path: root/Heat_Transfer_Principles_And_Applications_by_Dutta/ch9.ipynb

{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Chapter 9 : Evaporation and Evaporators"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 9.1 Page No : 391"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 18,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The rate at which heat must be supplied at 1 atm pressure is 2.441e+08 kj/ day\n",
      "The rate at which heat must be supplied at a pressure of 600 mm Hg is 2.373e+08 kj/day \n"
     ]
    }
   ],
   "source": [
    "# Variables\n",
    "ro = 1020.       \t\t\t# kg/m**3, density of feed\n",
    "sf = 4.1        \t\t\t#kj/kg C,specific heat of the feed\n",
    "sp = 3.9        \t\t\t#kj/kg C,specific heat of the product\n",
    "ci = 5.       \t\t\t#initial concentration\n",
    "cw = 100.-ci    \t\t\t#conc. of  water\n",
    "cf = 40.        \t\t\t#final conc.\n",
    "rate = 100.      \t\t\t#m**3/day, rate of conc. of aq. solution\n",
    "ft = 25.         \t\t\t# C, feed temp.\n",
    "\n",
    "#calculation and results\n",
    "#materiel balance\n",
    "Wf = rate*ro    \t\t\t#Kg. feed entering\n",
    "Ms = ro*ci      \t\t\t#Kg mass of solute\n",
    "Mw = ro*cw      \t\t\t#kg,mass of water\n",
    "fc = cw/ci      \t\t\t#kg,feed concentration\n",
    "pc = (100-cf)/cf \t\t\t# kg,product concentration\n",
    "wlwp = Ms*pc      \t\t\t#Kg, water leaving with the product\n",
    "Ws = Mw-wlwp     \t\t\t#kg,water evaporated\n",
    "Wp = wlwp+Ms     \t\t\t# kg, product\n",
    "#energy balance\n",
    "rt = 0.            \t\t\t#C reference temp.\n",
    "ef = sf*(ft-rt)   \t\t\t#kj/kg,enthlpy of the feed\n",
    "#case i\n",
    "Tp = 100.          \t\t\t#temp. of the product (because the solute has a 'high molecular wt' the boiling pt elevation is neglected)\n",
    "ip = sp*(Tp-rt)   \t\t\t#kj/kg, enthalpy of the product\n",
    "iv = 2680.         \t\t\t#kj/kg, enthalpy of the vapour generated at 100 C and 1 atm pr. from the steam table\n",
    "#refer to fig. 9.23\n",
    "#from energy balance eq. (Wf*if+qs = Wv*iv+Wp*ip)\n",
    "qs = Ws*iv+Wp*ip-Wf-ef  \t\t\t#Wv = Ws\n",
    "print \"The rate at which heat must be supplied at 1 atm pressure is %1.3e kj/ day\"%(qs)\n",
    "\n",
    "#case ii\n",
    "#650 mm Hg vaccum = 110 mmHg pressure\n",
    "bp = 53.5         \t\t\t#C, boiling point of water\n",
    "ip2 = sp*(bp-rt)  \t\t\t#kj/kg, enthalpy of the product\n",
    "es = 2604.         \t\t\t#kj/kg, enthalpy of the saturated steam (from steam table)\n",
    "#from energy balnce eq.\n",
    "qs2 = Wp*ip+Ws*es-Wf-ef\n",
    "print \"The rate at which heat must be supplied at a pressure of 600 mm Hg is %1.3e kj/day \"%(qs2)\n",
    "\n",
    "# note : rounding off error."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 9.2 Page No : 393"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 17,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The steam required is 1737 kg/h\n",
      "No. of tube are 102\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    "\n",
    "# Variables\n",
    "ci = 10.           \t\t\t#%,initial concentration\n",
    "cf = 40.           \t\t\t#%, final conc\n",
    "Wf = 2000.        \t\t\t#kg/h, feed rate\n",
    "ft = 30.           \t\t\t#C feed temp.\n",
    "rp = 0.33         \t\t\t#kg/cm**2, reduced pressure\n",
    "bt1 = 75.           \t\t\t#C,boiling point temp.\n",
    "sst = 115.          \t\t\t#C, saturated steam temp.\n",
    "l = 1.5             \t\t\t# m,height of calandria\n",
    "sh = 0.946        \t\t\t#kcal/kg C, specific heat of liquir\n",
    "lh = 556.5        \t\t\t#kcal/kg latent heat of steam\n",
    "bt2 = 345.         \t\t\t#K, boiling point of water \n",
    "h = 2150.          \t\t\t#kcal/h m**2 C, overall heat transfer coefficient\n",
    "si = 2000.*(ci/100)  \t\t\t#kg/h, solids in\n",
    "wi = 1800.          \t\t\t#kg/h,wate in\n",
    "\n",
    "# Calculations\n",
    "Wp = si/(cf/100)   \t\t\t#kg/h, product out\n",
    "Wv = Wf-Wp         \t\t\t#evaporation rate\n",
    "ef = sh*(ft-bt1)\n",
    "ip = 0\n",
    "lamda_s = 529.5   \t\t\t#kcal/kg, lamda_s = is-il\n",
    "bpe = (273+bt1)-345  \t\t\t#boiling point elevation.\n",
    "#from eergy balance eq.\n",
    "Ws = (Wp*ip+Wv*lh-Wf*ef)/lamda_s\n",
    "q = Ws*lamda_s       \t\t\t#kcal/h,rate of heat transfer\n",
    "A = q/(h*(sst-bt1))   \t\t\t# m**2\n",
    "di = 0.0221            \t\t\t#m,inside diameter\n",
    "At = math.pi*l*di        \t\t\t#m**2, area of a math.single tube\n",
    "N = A/At             \t\t\t#no. of tubes\n",
    "\n",
    "# Results\n",
    "print \"The steam required is %.0f kg/h\"%(Ws)\n",
    "print \"No. of tube are %d\"%(N)\n",
    "\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 9.3 Page No : 393"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 16,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The steam pressure to be used in the calandria is 2.15 barabs)\n",
      "The heat transfer rate required is 4.01e+06 Kj/h\n",
      "Rate of steam supply is 1833 kg/h\n"
     ]
    }
   ],
   "source": [
    "# Variables\n",
    "Wf = 2000.        \t\t\t#kg/h, feed rate\n",
    "ci = 8.          \t\t\t#% initial conc.\n",
    "cf = 40.          \t\t\t#% final conc.\n",
    "ft = 30.         \t\t\t#C, feed temp.\n",
    "vp = 660.        \t\t\t#mm Hg, vaccum pressure\n",
    "ssp = 8.         \t\t\t# bar absolute, saturated steam pr.\n",
    "\n",
    "#calculation\n",
    "sr = Wf*(ci/100)  \t\t\t#kg/h, solid rate\n",
    "Wp = sr/(cf/100)   \t\t\t#kg/h,concentrated product rate\n",
    "ap = 760-vp        \t\t\t#mm Hg, absolute pressure in the evaporator\n",
    "bt = 325.          \t\t\t#K,boiling temp. of water\n",
    "l_s = 2380.        \t\t\t#kj/kg, latent heat\n",
    "R = 8.303         \t\t\t#gas consmath.tant\n",
    "w = 40.             \t\t\t#g,mass of solute\n",
    "M = 18.           \t\t\t#g,molecular wt of solvent\n",
    "W = 60.           \t\t\t#g,mass of the solvent\n",
    "m = 2000.         \t\t\t#g,molecular wt of solute\n",
    "dtb = (R*bt**2*w*M)/(l_s*W*m)     \t\t\t#C, boiling point elevation\n",
    "bp = bt+dtb      \t\t\t#k,boiling point of 40% solution\n",
    "dt = 70.        \t\t\t#C, from given data flux becomes maximum at a temp. drop  = 70 C\n",
    "st = bp+dt     \t\t\t#K,saturation temp. of steam in the steam chest\n",
    "Sp = 2.15      \t\t\t# bar, from steam table, saturation lr. of steam at this temp.\n",
    "\n",
    "sh = 4.2       \t\t\t#kj/kg C, specific heat of product\n",
    "rt = 0.         \t\t\t#C reference teml.\n",
    "ef = sh*(ft-rt)  \t\t\t# kj/kg, enthalpy of the feed\n",
    "ip = sh*(54-rt)  \t\t\t#kj/kg, enthalpy of the product\n",
    "iv = 2607.        \t\t\t#kj/kg, enthalpy of vapour produced\n",
    "#from eq 9.6\n",
    "Wv = 1600.       \t\t\t#enthalpy of evaporation\n",
    "q = Wp*ip+Wv*iv-Wf*ef   \t\t\t#kj/h, heat transfe rate required\n",
    "hvp = 2188.      \t\t\t#kj/kg, heat of vaporization of saturated steam at 397 K\n",
    "rs = q/hvp      \t\t\t#kg/h, rate of steam supply\n",
    "\n",
    "# Results\n",
    "print \"The steam pressure to be used in the calandria is %.2f barabs)\"%(Sp);\n",
    "print \"The heat transfer rate required is %.2e Kj/h\"%(q);\n",
    "print \"Rate of steam supply is %.0f kg/h\"%(rs);\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 9.4 Page No : 402"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 15,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The evaporator area is 72 square metre \n",
      "Steam economy is 1.79\n"
     ]
    }
   ],
   "source": [
    "import math \n",
    "\n",
    "from numpy import array, linalg\n",
    "# Variables\n",
    "Wf = 6000.     \t\t\t#kg/h, feed rate\n",
    "ci = 2.        \t\t\t#%, initial concentration\n",
    "cf = 35.       \t\t\t#%, final conc.\n",
    "ft = 50.        \t\t\t#C,feed temp.\n",
    "ssp = 2.       \t\t\t#bar abs, saturated steaam pr.\n",
    "sep = 0.0139  \t\t\t#bar abs, maintained temp. in second effect\n",
    "h1 = 2000.     \t\t\t#W/m**2 K,overall heat transfer coeffcient in 1st effect\n",
    "h2 = 1500.     \t\t\t#W/m**2 K, overall heat transfer coefficient in 2nd effect\n",
    "cp = 4.1      \t\t\t#kj/kg k,specific heat\n",
    "\n",
    "#calculation\n",
    "si = Wf*(ci/100) \t\t\t#kg/h, solid in\n",
    "wi = 5880.      \t\t\t#kg/h, water in\n",
    "Wp = si/(cf/100)  \t\t\t#kg/h product out\n",
    "wo = Wp*(1-cf/100) \t\t\t#kg/h, water out with the product\n",
    "ter = wi-wo       \t\t\t#kg/h, total evaporation rate\n",
    "\n",
    "#boiling temp. in the first effect\n",
    "T1 = 120.       \t\t\t#C,Temprature\n",
    "l_s1 = 2200.    \t\t\t#kj/kg, latent heat\n",
    "T2 = 12.        \t\t\t#C,boiling point in second effect\n",
    "l_s2 = 2470.   \t\t\t# kj/kg  in second effect\n",
    "tatd = T1-T2    \t\t\t# C,tatd = dt1+dt2  = T1-T2  , total available temp. drop\n",
    "#from eq. 9.20\n",
    "#h1*dt1 = h2*dt2\n",
    "#solving above two equations by matrix\n",
    "A = array([[1,1],[2000,-1500]])\n",
    "C = array([108,0])\n",
    "X = linalg.solve(A,C)\n",
    "#X = inv(A)*C\n",
    "\n",
    "dt1 = X[0]\n",
    "dt2 = X[1]\n",
    "t1 = T1-dt1    \t\t\t#temp. of steam leaving the first effect\n",
    "t2 = T2-dt2    \t\t\t#temp. of steam leaving second effect\n",
    "#energy balance over the 1st effect, from eq.9.14\n",
    "rt1 = t1\n",
    "ef = cp*(ft-t1)   \t\t\t#kj/kg,enthalpy of feed\n",
    "i1 = 0\n",
    "lam_s1 = 2330.     \t\t\t#kj/kg\n",
    "is1 = lam_s1\n",
    "#Wf*ef+Ws*l_s = (Wf-Ws1)*i1+Ws1*is1\n",
    "#substituting we get,\n",
    "#Ws1 = 0.9442*Ws-253.4..........(1)\n",
    "#energy balance over second effect\n",
    "#from eq 9.15\n",
    "#(Wf-Ws1)*i1+Ws1*lam_s1 = (Wf-Ws1-Ws2)*i2+Ws2*is2\n",
    "rt2 = t2\n",
    "lam_s2 = 2470.\n",
    "is2 = lam_s2\n",
    "i2 = 0\n",
    "# substituting we get\n",
    "#Ws2 = 0.8404*Ws1+617.5............(2)\n",
    "#ter,Ws1+Ws2 = 5657...............(3)\n",
    "#solving by matrix method\n",
    "A = array([[0.9442,-1,0],[0,0.8404,-1],[0,1,1]])\n",
    "B = array([253.4,-617.5,5657])\n",
    "X = linalg.solve(A,B)\n",
    "#X = inv(A)*B\n",
    "Ws = X[0]\n",
    "Ws1 = X[1]\n",
    "Ws2 = X[2]\n",
    "\n",
    "#evaporator area\n",
    "A1 = Ws*l_s1/(h1*dt1)  \t\t\t#for 1st effect\n",
    "A2 = Ws1*lam_s1/(h2*dt2)  \t\t\t#for second effect\n",
    "\n",
    "#revised calculation\n",
    "#taking\n",
    "dt1_ = 48.\n",
    "dt2_ = 60.\n",
    "T1_ = T1-dt1_\n",
    "T2_ = T2-dt2_\n",
    "ls1_ = 2335.\n",
    "ls2_ = 2470.\n",
    "# energy balance over first effect gives\n",
    "#Ws1 = 0.9422Ws-231.8.........(4)\n",
    "#energy balance over second effect gives\n",
    "#Ws2 = 0.8457Ws1+579.5......(5)\n",
    "#solving eq 3,4,5\n",
    "P = array([[0.9422,-1,0],[0,0.8457,-1],[0,1,1]])\n",
    "Q = array([231.8,-579.5,5657])\n",
    "Y = linalg.solve(P,Q)\n",
    "#Y = inv(P)*Q\n",
    "Ws_ = Y[0]\n",
    "Ws1_ = Y[1]\n",
    "Ws2_ = Y[2]\n",
    "\n",
    "#eveporator area for 1st & 2nd effect in m**2\n",
    "A1_ = Ws_*l_s1/(h1*dt1_)\n",
    "A2_ = Ws1_*ls1_/(h2*dt2_)\n",
    "EA = (A1_+A2_)/2\n",
    "SE = (Ws1_+Ws2_)/Ws_\n",
    "\n",
    "# Results\n",
    "print \"The evaporator area is %.0f square metre \"%(EA);\n",
    "print \"Steam economy is %.2f\"%(SE);\n",
    "\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 9.5 Page No : 404"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Maximum no. of effects are 4\n"
     ]
    }
   ],
   "source": [
    "# Variables\n",
    "ssp = 3.32     \t\t\t#bar abs, saturated steam pr.\n",
    "rp = 0.195     \t\t\t# bar abs, residual pr. in the condenser\n",
    "tl = 41.      \t\t\t#K, sun of temp. losses because of BPE\n",
    "mt = 8.        \t\t\t#k,minimum available temp. driving force\n",
    "#calculation\n",
    "sst = 410.     \t\t\t#K,saturated steam temp.\n",
    "st = 333.        \t\t\t#K,corresponding saturation temp. when pressure in the last effect is 0.195 bar\n",
    "ttd = sst-st    \t\t\t#K,total temp. difference\n",
    "atd = ttd-tl    \t\t\t# K,available temp. drop across the unit\n",
    "n = atd/mt     \t\t\t#maximum no. of effect\n",
    "\n",
    "# Results\n",
    "print \"Maximum no. of effects are %.0f\"%(n);\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 9.6 Page No : 405"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 14,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The areas are now reasonably close  \n",
      "Steam Rate is 8854 Kg/h \n",
      "Steam economy is 1.93\n"
     ]
    }
   ],
   "source": [
    "# Variables\n",
    "fc = 9.5      \t\t\t#%,feed concentration\n",
    "pc = 50.     \t\t\t#%, product conc.\n",
    "ft = 40.     \t\t\t# C,feed temp.\n",
    "er = 2000.      \t\t\t#kg NaOH/h, evaporation rate\n",
    "vp = 714.     \t\t\t#mm Hg, vaccum pr. in last effect\n",
    "#heat transfer coefficients, W/m**2 C\n",
    "h1 = 6000.    \t\t\t#for first effect\n",
    "h2 = 3500.    \t\t\t#for second effect\n",
    "h3 = 2500.    \t\t\t#for third effect\n",
    "\n",
    "#calculatiin\n",
    "Wf = er/(fc/100)  \t\t\t#kg/h, 2 tons NaOH per hour, feed rate\n",
    "Wp = er/(pc/100)  \t\t\t#kg/h, product rate\n",
    "ter = Wf-Wp  \t\t\t#kg/h, total evaporation  rate\n",
    "#steam\n",
    "p = 3.3       \t\t\t#bar,assumed saturated\n",
    "#from steam table\n",
    "Ts = 137.      \t\t\t#C,temp.\n",
    "l_s = 2153.    \t\t\t#kj/kg, latent heat\n",
    "pl = 760.-vp   \t\t\t#mm Hg,pressure in the last effect\n",
    "bp = 37.       \t\t\t#C,boiling point of water\n",
    "#refer to fig. 9.24\n",
    "attd = Ts-bp  \t\t\t#C,apparent total temp. drop\n",
    "#let assume the following evaporation rate for three effects in kg/h\n",
    "ev1 = 5600.\n",
    "ev2 = 5680.\n",
    "ev3 = 5773.\n",
    "#conc. in three effects\n",
    "c1 = er/(Wf-ev1)\n",
    "c2 = er/(Wf-ev1-ev2)\n",
    "c3 = 0.5   \t\t\t# Variables\n",
    "#boiling point elevations in three effects in C\n",
    "bpe1 = 3.5\n",
    "bpe2 = 8.\n",
    "bpe3 = 39.\n",
    "attda = attd-(bpe1+bpe2+bpe3)  \t\t\t#actual total temp. drop available\n",
    "#temp. drop in three effects\n",
    "#from eq. 9.23\n",
    "dt1 = attda*((1/h1)/((1/h1)+(1/h2)+(1/h3)))\n",
    "dt2 = attda*((1/h2)/((1/h1)+(1/h2)+(1/h3)))\n",
    "dt3 = attda*((1/h3)/((1/h1)+(1/h2)+(1/h3)))\n",
    "\n",
    "#from table 9.4\n",
    "#enthalpy of solution in three effects in  kj/kg\n",
    "i1 = 486.\n",
    "i2 = 385.\n",
    "i3 = 460.\n",
    "#enthalpy of vapour generated for three effects in kj/kg\n",
    "is1 = 2729.\n",
    "is2 = 2691.\n",
    "is3 = 2646.\n",
    "#Enthalpy of condensate over effect 1,2,3 in kj/kg\n",
    "il1 = 0.\n",
    "il2 = 519.\n",
    "il3 = 418.\n",
    "#Enthalpy balance over effect 1\n",
    "ef = 145.      \t\t\t#kj/kg,enthalpy of feed\n",
    "#from energy balance eq.\n",
    "#Ws1 = 0.96Ws-3200......(1)\n",
    "#enthalpy balanc over effect 2\n",
    "#Ws2 = 0.9146Ws1+922...........(2)\n",
    "#enthalpy balanc over effet 3\n",
    "#Ws3 = 1.073Ws2+0.0343Ws1-722........(3)\n",
    "#ter = Ws1+Ws2+Ws3 = 17053..........(4)\n",
    "\n",
    "#Solving above four eqns by matrix\n",
    "A = array([[0.96,-1,0,0],[0,0.9146,-1,0],[0,0.0343,1.073,-1],[0,1,1,1]])\n",
    "B = array([3200,-922,722,17053])\n",
    "X = linalg.solve(A,B)\n",
    "#X = inv(A)*B\n",
    "Ws = X[0]\n",
    "Ws1 = X[1]\n",
    "Ws2 = X[2]\n",
    "Ws3 = X[3]\n",
    "\n",
    "#calculation of heat transfer areas iver effect 1, 2 ,3\n",
    "A1 = Ws*l_s*10**3/(h1*dt1*3600)\n",
    "A2 = Ws1*(is1-il2)*10**3/(h2*dt2*3600)\n",
    "A3 = Ws2*(is2-il3)*10**3/(h3*dt3*3600)\n",
    "\n",
    "#Revised dt\n",
    "avar = (A1+A2+A3)/3\n",
    "dt1_ = (A1/avar)*dt1\n",
    "dt2_ = (A2/avar)*dt2\n",
    "dt3_ = attda-dt1_-dt2_\n",
    "\n",
    "#from table 9.5\n",
    "#enthalpy of vapour generated over effect 1,2,3 in kj/kg\n",
    "is1_ = 2720.\n",
    "is2_ = 2685.\n",
    "is3_ = 2646.\n",
    "#enthalpy of soln on 1,2,3 in kj/kg\n",
    "i1_ = 470.\n",
    "i2_ = 380.\n",
    "i3_ = 460.\n",
    "#enthalpy of condensate over effect 1 ,2,3 in kj/kg\n",
    "il1_ = 0.\n",
    "il2_ = 513.\n",
    "il3_ = 412.\n",
    "#enthalpy balance ove effect 1,2,3 gives\n",
    "Ws_ = 8854.\n",
    "Ws1_ = 5432.\n",
    "Ws2_ = 5812.\n",
    "Ws3_ = 5809.\n",
    "#revised heat transfer areas for effect 1 ,2,3 in m**2\n",
    "A1_ = Ws_*l_s*1000/(h1*dt1_*3600)\n",
    "A2_ = Ws1_*(is1_-il2_)*10**3/(h2*dt2_*3600)\n",
    "A3_ = Ws2_*(is2_-il3_)*10**3/(h3*22.5*3600)\n",
    "avar_ = (A1_+A2_+A3_)/3\n",
    "SE = ter/Ws_\n",
    "\n",
    "# Results\n",
    "print \"The areas are now reasonably close  \"\n",
    "print \"Steam Rate is %.0f Kg/h \"%(Ws_)\n",
    "print \"Steam economy is %.2f\"%(SE)\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 9.7 Page No : 409"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 8,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The increase in evaporation capacity ic 113 percentage \n",
      " The percentage change in the math.cost of concentrating a ton of feed is 15 percentage\n"
     ]
    }
   ],
   "source": [
    "from numpy import array, linalg\n",
    "\n",
    "# Variables\n",
    "Wf = 3000.        \t\t\t#kg/h,feed\n",
    "fc = 8.        \t\t\t#%, feed concentration\n",
    "pc = 40.       \t\t\t#% product concentration\n",
    "si = Wf*(fc/100)      \t\t\t#kg,solid in\n",
    "pr = si/(40./100)      \t\t\t#g/h, product rate\n",
    "ft = 60.          \t\t\t#C,feed temp.\n",
    "er = Wf-pr           \t\t\t#kg/h, evaporation rate\n",
    "math.cost = 120000.   \t\t\t#total math.cost per year\n",
    "p1 = 4.5       \t\t\t#bar, low pressure steam\n",
    "scpt = 700.        \t\t\t#per ton.  math.cost of steam\n",
    "cp = 0.764        \t\t\t#  kcal/kg, specific heat\n",
    "\n",
    "#from table 9.6\n",
    "eep = 1.        \t\t\t#atm existing evaporator pressure \n",
    "oop = 400000.     \t\t\t# peryear ,other operatingmath.cost\n",
    "oop_ = 600000.     \t\t\t#per yr, for proposed condition\n",
    "wd = 300.         \t\t\t#days per year.working days\n",
    "wh = wd*24.        \t\t\t#working hr\n",
    "\n",
    "# Calculations\n",
    "#EXISTING OPERATING CONDITION \n",
    "rt = 0         \t\t\t#C,reference temp.\n",
    "ef = eep*(ft-rt)   \t\t\t#kcal/kg, enthalpy of feed\n",
    "pt = 100.     \t\t\t#C,product temp.\n",
    "i1 = cp*(pt-rt)   \t\t\t#kcal/kg, enthalpy of soln\n",
    "is1 = 639.       \t\t\t#kcal/kg,enthalpy of vapour generated at 1 atm (from steam table)\n",
    "l_s = 496.      \t\t\t#kcal/kg,latent heat of steam at 4.5 bar\n",
    "T = 425.      \t\t\t#K\n",
    "#heat balance\n",
    "Ws = (er*is1+pr*i1-Wf*ef)/l_s      \t\t\t#kg/h, steam required\n",
    "q = Ws*l_s       \t\t\t#ton/ hr,heat supplied\n",
    "x = q/(T-(pt+273))  \t\t\t#x = Ud*A\n",
    "#hourly math.cost\n",
    "sc = Ws/1000*(scpt)    \t\t\t# /perh, steam math.cost\n",
    "lc = 100.           \t\t\t#per h,labour math.cost\n",
    "oc = oop/(wh)     \t\t\t# per h,othe math.cost\n",
    "tc = sc+lc+oc  \t\t\t#total math.cost\n",
    "C = tc/(Wf/1000)     \t\t\t# per ton,math.cost per ton of feed\n",
    "\n",
    "#PROPOSED OPERATING CONDITION\n",
    "bpl = 320.        \t\t\t#K,boiling point of liquid\n",
    "dt = T-bpl\n",
    "q_ = x*dt     \t\t\t#kcal/h,rate of heat supply\n",
    "sr = q_/l_s   \t\t\t#steam rate ton per hr\n",
    "pt_ = 47.      \t\t\t#C,product temp .\n",
    "ep = cp*(pt_-rt)    \t\t\t#kcal/kg. enthalpy of product\n",
    "ev = 618.        \t\t\t#kcal/kg, enthalpy of vapour generated\n",
    "#heat balance\n",
    "#24Wf_-582Ws1_ = 2825000  ..........(1)\n",
    "#material balance\n",
    "# 4Wf_-5Ws1_ = 0  .............(2)\n",
    "#solving by matrix method\n",
    "a = array([[24,-582],[4,-5]])\n",
    "b = array([-2825000,0])\n",
    "x_ = linalg.solve(a,b)\n",
    "#x_ = inv(a)*b\n",
    "Wf_ = x_[0]\n",
    "Ws1_ = x_[1]\n",
    "ic = (Wf_-Wf)/Wf\n",
    "print \"The increase in evaporation capacity ic %d percentage \"%(ic*100)\n",
    "sr_ = Ws1_/1000    \t\t\t#ton per hr ,steam rate \n",
    "#hourly math.cost\n",
    "sc_ = Ws1_*scpt   \t\t\t#steam math.cost\n",
    "lc_ = 200.      \t\t\t#labour math.cost rs.200/ h\n",
    "oc_ = oop_/wh  \t\t\t# other math.cost\n",
    "tc_ = sc_/1000+lc_+oc_\n",
    "C_ = tc_/(Wf_/1000)  \t\t\t#math.cost per ton of feed\n",
    "ps = (C-C_)/C\n",
    "print \" The percentage change in the math.cost of concentrating a ton of feed is %.0f percentage\"%(ps*100)\n",
    "\n",
    "# rounding off error."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 9.8 Page No : 415"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 9,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Make up steam required is 1.302e+04 kg/day\n"
     ]
    }
   ],
   "source": [
    "# Variables\n",
    "q = 2200.         \t\t\t#kj/kg heat of condensation of steam \n",
    "#from example 9.1\n",
    "Qr = 2.337*10**8      \t\t\t#kj/day  rate of heat supply\n",
    "\n",
    "#calculation\n",
    "Rate = Qr/q          \t\t\t#kg/day steam supply rate\n",
    "Rate_ = 1.062*10**5   \t\t\t#approximate value\n",
    "E = 2800.             \t\t\t#kj/kg enthalpy of compressed vapour\n",
    "T = 175.7            \t\t\t#C, temprature\n",
    "Ts = 121.             \t\t\t#C Saturation temprature\n",
    "E1 = 2700.            \t\t\t#enthalpy at saturation temprature\n",
    "q1 = T-Ts            \t\t\t#Superheat of vapour\n",
    "T1 = 100.             \t\t\t#C hot water temprature\n",
    "E2 = 419.              \t\t\t#Enthalpy at hot water temp.\n",
    "x = (E-E1)/(E1-E2)    \t\t\t#water supplied per kg of superheated steam\n",
    "S = 1.044             \t\t\t#steam obtained after desuperheating\n",
    "R1 = 8.925*10**4       \t\t\t#kg/day rate of vapour generation \n",
    "R2 = S*R1             \t\t\t#Rate of recompressed sat. steam\n",
    "R2_ = 9.318*10**4       \t\t\t#approximate value\n",
    "SR = Rate_-R2_        \n",
    "\n",
    "# Results\n",
    "print \"Make up steam required is %.3e kg/day\"%(SR)\n",
    "\n"
   ]
  }
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