{
"cells": [
 {
		   "cell_type": "markdown",
	   "metadata": {},
	   "source": [
       "# Chapter 7: Liquid filtration"
	   ]
	},
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 7.1: volume_of_filtrate_collected_per_cycle.sci"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"printf('\n Example 7.1');\n",
"//In the leaf filter filtration is at const pressure from the start\n",
"//V^2 + 2ALV/v = 2(-deltaP)A^2t/(ruv)\n",
"\n",
"//In the filter press,a volume V1 of filtrate is obtained under const rate conditions in time t1,and filtration is then carried out at constant pressure.\n",
"//V1^2 + 2ALV1/v = 2(-deltaP)A^2t1/(ruv) \n",
"//and (V^2 − V1^2 ) + 2AL/υ(V − V1) = 2(−P)A^2/rμυ(t − t1)\n",
"\n",
"//for the leaf filter\n",
"t2 = 300;          //t2 is in secs\n",
"V2 = 2.5*10^(-4);   //V2 is in m^3\n",
"t3 = 600;           //t3 is in secs\n",
"V3 = 4*10^(-4);     //V3 is in m^3\n",
"A = 0.05;           //A is in m^2\n",
"deltaP = -7.13*10^(4); //it is in N/m^2\n",
"//putting these values in above eq\n",
"\n",
"a = [2*7.13*10^(4)*0.05^(2)*300 -2*0.05*2.5*10^(-4);2*7.13*10^(4)*0.05^(2)*600 -2*0.05*4*10^(-4)];\n",
"b = [(2.5*10^(-4))^2;(4*10^(-4))^2];\n",
"x = inv(a)*b;\n",
"y = [1/x(1);x(2)];\n",
"printf('\n L/υ=%f*10^(-3)   and rμυ = %f*10^(11)',y(2)*10^3,y(1)*10^(-11));\n",
"\n",
"//for the filter press\n",
"V1 = poly([0],'V1');\n",
"s = roots(V1^2 + (2.16*y(2)*V1)-(4*10^(5)*2.16^2)/y(1)*180);\n",
"printf('\n the value of V1 = %fm^3',s(2));\n",
"\n",
"//For a constant pressure period (t - t1)=900secs\n",
"//Calculting the total volume of filtrate\n",
"V = poly([0],'V');\n",
"d = roots((V^2-3.33*10^(-4))+(1.512*10^(-2)*(V-1.825*10^(-2))-5.235*10^(-6)*900));\n",
"printf('\n The value of V = %.3f m^3',d(2));\n",
"\n",
"f = (4*10^(5)*(2.16)^2)/(7.13*10^(11)*(6.15*10^(-2) + 2.16*3.5*10^(-3)));\n",
"printf('\n The final rate of filtration is %.2f*10^(-5) m^3/sec',f*10^(5));\n",
"\n",
"// Assuming  viscosity of the filtrate is the same as that of the wash-water\n",
"rw_400 = (0.25)*f;\n",
"printf('\n Rate of washing at 400 kN/m2 = %.1f*10^(-6) m^3/sec',rw_400*10^(6));\n",
"\n",
"rw_275 = rw_400*(275/400);\n",
"printf('\n Rate of washing at 275 kN/m^2 = %.1f*10^(-6) m^3/sec',rw_275*10^6);\n",
"printf('\n Thus the amount of wash-water passing in 600s = %.3f m^3',600*rw_275);\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 7.2: Effect_on_optimium_thickness_of_the_cake.sci"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"printf('\n Example 7.2');\n",
"//The slurry contains 100kg whiting/m^3 of water\n",
"printf('\n Volume of 100 kg whiting = %f m^3',100/3000);\n",
"printf('\n Volume of cake = %f m^3',0.0333/0.6);\n",
"printf('\n Volume of liquid in cake = %f m^3',0.05556*0.4);\n",
"printf('\n Volume of filtrate = %.3f m^3',(1-0.0222));\n",
"printf('\n volume of cake/volume of filtrate v = %f',0.0556/0.978);\n",
"A = 10^(-4);                 //area in sq meters\n",
"deltaP = -1.65*10^(5);       //P is in pascals\n",
"l = 0.01;                    //length is in meters\n",
"vol_flow_rate = 2*10^(-8);           //Volume flow rate is in m^3/sec\n",
"u = 10^(-3);                 //vicosity is in Ns/m^2\n",
"\n",
"r = poly([0],'r');\n",
"r1 = roots((10^4)*(2*10^(-8)*r)-1.65*10^(5)/(10^(-5)));\n",
"printf('\n r = %.2f*10^(13)/m^2',r1*10^(-13));\n",
"\n",
"function[Lopt]=optimum()\n",
"    Lopt = 1.161*10^(-3)*(900)^(0.5);  //t = 900 secs\n",
"    funcprot(0);\n",
"endfunction\n",
"printf('\n optimum frame thickness = %.1f mm',2*optimum()*1000);\n",
"\n",
"//total cycle time = 1.015L^2 + 900\n",
"//rate of cake production R = L/(1.015L^2 + 900)\n",
" \n",
" L = poly([0],'L');\n",
" L1 = roots(1.025*10^(6)*L^2 + 900 - 2.050*10^(6)*L^2);\n",
" printf('\n Frame thickness = %.2f mm',2*L1(1)*10^3);\n",
" \n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 7.3: Time_taken_to_produce_1_m3_of_filtrate_and_pressure_in_this_time.sci"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"printf('\n Example 7.3');\n",
"V = 0.094;         //volume in m^3\n",
"deltaP = -3530;    //P is in kN/m^2\n",
"\n",
"//At t = 1105 secs\n",
" V1 = 0.166;       //V is in m^3\n",
" deltaP1 = -5890;   //P is in kN/m^2\n",
"\n",
"a = [2.21*10^(6) -0.094;6.51*10^(6) -0.166];\n",
"b = [0.0088;0.0276];\n",
"x = inv(a)*b;\n",
"y = [x(2);x(1)];\n",
"printf('\n LA/v =%f       A^2/rμv = %f*10^(-7)',y(1),y(2)*10^7);\n",
"printf('\n For the full size plant:');\n",
"printf('\n LA/v = %f       A^2/rμv=%f*10^(-7)',10*y(1),y(2)*10^8);\n",
"\n",
"//Solving LHS of the integral\n",
"LHS = integrate('b+0.154+2.31','b',0,1);\n",
"//Equating LHS = RHS\n",
"t = LHS/(3.46*10^(-3));\n",
"printf('\n t = %d secs',t);\n",
"printf('\n deltaP = %dkN/m^2',(1+0.154)/(4.64*10^(-7)*857));\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 7.4: Speed_of_rotation_for_maximum_throughput.sci"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"printf('\n Example 7.4');\n",
"a = [2*84300*0.02^(2)*60 -2*0.02*0.0003;2*84300*0.02^(2)*120 -2*0.02*0.00044];\n",
"b = [0.0003^2;0.00044^2];\n",
"x = inv(a)*b;\n",
"y = [x(2);1/x(1)];\n",
"printf('\n L/v = %f         ruv = %f*10^(10)',y(1),y(2)*10^(-10));\n",
"printf('\n Area of filtering surface = %f m^2',4*(%pi));\n",
"printf('\n Bulk volume of cake deposited =%.3f m^3/revolution',4*(%pi)*0.005);\n",
"\n",
"V = sqrt(1*10^(-6)*143^2);\n",
"printf('\n V = %.3f m^3',V);\n",
"\n",
"t =poly([0],'t');\n",
"t1 = roots(0.141^2 +2*2.19*10^(-3)*0.141-2*84300*(4*(%pi))^(2)*t/(3.48*10^10));\n",
"printf('\n t = %f secs',t1);\n",
"printf('\n time for 1 revolution =%.1f secs',t1/0.4);\n",
"printf('\n speed = %.3fHz',0.4/t1);\n",
"printf('\n rate of filtrate production w = %.2f kg/sec',143/67.3)\n",
"printf('\n mass of slurry S =%.1f kg/sec',1.66*2.11);\n",
"\n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 7.5: Optimum_filtration_time_for_maximum_throughput.sci"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"printf('\n Example 7.5');\n",
"V1 = 0.00025;               //V is in m^3\n",
"t = 300;                    //t is in secs\n",
"a = [7.14*10^(-6) 2.86*10^(-4);11.42*10^(-6) 2.86*10^(-4)];\n",
"b = [1.2*10^(6);1*10^(6)];\n",
"x = inv(a)*b;\n",
"\n",
"//for the plate and frame filter\n",
"B1 = x(1)/(2*2.2^2*413*10^3);\n",
"B2 = x(2)/(2.2*413*1000);\n",
"\n",
"printf('\nrμv = %d\n',x(1));\n",
"printf('\n rμl = %d',x(2));\n",
"printf('\n B1= %f    B2= %f',B1,B2);\n",
"printf('\n the filtration time for maximum throughput is:');\n",
"t1 = 21.6*10^3;\n",
"t0= t1 +B2*(t1/B1)^(0.5);\n",
"printf('\n t = %f secs',t0);\n",
"V = (t1/B1)^(0.5);\n",
"printf('\n V= %f m^3',V);\n",
"printf('\nMean rate of filtration is: %.2f *10^-6 m^3/s',(V/(t1+t0))/10^-6);"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 7.6: Thickness_of_cake_produced.sci"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"printf('\n Example 7.6');\n",
"A=0.6*0.6*%pi;       //in m^2\n",
"rate=1.25*10^-4;     // in m^3/s\n",
"\n",
"v_w=0.2/(3*10^3);\n",
"v_f=10^-3-v_w;\n",
"\n",
"v=v_w/v_f;\n",
"v_rate=rate*v;\n",
"w=360*0.2;\n",
"\n",
"t=v_rate*w/A;\n",
"printf('\nThickness of cake produced is : %.1f mm',t/10^-4);\n",
"K = poly([0],'K');\n",
"K1 = roots((1.25*10^(-4)*360)^2-K*(6.5*10^(4)*(0.36*(%pi))^(2)*72));\n",
"printf('\n The value of K is %.2f*10^(-10)',K1*10^(10));\n",
"\n",
"//Filter press\n",
"//Using a filter press with n frames of thickness b m the total time, for one complete cycle of the press =(tf+120n+240),where tf is the time during which filtration is occurring\n",
"//overall rate of filtration = Vf/(tf + 120n + 240)\n",
"\n",
"// Vf = 0.3^(2)*n*b/0.143\n",
"//tf = 2.064*10^5 b^2\n",
"\n",
"b = poly([0],'b');\n",
"b1 = roots(b^2 - 0.0458*b - 0.001162);\n",
"printf('\n The thickness is %.4f m',b1(1));\n",
"\n",
"function[n]=number_of_plates()\n",
"    \n",
"    n = (0.030 + 25.8*b1(1)^2)/(0.629*b1(1)-0.015);\n",
"    funcprot(0);\n",
"endfunction\n",
"n = number_of_plates();\n",
"printf('\n The minimum number of plates required is %d',ceil(n));\n",
"\n",
"d = poly([0],'d');\n",
"d1 = roots(ceil(n)*(0.629*d-0.015)-0.030-25.8*d^2);\n",
"printf('\n The sizes of frames which will give exactly the required rate of filtration when six are used are %f mm',d1*10^3);\n",
"printf('\n\n\n  Thus any frame thickness between 47 and 99 mm will be satisfactory. In practice,50 mm (2 in) frames would probably be used.')\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 7.7: Increase_in_the_overall_throughput_of_the_press.sci"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"printf('\n Example 7.7');\n",
"//Case 1\n",
"\n",
"                 //dV/dt = A^2(-deltaP)/vru(V + AL/v) = a/V+b\n",
"                 //For constant rate filtration:\n",
"                 //Vo/to = a/Vo + b\n",
"                 //Vo^2 + bVo = ato\n",
"                 //For constant pressure filtration\n",
"                 //0.5(V^2 - Vo^2)+b(V-Vo)=a(t-to)\n",
"                 //to=600s,t-to=3600s,Vo=V/4\n",
"                 //V^2/16 +bV/4 = 600a\n",
"                 //o.5(V^2 - V^2/16)+b(V-V/4)=3600a\n",
"                 //3600a = (15/32)V^2 +3/4(bV) = 3/8(V^2) + 3/2(bV)\n",
"                 //b = V/8\n",
"                // a = (V^2/16 + V^2/32)/600 = (3/19200)V^2 \n",
"                //Total cycle time = 900 + 4200 = 5100secs\n",
"                //Filtration rate = V/5100 = 0.000196V\n",
"\n",
"//Case 2\n",
"                //V1/t1 = a/ (V1 +b/4)=Vo/to=a/(Vo+b)\n",
"                //0.5*(49/64V^2 - V1^2)+b/4(7/8V-V1)=a(t-t1)\n",
"                //V/2400 = (3/19200)V^2/(V1+V/32)\n",
"                //t1 = (to/Vo)V1\n",
"  t1 = 600/(1/4)*(11/32);\n",
"  printf('\n t1 = %dsecs',t1);\n",
"  //Substituting gives \n",
"  deltaT = (19200/3)*(784-121+34)/2048;\n",
"  printf('\n t -t1 = %d secs',deltaT); \n",
"  Cycle_time = 180+900+t1+deltaT;\n",
"  printf('\n cycle time = %d secs',Cycle_time);\n",
"  Increase = (0.000214 - 0.000196)/(0.000196)*100;\n",
"  printf('\n Increase in filtration rate is %.1f per cent',Increase);\n",
"  \n",
"  \n",
"  \n",
"  \n",
"  \n",
"  \n",
"  \n",
"  \n",
"  \n",
"  \n",
"  \n",
"  \n",
"  \n",
"  "
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   }
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