{ "cells": [ { "cell_type": "markdown", "metadata": {}, "source": [ "# Chapter 7 - Entropy" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 1: pg 159" ] }, { "cell_type": "code", "execution_count": 1, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Example 7.1\n", " The specific entropy of water is (kJ/kg K) = 1.304\n", " From table The accurate value of sf in this case is (kJ/kg K) = 1.307\n", "There is small error in book's final value of sf\n" ] } ], "source": [ "#pg 159\n", "print('Example 7.1');\n", "import math\n", "# aim : To determine\n", "# the specific enthalpy of water\n", "\n", "# Given values\n", "Tf = 273.+100;# Temperature,[K]\n", "\n", "# solution\n", "# from steam table\n", "cpl = 4.187;# [kJ/kg K]\n", "# using equation [8]\n", "sf = cpl*math.log(Tf/273.16);# [kJ/kg*K]\n", "print ' The specific entropy of water is (kJ/kg K) = ',round(sf,3)\n", "\n", "# using steam table\n", "sf = 1.307;# [kJ/kg K]\n", "print ' From table The accurate value of sf in this case is (kJ/kg K) = ',sf\n", "\n", "print \"There is small error in book's final value of sf\"\n", "\n", "# End\n" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 2: pg 160" ] }, { "cell_type": "code", "execution_count": 2, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Example 7.2\n", " (a) The specific entropy of wet steam is (kJ/kg K) = 5.52\n", " (b) The specific entropy using steam table is (kJ/kg K) = 5.559\n" ] } ], "source": [ "#pg 160\n", "print('Example 7.2');\n", "\n", "# aim : To determine\n", "# the specific entropy\n", "import math\n", "# Given values\n", "P = 2.;# pressure,[MN/m^2]\n", "x = .8;# dryness fraction\n", "\n", "# solution\n", "# from steam table at given pressure\n", "Tf = 485.4;# [K]\n", "cpl = 4.187;# [kJ/kg K]\n", "hfg = 1888.6;# [kJ/kg]\n", "\n", "# (a) finding entropy by calculation\n", "s = cpl*math.log(Tf/273.16)+x*hfg/Tf;# formula for entropy calculation\n", "\n", "print ' (a) The specific entropy of wet steam is (kJ/kg K) = ',round(s,2)\n", "\n", "# (b) calculation of entropy using steam table\n", "# from steam table at given pressure\n", "sf = 2.447;# [kJ/kg K]\n", "sfg = 3.89;# [kJ/kg K]\n", "# hence\n", "s = sf+x*sfg;# [kJ/kg K]\n", "\n", "print ' (b) The specific entropy using steam table is (kJ/kg K) = ',s\n", "\n", "# End\n" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 3: pg 161" ] }, { "cell_type": "code", "execution_count": 3, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Example 7.3\n", " (a) The specific entropy of steam is (kJ/kg K) = 6.822\n", " (b) The accurate value of specific entropy from steam table is (kJ/kg K) = 6.919\n" ] } ], "source": [ "#pg 161\n", "print('Example 7.3');\n", "import math\n", "# aim : To determine\n", "# the specific entropy of steam\n", "\n", "# Given values\n", "P = 1.5;#pressure,[MN/m^2]\n", "T = 273.+300;#temperature,[K]\n", "\n", "# solution\n", "\n", "# (a)\n", "# from steam table\n", "cpl = 4.187;# [kJ/kg K]\n", "Tf = 471.3;# [K]\n", "hfg = 1946.;# [kJ/kg]\n", "cpv = 2.093;# [kJ/kg K]\n", "\n", "# usung equation [2]\n", "s = cpl*math.log(Tf/273.15)+hfg/Tf+cpv*math.log(T/Tf);# [kJ/kg K]\n", "print ' (a) The specific entropy of steam is (kJ/kg K) = ',round(s,3)\n", "\n", "# (b)\n", "# from steam tables\n", "s = 6.919;# [kJ/kg K]\n", "print ' (b) The accurate value of specific entropy from steam table is (kJ/kg K) = ',s\n", "\n", "# End\n" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 4: pg 164" ] }, { "cell_type": "code", "execution_count": 4, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Example 7.4\n", " The final dryness fraction of steam is x2 = 0.989\n" ] } ], "source": [ "#pg 164\n", "print('Example 7.4');\n", "\n", "# aim : To determine\n", "# the dryness fraction of steam\n", "\n", "# Given values\n", "P1 = 2.;# initial pressure, [MN/m^2]\n", "t = 350.;# temperature, [C]\n", "P2 = .28;# final pressure, [MN/m^2]\n", "\n", "# solution\n", "# at 2 MN/m^2 and 350 C,steam is superheated because the saturation temperature is 212.4 C\n", "# From steam table\n", "s1 = 6.957;# [kJ/kg K]\n", "\n", "# for isentropic process\n", "s2 = s1;\n", "# also\n", "sf2 = 1.647;# [kJ/kg K]\n", "sfg2 = 5.368;# [kJ/kg K]\n", "\n", "# using\n", "# s2 = sf2+x2*sfg2, where x2 is dryness fraction of steam\n", "# hence\n", "x2 = (s2-sf2)/sfg2;\n", "print ' The final dryness fraction of steam is x2 = ',round(x2,3)\n", "\n", "# End\n" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 5: pg 165" ] }, { "cell_type": "code", "execution_count": 6, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Example 7.5\n", " (a) From steam table at .06 MN/m^2 steam is superheated and has temperature of 100 C with specific volume is (m^3/kg) = 2.83\n", " (b) The change in specific entropy during the hyperbolic process is (kJ/kg K) = 0.679\n" ] } ], "source": [ "#pg 165\n", "print('Example 7.5');\n", "\n", "# aim : To determine\n", "# the final condition of steam...\n", "# the change in specific entropy during hyperbolic process\n", "\n", "# Given values\n", "P1 = 2;# pressure, [MN/m^2]\n", "t = 250.;# temperature, [C]\n", "P2 = .36;# pressure, [MN/m^2]\n", "P3 = .06;# pressure, [MN/m^2]\n", "\n", "# solution\n", "\n", "# (a)\n", "# from steam table\n", "s1 = 6.545;# [kJ/kg K]\n", "# at .36 MN/m^2\n", "sg = 6.930;# [kJ/kg*K]\n", "\n", "sf2 = 1.738;# [kJ/kg K]\n", "sfg2 = 5.192;# [kJ/kg K]\n", "vg2 = .510;# [m^3]\n", "\n", "# so after isentropic expansion, steam is wet\n", "# hence, s2=sf2+x2*sfg2, where x2 is dryness fraction\n", "# also\n", "s2 = s1;\n", "# so\n", "x2 = (s2-sf2)/sfg2;\n", "# and\n", "v2 = x2*vg2;# [m^3]\n", "\n", "# for hyperbolic process\n", "# P2*v2=P3*v3\n", "# hence\n", "v3 = P2*v2/P3;# [m^3]\n", "\t\n", "print ' (a) From steam table at .06 MN/m^2 steam is superheated and has temperature of 100 C with specific volume is (m^3/kg) = ',round(v3,2)\n", "\n", "# (b)\n", "# at this condition\n", "s3 = 7.609;# [kJ/kg*K]\n", "# hence\n", "change_s23 = s3-sg;# change in specific entropy during the hyperblic process[kJ/kg*K]\n", "print ' (b) The change in specific entropy during the hyperbolic process is (kJ/kg K) = ',change_s23\n", "\n", "# In the book they have taken sg instead of s2 for part (b), so answer is not matching\n", "\n", "# End\n", "\n" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 6: pg 166" ] }, { "cell_type": "code", "execution_count": 7, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Example 7.6\n" ] }, { "data": { "image/png": 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"text/plain": [ "" ] }, "metadata": {}, "output_type": "display_data" }, { "name": "stdout", "output_type": "stream", "text": [ " (a) The heat transfer during the expansion is (kJ) (received) = 1224.986976\n", " (b) The work done during the expansion is (kJ) = 2705.234976\n" ] } ], "source": [ "#pg 166\n", "print('Example 7.6');\n", "\n", "# aim : To determine the\n", "# (a) heat transfer during the expansion and\n", "# (b) work done durind the expansion\n", "%matplotlib inline\n", "import matplotlib\n", "from matplotlib import pyplot\n", "# given values\n", "m = 4.5; # mass of steam,[kg]\n", "P1 = 3.; # initial pressure,[MN/m^2]\n", "T1 = 300.+273; # initial temperature,[K]\n", "\n", "P2 = .1; # final pressure,[MN/m^2]\n", "x2 = .96; # dryness fraction at final stage\n", "\n", "# solution\n", "# for state point 1,using steam table\n", "s1 = 6.541;# [kJ/kg/K]\n", "u1 = 2751;# [kJ/kg]\n", "\n", "# for state point 2\n", "sf2 = 1.303;# [kJ/kg/K]\n", "sfg2 = 6.056;# [kJ/kg/k]\n", "T2 = 273+99.6;# [K]\n", "hf2 = 417;# [kJ/kg]\n", "hfg2 = 2258;# [kJ/kg]\n", "vg2 = 1.694;# [m^3/kg]\n", "\n", "# hence\n", "s2 = sf2+x2*sfg2;# [kJ/kg/k]\n", "h2 = hf2+x2*hfg2;# [kJ/kg]\n", "u2 = h2-P2*x2*vg2*10**3;# [kJ/kg]\n", "\n", "# Diagram of example 7.6\n", "x = ([s1, s2]);\n", "y = ([T1, T2]);\n", "pyplot.plot(x,y);\n", "pyplot.title('Diagram for example 7.6(T vs s)');\n", "pyplot.xlabel('Entropy (kJ/kg K)');\n", "pyplot.ylabel('Temperature (K)');\n", "x = ([s1,s1]);\n", "y = ([0,T1]);\n", "pyplot.plot(x,y);\n", "x = ([s2,s2]);\n", "y = ([0,T2]);\n", "pyplot.plot(x,y);\n", "pyplot.show()\n", "# (a)\n", "# Q_rev is area of T-s diagram\n", "Q_rev = (T1+T2)/2*(s2-s1);# [kJ/kg]\n", "# so total heat transfer is\n", "Q_rev = m*Q_rev;# [kJ]\n", "\n", "# (b)\n", "del_u = u2-u1;# change in internal energy, [kJ/kg]\n", "# using 1st law of thermodynamics\n", "W = Q_rev-m*del_u;# [kJ]\n", "\n", "print ' (a) The heat transfer during the expansion is (kJ) (received) = ',Q_rev\n", "\n", "print ' (b) The work done during the expansion is (kJ) = ',W\n", "\n", "# End\n" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 7: pg 176" ] }, { "cell_type": "code", "execution_count": 8, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Example 7.7\n", " (a) The change of entropy is (kJ/K) = -0.046\n", " (b) The approximate change of entropy obtained by dividing the heat transferred by the gas by the mean absolute temperature during the compression (kJ/K) = -0.0448\n" ] } ], "source": [ "#pg 176\n", "print('Example 7.7');\n", "\n", "# aim : To determine the \n", "# (a) change of entropy\n", "# (b) The approximate change of entropy obtained by dividing the heat transferred by the gas by the mean absolute temperature during the compression\n", "import math\n", "# Given values\n", "P1 = 140.;# initial pressure,[kN/m^2]\n", "V1 = .14;# initial volume, [m^3]\n", "T1 = 273.+25;# initial temperature,[K]\n", "P2 = 1400.;# final pressure [kN/m^2]\n", "n = 1.25; # polytropic index\n", "cp = 1.041;# [kJ/kg K]\n", "cv = .743;# [kJ/kg K]\n", "\n", "# solution\n", "# (a)\n", "R = cp-cv;# [kJ/kg/K]\n", "# using ideal gas equation \n", "m = P1*V1/(R*T1);# mass of gas,[kg]\n", "# since gas is following law P*V^n=constant ,so \n", "V2 = V1*(P1/P2)**(1./n);# [m^3]\n", "\n", "# using eqn [9]\n", "del_s = m*(cp*math.log(V2/V1)+cv*math.log(P2/P1));# [kJ/K]\n", "print ' (a) The change of entropy is (kJ/K) = ',round(del_s,3)\n", "\n", "# (b)\n", "W = (P1*V1-P2*V2)/(n-1);# polytropic work,[kJ]\n", "Gamma = cp/cv;# heat capacity ratio\n", "Q = (Gamma-n)/(Gamma-1)*W;# heat transferred,[kJ]\n", "\n", "# Again using polytropic law\n", "T2 = T1*(V1/V2)**(n-1);# final temperature, [K]\n", "T_avg = (T1+T2)/2;# mean absolute temperature, [K]\n", "\n", "# so approximate change in entropy is\n", "del_s = Q/T_avg;# [kJ/K]\n", "\n", "print ' (b) The approximate change of entropy obtained by dividing the heat transferred by the gas by the mean absolute temperature during the compression (kJ/K) = ',round(del_s,4)\n", "\n", "# End\n" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 8: pg 179" ] }, { "cell_type": "code", "execution_count": 9, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Example 7.8\n", " The change of entropy in constant volume process is (kJ/kg K) = 0.149\n", " The change of entropy in constant pressure process is (kJ/kg K) = 0.332\n", "there is misprint in the book's result\n" ] } ], "source": [ "#pg 179\n", "print('Example 7.8');\n", "\n", "# aim : To determine\n", "# the change of entropy\n", "import math\n", "# Given values\n", "m = .3;# [kg]\n", "P1 = 350.;# [kN/m^2]\n", "T1 = 273.+35;# [K]\n", "P2 = 700.;# [kN/m^2]\n", "V3 = .2289;# [m^3]\n", "cp = 1.006;# [kJ/kg K]\n", "cv = .717;# [kJ/kg K]\n", "\n", "# solution\n", "# for constant volume process\n", "R = cp-cv;# [kJ/kg K]\n", "# using PV=mRT\n", "V1 = m*R*T1/P1;# [m^3]\n", "\n", "# for constant volume process P/T=constant,so\n", "T2 = T1*P2/P1;# [K]\n", "s21 = m*cv*math.log(P2/P1);# formula for entropy change for constant volume process\n", "print ' The change of entropy in constant volume process is (kJ/kg K) = ',round(s21,3)\n", "\n", "# 'For the above part result given in the book is wrong\n", "\n", "V2 = V1;\n", "# for constant pressure process\n", "T3 = T2*V3/V2;# [K]\n", "s32 = m*cp*math.log(V3/V2);# [kJ/kg K]\n", "\n", "print ' The change of entropy in constant pressure process is (kJ/kg K) = ',round(s32,3)\n", "\n", "print \"there is misprint in the book's result\"\n", "\n", "# End\n" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 9: pg 181" ] }, { "cell_type": "code", "execution_count": 11, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Example 7.9\n", " The change of entropy is (kJ/kg K) = 0.04151\n" ] } ], "source": [ "#pg 181\n", "print('Example 7.9');\n", "import math\n", "# aim : To determine\n", "# the change of entropy\n", "\n", "# Given values\n", "P1 = 700.;# initial pressure, [kN/m^2]\n", "T1 = 273.+150;# Temperature ,[K]\n", "V1 = .014;# initial volume, [m^3]\n", "V2 = .084;# final volume, [m^3]\n", "\n", "# solution\n", "# since process is isothermal so\n", "T2 = T1;\n", "# and using fig.7.10\n", "del_s = P1*V1*math.log(V2/V1)/T1 ;# [kJ/K]\n", "#results\n", "print ' The change of entropy is (kJ/kg K) = ',round(del_s,5)\n", "\n", "# End\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": "2.7.11" } }, "nbformat": 4, "nbformat_minor": 0 }