#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"
#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"
#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"ANSWER:Economoy pf evaporator is ",round(Eco,3)
print"Heat tarnsfer area to be provided = ",round(A,2),"m^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"
#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"
#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"
#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"
#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"
#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"Heat load is",round(Q),"W or J/s"
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"
#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"
#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"
#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"ANSWER:Area in each effect",round(A3,1),"sq m"
print"Steam economy is",round(eco,2)
print"Cooling water rate is",round(mw_dot/1000,2),"t/h"
#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"
#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"
#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)
#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"