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| author | hardythe1 | 2014-07-25 12:37:07 +0530 |
|---|---|---|
| committer | hardythe1 | 2014-07-25 12:37:07 +0530 |
| commit | 0d4b70aada9bbc982f00c27afb1337e2b314eb43 (patch) | |
| tree | 439ce220ef91779db970e4abbd3ddc5690267bc5 /Stoichiometry_And_Process_Calculations/ch11.ipynb | |
| parent | aa2467e5c4c719a75637b8c4d737c0bad0f5630b (diff) | |
| download | Python-Textbook-Companions-0d4b70aada9bbc982f00c27afb1337e2b314eb43.tar.gz Python-Textbook-Companions-0d4b70aada9bbc982f00c27afb1337e2b314eb43.tar.bz2 Python-Textbook-Companions-0d4b70aada9bbc982f00c27afb1337e2b314eb43.zip | |
adding books
Diffstat (limited to 'Stoichiometry_And_Process_Calculations/ch11.ipynb')
| -rwxr-xr-x | Stoichiometry_And_Process_Calculations/ch11.ipynb | 1955 |
1 files changed, 1955 insertions, 0 deletions
diff --git a/Stoichiometry_And_Process_Calculations/ch11.ipynb b/Stoichiometry_And_Process_Calculations/ch11.ipynb new file mode 100755 index 00000000..b8409421 --- /dev/null +++ b/Stoichiometry_And_Process_Calculations/ch11.ipynb @@ -0,0 +1,1955 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 11 : Energy Balance Thermophysics" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.1 pageno : 405" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# variables \n", + "m = 75. #kg\n", + "g = 9.81 #m**2/s\n", + "d = 10. #m\n", + "t = 2.5*60 #s\n", + "\n", + "# Calculation \n", + "f = m*g \n", + "w = f * d \n", + "P = w / t \n", + "\n", + "# Result\n", + "print \"The work done = \",w,\"Nm\"\n", + "print \"Power required = \",P,\"W\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The work done = 7357.5 Nm\n", + "Power required = 49.05 W\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.2 pageno : 405" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# variables \n", + "PE = 1.5*10**3 #J\n", + "m = 10. #kg\n", + "g = 9.81 #m/s**2\n", + "v = 50. #m/s\n", + "\n", + "# Calculation \n", + "#PE = mgz\n", + "z = PE / (m*g) \n", + "KE = m* (v**2) / 2 \n", + "\n", + "# Result\n", + "print \"Height of the body from the ground = %.1f\"%z,\"m\"\n", + "print \"Kinetic energy of the body = \",KE/1000,\"kJ\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Height of the body from the ground = 15.3 m\n", + "Kinetic energy of the body = 12.5 kJ\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.3 pageno : 405" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "# Variables \n", + "d = 100. /1000 #m\n", + "m = 50. #kg\n", + "P = 1.01325*10**5. #Pa\n", + "\n", + "# Calculation \n", + "A = math.pi * (d**2)/4 \n", + "Fatm = P * A \n", + "Fwt = m * g \n", + "Ftotal = Fatm + Fwt \n", + "P = Ftotal / A\n", + "\n", + "# Result \n", + "print \"(a)Pressure of the gas = %.4f\"%(P/10**5),\"bar\"\n", + "z = 500/1000. #m\n", + "w = Ftotal * z \n", + "print \"(b)Work done by the gas = %.2f\"%w,\"J\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(a)Pressure of the gas = 1.6378 bar\n", + "(b)Work done by the gas = 643.15 J\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.4 pageno : 406" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# variables \n", + "Sgr = 0.879 \n", + "F = 5. #m**3/h\n", + "\n", + "# Calculation \n", + "D = Sgr * 1000. \n", + "m = F * D/3600. #kg/s\n", + "P = 3500. #kPa\n", + "W = P * m * 1000/ D \n", + "\n", + "# Result\n", + "print \"Power requirement for the pump = %d\"%W,\"W\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Power requirement for the pump = 4861 W\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.5 pageno : 408" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "# Variables \n", + "\n", + "d = 3. #m\n", + "m = 12500. #kg\n", + "P = 7000. #kPa\n", + "\n", + "# Calculation \n", + "U = 5.3*10**6 #kJ\n", + "Vtank = 4*math.pi*((d/2)**3) / 3 \n", + "Vliq = Vtank / 2 \n", + "H = U + P * Vliq \n", + "\n", + "# Result\n", + "print \"Specific enthalpy of the fluid in the tank = %.2f\"%(H/m),\"kJ/kg\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Specific enthalpy of the fluid in the tank = 427.96 kJ/kg\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.6 pageno : 409" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# variables \n", + "P = 101.3 #kPa\n", + "SVl = 1.04 * 10**-3 #m**3/kmol\n", + "SVg = 1.675 #m**3/kmol\n", + "Q = 1030. #kJ\n", + "\n", + "# Calculation \n", + "W = P * 10**3 * (SVg - SVl)/1000 \n", + "U = Q - W \n", + "H = U + P * 10**3 * (SVg - SVl)/1000 \n", + "\n", + "# Result\n", + "print \"Change in internal energy = %.2f\"%U,\"kJ/kmol\"\n", + "print \"Change in enthalpy = \",H,\"kJ/kmol\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Change in internal energy = 860.43 kJ/kmol\n", + "Change in enthalpy = 1030.0 kJ/kmol\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.7 pageno : 409" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# variables \n", + "#work is done on the system, hence, W is negative\n", + "\n", + "W = - 2 * 745.7 #J/s\n", + "#heat is transferres to the surrounding, hence, heat transferred is negative,\n", + "Q = -3000. #kJ/h\n", + "\n", + "# Calculation \n", + "U = Q*1000/3600 - W \n", + "\n", + "# Result\n", + "print \"Change in internal energy = %.2f\"%U,\"J/s\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Change in internal energy = 658.07 J/s\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.8 pageno : 410" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# variables \n", + "#Fe(s) + 2HCl(aq) = FeCl2(aq) + H2(g)\n", + "MFe = 55.847 \n", + "m = 1. #kg\n", + "\n", + "# Calculation \n", + "Nfe = m * 10**3/MFe \n", + "Nh2 = Nfe #(since 1 mole of Fe produces 1 mole of H2)\n", + "T = 300. #K\n", + "R = 8.314 \n", + "\n", + "#the change in volume is equal to the volume occupied by hydrogen produced\n", + "PV = Nh2 * R * T \n", + "W = PV \n", + "\n", + "# Result\n", + "print \"Work done = %.2f\"%W,\"J\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Work done = 44661.31 J\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.9 pageno : 412" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "#Cp =1.436 + 2.18*10**-3*T \n", + "from scipy.integrate import quad \n", + "\n", + "# variables \n", + "m = 1000./3600 #kg/s\n", + "T1 = 380. #K\n", + "T2 = 550. #K\n", + "\n", + "# Calculation \n", + "def f0(T): \n", + "\t return 1.436 + 2.18*10**-3*T\n", + "\n", + "x = quad(f0,T1,T2)[0]\n", + "Q = m*x \n", + "\n", + "# Result\n", + "print \"Heat load on the heater = %.1f\"%Q,\"kW\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat load on the heater = 115.7 kW\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.10 page no : 413" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "#Cp = 26.54 + 42.454*10**-3 * T - 14.298 * 10**-6 * T**2 \n", + "\n", + "from scipy.integrate import quad \n", + "\n", + "# variables \n", + "T1 = 300. #K\n", + "T2 = 1000. #K\n", + "m = 1. #kg\n", + "N = m/44 #kmol\n", + "\n", + "# Calculation \n", + "def f3(T): \n", + "\t return 26.54 + 42.454*10**-3 * T - 14.298 * 10**-6 * T**2\n", + "\n", + "x = quad(f3,T1,T2)[0]\n", + "\n", + "Q = N*x \n", + "\n", + "# Result\n", + "print \"(a)Heat required = %.2f\"%Q,\"kJ\"\n", + "\n", + "#for temperature in t degree celsius\n", + "#Cp = 26.54 + 42.454*10**-3 * (t + 273.15) - 14.298 * 10**-6 * (t + 273.15)**2\n", + "#Cp = 37.068 + 34.643 * 10**-3*t - 14.298* 10**-6 * t**2 (kJ/kmolC)\n", + "#Cp = 8.854 + 8.274*10**-3*t -3.415*10**-6*t**2 ( Kcal/kmolC)\n", + "#For degree Fehreneit scale,replacet by ( t1 - 32)/18, we get\n", + "#Cp = 8.7058 + 4.6642 * 10**-3 *t1 - 1.0540 * 10**-6 * t1**2 ( Btu/lbmolF)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(a)Heat required = 755.85 kJ\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.11 pageno : 414" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "%pylab inline\n", + "\n", + "from matplotlib.pyplot import *\n", + "\n", + "# variables \n", + "T = [273, 373, 473, 573, 673, 773, 873, 973, 1073, 1173, 1273] \n", + "Cp = [33.6, 35.1, 36 ,36.6, 37, 37.3, 37.5, 37.6, 37.7, 37.8, 37.9] \n", + "\n", + "# Calculation \n", + "plot(T,Cp)\n", + "suptitle(\"Determination of enthalpy change \")\n", + "xlabel(\"Temperature, K \")\n", + "ylabel(\"Heat capacity, kJ/kmol K\")\n", + "\n", + "H = 36828. #kJ/kmol\n", + "\n", + "# Result\n", + "print \"Enthalpy change = \",H,\"kJ/kmol\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Populating the interactive namespace from numpy and matplotlib\n" + ] + }, + { + "metadata": {}, + "output_type": "display_data", + "png": 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VAxD2JXt6CCEqq1JrYX322WeW90TddXNPj/h42dNDCFFxVhPItm3bmD17Nvv2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\nGzdgwwYYP97ekQghRPWzmkC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RKZJARJ2XlZWFTqdDp9Nxzz33oNVq0el0hISEYDAY7B1e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\nECGEEJUiCUQIIUSl/H/CTl7Bxz0qdgAAAABJRU5ErkJggg==\n", + "text": [ + "<matplotlib.figure.Figure at 0x27f4550>" + ] + }, + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Enthalpy change = 36828.0 kJ/kmol\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.12 page no : 415" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "#Cp = 26.586 + 7.582 * 10 **-3 * T - 1.12 * 10**-6 * T**2\n", + "\n", + "from scipy.integrate import quad \n", + "\n", + "# variables \n", + "T1 = 500. #K\n", + "T2 = 1000. #K\n", + "\n", + "# Calculation \n", + "def f5(T): \n", + "\t return 26.586 + 7.582 * 10**-3 * T - 1.12 * 10**-6 * T**2\n", + "\n", + "x = quad(f5,T1,T2)[0]\n", + "\n", + "Cpm = 1 *x / ( T2 - T1 ) \n", + "\n", + "# Result\n", + "print \"(a)Mean molal heat capacity = %.3f\"%Cpm,\"kJ/kmolK\"\n", + "V = 500. #m**3 \n", + "N = V / 22.4143 \n", + "Q = N * Cpm * ( T2 - T1 ) \n", + "print \"(b)Heat to be supplied = %.3e\"%Q,\"kJ/h\"\n", + "T3 = 1500. #K\n", + "Q1 = Cpm * (T3 - T1) \n", + "\n", + "def f6(T): \n", + "\t return 26.586 + 7.582 * 10 **-3 * T - 1.12 * 10**-6 * T**2\n", + "\n", + "y = quad(f6,T1,T3)[0]\n", + "\n", + "Q2 = y \n", + "Perror = (Q2 - Q1) * 100 / Q2 \n", + "print \"(c)Percent error = %.1f\"%Perror,\"%\"\n", + "\n", + "# note : answer may vary because of rounding error.\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(a)Mean molal heat capacity = 31.619 kJ/kmolK\n", + "(b)Heat to be supplied = 3.527e+05 kJ/h\n", + "(c)Percent error = 4.1 %\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.13 page no : 416" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "# variables \n", + "T1 = 1500. #K\n", + "Tr = 273. #K\n", + "T2 = 400. #K\n", + "Cpm1 = 50. #kJ/kmol\n", + "Cpm2 = 35. #kJ/mol\n", + "\n", + "# Calculation \n", + "H = Cpm1 * ( T1 - Tr ) - Cpm2 * ( T2 - Tr ) \n", + "\n", + "# Result\n", + "print \"Enthalpy change = \",H,\"kJ/kmol\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Enthalpy change = 56905.0 kJ/kmol\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.14 pageno : 417" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from scipy.integrate import quad \n", + "\n", + "# variables \n", + "#CO, 26.586 + 7.582*10**-3*T - 1.12*10**-6*T**2\n", + "#CO2, 26.540 + 42.454*10**-3*T - 14.298*10**-6*T**2\n", + "#O2, 25.74 + 12.987*10**-3*T - 3.864*10**-6*T**2\n", + "#N2, 27.03 + 5.815*10**-3*T - 0.289*10**-6*T**2\n", + "#Cpmix = summation ( yi*Cpi ) = summation(yi*ai + yi*bi*T + yi*ci*T**2)\n", + "\n", + "xco2 = 0.09 \n", + "xco = 0.02 \n", + "xo2 = 0.07 \n", + "xn2 = 0.82 \n", + "T1 = 600. #K\n", + "T2 = 375. #K\n", + "\n", + "# Calculation \n", + "sumai = xco * 26.586 +xco2 * 26.540 + xo2 * 25.74 + xn2*27.03 \n", + "sumbi = xco * 7.582*10**-3 + xco2*42.454*10**-3+xo2*12.987*10**-3 + xn2*5.815*10**-3 \n", + "sumci = -(xco * 1.12*10**-6 + xco2*14.298*10**-6+xo2*3.864*10**-6+xn2*0.289*10**-6) \n", + "\n", + "def f8(T): \n", + "\t return sumai+sumbi*T+sumci*T**2\n", + "\n", + "H = quad(f8,T2,T1)[0]\n", + "\n", + "# Result\n", + "print \"Enthalpy change = %.2f\"%H,\"kJ/kmol\"\n", + "\n", + "# note : Calculating integration using quad may vary answer. Because Python use's its own methodology to calculate quad." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Enthalpy change = 7009.12 kJ/kmol\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.15 pageno : 420" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# variables \n", + "Hna = 26.04 #J/g-atomK\n", + "Hs = 22.6 #J/g-atomK\n", + "Ho = 16.8 #J/g-atomK\n", + "Hh = 9.6 #J/g-atomK\n", + "\n", + "# Calculation \n", + "Hna2so410h2o = 2*Hna + Hs + 14*Ho + 20*Hh \n", + "Hexp = 592.2 #J/molK\n", + "Deviation = (Hexp - Hna2so410h2o)*100/Hexp \n", + "\n", + "# Result\n", + "print \"Deviation in heat capacity = %.2f\"%Deviation,\"%\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Deviation in heat capacity = 15.25 %\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.16 pageno : 421" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# variables \n", + "P1 = 75. #kPa\n", + "T1 = 573. #K\n", + "Tvap = 365. #K\n", + "Tbasis = 273. #K\n", + "\n", + "#Since, the boiling point of water at 75kPa is 375K, the vapour at 573K is superheated \n", + "H1 = 3075. #kJ/kg\n", + "Cliq = 4.2 #kJ/kgK\n", + "Cvap = 1.97 # kJ/kg/K\n", + "m = 1. #kg\n", + "\n", + "\n", + "# Calculation \n", + "#let assume converting liq. water into superheated stream occurs in 3 steps,\n", + "\n", + "Hc1 = m*Cliq * ( Tvap - Tbasis ) \n", + "\n", + "\n", + "Hc3 = m*Cvap*(T1 - Tvap) \n", + "\n", + "#total enthalpy = 3075 = Hc1 + Hc2 + Hc3, therefore\n", + "Hc2 = H1 - Hc1 - Hc3 \n", + "\n", + "# Result\n", + "print \"Heat of vapourisation = \",Hc2,\"kJ/kg\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat of vapourisation = 2278.84 kJ/kg\n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.17 pageno : 422" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from scipy.integrate import quad\n", + "\n", + "# variables \n", + "T1 = 250. #K\n", + "T = 273.15 #K\n", + "T2 = 400. #K\n", + "Cice = 2.037 #kJ/kgK\n", + "T3 = 373.15 #K\n", + "Cliq = 75.726 #kJ/kmolK\n", + "\n", + "#Cp = 30.475 + 9.652*10**-3*T + 1.189*10**-6*T**2\n", + "Hfusion = 6012. #kJ/kmol\n", + "Hvap = 40608. #kJ/kmol\n", + "\n", + "\n", + "# Calculation \n", + "#1 - Heat for raising the temperature of ice, H1\n", + "H1 = Cice * (T - T1) \n", + "#2 - Latent heat of fusion of ice, Hf\n", + "Hf = Hfusion / 18.016 #kJ\n", + "\n", + "#3 - Sensible heat of raising the temperature of water, H2\n", + "H2 = Cliq * ( T3 - T)/18.016 \n", + "#4 - Latent heat of vaporization of water, Hv\n", + "Hv = Hvap / 18.016 \n", + "#5 - Sensible heat of raising the temperature of water vapou, H3\n", + "\n", + "def f4(T): \n", + "\t return 30.475 + 9.652*10.**-3*T + 1.189*10.**-6*T**2.\n", + "\n", + "H3 = 51.243883 #quadrature(f4,T3,T2)[0]\n", + "\n", + "Q = H1 + H2 + H3 + Hf + Hv \n", + "\n", + "# Result\n", + "print \"Heat required = %.1f\"%Q,\"kJ\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat required = 3106.4 kJ\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.18 pageno : 423" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# variables \n", + "#Cp = 0.16 + 4.78 * (10**-3) * T ( organic liquid )\n", + "#Cp = 0.7935 + 1.298 * (10**-4) * T ( CCL4 )\n", + "Tb = 349.9 #K\n", + "Hv = 195. #kJ/kg\n", + "Cp = 0.4693 #kJ/kgK\n", + "\n", + "# Calculation \n", + "#Let T be the final temperature\n", + "#integration(T - 650)(0.16 + 4.78 * (10**-3) * T)dt = integration(295 - T)(0.7935 + 1.298 * (10**-4) * T)dt\n", + "# the above equation yields, 2.4549*(10**-3)*T**2 + 0.9535*T - 1353.51 = 0, from this we get\n", + "T = 573.3 #K\n", + "\n", + "#since this temperature is above boiling point of CCl4,\n", + "#heat balance is, integration(T - 650)(0.16 + 4.78 * (10**-3) * T)dt = integration(295 - 349.9)(0.7935 + 1.298 * (10**-4) * T)dt + Hv + integration(349.9 - T)*0.4693*dT\n", + "#solving above equation, we get,\n", + "T1 = 540.1 #K\n", + "\n", + "# Result\n", + "print \"equilibrium temperature of the mixture = \",T1,\"K\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "equilibrium temperature of the mixture = 540.1 K\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.19 pageno : 427" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables \n", + "T1 = 363. #K\n", + "T2 = 373. #K\n", + "P1s = 70.11 #kPa\n", + "P2s = 101.3 #kPa\n", + "R = 8.314 #kJ/kmolK\n", + "\n", + "\n", + "# Calculation \n", + "# ln(P2s / P1s) = Hv / R * (1/T1 - 1/T2) \n", + "Hv = (math.log(P2s/P1s)*R)/(1/T1 - 1/T2) \n", + "Hv1 = Hv / (18) \n", + "\n", + "# Result\n", + "print \"Mean heat of vaporization = %.f\"%Hv1,\"kJ/kg\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Mean heat of vaporization = 2302 kJ/kg\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.20 pageno : 427" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# variables \n", + "T = 273.15 - 30 \n", + "R = 8.314 #K\n", + "\n", + "# Calculation \n", + "#lnPs = 14.2410 - 2137.72 / (T-26.72)\n", + "#dlnPs/dT = Hv / RT2\n", + "Hv = 2137.72 * R * T**2 / ( T - 26.72 )**2 \n", + "\n", + "# Result\n", + "print \"Heat of vaporization = %.2f\"%Hv,\"kJ/kmol\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat of vaporization = 22432.33 kJ/kmol\n" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.21 pageno : 427" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# variables \n", + "Hv1 = 2256. #kJ/kg\n", + "T1 = 373. #K\n", + "T2 = 473. #K\n", + "Tc = 647. #K\n", + "\n", + "# Calculation \n", + "Tr1 = T1 / Tc \n", + "Tr2 = T2 / Tc \n", + "#Hv2 / Hv1 = ((1-Tr2)/(1-Tr1))**0.38\n", + "Hv2 = Hv1*(((1-Tr2)/(1-Tr1))**0.38) \n", + "\n", + "# Result\n", + "print \"Latent heat of vaporization of water at 473K = %d\"%Hv2,\"kJ/kg\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Latent heat of vaporization of water at 473K = 1898 kJ/kg\n" + ] + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.22 pageno : 428" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "from scipy.integrate import quad \n", + "\n", + "\n", + "# Variables \n", + "#Cp = a + b*T\n", + "\n", + "T1 = 293.15 #K\n", + "Cp1 = 131.05 #J/molK\n", + "T2 = 323. #K\n", + "Cp2 = 138.04 #J/molK\n", + "\n", + "\n", + "# Calculation \n", + "#a + 293*b = 131.05\n", + "#a + 323*b = 138.04\n", + "b = (Cp1 - Cp2)/(T1 - T2) \n", + "a = Cp1 - b * T1 \n", + "\n", + "#Cp = 62.781 + 0.233*T\n", + "# Hvb / Tb = 36.63 + 8.31lnTb\n", + "Tb = 273.15 + 80.1 #K\n", + "\n", + "Hvb = (36.63 + 8.31*math.log(Tb)) * Tb \n", + "m = 100. #kg\n", + "\n", + "def f7(T): \n", + "\t return 62.781 + 0.233*T\n", + "\n", + "k = quad(f7,T1,Tb)[0]\n", + "\n", + "H = m*(10.**3) * ( k)/78.048 + m*(10.**3) * Hvb/78.048\n", + "\n", + "# Result\n", + "print \"Heat required = %.3e J\" %H\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat required = 4.928e+07 J\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.23 pageno : 430" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# variables \n", + "P = 10. #kPa\n", + "T1 = 323.15 #K\n", + "T2 = 373.15 #K\n", + "T = 358.15 #K\n", + "H1 = 2592.6 #kJ/kg\n", + "H2 = 2687.5 #kJ/kg\n", + "\n", + "\n", + "# Calculation \n", + "#H by interpolation,\n", + "H = H1 + ((H2 - H1)/(T2 - T1))*(T - T1) \n", + "Hl = 697.061 #kJ/kg\n", + "Hg = 2762. #kJ/kg\n", + "\n", + "#H = x*Hl + ( 1 - x )* Hg\n", + "x = (H - Hg)/(Hl - Hg) \n", + "Pmois = x*100 \n", + "Psteam = ( 1 - x )*100 \n", + "\n", + "# Result\n", + "print \"Percentage of moisture = %.f\"%Pmois,\"%\"\n", + "print \"Percentage of dry saturated steam = %.f\"%Psteam,\"%\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Percentage of moisture = 5 %\n", + "Percentage of dry saturated steam = 95 %\n" + ] + } + ], + "prompt_number": 27 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.24 pageno : 430" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# variables \n", + "P = 3500. #kPa\n", + "T = 673.15 #K\n", + "SV = 0.08453 #m**3/kg\n", + "Vcondensed = 1/2. \n", + "m = 100. #kg\n", + "\n", + "# Calculation \n", + "V = m * SV / (m/2) \n", + "#m*(Vl+Vg)*Vcondensed = m * SV\n", + "#But Vl is negligible,\n", + "Vg = m * SV / (m * Vcondensed) \n", + "#using steam table\n", + "T1 = 459.5 #K\n", + "P1 = 1158. #kPa\n", + "\n", + "#internal energy of superheated steam from steam table\n", + "I = 2928.4 #kJ/kg\n", + "U1 = m * I \n", + "Ul = 790. #kJ/kg\n", + "Ug = 2585.9 #kJ/kg\n", + "U2 = m*Vcondensed*Ul + m*(1-Vcondensed)*Ug \n", + "Q = U2 - U1 \n", + "\n", + "# Result\n", + "print \"The amount of heat removed fromt he system = \",Q,\"kJ\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The amount of heat removed fromt he system = -124045.0 kJ\n" + ] + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.25 pageno : 433" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# variables \n", + "m = 1000. #kg/h ( basis mass of 10% NaOH solution )\n", + "Pfeed = 10. #%\n", + "Ppro = 50. #(Percentage NaOH in product)\n", + "\n", + "\n", + "# Calculation \n", + "#Taking NaOH balance,P being the weight of the product\n", + "P = Pfeed * m / Ppro \n", + "\n", + "#W be the weight of water evaporized\n", + "W = m - P \n", + "\n", + "#step1 - cooling 1000kg/h of 10% solution from 305K to 298K\n", + "T1 = 305. #K\n", + "T2 = 298. #K\n", + "Cliq = 3.67 #kJ/kgK\n", + "H1 = m*Cliq * (T2 - T1) \n", + "\n", + "#step2 - separation into pure components\n", + "Hsolution = -42.85 #kJ/mol\n", + "H2 = -Pfeed * m *1000 *Hsolution/ (40*100) \n", + "\n", + "#step3 - W kg water is converted to water vapour\n", + "Hvap = 2442.5 #kJ/kg\n", + "H3 = W * Hvap \n", + "\n", + "#step4 - water vapour at 298K is heated to 373.15K\n", + "Cvap = 1.884 #kJ/kgK\n", + "T3 = 373.15 #K\n", + "H4 = W * Cvap * ( T3 - T2 ) \n", + "\n", + "#step5 - formation of 200kg of 50% NaOH solution at 298K\n", + "Hsolu = -25.89 #kJ/mol\n", + "H5 = Pfeed * m *1000 *Hsolu/ (40*100) \n", + "\n", + "#step6 - Heating the solution from 298K to 380K\n", + "Csolu = 3.34 #kJ/kg\n", + "T4 = 380. #K\n", + "H6 = P * Csolu * (T4 - T2) \n", + "Htotal = H1 + H2 + H3 + H4 + H5 + H6 \n", + "\n", + "# Result\n", + "print \"The enthalpy change accompanying the complete process = %d\"%Htotal,\"kJ\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The enthalpy change accompanying the complete process = 2138752 kJ\n" + ] + } + ], + "prompt_number": 28 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.26 page no : 435" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "# variables \n", + "Nwater = 0.8 #moles\n", + "Nethanol = 0.2 #moles\n", + "T = 323. #K\n", + "Cwater = 4.18*10**3 #J/kgK\n", + "Cethanol = 2.58*10**3 #J/kgK\n", + "Hmixing1 = -758. #J/mol ( at 298K )\n", + "Hmixing2 = -415. #J/mol ( at 323K )\n", + "T1 = 298. #K\n", + "T2 = 523. #K\n", + "\n", + "# Calculation \n", + "#step1 - 0.8 mol of water is cooled from 323 K to 298K\n", + "H1 = Nwater * 18 * Cwater * ( T1 - T )/ 1000 \n", + "\n", + "#step2 - 0.2 mol ethanol cooled from 323K to 298K\n", + "H2 = Nethanol * 46 * Cethanol * ( T1 - T )/1000 \n", + "\n", + "#step3 - 0.8 mol water and 0.2 mol ethanol are mixed together,\n", + "H3 = Hmixing1 \n", + "\n", + "#step4 solution is heated to 323K, H4 = Cpm * (T - T1)\n", + "#Hmixing2 = H1 + H2 + H3 + H4\n", + "H4 = Hmixing2 - H1 - H2 - H3 \n", + "Cpm = H4 / ( T - T1 ) \n", + "\n", + "# Result\n", + "print \"The mean heat capacity of a 20 percent solution = %.2f\"%Cpm,\"J/molK\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The mean heat capacity of a 20 percent solution = 97.65 J/molK\n" + ] + } + ], + "prompt_number": 30 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.27 page no : 437" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# variables \n", + "F = 1000. #kg/h\n", + "H1 = 116.3 #kJ/kg ( enthalpy of feed solution - 10% NaOH, 305 K )\n", + "H2 = 560.57 #kJ/kg ( enthalpy of thick liquor - 50% NaOH, 380 K )\n", + "Hsteam = 2676. #kJ/kg ( 1atm , 373.15K )\n", + "\n", + "#by doing material balances,\n", + "P = 200. #kg/h\n", + "mvap = 800. #kg/h\n", + "\n", + "# Calculation \n", + "#Enthalpy balance gives, F*H1 + Q = mvap*Hsteam + P*H2\n", + "Q = (mvap*Hsteam + P*H2)-F*H1 \n", + "\n", + "# Result\n", + "print \"Heat to be supplied = \",Q,\"kJ/h\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat to be supplied = 2136614.0 kJ/h\n" + ] + } + ], + "prompt_number": 31 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.28 pageno : 439" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# variables \n", + "U2 = 0.35*10**3 #kJ\n", + "U1 = 0.25*10**3 #kJ\n", + "#since the tank is rigid the volume does not change during heating, Under constant volume, the change in the internal energy is equal to the heat supplied\n", + "\n", + "# Calculation \n", + "Q = U2 - U1 \n", + "\n", + "# Result\n", + "print \"Heat transferred to the air = \",Q,\"kJ\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat transferred to the air = 100.0 kJ\n" + ] + } + ], + "prompt_number": 29 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.29 pageno : 439" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# variables \n", + "W = -2.25*745.7 #W ( work done on the system and 1hp = 745.7W)\n", + "Q = -3400. #kJ/h ( Heat transferred to the surrounding )\n", + "\n", + "# Calculation \n", + "U = Q*1000/3600 - W \n", + "\n", + "# Result\n", + "print \"Rise in the Internal energy of the system = %.3f\"%U,\"J/s\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Rise in the Internal energy of the system = 733.381 J/s\n" + ] + } + ], + "prompt_number": 32 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.30 pageno : 440" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# variables \n", + "#2Fe + 3/2O2 = Fe2O3\n", + "Hliberated = 831.08 #kJ\n", + "\n", + "# Calculation \n", + "Q = -Hliberated*1000 \n", + "\n", + "# Result\n", + "print \"Q = \",Q,\"J\"\n", + "\n", + "#P(V) = (n)RT\n", + "#W = P(V) = (n)RT\n", + "n = -1.5 \n", + "R = 8.314 \n", + "T = 298. #K\n", + "W = (n) * R * T \n", + "print \"W = %.1f\"%W,\"J\"\n", + "U = Q - W \n", + "print \"U = %.5e\"%U,\"J\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Q = -831080.0 J\n", + "W = -3716.4 J\n", + "U = -8.27364e+05 J\n" + ] + } + ], + "prompt_number": 34 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.31 pageno : 440" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# variables \n", + "Vgas = 0.09 #m**3\n", + "Vliq = 0.01 #m**3\n", + "SVliq = 1.061*10**-3 #m**3/kg\n", + "SVvap = 0.8857 #m**3/kg\n", + "\n", + "# Calculation \n", + "mvap = Vgas / SVvap \n", + "mliq = Vliq / SVliq \n", + "Ul = 504.5 #kJ/kg\n", + "Ug = 2529.5 #kJ/kg\n", + "U1 = Ul * mliq + Ug * mvap \n", + "SVtotal = (Vgas + Vliq)/(mvap + mliq) \n", + "#using steam table , these value of specific volume corresponds to\n", + "# pressure of 148.6bar and internal energy of 2464.6kJ/kg\n", + "U = 2464. #kJ/kg\n", + "Utotal = U * (mvap + mliq) \n", + "#Utotal - U1 = Q - W,but W = o, hence, \n", + "Q = Utotal - U1 \n", + "\n", + "# Result\n", + "print \"Heat to be added = %.f\"%Q,\"kJ\"\n", + "\n", + "\n", + "# note: answer may vary because of rounding error" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat to be added = 18462 kJ\n" + ] + } + ], + "prompt_number": 35 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.32 pageno : 443" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# variables \n", + "m = 10. #kg(air)\n", + "N = m / 29 #kmol\n", + "P1 = 100. #kPa\n", + "T1 = 300. #K\n", + "R = 8.314 \n", + "\n", + "# Calculation \n", + "V1 = N * R * T1 / P1 \n", + "V2 = V1 \n", + "T2 = 600. #K\n", + "Cv = 20.785 #kJ/kmolK\n", + "Cp = 29.099 #kJ/kmolK\n", + "U = N * Cv * (T2 - T1) \n", + "Q = U \n", + "W = Q - U \n", + "H = U + N * R * ( T2 - T1 ) \n", + "\n", + "# Result\n", + "print \"(a)Change in internal energy at constant volume = %d\"%U,\"kJ\"\n", + "print \"heat supplied at constant volume = %d\"%Q,\"kJ\"\n", + "print \"Work done at constant volume = \",W,\"kJ\"\n", + "print \"Change in Enthalpy at constant volume = %d\"%H,\"kJ\"\n", + "P2 = P1 \n", + "H2 = N * Cp * ( T2 - T1 ) \n", + "Q2 = H2 \n", + "U2 = H2 - N * R * (T2 - T1) \n", + "W2 = Q2 - U2 \n", + "print \"(b)Change in internal energy at constant Pressure = %d\"%U2,\"kJ\"\n", + "print \"heat supplied at constant Pressure = %d\"%Q2,\"kJ\"\n", + "print \"Work done at constant Pressure = %d\"%W2,\"kJ\"\n", + "print \"Change in Enthalpy at constant Pressure = %d\"%H2,\"kJ\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(a)Change in internal energy at constant volume = 2150 kJ\n", + "heat supplied at constant volume = 2150 kJ\n", + "Work done at constant volume = 0.0 kJ\n", + "Change in Enthalpy at constant volume = 3010 kJ\n", + "(b)Change in internal energy at constant Pressure = 2150 kJ\n", + "heat supplied at constant Pressure = 3010 kJ\n", + "Work done at constant Pressure = 860 kJ\n", + "Change in Enthalpy at constant Pressure = 3010 kJ\n" + ] + } + ], + "prompt_number": 36 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.33 pageno : 444" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# variables \n", + "Cp = 29.3 #kJ/kmol\n", + "R = 8.314 \n", + "Cv = Cp - R \n", + "T1 = 300. #K\n", + "P1 = 1. #bar\n", + "P2 = 2. #bar\n", + "\n", + "# Calculation \n", + "#step1 - Volume remains constant, therefore the work done is\n", + "# zero and heat supplied is Cv, Also T2/T1 = P2/P1\n", + "T2 = P2 * T1 / P1 \n", + "Q1 = Cv * ( T2 - T1 ) \n", + "W1 = 0 \n", + "\n", + "# Result\n", + "print \"Work done at constant volume = \",W1,\"kJ\"\n", + "print \"Heat supplied at constant volume = \",Q1,\"kJ\"\n", + "\n", + "#step2 - Process is abdiabatic\n", + "Q2 = 0 \n", + "r = 1.4 \n", + "T3 = T2 * (( P1 / P2 )**((r - 1)/r)) \n", + "W2 = Cv * ( T2 - T3 ) \n", + "\n", + "print \"Work done in adiabatic process = %.1f\"%W2,\"kJ\"\n", + "print \"Heat supplied in adiabatic process = \",Q2,\"kJ\"\n", + "\n", + "#step3 - process is isobaric\n", + "Q3 = Cp * (T1 - T3) \n", + "U3 = Cv * (T1 - T3) \n", + "W3 = Q3 - U3 \n", + "print \"Work done at constant pressure = %.2f\"%W3,\"kJ\"\n", + "print \"Heat supplied at constant pressure = %.1f\"%Q3,\"kJ\"\n", + "\n", + "\n", + "# Note : Answers in book is wrong. while calculating heat supplied i.e. Cv(T2-T1), value of Cv is been taken wrongly." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Work done at constant volume = 0 kJ\n", + "Heat supplied at constant volume = 6295.8 kJ\n", + "Work done in adiabatic process = 2262.3 kJ\n", + "Heat supplied in adiabatic process = 0 kJ\n", + "Work done at constant pressure = -1597.96 kJ\n", + "Heat supplied at constant pressure = -5631.5 kJ\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.34 pageno : 445" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables \n", + "P1 = 5. #bar\n", + "P2 = 4. #bar\n", + "T1 = 600. #K\n", + "V = 0.1 #m**3\n", + "T2 = 400. #K\n", + "T = 298. #K\n", + "Cp = 30. #J/molK\n", + "\n", + "#step1 - isothermal condition\n", + "U1 = 0 \n", + "H1 = 0 \n", + "P = 1. #bar\n", + "R = 8.314 \n", + "\n", + "# Calculation \n", + "W1 = R*T1*math.log(P1/P2) \n", + "Q1 = W1 \n", + "\n", + "# Result\n", + "print \"(a)Change in the internal energy in isothermal condition = \",U1,\"kJ/kmol\"\n", + "print \"Change in the enthalpy energy in isothermal condition = \",H1,\"kJ/kmol\"\n", + "print \"Work done in isothermal condition = %.2f\"%W1,\"kJ/kmol\"\n", + "print \"Heat supplied in isothermal condition = %.2f\"%Q1,\"kJ/kmol\"\n", + "N = round(P * (1.01325 * 10**5) * V / ( R * T ),2) # answer slightly different because of rouding error.\n", + "Cv = Cp - R \n", + "U2 = Cv * (T2 - T)*N #Answer differes due to slighlty different value of N\n", + "H2 = Cp * (T2 - T)*N #Answer differes due to slighlty different value of N\n", + "W2 = 0 \n", + "Q2 = U2 + W2 \n", + "print \n", + "print \"\\n(b)Change in the internal energy at constant volume condition = %d\"%U2,\"J\"\n", + "print \"Change in the enthalpy energy at constant volume condition = %d\"%H2,\"kJ/kmol\"\n", + "print \"Work done at constant volume condition = \",W2,\"kJ/kmol\"\n", + "print \"Heat supplied at constant volume condition = %d\"%Q2,\"kJ/kmol\"\n", + "\n", + "# note : answer varies because of rounding error." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(a)Change in the internal energy in isothermal condition = 0 kJ/kmol\n", + "Change in the enthalpy energy in isothermal condition = 0 kJ/kmol\n", + "Work done in isothermal condition = 1113.13 kJ/kmol\n", + "Heat supplied in isothermal condition = 1113.13 kJ/kmol\n", + "\n", + "\n", + "(b)Change in the internal energy at constant volume condition = 9046 J\n", + "Change in the enthalpy energy at constant volume condition = 12515 kJ/kmol\n", + "Work done at constant volume condition = 0 kJ/kmol\n", + "Heat supplied at constant volume condition = 9046 kJ/kmol\n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.35 pageno : 448" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# variables \n", + "m = 1. #kg\n", + "u2 = 0.5 #m/s\n", + "u1 = 60. #m/s\n", + "H = -3000. #kJ/kg\n", + "\n", + "# Calculation \n", + "#KE = (u**2)/2\n", + "KE = ((u2 ** 2) - (u1**2))/2000 \n", + "g = 9.81 #m/s**2\n", + "Z1 = 7.5 #m\n", + "Z2 = 2. #m\n", + "\n", + "#PE = g * (Z)\n", + "PE = g * (Z2 - Z1)/1000 \n", + "W = 800. #kJ/kg\n", + "Q = H + PE + KE + W \n", + "\n", + "# Result\n", + "print \"Heat removed from the fluid = %.2f\"%Q,\"kJ/kg\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat removed from the fluid = -2201.85 kJ/kg\n" + ] + } + ], + "prompt_number": 39 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.37 pageno : 449" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# variables \n", + "g = 9.81 #m/s**2\n", + "z = 55. \n", + "PE = g * z \n", + "KE = 0. \n", + "T2 = 288. #K\n", + "f = 1.5*10**-2 #m**3/min\n", + "D = 1000. #kg/m**3\n", + "m = f * D \n", + "Qsupp = 500. #kJ/min\n", + "Qlost = 400. #kJ/min\n", + "\n", + "# Calculation \n", + "Qnet = (Qsupp - Qlost) * D / m \n", + "W = 2*745.7 #W\n", + "Ws = -W * 0.6 / (m/60) \n", + "H = Qnet - Ws - PE - KE \n", + "Cp = 4200. \n", + "T1 = H / Cp \n", + "T = T1 + T2 \n", + "\n", + "# Result\n", + "print \"The temperature of exit water = %.2f\"%T,\"K\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The temperature of exit water = 290.31 K\n" + ] + } + ], + "prompt_number": 40 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.38 pageno : 450" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# variables \n", + "m = 1000. #kg/h (dried product)\n", + "\n", + "# S be the amount of dry solid in the product stream \n", + "Pmoisture1 = 4. #%\n", + "Pmoisture2 = 0.2 #%\n", + "P = 1.\n", + "S = m *(1 - P/1000) \n", + "\n", + "# Calculation \n", + "X1 = Pmoisture1/(100 - Pmoisture1) \n", + "X2 = Pmoisture2/(100 - Pmoisture2) \n", + "\n", + "#let G be the weight of dry air in the air stream \n", + "Y1 = 0.01 #kg water/kg dry solid\n", + "Cp = 1.507 \n", + "Cw = 4.2 \n", + "T1 = 298. #K\n", + "T = 273. #K\n", + "T2 = 333. #K\n", + "Tg1 = 363. #K\n", + "Tg2 = 305. #K\n", + "\n", + "Hs1 = (Cp + X1 * Cw) * (T1 - T) \n", + "Hs2 = (Cp + X2 * Cw) * (T2 - T) \n", + "#Hg = Cs(Tg - To) + Y*L\n", + "#Cs = 1.005 + 1.884*Y\n", + "L = 2502.3 #kJ/kg dry air\n", + "Hg1 = (1.005 + 1.884 * Y1)*(Tg1 - T) + Y1 * L \n", + "Q = -40000. #kJ/h\n", + "\n", + "#Calculating for T2, Hg2 = 32.16 + 2562.59*Y \n", + "#change in enthalpy = Q\n", + "#H1 = S * Hs1 + G * HG1 = 37814.22 + 117.17G\n", + "#H2 = 100728.14 + G* (32.16 + 2561.59*Y)\n", + "#change in enthalpy = Q\n", + "#62913.92 + G *(-85.01 + 2561.59*Y) + 40000 = 0\n", + "#102913.92 + G *(-85.01 + 2561.59*Y) = 0 (1)\n", + "#moisture balance, S*X1 + G*Y1 = S*X2 + G*Y2\n", + "#G*(Y-0.01) = 39.62 (2)\n", + "#solving simultaneously ( 1 ) and ( 2 ), \n", + "Gdry = 3443. #kg/h\n", + "G = Gdry*(1 + Y1) \n", + "\n", + "# Result\n", + "print \"Air requirement = \",G,\"kg/h\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Air requirement = 3477.43 kg/h\n" + ] + } + ], + "prompt_number": 39 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.39 pageno : 452" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# variables \n", + "m = 1000. #kg/h ( feed solution )\n", + "\n", + "#F - mass of feed distilled, W - mass of the bottom product, D - mass of the distillate, xf, xd and xw - weight fraction of actone in feed, distillate and residue resp_\n", + "#total balance, F = D + W\n", + "#Acetone balance, F*xf = D*xd + w*xw\n", + "F = 1000 \n", + "xf = 0.10 \n", + "xd = 0.9 \n", + "xw = 0.01 \n", + "\n", + "# Calculation \n", + "#substituting in above equations,\n", + "D = F * (xf - xw) / (xd - xw) \n", + "W = F - D \n", + "R = 8 \n", + "L = R * D \n", + "#material balance around the condenser,G vapour reaching the condenser\n", + "G = L + D \n", + "Td = 332. #K\n", + "T2 = 300. #K\n", + "Tw = 370. #K\n", + "Tf = 340. #K\n", + "Lacetone1 = 620. #kJ/kg\n", + "Lwater1 = 2500. #kJ/kg\n", + "Ld = xd * Lacetone1 + (1 - xd) * Lwater1 \n", + "Cpacetone = 2.2 #kJ/kgK\n", + "Cpwater = 4.2 #kJ/kgK\n", + "Cp = xd * Cpacetone + (1-xd)*Cpwater \n", + "H = Ld + Cp * ( Td - T2 ) \n", + "Cpc = 4.2 #kJ/kg\n", + "Tc = 30. #K ( change in temperature allowable for cooling water )\n", + "m = G * H / ( Cpc * Tc ) \n", + "\n", + "# Result\n", + "print \"(a)The circulation rate of cooling water = %2f\"%m,\"kg/h\"\n", + "Qc = G * H \n", + "Hd = 0. \n", + "Hw = (xw * Cpacetone + (1-xw)*Cpwater)*(Tw - T2) \n", + "Hf = (xf * Cpacetone + (1-xf)*Cpwater)*(Tf - T2) \n", + "Qb = D * Hd + W * Hw + Qc - F * Hf \n", + "Hcondensation = 2730. #kJ/kg\n", + "msteam = Qb/Hcondensation \n", + "print \"(b)Amount of steam supplied = %.2f\"%msteam,\"kg/h\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(a)The circulation rate of cooling water = 6391.011236 kg/h\n", + "(b)Amount of steam supplied = 332.70 kg/h\n" + ] + } + ], + "prompt_number": 41 + } + ], + "metadata": {} + } + ] +}
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