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diff --git a/Heat_Transfer_Principles_And_Applications_by_Dutta/ch9.ipynb b/Heat_Transfer_Principles_And_Applications_by_Dutta/ch9.ipynb new file mode 100644 index 00000000..6a4e4ff0 --- /dev/null +++ b/Heat_Transfer_Principles_And_Applications_by_Dutta/ch9.ipynb @@ -0,0 +1,692 @@ +{ + "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" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} |