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authorTrupti Kini2016-09-09 23:30:25 +0600
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A Heat_Transfer_Principles_And_Applications_by_Dutta/README.txt A Heat_Transfer_Principles_And_Applications_by_Dutta/ch10.ipynb A Heat_Transfer_Principles_And_Applications_by_Dutta/ch11.ipynb A Heat_Transfer_Principles_And_Applications_by_Dutta/ch2.ipynb A Heat_Transfer_Principles_And_Applications_by_Dutta/ch3.ipynb A Heat_Transfer_Principles_And_Applications_by_Dutta/ch4.ipynb A Heat_Transfer_Principles_And_Applications_by_Dutta/ch5.ipynb A Heat_Transfer_Principles_And_Applications_by_Dutta/ch6.ipynb A Heat_Transfer_Principles_And_Applications_by_Dutta/ch7.ipynb A Heat_Transfer_Principles_And_Applications_by_Dutta/ch8.ipynb A Heat_Transfer_Principles_And_Applications_by_Dutta/ch9.ipynb A Heat_Transfer_Principles_And_Applications_by_Dutta/screenshots/10.png A Heat_Transfer_Principles_And_Applications_by_Dutta/screenshots/5.png A Heat_Transfer_Principles_And_Applications_by_Dutta/screenshots/51.png A Heat_Transfer_in_SI_units_by_Holman/Chapter1.ipynb A Heat_Transfer_in_SI_units_by_Holman/Chapter10.ipynb A Heat_Transfer_in_SI_units_by_Holman/Chapter11.ipynb A Heat_Transfer_in_SI_units_by_Holman/Chapter2.ipynb A Heat_Transfer_in_SI_units_by_Holman/Chapter3.ipynb A Heat_Transfer_in_SI_units_by_Holman/Chapter4.ipynb A Heat_Transfer_in_SI_units_by_Holman/Chapter5.ipynb A Heat_Transfer_in_SI_units_by_Holman/Chapter6.ipynb A Heat_Transfer_in_SI_units_by_Holman/Chapter7.ipynb A Heat_Transfer_in_SI_units_by_Holman/Chapter8.ipynb A Heat_Transfer_in_SI_units_by_Holman/Chapter9.ipynb A Heat_Transfer_in_SI_units_by_Holman/README.txt A Heat_Transfer_in_SI_units_by_Holman/screenshots/9.1.png A Heat_Transfer_in_SI_units_by_Holman/screenshots/9.2.png A Heat_Transfer_in_SI_units_by_Holman/screenshots/9.4.png A Power_Electronics_Principles_and_Applications_by_Jacob/Chapter1.ipynb A Power_Electronics_Principles_and_Applications_by_Jacob/Chapter2.ipynb A Power_Electronics_Principles_and_Applications_by_Jacob/Chapter3.ipynb A Power_Electronics_Principles_and_Applications_by_Jacob/Chapter4.ipynb A Power_Electronics_Principles_and_Applications_by_Jacob/Chapter5.ipynb A Power_Electronics_Principles_and_Applications_by_Jacob/Chapter6.ipynb A Power_Electronics_Principles_and_Applications_by_Jacob/Chapter7.ipynb A Power_Electronics_Principles_and_Applications_by_Jacob/Chapter8.ipynb A Power_Electronics_Principles_and_Applications_by_Jacob/Chapter9.ipynb A Power_Electronics_Principles_and_Applications_by_Jacob/README.txt A Power_Electronics_Principles_and_Applications_by_Jacob/screenshots/4.png A Power_Electronics_Principles_and_Applications_by_Jacob/screenshots/5.png A Power_Electronics_Principles_and_Applications_by_Jacob/screenshots/6.png A sample_notebooks/AviralYadav/Chapter5.ipynb
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+{
+ "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"
+ }
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