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{
"metadata": {
"name": ""
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 1:Water"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no:1,Page no:1.7"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"W1=219.0 #amount of Mg(HCO3)2 in water in ppm#\n",
"W2=36.0 #amount of Mg2+ in water in ppm#\n",
"W3=18.3 #amount of (HCO3)- in water in ppm#\n",
"W4=1.5 #amount of H+_in water in ppm#\n",
"M1=100/146.0 #multiplication factor of Mg(HCO3)2#\n",
"M2=100/24.0 #multiplication factor of Mg(HCO3)2#\n",
"M3=100/122.0 #multiplication factor of Mg(HCO3)2#\n",
"M4=100/2.0 #multiplication factor of Mg(HCO3)2#\n",
"\n",
"#Calculation\n",
"P1=W1*M1 #in terms of CaCO3#\n",
"P2=W2*M2 #in terms of CaCO3#\n",
"P3=W3*M3 #in terms of CaCO3#\n",
"P4=W4*M4 #in terms of CaCO3#\n",
"L=0.74*((2*P1)+P2+P3+P4) \n",
"\n",
"R=1.0 #water supply rate in m**3/s#\n",
"D=R*60.0*60.0*24.0*L \n",
"K=D*1000.0 #in lit/day#\n",
"T=K/10.0**9 #in tonnes#\n",
"S=1.06*(P2+P4-P3) \n",
"D2=R*60*60*24*S \n",
"A=D2*1000 #in lit/day#\n",
"B=A/10.0**9 #in tonnes#\n",
"J1=90/100.0 #purity of lime#\n",
"J2=95/100.0 #purity of soda#\n",
"C1=500.0 #cost of one tonne lime#\n",
"C2=7000.0 #cost of one tonne soda#\n",
"CL=round(T,1)*C1/J1 \n",
"print\"\\ncost of lime is\",CL,\"Rs\"\n",
"CS=round(B,1)*C2/J2 \n",
"print\"\\ncost of soda is \",CS,\"Rs\"\n",
"C=CL+CS \n",
"\n",
"#Result\n",
"print\"\\ntotal cost is \",round(C) ,\"Rs\"\n",
"\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"cost of lime is 19166.6666667 Rs\n",
"\n",
"cost of soda is 141473.684211 Rs\n",
"\n",
"total cost is 160640.0 Rs\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no:15,Page no:1.25"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"W1=40.0 #amount of Ca2+ in water in mg/l#\n",
"W2=24.0 #amount of Mg2+ in water in mg/l#\n",
"W3=8.05 #amount of Na+ in water in mg/l#\n",
"W4=183.0 #amount of (HCO3)- in water in mg/l#\n",
"W5=55.68 #amount of (SO4)2- in water in mg/l#\n",
"W6=6.74 #amount of Cl- in water in mg/l#\n",
"M1=100/40.0 #multiplication factor of Ca2+#\n",
"M2=100/24.0 #multiplication factor of Mg2+#\n",
"M3=100/(23.0*2) #multiplication factor of Na+#\n",
"M4=100/(61.0*2) #multiplication factor of (HCO3)-#\n",
"M5=100/96.0 #multiplication factor of (SO4)2-#\n",
"M6=100/(35.5*2) #multiplication factor of Cl-#\n",
"\n",
"#Calculation\n",
"P1=W1*M1 #in terms of CaCO3#\n",
"P2=W2*M2 #in terms of CaCO3#\n",
"P3=W3*M3 #in terms of CaCO3#\n",
"P4=W4*M4 #in terms of CaCO3#\n",
"P5=W5*M5 #in terms of CaCO3#\n",
"P6=W6*M6 #in terms of CaCO3#\n",
"\n",
"\n",
"#Result\n",
"print\"\\nCalcium alkalinity =\",P1,\"ppm\" \n",
"print\"\\nMagnesium alkalinity =\",P4-P1,\"ppm\"\n",
"print\"\\n total alkalinity = \",P1+P4-P1,\"ppm\"\n",
"print\"\\n total hardness = \",P1+P2,\"ppm\"\n",
"print\"\\nCa temporary hardness = \",P1,\"ppm\"\n",
"print\"\\nMg temporary hardness = \",P4-P1,\"ppm\"\n",
"print\"\\nMg permanant hardness = \",P2-(P4-P1),\"ppm\"\n",
"print\"\\nSalts are:\"\n",
"print\"\\nCa(HCO3)2 salt = \",P1,\"ppm\"\n",
"print\"\\nMg(HCO3)2 salt = \",P4-P1,\"ppm\"\n",
"print\"\\nMgSO4 salt = \",P2-(P4-P1),\"ppm\"\n",
"print\"\\nNaCl salt = \",P6,\"ppm\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"Calcium alkalinity = 100.0 ppm\n",
"\n",
"Magnesium alkalinity = 50.0 ppm\n",
"\n",
" total alkalinity = 150.0 ppm\n",
"\n",
" total hardness = 200.0 ppm\n",
"\n",
"Ca temporary hardness = 100.0 ppm\n",
"\n",
"Mg temporary hardness = 50.0 ppm\n",
"\n",
"Mg permanant hardness = 50.0 ppm\n",
"\n",
"Salts are:\n",
"\n",
"Ca(HCO3)2 salt = 100.0 ppm\n",
"\n",
"Mg(HCO3)2 salt = 50.0 ppm\n",
"\n",
"MgSO4 salt = 50.0 ppm\n",
"\n",
"NaCl salt = 9.49295774648 ppm\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no:19,Page no:1.39"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"P=4.7 #required HCl in ml using HpH indicator #\n",
"H=10.5 #required HCl im ml using MeOH indicator#\n",
"M=P+H \n",
"N=0.02 #normality of HCl#\n",
"\n",
"print\"\\nSince P<0.5*M sample contain (CO3)2- and (HCO3)- alkalinity\"\n",
"C=50 #equivalent of CaCO3 in mg for 1ml 1N HCl#\n",
"\n",
"#Calculation\n",
"A=C*(2*P)*N #amount of (CO3)2- alkalinity in mg in 100 ml of water#\n",
"B=A*1000/100 \n",
"D=C*(M-2*P)*N #the amount of (HCO3)- alkalinity in mg in 100 ml of water#\n",
"E=D*1000/100 \n",
"T=B+E \n",
"\n",
"#Result\n",
"print\"\\nTotal alkalinity is \",T,\"mg/l or ppm\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"Since P<0.5*M sample contain (CO3)2- and (HCO3)- alkalinity\n",
"\n",
"Total alkalinity is 152.0 mg/l or ppm\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no:25,Page no:1.56"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"W1=160.0 #amount of Ca2+ in ppm#\n",
"W2=88.0 #amount of Mg2+ in ppm#\n",
"W3=72.0 #amount of CO2 in ppm#\n",
"W4=488.0 #amount of (HCO3)- in ppm#\n",
"W5=139.0 #amount of (FeSO4).7H2O in ppm#\n",
"M1=100/40.0 #multiplication factor of Ca2+#\n",
"M2=100/24.0 #multiplication factor of Mg2+#\n",
"M3=100/44.0 #multiplication factor of CO2#\n",
"M4=100/(61.0*2.0) #multiplication factor of (HCO3)-#\n",
"M5=100/278.0 #multiplication factor of (FeSO4).7H2O#\n",
"\n",
"P1=400 #in terms of CaCO3#\n",
"P2=300 #in terms of CaCO3#\n",
"P3=200 #in terms of CaCO3#\n",
"P4=400 #in terms of CaCO3#\n",
"P5=50 #in terms of CaCO3#\n",
"V=100000.0 #volume of water in litres#\n",
"\n",
"\n",
"#Calculation\n",
"L=0.74*(P2+P3+P4+P5)*V #lime required in mg#\n",
"L=L/10.0**6 \n",
"S=1.06*(P1+P2+P5-P4)*V #soda required in mg#\n",
"S=S/10.0**6 \n",
"\n",
"#Result\n",
"print\"Lime required is \",L,\"kg\"\n",
"print\"\\nSoda required is \",S,\"kg\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Lime required is 70.3 kg\n",
"\n",
"Soda required is 37.1 kg\n"
]
}
],
"prompt_number": 4
},
],
"metadata": {}
}
]
}
|
