate of steam supply

# Results
print "The steam pressure to be used in the calandria is %.2f barabs)"%(Sp);
print "The heat transfer rate required is %.2e Kj/h"%(q);
print "Rate of steam supply is %.0f kg/h"%(rs);

The steam pressure to be used in the calandria is 2.15 barabs)
The heat transfer rate required is 4.01e+06 Kj/h
Rate of steam supply is 1833 kg/h


## Example 9.4 Page No : 402¶

In :
import math

from numpy import array, linalg
# Variables
Wf = 6000.     			#kg/h, feed rate
ci = 2.        			#%, initial concentration
cf = 35.       			#%, final conc.
ft = 50.        			#C,feed temp.
ssp = 2.       			#bar abs, saturated steaam pr.
sep = 0.0139  			#bar abs, maintained temp. in second effect
h1 = 2000.     			#W/m**2 K,overall heat transfer coeffcient in 1st effect
h2 = 1500.     			#W/m**2 K, overall heat transfer coefficient in 2nd effect
cp = 4.1      			#kj/kg k,specific heat

#calculation
si = Wf*(ci/100) 			#kg/h, solid in
wi = 5880.      			#kg/h, water in
Wp = si/(cf/100)  			#kg/h product out
wo = Wp*(1-cf/100) 			#kg/h, water out with the product
ter = wi-wo       			#kg/h, total evaporation rate

#boiling temp. in the first effect
T1 = 120.       			#C,Temprature
l_s1 = 2200.    			#kj/kg, latent heat
T2 = 12.        			#C,boiling point in second effect
l_s2 = 2470.   			# kj/kg  in second effect
tatd = T1-T2    			# C,tatd = dt1+dt2  = T1-T2  , total available temp. drop
#from eq. 9.20
#h1*dt1 = h2*dt2
#solving above two equations by matrix
A = array([[1,1],[2000,-1500]])
C = array([108,0])
X = linalg.solve(A,C)
#X = inv(A)*C

dt1 = X
dt2 = X
t1 = T1-dt1    			#temp. of steam leaving the first effect
t2 = T2-dt2    			#temp. of steam leaving second effect
#energy balance over the 1st effect, from eq.9.14
rt1 = t1
ef = cp*(ft-t1)   			#kj/kg,enthalpy of feed
i1 = 0
lam_s1 = 2330.     			#kj/kg
is1 = lam_s1
#Wf*ef+Ws*l_s = (Wf-Ws1)*i1+Ws1*is1
#substituting we get,
#Ws1 = 0.9442*Ws-253.4..........(1)
#energy balance over second effect
#from eq 9.15
#(Wf-Ws1)*i1+Ws1*lam_s1 = (Wf-Ws1-Ws2)*i2+Ws2*is2
rt2 = t2
lam_s2 = 2470.
is2 = lam_s2
i2 = 0
# substituting we get
#Ws2 = 0.8404*Ws1+617.5............(2)
#ter,Ws1+Ws2 = 5657...............(3)
#solving by matrix method
A = array([[0.9442,-1,0],[0,0.8404,-1],[0,1,1]])
B = array([253.4,-617.5,5657])
X = linalg.solve(A,B)
#X = inv(A)*B
Ws = X
Ws1 = X
Ws2 = X

#evaporator area
A1 = Ws*l_s1/(h1*dt1)  			#for 1st effect
A2 = Ws1*lam_s1/(h2*dt2)  			#for second effect

#revised calculation
#taking
dt1_ = 48.
dt2_ = 60.
T1_ = T1-dt1_
T2_ = T2-dt2_
ls1_ = 2335.
ls2_ = 2470.
# energy balance over first effect gives
#Ws1 = 0.9422Ws-231.8.........(4)
#energy balance over second effect gives
#Ws2 = 0.8457Ws1+579.5......(5)
#solving eq 3,4,5
P = array([[0.9422,-1,0],[0,0.8457,-1],[0,1,1]])
Q = array([231.8,-579.5,5657])
Y = linalg.solve(P,Q)
#Y = inv(P)*Q
Ws_ = Y
Ws1_ = Y
Ws2_ = Y

#eveporator area for 1st & 2nd effect in m**2
A1_ = Ws_*l_s1/(h1*dt1_)
A2_ = Ws1_*ls1_/(h2*dt2_)
EA = (A1_+A2_)/2
SE = (Ws1_+Ws2_)/Ws_

# Results
print "The evaporator area is %.0f square metre "%(EA);
print "Steam economy is %.2f"%(SE);

The evaporator area is 72 square metre
Steam economy is 1.79


## Example 9.5 Page No : 404¶

In :
# Variables
ssp = 3.32     			#bar abs, saturated steam pr.
rp = 0.195     			# bar abs, residual pr. in the condenser
tl = 41.      			#K, sun of temp. losses because of BPE
mt = 8.        			#k,minimum available temp. driving force
#calculation
sst = 410.     			#K,saturated steam temp.
st = 333.        			#K,corresponding saturation temp. when pressure in the last effect is 0.195 bar
ttd = sst-st    			#K,total temp. difference
atd = ttd-tl    			# K,available temp. drop across the unit
n = atd/mt     			#maximum no. of effect

# Results
print "Maximum no. of effects are %.0f"%(n);

Maximum no. of effects are 4


## Example 9.6 Page No : 405¶

In :
# Variables
fc = 9.5      			#%,feed concentration
pc = 50.     			#%, product conc.
ft = 40.     			# C,feed temp.
er = 2000.      			#kg NaOH/h, evaporation rate
vp = 714.     			#mm Hg, vaccum pr. in last effect
#heat transfer coefficients, W/m**2 C
h1 = 6000.    			#for first effect
h2 = 3500.    			#for second effect
h3 = 2500.    			#for third effect

#calculatiin
Wf = er/(fc/100)  			#kg/h, 2 tons NaOH per hour, feed rate
Wp = er/(pc/100)  			#kg/h, product rate
ter = Wf-Wp  			#kg/h, total evaporation  rate
#steam
p = 3.3       			#bar,assumed saturated
#from steam table
Ts = 137.      			#C,temp.
l_s = 2153.    			#kj/kg, latent heat
pl = 760.-vp   			#mm Hg,pressure in the last effect
bp = 37.       			#C,boiling point of water
#refer to fig. 9.24
attd = Ts-bp  			#C,apparent total temp. drop
#let assume the following evaporation rate for three effects in kg/h
ev1 = 5600.
ev2 = 5680.
ev3 = 5773.
#conc. in three effects
c1 = er/(Wf-ev1)
c2 = er/(Wf-ev1-ev2)
c3 = 0.5   			# Variables
#boiling point elevations in three effects in C
bpe1 = 3.5
bpe2 = 8.
bpe3 = 39.
attda = attd-(bpe1+bpe2+bpe3)  			#actual total temp. drop available
#temp. drop in three effects
#from eq. 9.23
dt1 = attda*((1/h1)/((1/h1)+(1/h2)+(1/h3)))
dt2 = attda*((1/h2)/((1/h1)+(1/h2)+(1/h3)))
dt3 = attda*((1/h3)/((1/h1)+(1/h2)+(1/h3)))

#from table 9.4
#enthalpy of solution in three effects in  kj/kg
i1 = 486.
i2 = 385.
i3 = 460.
#enthalpy of vapour generated for three effects in kj/kg
is1 = 2729.
is2 = 2691.
is3 = 2646.
#Enthalpy of condensate over effect 1,2,3 in kj/kg
il1 = 0.
il2 = 519.
il3 = 418.
#Enthalpy balance over effect 1
ef = 145.      			#kj/kg,enthalpy of feed
#from energy balance eq.
#Ws1 = 0.96Ws-3200......(1)
#enthalpy balanc over effect 2
#Ws2 = 0.9146Ws1+922...........(2)
#enthalpy balanc over effet 3
#Ws3 = 1.073Ws2+0.0343Ws1-722........(3)
#ter = Ws1+Ws2+Ws3 = 17053..........(4)

#Solving above four eqns by matrix
A = array([[0.96,-1,0,0],[0,0.9146,-1,0],[0,0.0343,1.073,-1],[0,1,1,1]])
B = array([3200,-922,722,17053])
X = linalg.solve(A,B)
#X = inv(A)*B
Ws = X
Ws1 = X
Ws2 = X
Ws3 = X

#calculation of heat transfer areas iver effect 1, 2 ,3
A1 = Ws*l_s*10**3/(h1*dt1*3600)
A2 = Ws1*(is1-il2)*10**3/(h2*dt2*3600)
A3 = Ws2*(is2-il3)*10**3/(h3*dt3*3600)

#Revised dt
avar = (A1+A2+A3)/3
dt1_ = (A1/avar)*dt1
dt2_ = (A2/avar)*dt2
dt3_ = attda-dt1_-dt2_

#from table 9.5
#enthalpy of vapour generated over effect 1,2,3 in kj/kg
is1_ = 2720.
is2_ = 2685.
is3_ = 2646.
#enthalpy of soln on 1,2,3 in kj/kg
i1_ = 470.
i2_ = 380.
i3_ = 460.
#enthalpy of condensate over effect 1 ,2,3 in kj/kg
il1_ = 0.
il2_ = 513.
il3_ = 412.
#enthalpy balance ove effect 1,2,3 gives
Ws_ = 8854.
Ws1_ = 5432.
Ws2_ = 5812.
Ws3_ = 5809.
#revised heat transfer areas for effect 1 ,2,3 in m**2
A1_ = Ws_*l_s*1000/(h1*dt1_*3600)
A2_ = Ws1_*(is1_-il2_)*10**3/(h2*dt2_*3600)
A3_ = Ws2_*(is2_-il3_)*10**3/(h3*22.5*3600)
avar_ = (A1_+A2_+A3_)/3
SE = ter/Ws_

# Results
print "The areas are now reasonably close  "
print "Steam Rate is %.0f Kg/h "%(Ws_)
print "Steam economy is %.2f"%(SE)

The areas are now reasonably close
Steam Rate is 8854 Kg/h
Steam economy is 1.93


## Example 9.7 Page No : 409¶

In :
from numpy import array, linalg

# Variables
Wf = 3000.        			#kg/h,feed
fc = 8.        			#%, feed concentration
pc = 40.       			#% product concentration
si = Wf*(fc/100)      			#kg,solid in
pr = si/(40./100)      			#g/h, product rate
ft = 60.          			#C,feed temp.
er = Wf-pr           			#kg/h, evaporation rate
math.cost = 120000.   			#total math.cost per year
p1 = 4.5       			#bar, low pressure steam
scpt = 700.        			#per ton.  math.cost of steam
cp = 0.764        			#  kcal/kg, specific heat

#from table 9.6
eep = 1.        			#atm existing evaporator pressure
oop = 400000.     			# peryear ,other operatingmath.cost
oop_ = 600000.     			#per yr, for proposed condition
wd = 300.         			#days per year.working days
wh = wd*24.        			#working hr

# Calculations
#EXISTING OPERATING CONDITION
rt = 0         			#C,reference temp.
ef = eep*(ft-rt)   			#kcal/kg, enthalpy of feed
pt = 100.     			#C,product temp.
i1 = cp*(pt-rt)   			#kcal/kg, enthalpy of soln
is1 = 639.       			#kcal/kg,enthalpy of vapour generated at 1 atm (from steam table)
l_s = 496.      			#kcal/kg,latent heat of steam at 4.5 bar
T = 425.      			#K
#heat balance
Ws = (er*is1+pr*i1-Wf*ef)/l_s      			#kg/h, steam required
q = Ws*l_s       			#ton/ hr,heat supplied
x = q/(T-(pt+273))  			#x = Ud*A
#hourly math.cost
sc = Ws/1000*(scpt)    			# /perh, steam math.cost
lc = 100.           			#per h,labour math.cost
oc = oop/(wh)     			# per h,othe math.cost
tc = sc+lc+oc  			#total math.cost
C = tc/(Wf/1000)     			# per ton,math.cost per ton of feed

#PROPOSED OPERATING CONDITION
bpl = 320.        			#K,boiling point of liquid
dt = T-bpl
q_ = x*dt     			#kcal/h,rate of heat supply
sr = q_/l_s   			#steam rate ton per hr
pt_ = 47.      			#C,product temp .
ep = cp*(pt_-rt)    			#kcal/kg. enthalpy of product
ev = 618.        			#kcal/kg, enthalpy of vapour generated
#heat balance
#24Wf_-582Ws1_ = 2825000  ..........(1)
#material balance
# 4Wf_-5Ws1_ = 0  .............(2)
#solving by matrix method
a = array([[24,-582],[4,-5]])
b = array([-2825000,0])
x_ = linalg.solve(a,b)
#x_ = inv(a)*b
Wf_ = x_
Ws1_ = x_
ic = (Wf_-Wf)/Wf
print "The increase in evaporation capacity ic %d percentage "%(ic*100)
sr_ = Ws1_/1000    			#ton per hr ,steam rate
#hourly math.cost
sc_ = Ws1_*scpt   			#steam math.cost
lc_ = 200.      			#labour math.cost rs.200/ h
oc_ = oop_/wh  			# other math.cost
tc_ = sc_/1000+lc_+oc_
C_ = tc_/(Wf_/1000)  			#math.cost per ton of feed
ps = (C-C_)/C
print " The percentage change in the math.cost of concentrating a ton of feed is %.0f percentage"%(ps*100)

# rounding off error.

The increase in evaporation capacity ic 113 percentage
The percentage change in the math.cost of concentrating a ton of feed is 15 percentage


## Example 9.8 Page No : 415¶

In :
# Variables
q = 2200.         			#kj/kg heat of condensation of steam
#from example 9.1
Qr = 2.337*10**8      			#kj/day  rate of heat supply

#calculation
Rate = Qr/q          			#kg/day steam supply rate
Rate_ = 1.062*10**5   			#approximate value
E = 2800.             			#kj/kg enthalpy of compressed vapour
T = 175.7            			#C, temprature
Ts = 121.             			#C Saturation temprature
E1 = 2700.            			#enthalpy at saturation temprature
q1 = T-Ts            			#Superheat of vapour
T1 = 100.             			#C hot water temprature
E2 = 419.              			#Enthalpy at hot water temp.
x = (E-E1)/(E1-E2)    			#water supplied per kg of superheated steam
S = 1.044             			#steam obtained after desuperheating
R1 = 8.925*10**4       			#kg/day rate of vapour generation
R2 = S*R1             			#Rate of recompressed sat. steam
R2_ = 9.318*10**4       			#approximate value
SR = Rate_-R2_

# Results
print "Make up steam required is %.3e kg/day"%(SR)

Make up steam required is 1.302e+04 kg/day