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{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 8 : Filtration"
]
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"example 8.1 page no : 145"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"import math \n",
"from numpy import *\n",
"\n",
"# Initialization of Variable\n",
"\n",
"a = 78./1000 #dV/dt\n",
"rho = 998. #density of water\n",
"rhoc = 2230. #density of china clay\n",
"rhod = 1324. #density of cowdung cake\n",
"mu = 1.003/1000\n",
"P2 = 3.23*1000 #pressure after 2 min.\n",
"P5 = 6.53*1000 #pressure after 5 min.\n",
"t = 30*60.\n",
"b = array([[P2],[P5]])\n",
"A = array([[(a**2)*120., a],[(a**2)*300., a]])\n",
"x = linalg.solve(A, b)\n",
"P = x[0]*a**2*t+x[1]*a\n",
"print \"pressure drop at t = 30min in (kN/m**2):\",P/1000\n",
"\n",
"#part2\n",
"J = 0.0278 #mass fraction\n",
"l = 1.25\n",
"b1 = 0.7\n",
"A1 = l*b1*17*2 #area of filtering\n",
"V = a*30*60. #volume of filterate\n",
"e = 1-rhod/rhoc\n",
"nu = J*rho/((1-J)*(1-e)*rhoc-J*e*rho)\n",
"l1 = nu*V/A1\n",
"print \"the thickness of filtercake formed after 30 min in (m): %.4f\"%l1\n",
"\n",
"#part3\n",
"r = x[0][0]/mu/nu*A1**2\n",
"L = x[1][0]*A1/r/mu\n",
"print \"thickness of cake required in (m): %.4f\"%L\n",
"\n",
"#part 4\n",
"S = math.sqrt(r*e**3./5/(1-e)**2)\n",
"d = 6./S\n",
"print \"average particle diameter in(10**-6m): %.4f\"%(d*10**6)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"pressure drop at t = 30min in (kN/m**2): [ 34.03]\n",
"the thickness of filtercake formed after 30 min in (m): 0.1026\n",
"thickness of cake required in (m): 0.0032\n",
"average particle diameter in(10**-6m): 87.9625\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"example 8.2 pageno :148"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"import math \n",
"from numpy import *\n",
"\n",
"\n",
"# Initialization of Variable\n",
"P1 = 5.34*1000 #pressure after 3 min.\n",
"P2 = 9.31*1000 #pressure after 8 min.\n",
"a = 240./1000000 #dV/dt\n",
"P3 = 15.*10**3 #final pressure\n",
"\n",
"#calculation\n",
"b = array([[P1],[P2]])\n",
"A = array([[a**2*180, a],[a**2*480, a]])\n",
"x = linalg.solve(A,b)\n",
"\n",
"#part1\n",
"t = (P3-x[1][0]*a)/x[0][0]/a**2\n",
"\n",
"print \"time at which the required pressure drop have taken place in (s): %.4f\"%t\n",
"\n",
"#part 2\n",
"V1 = a*t\n",
"print \"volume of filterate in (m**3): %.4f\"%V1\n",
"\n",
"#part 3\n",
"V2 = 0.75\n",
"t2 = t+x[0][0]/2/P3*(V2**2-V1**2)+x[1][0]/P3*(V2-V1)\n",
"print \"the time required to collect 750dm**3 of filterate in (s): %.4f\"%t2\n",
"\n",
"#part 4\n",
"P4 = 12.*10**3\n",
"a = P4/(x[0][0]*V2+x[1][0])\n",
"t = 10./1000/a\n",
"print \"time required to pass 10dm**3 volume in (s): %.4f\"%t\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"time at which the required pressure drop have taken place in (s): 909.9748\n",
"volume of filterate in (m**3): 0.2184\n",
"the time required to collect 750dm**3 of filterate in (s): 5289.2396\n",
"time required to pass 10dm**3 volume in (s): 153.8617\n"
]
}
],
"prompt_number": 15
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"example 8.3 pageno : 150"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"import math \n",
"from numpy import *\n",
"\n",
"# Initialization of Variable\n",
"a = 16./1000 #dV/dt\n",
"J = 0.0876 #mass fraction\n",
"rho = 999. #density of water\n",
"rhoc = 3470. #density of slurry\n",
"mu = 1.12/1000\n",
"rhos = 1922. #density of dry filter cake\n",
"t1 = 3*60.\n",
"t2 = 8*60.\n",
"V1 = 33.8/1000 #volume at t1\n",
"V2 = 33.8/1000+23.25/1000 #volume at t2\n",
"P = 12*1000. #pressure difference\n",
"Ap = 70.**2./10000*2*9\n",
"As = 650/10000.\n",
"\n",
"#calculation\n",
"\n",
"b = array([t1,t2])\n",
"A = array([[V1**2/2/P, V1/P],[V2**2/2/P, V2/P]])\n",
"x = linalg.solve(A, b)\n",
"K1p = x[0]*As**2/Ap**2\n",
"K2p = x[1]*As/Ap\n",
"P2 = 15*1000. #final pressure drop\n",
"t = (P2-K2p*a)/K1p/a**2 #time for filterate\n",
"V = a*t #volume of filterate\n",
"e = 1-rhos/rhoc\n",
"nu = J*rho/((1-J)*(1-e)*rhoc-J*e*rho)\n",
"l = (11.-1)/200.\n",
"Vf = Ap*l/nu\n",
"tf = t+K1p/2/P2*(Vf**2-V**2)+K2p/P2*(Vf-V)\n",
"r = K1p/mu/nu*Ap**2\n",
"L = K2p*Ap/r/mu\n",
"print \"the thickness of filter which has resistance equal to resistance of filter medium in (m):%.5f\"%L\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"the thickness of filter which has resistance equal to resistance of filter medium in (m):0.00247\n"
]
}
],
"prompt_number": 8
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"example 8.4 page no : 154"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"import math \n",
"from numpy import *\n",
"\n",
"# Initialization of Variable\n",
"\n",
"t1 = 3*60. #time 3min\n",
"t2 = 12*60. #time 12min\n",
"t3 = 5*60. #time 5min\n",
"P = 45*1000. #pressure at t1&t2\n",
"P2 = 85*1000. #pres. at t3\n",
"a = 1.86 #area\n",
"mu = 1.29/1000.\n",
"c = 11.8\n",
"V1 = 5.21/1000 #volume at t1\n",
"V2 = 17.84/1000 #volume at t2\n",
"V3 = 10.57/1000 #volume at t3\n",
"\n",
"#calculation\n",
"b = array([t1,t2])\n",
"A = array([[mu*c/2/a**2/P*V1**2, V1/P],[mu*c/2/a**2/P*V2**2, V2/P]])\n",
"x = linalg.solve(A,b)\n",
"r45 = x[0]\n",
"r85 = (t3-x[1]*V3/P2)*2*a**2*P2/V3**2/mu/c\n",
"n = math.log(r45/r85)/math.log(45./85)\n",
"rbar = r45/(1-n)/(45.*1000)**n\n",
"r78 = rbar*(1-n)*(78.*1000)**n\n",
"\n",
"#part1\n",
"#polynomial in V as a1x**2+bx+c1 = 0\n",
"c1 = 90.*60 #time at 90 \n",
"Pt = 78*1000. #Pt = pressure at time t = 90\n",
"r78 = round(r78/10.**12)*10.**12\n",
"a1 = r78*mu/a**2/Pt*c/2.\n",
"b = x[1]/Pt\n",
"y = poly1d([a1,b,-c1],False)\n",
"V1 = roots(y)\n",
"print \"Volume at P = 90kPa in (m**3): %.4f\"%V1[1]\n",
"\n",
"#part2\n",
"Pt = 45.*1000\n",
"c1 = 90.*60\n",
"a1 = r45*mu/a**2/Pt*c/2\n",
"b = x[1]/Pt\n",
"y = poly1d([a1,b,-c1],False)\n",
"V1 = roots(y)\n",
"print \"Volume at p = 45kPa in (m**3): %.4f\"%V1[1]\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Volume at P = 90kPa in (m**3): 0.0660\n",
"Volume at p = 45kPa in (m**3): 0.0789\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"example 8.5 page no : 157"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"from numpy import *\n",
"import math \n",
"\n",
"\n",
"# Initialization of Variable\n",
"t = 60*0.3/0.5 #time of 1 revollution\n",
"d = 34/1000000.\n",
"S = 6./d\n",
"e = 0.415\n",
"J = 0.154\n",
"P = 34.8*1000\n",
"mu = 1.17/1000\n",
"L = 2.35/1000\n",
"rho = 999. #density of water\n",
"rhos = 4430. #density of barium carbonate\n",
"\n",
"#calculation\n",
"#part1\n",
"nu = J*rho/((1-J)*(1-e)*rhos-J*e*rho)\n",
"r = 5*S**2*(1-e)**2/e**3\n",
"\n",
"#quadratic in l\n",
"#in the form of ax**2+bx+c = 0\n",
"c = -t\n",
"b = r*mu*L/nu/P\n",
"a = r*mu/2/nu/P\n",
"y = poly1d([a,b,c],False)\n",
"l = roots(y)\n",
"print \"thickness of filter cake in (m): %.4f\"%l[1]\n",
"\n",
"#part2\n",
"d = 1.2\n",
"l1 = 2.6\n",
"pi = 3.1428\n",
"u = pi*d*0.5/60\n",
"Q = u*l1*l[1]\n",
"mnet = Q*(1-e)*rhos+Q*e*rho\n",
"print \"rate at which wet cake will be scrapped in (kg/s): %.4f\"%mnet\n",
"\n",
"#part3\n",
"md = Q*(1-e)*rhos #rate at which solid scrapped from the drum\n",
"r = md/0.154\n",
"print \"rate of which slurry is treated is (kg/h): %.4f\"%(r*3600)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"thickness of filter cake in (m): 0.0122\n",
"rate at which wet cake will be scrapped in (kg/s): 3.0088\n",
"rate of which slurry is treated is (kg/h): 60635.4180\n"
]
}
],
"prompt_number": 7
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"example 8.6 page no : 159"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"import math \n",
"\n",
"# Initialization of Variable\n",
"mu = 0.224\n",
"rho = 1328.\n",
"K = 5.\n",
"b = 3*.5 #radius\n",
"h = 2.5\n",
"pi = 3.1428\n",
"x = 2.1*.5\n",
"rhos = 1581. #density of sucrose\n",
"e = 0.435 #void ratio\n",
"J = 0.097 #mass fraction\n",
"m = 3500. #mass flowing\n",
"a = 85/10.**6 #side length\n",
"L = 48./1000 #thickness\n",
"omega = 2*pi*325./60.\n",
"\n",
"#calculation\n",
"bi = b**2-m/pi/h/(1-e)/rhos #inner radius\n",
"bi = math.sqrt(bi)\n",
"bi = round(bi*1000)/1000.\n",
"nu = J*rho/((1-J)*(1-e)*rhos-J*e*rho)\n",
"S = 6./a\n",
"r = 5*S**2*(1-e)**2/e**3\n",
"t = ((b**2-bi**2)*(1+2*L/b)+2*bi**2*math.log(bi/b))/(2*nu*rho*omega**2/r/mu*(b**2-x**2))\n",
"print \"time taken to collect sucrose crystal in (s): %.4f\"%t\n",
"\n",
"#part2\n",
"vl = pi*(b**2-bi**2)*h*e\n",
"vs = pi*(b**2-bi**2)*h/nu-vl\n",
"print \"volume of liquid separated as filterate i (m**3): %.4f\"%vs\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"time taken to collect sucrose crystal in (s): 3287.3308\n",
"volume of liquid separated as filterate i (m**3): 21.1677\n"
]
}
],
"prompt_number": 8
},
{
"cell_type": "code",
"collapsed": false,
"input": [],
"language": "python",
"metadata": {},
"outputs": []
}
],
"metadata": {}
}
]
}