{
"metadata": {
"name": ""
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Evaporation"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no:6.1,Page no:6.19"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Boiling point Elevation\n",
"#Variable declaration\n",
"T=380 #B.P of solution[K]\n",
"T_dash=373 #B.P of water [K]\n",
"Ts=399 #Saturating temperature in [K]\n",
"#Calculation\n",
"BPE=T-T_dash #Boiling point elevation in [K]\n",
"DF=Ts-T #Driving force in [K]\n",
"#Result\n",
"print\"Boiling point of elevation of the solution is\",BPE,\"K\"\n",
"print\"Driving forve for heat transfer is\",DF,\"K\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Boiling point of elevation of the solution is 7 K\n",
"Driving forve for heat transfer is 19 K\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no:6.2 ,Page no:6.20"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Capacity of evaporator\n",
"#Variable declaration\n",
"m_dot=10000 #Weak liquor entering in [kg/h]\n",
"fr_in=0.04 #Fraciton of caustic soda IN i.e 4%\n",
"fr_out=0.25 #Fraciton of caustic soda OUT i.e 25%\n",
"#Let mdash_dot be the kg/h of thick liquor leaving\n",
"\n",
"#Calculation\n",
"mdash_dot=fr_in*m_dot/fr_out #[kg/h]\n",
"\n",
"#Overall material balance\n",
"#kg/h of feed=kg/h of water evaporated +kg/h of thick liquor\n",
"#we=water evaporated in kg/h\n",
"#Therefore\n",
"we=m_dot-mdash_dot #[kg/h]\n",
"\n",
"#Result\n",
"\n",
"print\"Capacity of evaporator is\",we,\"kg/h\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Capacity of evaporator is 8400.0 kg/h\n"
]
}
],
"prompt_number": 38
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no: 6.3,Page no:6.20"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Economy of Evaporator\n",
"#Variable declaration\n",
"ic=0.05 #Initial concentration (5%)\n",
"fc=0.2 #Final concentration (20%)\n",
"T_dash=373 #B.P of water in [K]\n",
"bpe=5 #Boiling point elevation[K]\n",
"mf_dot=5000 #[Basis] feed to evaporator in [kg/h]\n",
"\n",
"#Calculation\n",
"\n",
"#Material balance of solute\n",
"mdash_dot=ic*mf_dot/fc #[kg/h]\n",
"#Overall material balance\n",
"mv_dot=mf_dot-mdash_dot #Water evaporated [kg/h]\n",
"lambda_s=2185 #Latent heat of condensation of steam[kJ/kg]\n",
"lambda_v=2257 #Latent heat of vaporisation of water [kJ/kg]\n",
"lambda1=lambda_v #[kJ/kg]\n",
"T=T_dash+bpe #Temperature of thick liquor[K]\n",
"Tf=298 #Temperature of feed [K]\n",
"Cpf=4.187 #Sp. heat of feed in [kJ/kg.K]\n",
"#Heat balance over evaporator=ms_dot\n",
"ms_dot=(mf_dot*Cpf*(T-Tf)+mv_dot*lambda1)/lambda_s #Steam consumption [kg/h]\n",
"Eco=mv_dot/ms_dot #Economy of evaporator\n",
"Ts=399 #Saturation temperature of steam in [K]\n",
"dT=Ts-T #Temperature driving force [K] \n",
"U=2350 #[W/sq m.K]\n",
"Q=ms_dot*lambda_s #Rate of heat transfer in [kJ/kg]\n",
"Q=Q*1000/3600 #[J/s]=[W]\n",
"A=Q/(U*dT) #Heat transfer area in [sq m]\n",
"\n",
"#Result\n",
"print\"ANSWER:Economoy pf evaporator is \",round(Eco,3)\n",
"print\"Heat tarnsfer area to be provided = \",round(A,2),\"m^2\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"ANSWER:Economoy pf evaporator is 0.808\n",
"Heat tarnsfer area to be provided = 57.07 m^2\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no: 6.4,Page no:6.22"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Steam economy\n",
"\n",
"#Variable declaration\n",
"\n",
"Cpf=3.98 #Specific heat of feed in kJ/(kg.K)\n",
"lambda_s=2202 #Latent heat of conds of heat at 0.2MPa in [kJ/kg]\n",
"lambda1=2383 #Latent heat of vaporisation of water aty 323 [kJ/kg\n",
"ic=0.1 #Initial concentration of soilds in [%]\n",
"fc=0.5 #Final concentration\n",
"m_dot=30000 #Feed to evaporator in [kg/h]\n",
"\n",
"#Calculation\n",
"\n",
"mdash_dot=ic* m_dot/fc #Mass flow rate of thick liquor in [kg/h]\n",
"mv_dot=m_dot-mdash_dot #Water evaporated in [kg/h]\n",
"\n",
"#Case 1: Feed at 293K\n",
"mf_dot=30000 #[kg/h]\n",
"mv_dot=24000 #[kg/h]\n",
"Cpf=3.98 #[kJ/(kg.K)]\n",
"Ts=393 #Saturation temperature of steam in [K]\n",
"T=323 #Boiling point of solution [K]\n",
"lambda_s=2202 #Latent heat of condensation [kJ/kg]\n",
"lambda1=2383 #Latent heat of vaporisation[kJ/kg]\n",
"Tf=293 #Feed temperature\n",
"#Enthalpy balance over the evaporator:\n",
"ms_dot=(mf_dot*Cpf*(T-Tf)+mv_dot*lambda1)/lambda_s #Steam consumption[kg/h]\n",
"eco=(mv_dot/ms_dot) #Steam economy\n",
"print\"When Feed introduced at 293 K ,Steam economy is \",round(eco,2) \n",
"dT=Ts-T #[K]\n",
"U=2900 #[W/sq m.K]\n",
"Q=ms_dot*lambda_s #Heat load =Rate of heat transfer in [kJ/h]\n",
"Q=Q*1000/3600 #[J/s]\n",
"A=Q/(U*dT) #Heat transfer area required [sq m]\n",
"\n",
"#Result\n",
"print\"ANSWER-(i) At 293 K,Heat transfer area required is\",round(A,2),\"m^2\"\n",
"\n",
"#Case2: Feed at 308K\n",
"Tf=308 #[Feed temperature][K]\n",
"\n",
"#Calculation\n",
"ms_dot=(mf_dot*Cpf*(T-Tf)+mv_dot*lambda1)/lambda_s #Steam consumption in [kg/h]\n",
"eco=mv_dot/ms_dot #Economy of evaporator\n",
"Q=ms_dot*lambda_s #[kJ/h]\n",
"Q=Q*1000/3600 #[J/s]\n",
"A=Q/(U*dT) #Heat transfer area required [sq m]\n",
"#Result\n",
"print\"ANSWER-(ii) When T=308 K,Economy of evaporator is \",round(eco,3)\n",
"print\"ANSWER-(iii) When T=308 K,Heat transfer Area required is \",round(A,2),\"m^2\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"When Feed introduced at 293 K ,Steam economy is 0.87\n",
"ANSWER-(i) At 293 K,Heat transfer area required is 83.16 m^2\n",
"ANSWER-(ii) When T=308 K,Economy of evaporator is 0.896\n",
"ANSWER-(iii) When T=308 K,Heat transfer Area required is 80.71 m^2\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no: 6.5,Page no:6.24"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Evaporator economy\n",
"#Variable declaration\n",
"m_dot=5000 #Feed to the evaporator [kg/h]\n",
"Cpf=4.187 #Cp of feed in [kJ/kg.K]\n",
"ic=0.10 #Initial concentration\n",
"fc=0.4 #Final concentration\n",
"lambda_s=2162 #Latent heat of condensing steam [kJ/kg]\n",
"P=101.325 #Pressure in the evaporator[kPa]\n",
"bp=373 #[K]\n",
"Hv=2676 #Enthalpy of water vapor [kJ/kg]\n",
"H_dash=419 #[kJ/kg]\n",
"Hf=170 #[kJ/kg]\n",
"U=1750 #[W/sq m.K]\n",
"dT=34 #[K]\n",
"#Calculation\n",
"mdash_dot=m_dot*ic/fc #[kg/h] of thick liquor\n",
"mv_dot=m_dot-mdash_dot #Water evaporated in[kg/h]\n",
"ms_dot=(mv_dot*Hv+mdash_dot*H_dash-m_dot*Hf)/lambda_s #Steam consumption in [kg/h]\n",
"eco=mv_dot/ms_dot #Steam economy of evaporator\n",
"Q=ms_dot*lambda_s #[kJ/h]\n",
"Q=Q*1000/3600 #[J/s]\n",
"A=Q/(U*dT) #[sq m]\n",
"#Result\n",
"print\"Heat transfer area to be provided is\",round(A,2),\"m^2\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Heat transfer area to be provided is 45.33 m^2\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no:6.6 ,Page no:6.26"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Single effect Evaporator\n",
"#Variable declaration\n",
"mf_dot=5000 #[kg/h]\n",
"ic=0.01 #Initial concentration [kg/h]\n",
"fc=0.02 #Final concentration [kg/h]\n",
"T=373 #Boiling pt of saturation in [K]\n",
"Ts=383 #Saturation temperature of steam in [K] \n",
"Hf=125.79 #[kJ/kg]\n",
"Hdash=419.04 #[kJ/kg]\n",
"Hv=2676.1 #[kJ/kg]\n",
"lambda_s=2230.2 #[kJ/kg]\n",
"#Calculation\n",
"mdash_dot=ic*mf_dot/fc #[kg/h]\n",
"mv_dot=mf_dot-mdash_dot #Water evaporated in [kg/h]\n",
"ms_dot=(mdash_dot*Hdash+mv_dot*Hv-mf_dot*Hf)/lambda_s #Steam flow rate in [kg/h]\n",
"eco=mv_dot/ms_dot #Steam economy\n",
"Q=ms_dot*lambda_s #Rate of heat transfer in [kJ/h]\n",
"Q=Q*1000/3600 #[J/s]\n",
"dT=Ts-T #[K]\n",
"\n",
"A=69 #Heating area of evaporator in [sq m]\n",
"U=Q/(A*dT) #Overall heat transfer coeff in [W/sq m.K]\n",
"\n",
"#Result\n",
"print\"Steam economy is\",round(eco,3)\n",
"print\"Overall heat transfer coefficient is\",round(U),\"W/m^2.K\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Steam economy is 0.784\n",
"Overall heat transfer coefficient is 2862.0 W/m^2.K\n"
]
}
],
"prompt_number": 54
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no: 6.7,Page no:6.27"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Single efect evaporator reduced pressure\n",
"#From previous example:\n",
"#Variable declaration\n",
"mf_dot=5000 #[kg/h]\n",
"Hf=125.79 #[kJ/kg]\n",
"lambda_s=2230.2 #[kJ/kg]\n",
"mdash_dot=2500 #[kg/h]\n",
"Hdash=313.93 #[kJ/kg]\n",
"mv_dot=2500 #[kg/h]\n",
"Hv=2635.3 #[kJ/kg]\n",
"U=2862 #[W/sq m.K]\n",
"dT=35 #[K]\n",
"#Calculation\n",
"ms_dot=(mdash_dot*Hdash+mv_dot*Hv-mf_dot*Hf)/lambda_s #Steam flow rate in [kg/h]\n",
"Q=ms_dot*lambda_s #[kJ/h]\n",
"Q=Q*1000/3600 #[W]\n",
"A=Q/(U*dT) #[sq m]\n",
"#Result\n",
"print\"The heat transfer area in this case is\",round(A,2),\"m^2\"\n",
"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\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The heat transfer area in this case is 18.7 m^2\n",
"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\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no: 6.8,Page no:6.27"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Mass flow rate\n",
"#Variable declaration\n",
"mf_dot=6000 #Feed rate in [kg/h]\n",
"#Taking the given values from previous example(6.6)\n",
"Hf=125.79 #[kJ/kg]\n",
"ms_dot=3187.56 #[kg/h]\n",
"lambda_s=2230.2 #[kJ/kg]\n",
"Hdash=419.04 #[kJ/kg]\n",
"Hv=2676.1 #[kJ/kg]\n",
"#Calculation\n",
"mv_dot=(mf_dot*Hf+ms_dot*lambda_s-6000*Hdash)/(Hv-Hdash) #Water evaporated in [kg/h]\n",
"mdash_dot=6000-mv_dot #Mass flow rate of product [kg/h]\n",
"x=(0.01*mf_dot)*100/mdash_dot #Wt % of solute in products\n",
"#Result\n",
"print\"Mass flow rate of product is\",round(mdash_dot,1),\"kg/h\"\n",
"print\"The product concentration is\",round(x,3),\"% by weight\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Mass flow rate of product is 3629.9 kg/h\n",
"The product concentration is 1.653 % by weight\n"
]
}
],
"prompt_number": 44
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no:6.9 ,Page no:6.28"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Heat load in single effect evaporator\n",
"#Variable declaration\n",
"Tf=298 #Feed temperature in [K]\n",
"T_dash=373 #[K]\n",
"Cpf=4 #[kJ/kg.K]\n",
"fc=0.2 #Final concentration of salt\n",
"ic=0.05 #Initial concentration\n",
"mf_dot=20000 #[kg/h] Feed to evaporator\n",
"#Calculation\n",
"mdash_dot=ic*mf_dot/fc #Thick liquor [kg/h]\n",
"mv_dot=mf_dot-mdash_dot #Water evaporated in [kg/h]\n",
"lambda_s=2185 #[kJ/kg]\n",
"lambda1=2257 #[kJ/kg]\n",
"bpr=7 #Boiling point rise[K]\n",
"T=T_dash+bpr #Boiling point of solution in[K]\n",
"Ts=39 #Temperature of condensing steam in [K]\n",
"ms_dot=(mf_dot*Cpf*(T-Tf)+mv_dot*lambda1)/lambda_s #Steam consumption in [kg/h]\n",
"eco=mv_dot/ms_dot #Economy of evaporator \n",
"Q=ms_dot*lambda_s #[kJ/h]\n",
"Q=Q*1000/3600 #[J/s]\n",
"#Result\n",
"print\"Heat load is\",round(Q),\"W or J/s\"\n",
"print\"Economy of evaporator is \",round(eco,3)\n",
"print\"NOTE:Again there is a calcualtion mistake in book at line 19 of code,it is written as 4041507.1 instead of 40415071\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Heat load is 11226389.0 W or J/s\n",
"Economy of evaporator is 0.811\n",
"NOTE:Again there is a calcualtion mistake in book at line 19 of code,it is written as 4041507.1 instead of 40415071\n"
]
}
],
"prompt_number": 45
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no:6.10 ,Page no:6.32"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Triple efect evaporator\n",
"#Variable declaration\n",
"Ts=381.3 #[K]\n",
"dT=56.6 #[K]\n",
"U1=2800.0 #Overall heat transfer coeff in first effect\n",
"U2=2200.0 #Overall heat transfer coeff in first effect\n",
"U3=1100.0 #Overall heat transfer coeff in first effect\n",
"#Calculation\n",
"dT1=dT/(1+(U1/U2)+(U1/U3)) #/[K]\n",
"dT2=dT/(1+(U2/U1)+(U2/U3)) #/[K]\n",
"dT3=dT-(dT1+dT2) #[K]\n",
"#dT1=Ts-T1_dash #[K]\n",
"T1dash=Ts-dT1\n",
"#dT2=T1_dash-T2_dash #[K]\n",
"T2_dash=T1dash-dT2 #[K]\n",
"#Result\n",
"print\"Boiling point of solution in first effect =\",round(T1dash,2),\"K\"\n",
"print\"Boiling point of solution in second effect =\",round(T2_dash,1),\"K\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Boiling point of solution in first effect = 369.55 K\n",
"Boiling point of solution in second effect = 354.6 K\n"
]
}
],
"prompt_number": 46
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no:6.11,Page no:6.33"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Double effect evaporator\n",
"#Variable declaration\n",
"mf_dot=10000.0 #[kg/h] of feed\n",
"ic=0.09 #Initial concentration \n",
"fc=0.47 #Final concentration\n",
"m1dot_dash=ic*mf_dot/fc #[kg/h]\n",
"Ps=686.616 #Steam pressure [kPa.g]\n",
"Ps=Ps+101.325 #[kPa]\n",
"Ts=442.7 #Saturation temperature in [K]\n",
"P2=86.660 #Vacuum in second effect in [kPa]\n",
"U1=2326.0 #Overall heat transfer in first effect [W/sq m.K]\n",
"U2=1744.5 #Overall heat transfer in 2nd effect [W/sqm.K]\n",
"P2_abs=101.325-P2 #Absolute pressure in second effect[kPa]\n",
"T2=326.3 #Temperature in 2nd effect in [K]\n",
"dT=Ts-T2 #[K]\n",
"Tf=309.0 #Feed temperature in[K]\n",
"T=273.0 #[K]\n",
"Cpf=3.77 #kJ/kg.K Specific heat for all caustic streams\n",
"#Q1=Q2\n",
"#U1*A1*dT1=U2*A2*dT2\n",
"#Calculation\n",
"dT2=dT/1.75 #[K]\n",
"dT1=(U2/U1)*dT2 #[K]\n",
"#Since there is no B.P.R\n",
"Tv1=Ts-dT1 #Temperature in vapor space of first effect in [K]\n",
"Tv2=Tv1-dT2 #Second effect [K]\n",
"Hf=Cpf*(Tf-T) #Feed enthalpy[kJ/kg]\n",
"H1dash=Cpf*(Tv1-T) #Enthalpy of final product[kJ/kg]\n",
"H2dash=Cpf*(Tv2-T) #kJ/kg\n",
"#For steam at 442.7 K\n",
"lambda_s=2048.7 #[kJ/kg]\n",
"#For vapour at 392.8 K\n",
"Hv1=2705.22 #[kJ/kg]\n",
"lambda_v1=2202.8 #[kJ/kg]\n",
"#for vapour at 326.3 K:\n",
"Hv2=2597.61 #[kJ/kg]\n",
"lambda_v2=2377.8 #[kJ/kg]\n",
"\n",
"#Overall material balance:\n",
"mv_dot=mf_dot-m1dot_dash #[kg/h]\n",
"\n",
"#Equation 4 becomes:\n",
"#mv1_dot*lambda_v1+mf_dot*Hf=(mv_dot-mv1_dot)*Hv2+(mf_dot-mv2_dot)*H2_dash\n",
"mv1_dot=(H2dash*(mf_dot-mv_dot)-mf_dot*Hf+mv_dot*Hv2)/(Hv2+lambda_v1-H2dash) \n",
"mv2_dot=mv_dot-mv1_dot #[kg/h]\n",
"\n",
"#From equation 2\n",
"\n",
"m2dot_dash=m1dot_dash+mv1_dot #First effect material balance[kg/h]\n",
"ms_dot=(mv1_dot*Hv1+m1dot_dash*H1dash-m2dot_dash*H2dash)/lambda_s #[kg/h]\n",
"\n",
"\n",
"#Heat transfer Area\n",
"#First effect\n",
"A1=ms_dot*lambda_s*(10.0**3.0)/(3600.0*U1*dT1) #[sq m]\n",
"\n",
"#Second effect\n",
"lambda_v1=lambda_v1*(10**3.0)/3600.0\n",
"A2=mv1_dot*lambda_v1/(U2*dT2) #[sq m]\n",
"\n",
"#Since A1 not= A2\n",
"\n",
"#SECOND TRIAL\n",
"Aavg=(A1+A2)/2 #[sq m]\n",
"dT1_dash=dT1*A1/Aavg #[K]\n",
"dT2_dash=dT-dT1 #/[K]\n",
"\n",
"#Temperature distribution\n",
"Tv1=Ts-dT1_dash #[K]\n",
"Tv2=Tv1-dT2_dash #[K]\n",
"Hf=135.66 #[kJ/kg]\n",
"H1dash=Cpf*(Tv1-T) #[kJ/kg]\n",
"H2dash=200.83 #[kJ/kg]\n",
"\n",
"#Vapour at 388.5 K\n",
"Hv1=2699.8 #[kJ/kg]\n",
"lambda_v1=2214.92 #[kJ/kg]\n",
"mv1_dot=(H2dash*(mf_dot-mv_dot)-mf_dot*Hf+mv_dot*Hv2)/(Hv2+lambda_v1-H2dash) \n",
"mv2_dot=mv_dot-mv1_dot #[kg/h]\n",
"\n",
"#First effect Energy balance\n",
"ms_dot=((mv1_dot*Hv1+m1dot_dash*H1dash)-(mf_dot-mv2_dot)*H2dash)/lambda_s #[kg/h]\n",
"\n",
"#Area of heat transfer\n",
"lambda_s=lambda_s*1000.0/3600.0 \n",
"A1=ms_dot*lambda_s/(U1*dT1_dash) #[sq m]\n",
"\n",
"#Second effect:\n",
"A2=(mv1_dot*lambda_v1*1000)/(3600.0*U2*dT2_dash) #[sq m]\n",
"\n",
"#Result\n",
"\n",
"print\"A1(\",round(A1,1),\")=A2(\",round(A2),\"),So the area in each effect can be\",round(A1,1),\"m^2\"\n",
"print\"Heat transfer surface in each effect is\",round(A1,1),\"m^2\"\n",
"print\"Steam consumption=\",round(ms_dot),\"(approx)kg/h\"\n",
"print\"Evaporation in the first effect is\",round(mv1_dot),\"kg/h\"\n",
"print\"Evaporation in 2nd effect is\",round(mv2_dot),\"kg/h\" \n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"A1( 24.9 )=A2( 23.0 ),So the area in each effect can be 24.9 m^2\n",
"Heat transfer surface in each effect is 24.9 m^2\n",
"Steam consumption= 5517.0 (approx)kg/h\n",
"Evaporation in the first effect is 4343.0 kg/h\n",
"Evaporation in 2nd effect is 3742.0 kg/h\n"
]
}
],
"prompt_number": 12
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no:6.12 ,Page no:6.37"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#lye in Triple effect evaporator\n",
"#Variable declaration\n",
"Tf=353.0 #[K]\n",
"T=273.0 #[K]\n",
"mf_dot=10000.0 #Feed [kg/h]\n",
"ic=0.07 #Initial conc of glycerine \n",
"fc=0.4 #FinaL CONC OF GLYCERINE\n",
"#Overall glycerine balance\n",
"P=313.0 #Steam pressure[kPa]\n",
"Ts=408.0 #[from steam table][K]\n",
"P1=15.74 #[Pressure in last effect][kPa]\n",
"Tv3=328.0 #[Vapour temperature]\n",
"#Calculation\n",
"m3dot_dash=(ic/fc)*mf_dot #[kg/h]\n",
"mv_dot=mf_dot-m3dot_dash #/[kg/h]\n",
"dT=Ts-Tv3 #Overall apparent [K]\n",
"bpr1=10.0 #[K]\n",
"bpr2=bpr1 \n",
"bpr3=bpr2 \n",
"sum_bpr=bpr1+bpr2+bpr3 #[K]\n",
"dT=dT-sum_bpr #True_Overall\n",
"dT1=14.5 #[K]\n",
"dT2=16.0 #[K]\n",
"dT3=19.5 #[K]\n",
"Cpf=3.768 #[kJ/(kg.K)]\n",
"#Enthalpies of various streams\n",
"Hf=Cpf*(Tf-T) #[kJ/kg]\n",
"H1=Cpf*(393.5-T) #[kJ/kg]\n",
"H2=Cpf*(367.5-T) #[kJ/kg]\n",
"H3=Cpf*(338.0-T) #[kJ/kg]\n",
"#For steam at 40K\n",
"lambda_s=2160.0 #[kJ/kg]\n",
"Hv1=2692.0 #[kJ/kg]\n",
"lambda_v1=2228.3 #[kJ/kg]\n",
"Hv2=2650.8 #[kJ/kg]\n",
"lambda_v2=2297.4 #[kJ/kg]\n",
"Hv3=2600.5 #[kJ/kg]\n",
"lambda_v3=2370.0 #[kJ/kg]\n",
"\n",
"#MATERIAL AND EBERGY BALANCES\n",
"#First effect\n",
"#Material balance\n",
"\n",
"#m1dot_dash=mf_dot-mv1_dot\n",
"#m1dot_dash=1750+mv2_dot+mv3_dot \n",
"\n",
"#Energy balance\n",
"#ms_dot*lambda_s+mf_Dot*hf=mv1_dot*Hv1+m1dot_dash*H1\n",
"#2160*ms_dot+2238*(mv2_dot+mv3_dot)=19800500\n",
"\n",
"#Second effect\n",
"#Energy balance:\n",
"#mv3_dot=8709.54-2.076*mv2_dot\n",
"\n",
"#Third effect:\n",
"#m2dot_dash=mv3_dot+m3dot_dash\n",
"#m2dot_dash=mv3_dot+1750\n",
"#From eqn 8 we get\n",
"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)\n",
"#From eqn 8:\n",
"mv3_dot=8709.54-2.076*mv2_dot #[kg/h]\n",
"mv1_dot=mv_dot-(mv2_dot+mv3_dot) #[kg/h]\n",
"#From equation 4:\n",
"#m1dot_dash=mf_dot-mv1_dot\n",
"#ms_dot=(mv1_dot*Hv1+m1dot_dash*H1-mf_dot*Hf)/lambda_s #[kg/h]\n",
"ms_dot=(19800500.0-2238.0*(mv2_dot+mv3_dot))/2160.0 #[kg/h]\n",
"\n",
"#Heat transfer Area is\n",
"U1=710.0 #[W/sq m.K]\n",
"U2=490.0 #[W/sq m.K]\n",
"U3=454.0 #[W/sq m.K]\n",
"A1=(ms_dot*lambda_s*1000.0)/(3600.0*U1*dT1) #[sq m]\n",
"A2=mv1_dot*lambda_v1*1000.0/(3600.0*U2*dT2) #[sq m]\n",
"A3=mv2_dot*lambda_v2*1000.0/(3600.0*U3*dT3) #[sq m]\n",
"#The deviaiton is within +-10%\n",
"#Hence maximum A1 area can be recommended\n",
"\n",
"eco=(mv_dot/ms_dot) #[Steam economy]\n",
"\n",
"Qc=mv3_dot*lambda_v3 #[kJ/h]\n",
"dT=25.0 #Rise in water temperature\n",
"Cp=4.187\n",
"mw_dot=Qc/(Cp*dT)\n",
"#Result\n",
"print\"ANSWER:Area in each effect\",round(A3,1),\"sq m\" \n",
"print\"Steam economy is\",round(eco,2) \n",
"print\"Cooling water rate is\",round(mw_dot/1000,2),\"t/h\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"ANSWER:Area in each effect 200.2 sq m\n",
"Steam economy is 2.55\n",
"Cooling water rate is 66.63 t/h\n"
]
}
],
"prompt_number": 48
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no:6.13 ,Page no:6.42"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Triple effect unit\n",
"#Variable declaration\n",
"Cpf=4.18 #[kJ/kg.K]\n",
"dT1=18 #[K]\n",
"dT2=17 #[K]\n",
"dT3=34 #[K]\n",
"mf_dot=4 #[kg/s]\n",
"Ts=394 #[K]\n",
"bp=325 #Bp of water at 13.172 kPa [K]\n",
"dT=Ts-bp #[K]\n",
"lambda_s=2200 #[kJ/kg]\n",
"T1=Ts-dT1 #[K]\n",
"lambda1=2249 #[kJ/kg]\n",
"lambda_v1=lambda1 #[kJ/kg]\n",
"#Calculation\n",
"T2=T1-dT2 #[K]\n",
"lambda2=2293 #[kJ/kg]\n",
"lambda_v2=lambda2 #[kJ/kg]\n",
"\n",
"T3=T2-dT3 #[K]\n",
"lambda3=2377 #[kJ/kg]\n",
"lambda_v3=lambda3 #[kJ/kg]\n",
"\n",
"ic=0.1 #Initial conc of solids\n",
"fc=0.5 #Final conc of solids\n",
"m3dot_dash=(ic/fc)*mf_dot #[kg/s]\n",
"mv_dot=mf_dot-m3dot_dash #Total evaporation in [kg/s]\n",
"#Material balance over first effect\n",
"#mf_dot=mv1_dot_m1dot_dash\n",
"#Energy balance:\n",
"#ms_dot*lambda_s=mf_dot*(Cpf*(T1-Tf)+mv1_dot*lambda_v1)\n",
"\n",
"#Material balance over second effect\n",
"#m1dot_dash=mv2_dot+m2dot_dash\n",
"#Enthalpy balance:\n",
"#mv1_dot*lambda_v1+m1dot_dash(cp*(T1-T2)=mv2_dot*lambda_v2)\n",
"\n",
"#Material balance over third effect\n",
"#m2dot_dash=mv3_dot+m3dot+dash\n",
"\n",
"#Enthalpy balance:\n",
"#mv2_lambda_v2+m2dot_dash*cp*(T2-T3)=mv3_dot*lambda_v3\n",
"294\n",
"mv2_dot=3.2795/3.079 #[kg/s]\n",
"mv1_dot=1.053*mv2_dot-0.1305 #[kg/s]\n",
"mv3_dot=1.026*mv2_dot+0.051 #[kg/s]\n",
"ms_dot=(mf_dot*Cpf*(T1-294)+mv1_dot*lambda_v1)/lambda_s #[kg/s]\n",
"eco=mv_dot/ms_dot #Steam economy \n",
"eco=round(eco)\n",
"U1=3.10 #[kW/sq m.K]\n",
"U2=2 #[kW/sq m.K]\n",
"U3=1.10 #[kW/sq m.K]\n",
"#First effect:\n",
"A1=ms_dot*lambda_s/(U1*dT1) #[sq m]\n",
"A2=mv1_dot*lambda_v1/(U2*dT2) #[sq m]\n",
"A3=mv2_dot*lambda_v2/(U3*dT3) #[sq m]\n",
"#Areas are calculated witha deviation of +-10%\n",
"#Result\n",
"print\"Steam economy is\",eco \n",
"print\"Area pf heat transfer in each effect is\",round(A3,1),\"m^2\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Steam economy is 2.0\n",
"Area pf heat transfer in each effect is 65.3 m^2\n"
]
}
],
"prompt_number": 9
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no: 6.14,Page no:6.45"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Quadruple effect evaporator\n",
"#Variable declaration\n",
"mf_dot=1060 #[kg/h]\n",
"ic=0.04 #Initial concentration\n",
"fc=0.25 #Final concentration\n",
"m4dot_dash=(ic/fc)*mf_dot #[kg/h]\n",
"#Total evaporation=\n",
"mv_dot=mf_dot-m4dot_dash #[kg/h]\n",
"\n",
"#Fromsteam table:\n",
"P1=370 #[kPa.g]\n",
"T1=422.6 #[K]\n",
"lambda1=2114.4 #[kJ/kg]\n",
"\n",
"P2=235 #[kPa.g]\n",
"T2=410.5 #[K]\n",
"lambda2=2151.5 #[kJ/kg]\n",
"\n",
"P3=80 #[kPa.g]\n",
"T3=390.2 #[K]\n",
"lambda3=2210.2 #[kJ/kg]\n",
"\n",
"P4=50.66 #[kPa.g]\n",
"T4=354.7 #[K]\n",
"lambda4=2304.6 #[kJ/kg]\n",
"\n",
"P=700 #Latent heat of steam[kPa .g]\n",
"lambda_s=2046.3 #[kJ/kg]\n",
"\n",
"#Calculation\n",
"#FIRST EFFECT\n",
"#Enthalpy balance:\n",
"#ms_dot=mf_dot*Cpf*(T1-Tf)+mv1_dot*lambda1\n",
"#ms_dot=1345.3-1.033*m1dot_dash\n",
"\n",
"#SECOND EFFECT\n",
"#m1dot_dash=m2dot_dash+mdot_v2\n",
"#Enthalpy balance:\n",
"#m1dot_dash=531.38+0.510*m2dot_dash\n",
"\n",
"#THIRD EFFECT\n",
"#Material balance:\n",
"#m2dot_dash-m3dot_dash+mv3_dot\n",
"\n",
"#FOURTH EFFECT\n",
"#m3dot_dash=m4dot_dash+mv4_dot\n",
"mv4dot_dash=169.6 #[kg/h]\n",
"m3dot_dash=416.7 #[kg/h]\n",
"\n",
"#From eq n 4:\n",
"m2dot_dash=-176.84+1.98*m3dot_dash #[kg/h]\n",
"\n",
"#From eqn 2:\n",
"m1dot_dash=531.38+0.510*m2dot_dash #[kg/h]\n",
"\n",
"#From eqn 1:\n",
"ms_dot=1345.3-1.033*m1dot_dash\n",
"eco=mv_dot/ms_dot #[kg evaporation /kg steam]\n",
"#Result\n",
"print\"Steam economy is\",round(eco,3),\"evaporation/kg steam\" \n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Steam economy is 1.957 evaporation/kg steam\n"
]
}
],
"prompt_number": 50
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no:6.15 ,Page no:6.48"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Single effect Calendria\n",
"import math\n",
"#Variable declaration\n",
"m1_dot=5000 #[kg/h]\n",
"ic=0.1 #Initial concentration\n",
"fc=0.5 #Final concentration\n",
"mf_dot=(fc/ic)*m1_dot #[kg/h]\n",
"mv_dot=mf_dot-m1_dot #Water evaporated[kg/h]\n",
"P=357 #Steam pressure[kN/sq m]\n",
"Ts=412 #[K]\n",
"H=2732 #[kJ/kg]\n",
"lambda1=2143 #[kJ/kg]\n",
"bpr=18.5 #[K]\n",
"T_dash=352+bpr #[K]\n",
"Hf=138 #[kJ/kg]\n",
"lambda_s=2143 #[kJ/kg]\n",
"Hv=2659 #[kJ/kg]\n",
"H1=568 #[kJ/kg]\n",
"#Calculation\n",
"ms_dot=(mv_dot*Hv+m1_dot*H1-mf_dot*Hf)/lambda_s #Steam consumption in kg/h\n",
"eco=mv_dot/ms_dot #Economy\n",
"dT=Ts-T_dash #[K]\n",
"hi=4500 #[W/sq m.K]\n",
"ho=9000 #[W/sq m.K]\n",
"Do=0.032 #[m]\n",
"Di=0.028 #[m]\n",
"x1=(Do-Di)/2 #[m]\n",
"Dw=(Do-Di)/math.log(32.0/28.0) #[m]\n",
"x2=0.25*10**-3 #[m]\n",
"L=2.5 #Length [m]\n",
"hio=hi*(Di/Do) #[W/sq m.K]\n",
"print\"NOTE:In textbook this value of hio is wrongly calculated as 3975.5..So we will take this\"\n",
"hio=3975.5\n",
"k1=45.0 #Tube material in [W/sq m.K]\n",
"k2=2.25 #For scale[W/m.K]\n",
"Uo=1.0/(1.0/ho+1.0/hio+(x1*Dw)/(k1*Do)+(x2/k2)) #Overall heat transfer coeff in W/sq m.K\n",
"Q=ms_dot*lambda_s #[kJ/h]\n",
"Q=Q*1000.0/3600.0 #[W]\n",
"\n",
"A=Q/(Uo*dT) #[sq m]\n",
"n=A/(math.pi*Do*L) #from A=n*math.pi*Do*L \n",
"#Result\n",
"print\"Steam consumption is\",round(ms_dot),\"kg/h\" \n",
"print\"Capacity is\",round(mv_dot),\"kg/h\"\n",
"print\"Steam economy is \",round(eco,3)\n",
"print\" No. of tubes required is \",round(n)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"NOTE:In textbook this value of hio is wrongly calculated as 3975.5..So we will take this\n",
"Steam consumption is 24531.0 kg/h\n",
"Capacity is 20000.0 kg/h\n",
"Steam economy is 0.815\n",
" No. of tubes required is 722.0\n"
]
}
],
"prompt_number": 10
},
{
"cell_type": "heading",
"level": 2,
"metadata": {
"slideshow": {
"slide_type": "subslide"
}
},
"source": [
"Example no:6.16 ,Page no:6.50"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Single effect evaporator\n",
"#Variable declaration\n",
"bpr=40.6 #[K]\n",
"Cpf=1.88 #[kJ/kg.K]\n",
"Hf=214 #[kJ/kg]\n",
"H1=505 #[kJ/kg]\n",
"mf_dot=4536 #[kg/h] of feed solution\n",
"ic=0.2 #Initial conc\n",
"fc=0.5 #Final concentration\n",
"m1dot_dash=(ic/fc)*mf_dot #Thisck liquor flow arte[kg/h]\n",
"mv_dot=mf_dot-m1dot_dash #[kg/H]\n",
"Ts=388.5 #Saturation temperature of steam in [K]\n",
"bp=362.5 #b.P of solution in [K]\n",
"lambda_s=2214 #[kJ/kg]\n",
"P=21.7 #Vapor space in [kPa]\n",
"Hv=2590.3 #[kJ/kg]\n",
"\n",
"#Calculation\n",
"#Enthalpy balance over evaporator\n",
"ms_dot=(m1dot_dash*H1+mv_dot*Hv-mf_dot*Hf)/lambda_s #[kg/h\n",
"print\"Steam consumption is\",round(ms_dot,1),\"kg/h\" \n",
"dT=Ts-bp #[K]\n",
"U=1560 #[W/sq m.K]\n",
"Q=ms_dot*lambda_s #[kJ/h]\n",
"Q=Q*1000/3600 #[W]\n",
"A=Q/(U*dT) #[sq m]\n",
"print\"Heat transfer area is\",round(A,2),\"m^2\"\n",
"\n",
"#Calculations considering enthalpy of superheated vapour\n",
"\n",
"Hv=Hv+Cpf*bpr #[kJ/kg]\n",
"ms_dot=(m1dot_dash*H1+mv_dot*Hv-mf_dot*Hf)/lambda_s #[kg/h]\n",
"print\" Now,Steam consumption is\",round(ms_dot,2),\"kg/h\" \n",
"eco=mv_dot/ms_dot #Steam economy\n",
"print\"Economy of evaporator \",round(eco,2)\n",
"Q=ms_dot*lambda_s #[kJ/h]\n",
"Q=Q*1000.0/3600.0 #[w]\n",
"A2=Q/(U*dT) #Area\n",
"print\"Now,Area is\",round(A2,2) \n",
"perc=(A2-A)*100/A #%error in the heat transfer area \n",
"#Result\n",
"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\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Steam consumption is 3159.6 kg/h\n",
"Heat transfer area is 47.91 m^2\n",
" Now,Steam consumption is 3253.42 kg/h\n",
"Economy of evaporator 0.84\n",
"Now,Area is 49.33\n",
"If enthalpy of water vapour Hv were based on the saturated vapour at the pressure\n",
"the error introduced is only 2.97 percent\n"
]
}
],
"prompt_number": 11
}
],
"metadata": {}
}
]
}