# Chapter6: Evaporation¶

## Example no:6.1,Page no:6.19¶

In [1]:
#Boiling point Elevation
#Variable declaration
T=380       #B.P of solution[K]
T_dash=373      #B.P of water [K]
Ts=399      #Saturating temperature in [K]
#Calculation
BPE=T-T_dash        #Boiling point elevation in [K]
DF=Ts-T     #Driving force in [K]
#Result
print"Boiling point of elevation of the solution is",BPE,"K"
print"Driving forve for heat transfer is",DF,"K"

Boiling point of elevation of the solution is 7 K
Driving forve for heat transfer is 19 K


## Example no:6.2 ,Page no:6.20¶

In [38]:
#Capacity of evaporator
#Variable declaration
m_dot=10000     #Weak liquor entering in [kg/h]
fr_in=0.04       #Fraciton of caustic soda IN i.e 4%
fr_out=0.25    #Fraciton of caustic soda OUT i.e 25%
#Let mdash_dot be the kg/h of thick liquor leaving

#Calculation
mdash_dot=fr_in*m_dot/fr_out        #[kg/h]

#Overall material balance
#kg/h of feed=kg/h of water evaporated +kg/h of thick liquor
#we=water evaporated in kg/h
#Therefore
we=m_dot-mdash_dot      #[kg/h]

#Result

print"Capacity of evaporator is",we,"kg/h"

Capacity of evaporator is 8400.0 kg/h


## Example no: 6.3,Page no:6.20¶

In [1]:
#Economy of Evaporator
#Variable declaration
ic=0.05     #Initial concentration (5%)
fc=0.2      #Final concentration   (20%)
T_dash=373      #B.P of water in [K]
bpe=5       #Boiling point elevation[K]
mf_dot=5000       #[Basis] feed to evaporator in [kg/h]

#Calculation

#Material balance of solute
mdash_dot=ic*mf_dot/fc       #[kg/h]
#Overall material balance
mv_dot=mf_dot-mdash_dot      #Water evaporated [kg/h]
lambda_s=2185        #Latent heat of condensation of steam[kJ/kg]
lambda_v=2257       #Latent heat of vaporisation of water [kJ/kg]
lambda1=lambda_v     #[kJ/kg]
T=T_dash+bpe        #Temperature of thick liquor[K]
Tf=298      #Temperature of feed [K]
Cpf=4.187       #Sp. heat of feed in [kJ/kg.K]
#Heat balance over evaporator=ms_dot
ms_dot=(mf_dot*Cpf*(T-Tf)+mv_dot*lambda1)/lambda_s   #Steam consumption [kg/h]
Eco=mv_dot/ms_dot       #Economy of evaporator
Ts=399      #Saturation temperature of steam in [K]
dT=Ts-T     #Temperature driving force [K]
U=2350      #[W/sq m.K]
Q=ms_dot*lambda_s       #Rate of heat transfer in [kJ/kg]
Q=Q*1000/3600           #[J/s]=[W]
A=Q/(U*dT)              #Heat transfer area in [sq m]

#Result
print"Heat tarnsfer area to be provided = ",round(A,2),"m^2"

ANSWER:Economoy pf evaporator is  0.808
Heat tarnsfer area to be provided =  57.07 m^2


## Example no: 6.4,Page no:6.22¶

In [2]:
#Steam economy

#Variable declaration

Cpf=3.98        #Specific heat of feed in kJ/(kg.K)
lambda_s=2202    #Latent heat of conds of heat at 0.2MPa in [kJ/kg]
lambda1=2383     #Latent heat of vaporisation of water aty 323 [kJ/kg
ic=0.1          #Initial concentration of soilds in [%]
fc=0.5          #Final concentration
m_dot=30000     #Feed to evaporator in [kg/h]

#Calculation

mdash_dot=ic* m_dot/fc  #Mass flow rate of thick liquor in [kg/h]
mv_dot=m_dot-mdash_dot      #Water evaporated in [kg/h]

#Case 1: Feed at 293K
mf_dot=30000        #[kg/h]
mv_dot=24000        #[kg/h]
Cpf=3.98        #[kJ/(kg.K)]
Ts=393      #Saturation temperature of steam in [K]
T=323       #Boiling point of solution [K]
lambda_s=2202       #Latent heat of condensation [kJ/kg]
lambda1=2383     #Latent heat of vaporisation[kJ/kg]
Tf=293          #Feed temperature
#Enthalpy balance over the evaporator:
ms_dot=(mf_dot*Cpf*(T-Tf)+mv_dot*lambda1)/lambda_s       #Steam consumption[kg/h]
eco=(mv_dot/ms_dot)         #Steam economy
print"When Feed introduced at 293 K ,Steam economy is ",round(eco,2)
dT=Ts-T                     #[K]
U=2900          #[W/sq m.K]
Q=ms_dot*lambda_s           #Heat load =Rate of heat transfer in [kJ/h]
Q=Q*1000/3600               #[J/s]
A=Q/(U*dT)              #Heat transfer area required [sq m]

#Result
print"ANSWER-(i) At 293 K,Heat transfer area required is",round(A,2),"m^2"

#Case2: Feed at 308K
Tf=308      #[Feed temperature][K]

#Calculation
ms_dot=(mf_dot*Cpf*(T-Tf)+mv_dot*lambda1)/lambda_s           #Steam consumption in [kg/h]
eco=mv_dot/ms_dot               #Economy of evaporator
Q=ms_dot*lambda_s               #[kJ/h]
Q=Q*1000/3600                   #[J/s]
A=Q/(U*dT)                      #Heat transfer area required [sq m]
#Result
print"ANSWER-(ii) When T=308 K,Economy of evaporator is ",round(eco,3)
print"ANSWER-(iii) When T=308 K,Heat transfer Area required is ",round(A,2),"m^2"

When Feed introduced at 293 K ,Steam economy is  0.87
ANSWER-(i) At 293 K,Heat transfer area required is 83.16 m^2
ANSWER-(ii) When T=308 K,Economy of evaporator is  0.896
ANSWER-(iii) When T=308 K,Heat transfer Area required is  80.71 m^2


## Example no: 6.5,Page no:6.24¶

In [4]:
#Evaporator economy
#Variable declaration
m_dot=5000  #Feed to the evaporator [kg/h]
Cpf=4.187           #Cp of feed in [kJ/kg.K]
ic=0.10             #Initial concentration
fc=0.4              #Final concentration
lambda_s=2162           #Latent heat of condensing steam [kJ/kg]
P=101.325       #Pressure in the evaporator[kPa]
bp=373          #[K]
Hv=2676     #Enthalpy of water vapor [kJ/kg]
H_dash=419          #[kJ/kg]
Hf=170          #[kJ/kg]
U=1750          #[W/sq m.K]
dT=34           #[K]
#Calculation
mdash_dot=m_dot*ic/fc           #[kg/h] of thick liquor
mv_dot=m_dot-mdash_dot          #Water evaporated in[kg/h]
ms_dot=(mv_dot*Hv+mdash_dot*H_dash-m_dot*Hf)/lambda_s          #Steam consumption in [kg/h]
eco=mv_dot/ms_dot           #Steam economy of evaporator
Q=ms_dot*lambda_s           #[kJ/h]
Q=Q*1000/3600               #[J/s]
A=Q/(U*dT)                  #[sq m]
#Result
print"Heat transfer area to be provided is",round(A,2),"m^2"

Heat transfer area to be provided is 45.33 m^2


## Example no:6.6 ,Page no:6.26¶

In [54]:
#Single effect Evaporator
#Variable declaration
mf_dot=5000           #[kg/h]
ic=0.01             #Initial concentration [kg/h]
fc=0.02             #Final concentration  [kg/h]
T=373               #Boiling pt of saturation in [K]
Ts=383              #Saturation temperature of steam in [K]
Hf=125.79           #[kJ/kg]
Hdash=419.04            #[kJ/kg]
Hv=2676.1           #[kJ/kg]
lambda_s=2230.2     #[kJ/kg]
#Calculation
mdash_dot=ic*mf_dot/fc   #[kg/h]
mv_dot=mf_dot-mdash_dot      #Water evaporated in [kg/h]
ms_dot=(mdash_dot*Hdash+mv_dot*Hv-mf_dot*Hf)/lambda_s   #Steam flow rate in [kg/h]
eco=mv_dot/ms_dot           #Steam economy
Q=ms_dot*lambda_s           #Rate of heat transfer in [kJ/h]
Q=Q*1000/3600               #[J/s]
dT=Ts-T                     #[K]

A=69            #Heating area of evaporator in [sq m]
U=Q/(A*dT)      #Overall heat transfer coeff in [W/sq m.K]

#Result
print"Steam economy is",round(eco,3)
print"Overall heat transfer coefficient is",round(U),"W/m^2.K"

Steam economy is 0.784
Overall heat transfer coefficient is 2862.0 W/m^2.K


## Example no: 6.7,Page no:6.27¶

In [6]:
#Single efect evaporator reduced pressure
#From previous example:
#Variable declaration
mf_dot=5000         #[kg/h]
Hf=125.79           #[kJ/kg]
lambda_s=2230.2     #[kJ/kg]
mdash_dot=2500   #[kg/h]
Hdash=313.93         #[kJ/kg]
mv_dot=2500         #[kg/h]
Hv=2635.3           #[kJ/kg]
U=2862              #[W/sq m.K]
dT=35       #[K]
#Calculation
ms_dot=(mdash_dot*Hdash+mv_dot*Hv-mf_dot*Hf)/lambda_s   #Steam flow rate in [kg/h]
Q=ms_dot*lambda_s           #[kJ/h]
Q=Q*1000/3600           #[W]
A=Q/(U*dT)      #[sq m]
#Result
print"The heat transfer area in this case is",round(A,2),"m^2"
print"NOTE :There is a calculation mistake in the book at the line12 of this code,ms_dot value is written as 2320.18,which is wrong"

The heat transfer area in this case is 18.7 m^2
NOTE :There is a calculation mistake in the book at the line12 of this code,ms_dot value is written as 2320.18,which is wrong


## Example no: 6.8,Page no:6.27¶

In [44]:
#Mass flow rate
#Variable declaration
mf_dot=6000         #Feed rate in [kg/h]
#Taking the given values from previous example(6.6)
Hf=125.79           #[kJ/kg]
ms_dot=3187.56      #[kg/h]
lambda_s=2230.2         #[kJ/kg]
Hdash=419.04        #[kJ/kg]
Hv=2676.1           #[kJ/kg]
#Calculation
mv_dot=(mf_dot*Hf+ms_dot*lambda_s-6000*Hdash)/(Hv-Hdash)  #Water evaporated in [kg/h]
mdash_dot=6000-mv_dot       #Mass flow rate of product [kg/h]
x=(0.01*mf_dot)*100/mdash_dot       #Wt % of solute in products
#Result
print"Mass flow rate of product is",round(mdash_dot,1),"kg/h"
print"The product concentration is",round(x,3),"% by weight"

Mass flow rate of product is 3629.9 kg/h
The product concentration is 1.653 % by weight


## Example no:6.9 ,Page no:6.28¶

In [45]:
#Heat load in single effect evaporator
#Variable declaration
Tf=298      #Feed temperature in [K]
T_dash=373      #[K]
Cpf=4       #[kJ/kg.K]
fc=0.2      #Final concentration of salt
ic=0.05     #Initial concentration
mf_dot=20000    #[kg/h] Feed to evaporator
#Calculation
mdash_dot=ic*mf_dot/fc      #Thick liquor [kg/h]
mv_dot=mf_dot-mdash_dot     #Water evaporated in [kg/h]
lambda_s=2185       #[kJ/kg]
lambda1=2257       #[kJ/kg]
bpr=7       #Boiling point rise[K]
T=T_dash+bpr     #Boiling point of solution in[K]
Ts=39       #Temperature of condensing steam in [K]
ms_dot=(mf_dot*Cpf*(T-Tf)+mv_dot*lambda1)/lambda_s   #Steam consumption in [kg/h]
eco=mv_dot/ms_dot           #Economy of evaporator
Q=ms_dot*lambda_s               #[kJ/h]
Q=Q*1000/3600           #[J/s]
#Result
print"Economy of evaporator is ",round(eco,3)
print"NOTE:Again there is a calcualtion mistake in book at line 19 of code,it is written as 4041507.1 instead of 40415071"

Heat load is 11226389.0 W or J/s
Economy of evaporator is  0.811
NOTE:Again there is a calcualtion mistake in book at line 19 of code,it is written as 4041507.1 instead of 40415071


## Example no:6.10 ,Page no:6.32¶

In [5]:
#Triple efect evaporator
#Variable declaration
Ts=381.3        #[K]
dT=56.6      #[K]
U1=2800.0  #Overall heat transfer coeff in first effect
U2=2200.0  #Overall heat transfer coeff in first effect
U3=1100.0  #Overall heat transfer coeff in first effect
#Calculation
dT1=dT/(1+(U1/U2)+(U1/U3))  #/[K]
dT2=dT/(1+(U2/U1)+(U2/U3))  #/[K]
dT3=dT-(dT1+dT2)             #[K]
#dT1=Ts-T1_dash      #[K]
T1dash=Ts-dT1
#dT2=T1_dash-T2_dash         #[K]
T2_dash=T1dash-dT2             #[K]
#Result
print"Boiling point of solution in first effect =",round(T1dash,2),"K"
print"Boiling point of solution in second effect =",round(T2_dash,1),"K"

Boiling point of solution in first effect = 369.55 K
Boiling point of solution in second effect = 354.6 K


## Example no:6.11,Page no:6.33¶

In [4]:
#Double effect evaporator
#Variable declaration
mf_dot=10000.0        #[kg/h] of feed
ic=0.09     #Initial concentration
fc=0.47     #Final concentration
m1dot_dash=ic*mf_dot/fc     #[kg/h]
Ps=686.616      #Steam pressure [kPa.g]
Ps=Ps+101.325       #[kPa]
Ts=442.7        #Saturation temperature in [K]
P2=86.660       #Vacuum in second effect in [kPa]
U1=2326.0     #Overall heat transfer in first effect [W/sq m.K]
U2=1744.5   #Overall heat transfer in 2nd effect [W/sqm.K]
P2_abs=101.325-P2   #Absolute pressure in second effect[kPa]
T2=326.3        #Temperature in 2nd effect in [K]
dT=Ts-T2        #[K]
Tf=309.0      #Feed temperature in[K]
T=273.0       #[K]
Cpf=3.77        #kJ/kg.K  Specific heat for all caustic streams
#Q1=Q2
#U1*A1*dT1=U2*A2*dT2
#Calculation
dT2=dT/1.75     #[K]
dT1=(U2/U1)*dT2     #[K]
#Since there is no B.P.R
Tv1=Ts-dT1      #Temperature in vapor space of first effect in [K]
Tv2=Tv1-dT2     #Second effect [K]
Hf=Cpf*(Tf-T)       #Feed enthalpy[kJ/kg]
H1dash=Cpf*(Tv1-T)      #Enthalpy of final product[kJ/kg]
H2dash=Cpf*(Tv2-T)      #kJ/kg
#For steam at 442.7 K
lambda_s=2048.7     #[kJ/kg]
#For vapour at 392.8 K
Hv1=2705.22     #[kJ/kg]
lambda_v1=2202.8        #[kJ/kg]
#for vapour at 326.3 K:
Hv2=2597.61     #[kJ/kg]
lambda_v2=2377.8        #[kJ/kg]

#Overall material balance:
mv_dot=mf_dot-m1dot_dash        #[kg/h]

#Equation 4 becomes:
#mv1_dot*lambda_v1+mf_dot*Hf=(mv_dot-mv1_dot)*Hv2+(mf_dot-mv2_dot)*H2_dash
mv1_dot=(H2dash*(mf_dot-mv_dot)-mf_dot*Hf+mv_dot*Hv2)/(Hv2+lambda_v1-H2dash)
mv2_dot=mv_dot-mv1_dot              #[kg/h]

#From equation 2

m2dot_dash=m1dot_dash+mv1_dot           #First effect material balance[kg/h]
ms_dot=(mv1_dot*Hv1+m1dot_dash*H1dash-m2dot_dash*H2dash)/lambda_s     #[kg/h]

#Heat transfer Area
#First effect
A1=ms_dot*lambda_s*(10.0**3.0)/(3600.0*U1*dT1)     #[sq m]

#Second effect
lambda_v1=lambda_v1*(10**3.0)/3600.0
A2=mv1_dot*lambda_v1/(U2*dT2)       #[sq m]

#Since A1 not= A2

#SECOND TRIAL
Aavg=(A1+A2)/2          #[sq m]
dT1_dash=dT1*A1/Aavg        #[K]
dT2_dash=dT-dT1         #/[K]

#Temperature distribution
Tv1=Ts-dT1_dash         #[K]
Tv2=Tv1-dT2_dash            #[K]
Hf=135.66       #[kJ/kg]
H1dash=Cpf*(Tv1-T)      #[kJ/kg]
H2dash=200.83          #[kJ/kg]

#Vapour at 388.5 K
Hv1=2699.8      #[kJ/kg]
lambda_v1=2214.92       #[kJ/kg]
mv1_dot=(H2dash*(mf_dot-mv_dot)-mf_dot*Hf+mv_dot*Hv2)/(Hv2+lambda_v1-H2dash)
mv2_dot=mv_dot-mv1_dot  #[kg/h]

#First effect Energy balance
ms_dot=((mv1_dot*Hv1+m1dot_dash*H1dash)-(mf_dot-mv2_dot)*H2dash)/lambda_s   #[kg/h]

#Area of heat transfer
lambda_s=lambda_s*1000.0/3600.0
A1=ms_dot*lambda_s/(U1*dT1_dash)        #[sq m]

#Second effect:
A2=(mv1_dot*lambda_v1*1000)/(3600.0*U2*dT2_dash)        #[sq m]

#Result

print"A1(",round(A1,1),")=A2(",round(A2),"),So the area in each effect can be",round(A1,1),"m^2"
print"Heat transfer surface in each effect is",round(A1,1),"m^2"
print"Steam consumption=",round(ms_dot),"(approx)kg/h"
print"Evaporation in the first effect is",round(mv1_dot),"kg/h"
print"Evaporation in  2nd effect is",round(mv2_dot),"kg/h"

A1( 24.9 )=A2( 23.0 ),So the area in each effect can be 24.9 m^2
Heat transfer surface in each effect is 24.9 m^2
Steam consumption= 5517.0 (approx)kg/h
Evaporation in the first effect is 4343.0 kg/h
Evaporation in  2nd effect is 3742.0 kg/h


## Example no:6.12 ,Page no:6.37¶

In [4]:
#lye in Triple effect evaporator
#Variable declaration
Tf=353.0           #[K]
T=273.0           #[K]
mf_dot=10000.0          #Feed [kg/h]
ic=0.07          #Initial conc of glycerine
fc=0.4           #FinaL CONC OF GLYCERINE
#Overall glycerine balance
P=313.0            #Steam pressure[kPa]
Ts=408.0       #[from steam table][K]
P1=15.74         #[Pressure in last effect][kPa]
Tv3=328.0          #[Vapour temperature]
#Calculation
m3dot_dash=(ic/fc)*mf_dot           #[kg/h]
mv_dot=mf_dot-m3dot_dash            #/[kg/h]
dT=Ts-Tv3      #Overall apparent [K]
bpr1=10.0          #[K]
bpr2=bpr1
bpr3=bpr2
sum_bpr=bpr1+bpr2+bpr3      #[K]
dT=dT-sum_bpr               #True_Overall
dT1=14.5             #[K]
dT2=16.0               #[K]
dT3=19.5             #[K]
Cpf=3.768            #[kJ/(kg.K)]
#Enthalpies of various streams
Hf=Cpf*(Tf-T)           #[kJ/kg]
H1=Cpf*(393.5-T)           #[kJ/kg]
H2=Cpf*(367.5-T)           #[kJ/kg]
H3=Cpf*(338.0-T)           #[kJ/kg]
#For steam at 40K
lambda_s=2160.0       #[kJ/kg]
Hv1=2692.0        #[kJ/kg]
lambda_v1=2228.3        #[kJ/kg]
Hv2=2650.8          #[kJ/kg]
lambda_v2=2297.4        #[kJ/kg]
Hv3=2600.5              #[kJ/kg]
lambda_v3=2370.0      #[kJ/kg]

#MATERIAL AND EBERGY BALANCES
#First effect
#Material balance

#m1dot_dash=mf_dot-mv1_dot
#m1dot_dash=1750+mv2_dot+mv3_dot

#Energy balance
#ms_dot*lambda_s+mf_Dot*hf=mv1_dot*Hv1+m1dot_dash*H1
#2160*ms_dot+2238*(mv2_dot+mv3_dot)=19800500

#Second effect
#Energy balance:
#mv3_dot=8709.54-2.076*mv2_dot

#Third effect:
#m2dot_dash=mv3_dot+m3dot_dash
#m2dot_dash=mv3_dot+1750
#From eqn 8 we get
mv2_dot=(8709.54*2600.5+1750*244.92-8790.54*356.1-356.1*1750)/(-2.076*356.1+2297.4+2600.5*2.076)
#From eqn 8:
mv3_dot=8709.54-2.076*mv2_dot           #[kg/h]
mv1_dot=mv_dot-(mv2_dot+mv3_dot)        #[kg/h]
#From equation 4:
#m1dot_dash=mf_dot-mv1_dot
#ms_dot=(mv1_dot*Hv1+m1dot_dash*H1-mf_dot*Hf)/lambda_s   #[kg/h]
ms_dot=(19800500.0-2238.0*(mv2_dot+mv3_dot))/2160.0           #[kg/h]

#Heat transfer Area is
U1=710.0          #[W/sq m.K]
U2=490.0          #[W/sq m.K]
U3=454.0          #[W/sq m.K]
A1=(ms_dot*lambda_s*1000.0)/(3600.0*U1*dT1)     #[sq m]
A2=mv1_dot*lambda_v1*1000.0/(3600.0*U2*dT2)   #[sq m]
A3=mv2_dot*lambda_v2*1000.0/(3600.0*U3*dT3)   #[sq m]
#The deviaiton is within +-10%
#Hence maximum A1 area can be recommended

eco=(mv_dot/ms_dot)     #[Steam economy]

Qc=mv3_dot*lambda_v3        #[kJ/h]
dT=25.0           #Rise in water temperature
Cp=4.187
mw_dot=Qc/(Cp*dT)
#Result
print"Steam economy is",round(eco,2)
print"Cooling water rate is",round(mw_dot/1000,2),"t/h"

ANSWER:Area in each effect 200.2 sq m
Steam economy is 2.55
Cooling water rate is 66.63 t/h


## Example no:6.13 ,Page no:6.42¶

In [9]:
#Triple effect unit
#Variable declaration
Cpf=4.18        #[kJ/kg.K]
dT1=18          #[K]
dT2=17          #[K]
dT3=34          #[K]
mf_dot=4        #[kg/s]
Ts=394          #[K]
bp=325      #Bp of water at 13.172 kPa [K]
dT=Ts-bp        #[K]
lambda_s=2200           #[kJ/kg]
T1=Ts-dT1           #[K]
lambda1=2249        #[kJ/kg]
lambda_v1=lambda1       #[kJ/kg]
#Calculation
T2=T1-dT2           #[K]
lambda2=2293        #[kJ/kg]
lambda_v2=lambda2       #[kJ/kg]

T3=T2-dT3           #[K]
lambda3=2377        #[kJ/kg]
lambda_v3=lambda3       #[kJ/kg]

ic=0.1      #Initial conc of solids
fc=0.5      #Final conc of solids
m3dot_dash=(ic/fc)*mf_dot       #[kg/s]
mv_dot=mf_dot-m3dot_dash        #Total evaporation in [kg/s]
#Material balance over first effect
#mf_dot=mv1_dot_m1dot_dash
#Energy balance:
#ms_dot*lambda_s=mf_dot*(Cpf*(T1-Tf)+mv1_dot*lambda_v1)

#Material balance over second effect
#m1dot_dash=mv2_dot+m2dot_dash
#Enthalpy balance:
#mv1_dot*lambda_v1+m1dot_dash(cp*(T1-T2)=mv2_dot*lambda_v2)

#Material balance over third effect
#m2dot_dash=mv3_dot+m3dot+dash

#Enthalpy balance:
#mv2_lambda_v2+m2dot_dash*cp*(T2-T3)=mv3_dot*lambda_v3
294
mv2_dot=3.2795/3.079        #[kg/s]
mv1_dot=1.053*mv2_dot-0.1305     #[kg/s]
mv3_dot=1.026*mv2_dot+0.051     #[kg/s]
ms_dot=(mf_dot*Cpf*(T1-294)+mv1_dot*lambda_v1)/lambda_s      #[kg/s]
eco=mv_dot/ms_dot           #Steam economy
eco=round(eco)
U1=3.10     #[kW/sq m.K]
U2=2     #[kW/sq m.K]
U3=1.10     #[kW/sq m.K]
#First effect:
A1=ms_dot*lambda_s/(U1*dT1)         #[sq m]
A2=mv1_dot*lambda_v1/(U2*dT2)        #[sq m]
A3=mv2_dot*lambda_v2/(U3*dT3)        #[sq m]
#Areas are calculated witha  deviation of +-10%
#Result
print"Steam economy is",eco
print"Area pf heat transfer in each effect is",round(A3,1),"m^2"

Steam economy is 2.0
Area pf heat transfer in each effect is 65.3 m^2


## Example no: 6.14,Page no:6.45¶

In [2]:
#Quadruple effect evaporator
#Variable declaration
mf_dot=1060     #[kg/h]
ic=0.04     #Initial concentration
fc=0.25         #Final concentration
m4dot_dash=(ic/fc)*mf_dot       #[kg/h]
#Total evaporation=
mv_dot=mf_dot-m4dot_dash        #[kg/h]

#Fromsteam table:
P1=370      #[kPa.g]
T1=422.6        #[K]
lambda1=2114.4      #[kJ/kg]

P2=235      #[kPa.g]
T2=410.5        #[K]
lambda2=2151.5      #[kJ/kg]

P3=80      #[kPa.g]
T3=390.2        #[K]
lambda3=2210.2      #[kJ/kg]

P4=50.66      #[kPa.g]
T4=354.7        #[K]
lambda4=2304.6      #[kJ/kg]

P=700       #Latent heat of steam[kPa .g]
lambda_s=2046.3         #[kJ/kg]

#Calculation
#FIRST EFFECT
#Enthalpy balance:
#ms_dot=mf_dot*Cpf*(T1-Tf)+mv1_dot*lambda1
#ms_dot=1345.3-1.033*m1dot_dash

#SECOND EFFECT
#m1dot_dash=m2dot_dash+mdot_v2
#Enthalpy balance:
#m1dot_dash=531.38+0.510*m2dot_dash

#THIRD EFFECT
#Material balance:
#m2dot_dash-m3dot_dash+mv3_dot

#FOURTH EFFECT
#m3dot_dash=m4dot_dash+mv4_dot
mv4dot_dash=169.6           #[kg/h]
m3dot_dash=416.7        #[kg/h]

#From eq n 4:
m2dot_dash=-176.84+1.98*m3dot_dash      #[kg/h]

#From eqn 2:
m1dot_dash=531.38+0.510*m2dot_dash      #[kg/h]

#From eqn 1:
ms_dot=1345.3-1.033*m1dot_dash
eco=mv_dot/ms_dot           #[kg evaporation /kg steam]
#Result
print"Steam economy is",round(eco,3),"evaporation/kg steam"

Steam economy is 1.957 evaporation/kg steam


## Example no:6.15 ,Page no:6.48¶

In [10]:
#Single effect Calendria
import math
#Variable declaration
m1_dot=5000     #[kg/h]
ic=0.1      #Initial concentration
fc=0.5      #Final concentration
mf_dot=(fc/ic)*m1_dot       #[kg/h]
mv_dot=mf_dot-m1_dot        #Water evaporated[kg/h]
P=357       #Steam pressure[kN/sq m]
Ts=412       #[K]
H=2732      #[kJ/kg]
lambda1=2143     #[kJ/kg]
bpr=18.5            #[K]
T_dash=352+bpr      #[K]
Hf=138      #[kJ/kg]
lambda_s=2143       #[kJ/kg]
Hv=2659     #[kJ/kg]
H1=568      #[kJ/kg]
#Calculation
ms_dot=(mv_dot*Hv+m1_dot*H1-mf_dot*Hf)/lambda_s         #Steam consumption in kg/h
eco=mv_dot/ms_dot       #Economy
dT=Ts-T_dash        #[K]
hi=4500     #[W/sq m.K]
ho=9000     #[W/sq m.K]
Do=0.032        #[m]
Di=0.028        #[m]
x1=(Do-Di)/2        #[m]
Dw=(Do-Di)/math.log(32.0/28.0)   #[m]
x2=0.25*10**-3      #[m]
L=2.5       #Length [m]
hio=hi*(Di/Do)        #[W/sq m.K]
print"NOTE:In textbook this value of hio is wrongly calculated as 3975.5..So we will take this"
hio=3975.5
k1=45.0       #Tube material in [W/sq m.K]
k2=2.25     #For scale[W/m.K]
Uo=1.0/(1.0/ho+1.0/hio+(x1*Dw)/(k1*Do)+(x2/k2))     #Overall heat transfer coeff in W/sq m.K
Q=ms_dot*lambda_s       #[kJ/h]
Q=Q*1000.0/3600.0           #[W]

A=Q/(Uo*dT)             #[sq m]
n=A/(math.pi*Do*L)          #from A=n*math.pi*Do*L
#Result
print"Steam consumption is",round(ms_dot),"kg/h"
print"Capacity is",round(mv_dot),"kg/h"
print"Steam economy is ",round(eco,3)
print" No. of tubes required is ",round(n)

NOTE:In textbook this value of hio is wrongly calculated as 3975.5..So we will take this
Steam consumption is 24531.0 kg/h
Capacity is 20000.0 kg/h
Steam economy is  0.815
No. of tubes required is  722.0


## Example no:6.16 ,Page no:6.50¶

In [3]:
#Single effect evaporator
#Variable declaration
bpr=40.6         #[K]
Cpf=1.88        #[kJ/kg.K]
Hf=214      #[kJ/kg]
H1=505      #[kJ/kg]
mf_dot=4536     #[kg/h] of feed solution
ic=0.2       #Initial conc
fc=0.5       #Final concentration
m1dot_dash=(ic/fc)*mf_dot       #Thisck liquor flow arte[kg/h]
mv_dot=mf_dot-m1dot_dash        #[kg/H]
Ts=388.5         #Saturation temperature of steam in [K]
bp=362.5        #b.P of solution in [K]
lambda_s=2214      #[kJ/kg]
P=21.7       #Vapor space in [kPa]
Hv=2590.3        #[kJ/kg]

#Calculation
#Enthalpy balance over evaporator
ms_dot=(m1dot_dash*H1+mv_dot*Hv-mf_dot*Hf)/lambda_s     #[kg/h
print"Steam consumption is",round(ms_dot,1),"kg/h"
dT=Ts-bp        #[K]
U=1560      #[W/sq m.K]
Q=ms_dot*lambda_s           #[kJ/h]
Q=Q*1000/3600               #[W]
A=Q/(U*dT)      #[sq m]
print"Heat transfer area is",round(A,2),"m^2"

#Calculations considering enthalpy of superheated vapour

Hv=Hv+Cpf*bpr   #[kJ/kg]
ms_dot=(m1dot_dash*H1+mv_dot*Hv-mf_dot*Hf)/lambda_s     #[kg/h]
print" Now,Steam consumption is",round(ms_dot,2),"kg/h"
eco=mv_dot/ms_dot       #Steam economy
print"Economy of evaporator ",round(eco,2)
Q=ms_dot*lambda_s       #[kJ/h]
Q=Q*1000.0/3600.0           #[w]
A2=Q/(U*dT)              #Area
print"Now,Area is",round(A2,2)
perc=(A2-A)*100/A           #%error in the heat transfer area
#Result
print"If enthalpy of water vapour Hv were based on the saturated vapour at the pressure\nthe error introduced is only",round(perc,2),"percent"

Steam consumption is 3159.6 kg/h
Heat transfer area is 47.91 m^2
Now,Steam consumption is 3253.42 kg/h
Economy of evaporator  0.84
Now,Area is 49.33
If enthalpy of water vapour Hv were based on the saturated vapour at the pressure
the error introduced is only 2.97 percent