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diff --git a/Basic_Engineering_Thermodynamics_by_Rayner_Joel/Chapter4_1.ipynb b/Basic_Engineering_Thermodynamics_by_Rayner_Joel/Chapter4_1.ipynb new file mode 100644 index 00000000..72f19875 --- /dev/null +++ b/Basic_Engineering_Thermodynamics_by_Rayner_Joel/Chapter4_1.ipynb @@ -0,0 +1,1296 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 4: Steam and two phase systems" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1: pg 61" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 4.1\n", + "The specific liquid enthalpy (kJ/kg) = 640.1\n", + "The specific enthalpy of evaporation (kJ/kg) = 2107.4\n", + "The specific enthalpy of dry saturated steam (kJ/kg) = 2747.5\n" + ] + } + ], + "source": [ + "#pg 61\n", + "#calculate the specific liquid enthalpy of evaporation and dry saturated steam\n", + "print('Example 4.1');\n", + "\n", + "# aim : To determine\n", + "# the enthalpy\n", + "\n", + "# Given values\n", + "P = .50;# Pressure, [MN/m^2]\n", + "\n", + "# solution\n", + "\n", + "# From steam tables, at given pressure\n", + "hf = 640.1;# specific liquid enthalpy ,[kJ/kg]\n", + "hfg = 2107.4;# specific enthalpy of evaporation ,[kJ/kg]\n", + "hg = 2747.5; # specific enthalpy of dry saturated steam ,[kJ/kg]\n", + "tf = 151.8; # saturation temperature,[C]\n", + "#results\n", + "print 'The specific liquid enthalpy (kJ/kg) = ',hf\n", + "print 'The specific enthalpy of evaporation (kJ/kg) = ',hfg\n", + "print 'The specific enthalpy of dry saturated steam (kJ/kg) = ',hg\n", + "\t\n", + "# End\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2: pg 61" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 4.2\n", + "The Saturation temperature (C) = 212.4\n", + "The Specific liquid enthalpy (kJ/kg) = 908.6\n", + "The Specific enthalpy of evaporation (kJ/kg) = 1888.6\n", + "The Specific enthalpy of dry saturated steam (kJ/kg) = 2797.2\n", + "The answers in the textbook are a bit different due to rounding off error\n" + ] + } + ], + "source": [ + "#pg 61\n", + "#calculate the saturation temperature, specific enthalpy\n", + "print('Example 4.2');\n", + "import numpy\n", + "# aim : To determine \n", + "# saturation temperature and enthalpy\n", + "\n", + "# Given values\n", + "P = 2.04;# pressure, [MN/m^2]\n", + "\n", + "# solution\n", + "# since in the steam table values of enthalpy and saturation temperature at 2 and 2.1 MN?m^2 are given, so for knowing required values at given pressure,there is need to do interpolation\n", + "\n", + "# calculation of saturation temperature and results\n", + "# from steam table\n", + "# P in [MN/m^2] and tf in [C]\n", + "Table_P_tf_x=[2.1,2.0]\n", + "Table_P_tf_y=[214.9,212.4]\n", + "# using interpolation\n", + "tf = numpy.interp(P,Table_P_tf_x,Table_P_tf_y);# saturation temperature at given condition\n", + "print 'The Saturation temperature (C) = ',tf\n", + "\n", + "# calculation of specific liquid enthalpy\n", + "# from steam table\n", + "Table_P_hf_y = [920.0,908.6];# P in [MN/m^2] and hf in [kJ/kg]\n", + "Table_P_hf_x=[2.1,2.0]\n", + "# using interpolation\n", + "hf = numpy.interp(P,Table_P_hf_x,Table_P_hf_y); # enthalpy at given condition, [kJ/kg]\n", + "print 'The Specific liquid enthalpy (kJ/kg) = ',hf\n", + "\n", + "# calculation of specific enthalpy of evaporation\n", + "# from steam table\n", + "Table_P_hfg_x = [2.1,2.0];# P in [MN/m^2] and hfg in [kJ/kg]\n", + "Table_P_hfg_y=[1878.2,1888.6]\n", + "# using interpolation \n", + "hfg = numpy.interp(P,Table_P_hfg_x,Table_P_hfg_y); # enthalpy at given condition, [kJ/kg]\n", + "print 'The Specific enthalpy of evaporation (kJ/kg) = ',hfg\n", + "\n", + "# calculation of specific enthalpy of dry saturated steam\n", + "# from steam table\n", + "Table_P_hg_y = [2798.2,2797.2];#P in [MN/m^2] and hg in [kJ/kg]\n", + "Table_P_hg_x=[2.1,2.0]\n", + "# using interpolation\n", + "hg = numpy.interp(P,Table_P_hg_x,Table_P_hg_y); # enthalpy at given condition, [kJ/kg]\n", + "print 'The Specific enthalpy of dry saturated steam (kJ/kg) = ',hg\n", + "\n", + "print'The answers in the textbook are a bit different due to rounding off error'\n", + "# End\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3: pg 63" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 4.3\n", + "The specific enthalpy of steam at 2 MN/m^2 with temperature 250 C (kJ/kg) = 2902\n", + "The specific enthalpy at given T and P by alternative path (kJ/kg) = 2875.9\n" + ] + } + ], + "source": [ + "#pg 63\n", + "#calculate the specific enthalpy\n", + "print('Example 4.3');\n", + "\n", + "# aim : To determine\n", + "# the specific enthalpy\n", + "\n", + "# given values\n", + "P = 2; # pressure ,[MN/m^2]\n", + "t = 250; # Temperature, [C]\n", + "cp = 2.0934; # average value of specific heat capacity, [kJ/kg K]\n", + "\n", + "# solution\n", + "\n", + "# looking up steam table it shows that at given pressure saturation temperature is 212.4 C,so\n", + "tf = 212.4; # [C]\n", + "# hence,\n", + "Degree_of_superheat = t-tf;# [C]\n", + "# from table at given temperature 250 C\n", + "h = 2902; # specific enthalpy of steam at 250 C ,[kJ/kg]\n", + "\n", + "# Also from steam table enthalpy at saturation temperature is\n", + "hf = 2797.2 ;# [kJ/kg]\n", + "# so enthalpy at given temperature is\n", + "h2 = hf+cp*(t-tf);# [kJ/kg]\n", + "#results\n", + "print 'The specific enthalpy of steam at 2 MN/m^2 with temperature 250 C (kJ/kg) = ',h\n", + "print 'The specific enthalpy at given T and P by alternative path (kJ/kg) = ',round(h2,1)\n", + "\n", + "# End\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4: pg 63" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 4.4\n", + " The estimated specific enthalpy (kJ/kg) = 3002.08\n", + " The accurate specific enthalpy of steam at pressure of 2.5 MN/m^2 and with a temperature 320 C (kJ/kg) = 3025.0\n", + " The answer is a bit different from textbook due to rounding off error\n" + ] + } + ], + "source": [ + "#pg 63\n", + "#calculate the estimated and accurate specific enthalpy\n", + "print('Example 4.4');\n", + "import numpy\n", + "# aim : To determine\n", + "# the specific enthalpy of steam\n", + "\n", + "# Given values\n", + "P = 2.5;# pressure, [MN/m^2]\n", + "t = 320; # temperature, [C]\n", + "\n", + "# solution\n", + "# from steam table at given condition the saturation temperature of steam is 223.9 C, therefore steam is superheated\n", + "tf = 223.9;# [C]\n", + "\n", + "# first let's calculate estimated enthalpy\n", + "# again from steam table \n", + "\n", + "hg = 2800.9;# enthalpy at saturation temp, [kJ/kg]\n", + "cp =2.0934;# specific heat capacity of steam,[kJ/kg K]\n", + "\n", + "# so enthalpy at given condition is\n", + "h = hg+cp*(t-tf);# [kJ/kg]\n", + "print ' The estimated specific enthalpy (kJ/kg) = ',round(h,2)\n", + "\n", + "# calculation of accurate specific enthalpy\n", + "# we need double interpolation for this\n", + "\n", + "# first interpolation w.r.t. to temperature\n", + "# At 2 MN/m^2\n", + "Table_t_h_x = [325.,300.];# where, t in [C] and h in [kJ/kg]\n", + "Table_t_h_y=[3083.,3025.]\n", + "h1 = numpy.interp(320.,Table_t_h_x,Table_t_h_y); # [kJ/kg]\n", + "\n", + "# at 4 MN/m^2\n", + "Table_t_h_x = [325.,300.]; # t in [C] and h in [kJ/kg]\n", + "Table_t_h_y=[3031.,2962.]\n", + "h2 = numpy.interp(320.,Table_t_h_x,Table_t_h_y); # [kJ/kg]\n", + "\n", + "# now interpolation w.r.t. pressure\n", + "Table_P_h_x = [4.,2.]; # where P in NM/m^2 and h1,h2 in kJ/kg\n", + "Table_P_h_y=[h2,h1]\n", + "h = numpy.interp(2.5,Table_P_h_x,Table_P_h_y);# [kJ/kg]\n", + "print ' The accurate specific enthalpy of steam at pressure of 2.5 MN/m^2 and with a temperature 320 C (kJ/kg) = ',h\n", + "\n", + "# End\n", + "print ' The answer is a bit different from textbook due to rounding off error'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5: pg 65" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 4.5\n", + " The specific enthalpy of wet steam (kJ/kg) = 2317.605\n" + ] + } + ], + "source": [ + "#pg 65\n", + "#calculate the specific enthalpy\n", + "print('Example 4.5');\n", + "\n", + "# aim : To determine \n", + "# the specific enthalpy \n", + "\n", + "# Given values\n", + "P = 70; # pressure, [kn/m^2]\n", + "x = .85; # Dryness fraction\n", + "\n", + "# solution\n", + "\n", + "# from steam table, at given pressure \n", + "hf = 376.8;# [kJ/kg]\n", + "hfg = 2283.3;# [kJ/kg]\n", + "\n", + "# now using equation [2]\n", + "h = hf+x*hfg;# specific enthalpy of wet steam,[kJ/kg]\n", + "\n", + "#results\n", + "print ' The specific enthalpy of wet steam (kJ/kg) = ',h\n", + "\n", + "# There is minor variation in the book's answer\n", + "\n", + "# End\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8: pg 68" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 4.8\n", + "The specific volume of wet steam (m^3/kg) = 0.14121\n" + ] + } + ], + "source": [ + "#pg 68\n", + "#calculate the specific volume\n", + "print('Example 4.8');\n", + "\n", + "# aim : To determine \n", + "# the specific volume of wet steam\n", + "\n", + "# Given values\n", + "P = 1.25; # pressure, [MN/m^2]\n", + "x = .9; # dry fraction\n", + "\n", + "# solution\n", + "# from steam table at given pressure\n", + "vg = .1569;# [m^3/kg]\n", + "# hence\n", + "v = x*vg; # [m^3/kg]\n", + "#results\n", + "print 'The specific volume of wet steam (m^3/kg) = ',v\n", + "\n", + "# End\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9: pg 69" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 4.9\n", + " The specific volume of steam at pressure of 2 MN/m^2 and with temperature 325 C (m^3/kg) = 0.1321\n", + " The degree of superheat (C) = 112.6\n" + ] + } + ], + "source": [ + "#pg 69\n", + "#calculate the specific volume and degree of superheat\n", + "print('Example 4.9');\n", + "\n", + "# aim : To determine\n", + "# the specific volume \n", + "\n", + "# Given values\n", + "t = 325; # temperature, [C]\n", + "P = 2; # pressure, [MN/m^2]\n", + "\n", + "# solution\n", + "# from steam table at given t and P\n", + "vf = .1321; # [m^3/kg]\n", + "tf = 212.4; # saturation temperature, [C]\n", + "doh= t-tf; # degree of superheat, [C]\n", + "#results\n", + "print ' The specific volume of steam at pressure of 2 MN/m^2 and with temperature 325 C (m^3/kg) = ',vf\n", + "print ' The degree of superheat (C) = ',doh\n", + "\n", + "# End \n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10: pg 69" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example .10\n", + " (a) The mass of steam entering the heater (kg/h) = 81.8\n", + " (b) The mass of water entering the heater (kg/h) = 628.23\n" + ] + } + ], + "source": [ + "#pg 69\n", + "#calculate the mass of steam and water\n", + "print('Example .10');\n", + "import math\n", + "# aim : To determine\n", + "# (a) the mass of steam entering the heater\n", + "# (b) the mass of water entering the heater\n", + "\n", + "# Given values\n", + "x = .95;# Dryness fraction\n", + "P = .7;# pressure,[MN/m**2]\n", + "d = 25;# internal diameter of heater,[mm]\n", + "C = 12; # steam velocity in the pipe,[m/s]\n", + "\n", + "# solution\n", + "# from steam table at .7 MN/m**2 pressure\n", + "hf = 697.1;# [kJ/kg]\n", + "hfg = 2064.9;# [kJ/kg]\n", + "hg = 2762.0; # [kJ/kg]\n", + "vg = .273; # [m**3/kg]\n", + "\n", + "# (a)\n", + "v = x*vg; # [m**3/kg]\n", + "ms_dot = math.pi*(d*10**-3)**2*C*3600/(4*v);# mass of steam entering, [kg/h]\n", + "\n", + "# (b)\n", + "h = hf+x*hfg;# specific enthalpy of steam entering heater,[kJ/kg]\n", + "# again from steam tables\n", + "hf1 = 376.8;# [kJ/kg] at 90 C\n", + "hf2 = 79.8;# [kJ/kg] at 19 C\n", + "\n", + "# using energy balance,mw_dot*(hf1-hf2)=ms_dot*(h-hf1)\n", + "mw_dot = ms_dot*(h-hf1)/(hf1-hf2);# mass of water entering to heater,[kg/h]\n", + "#results\n", + "print ' (a) The mass of steam entering the heater (kg/h) = ',round(ms_dot,1)\n", + "print ' (b) The mass of water entering the heater (kg/h) = ',round(mw_dot,2)\n", + "\n", + "# End\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11: pg 72" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 4.11\n", + " The change in internal energy (kJ) = -486.753\n" + ] + } + ], + "source": [ + "#pg 72\n", + "#calculate the change in internal energy\n", + "print('Example 4.11');\n", + "\n", + "# aim: To determine\n", + "# the change of internal energy\n", + "\n", + "# Given values\n", + "m = 1.5;# mass of steam,[kg]\n", + "P1 = 1;# initial pressure, [MN/m**2]\n", + "t = 225;# temperature, [C]\n", + "P2 = .28;# final pressure, [MN/m**2]\n", + "x = .9;# dryness fraction of steam at P2\n", + "\n", + "# solution\n", + "\n", + "# from steam table at P1\n", + "h1 = 2886;# [kJ/kg]\n", + "v1 = .2198; # [m**3/kg]\n", + "# hence\n", + "u1 = h1-P1*v1*10**3;# internal energy [kJ/kg]\n", + "\n", + "# at P2\n", + "hf2 = 551.4;# [kJ/kg]\n", + "hfg2 = 2170.1;# [kJ/kg]\n", + "vg2 = .646; # [m**3/kg]\n", + "# so\n", + "h2 = hf2+x*hfg2;# [kj/kg]\n", + "v2 = x*vg2;# [m**3/kg]\n", + "\n", + "# now\n", + "u2 = h2-P2*v2*10**3;# [kJ/kg]\n", + "\n", + "# hence change in specific internal energy is\n", + "del_u = u2-u1;# [kJ/kg]\n", + "\n", + "del_u = m*del_u;# [kJ];\n", + "#results\n", + "print ' The change in internal energy (kJ) = ',del_u\n", + "\n", + "# End\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12: pg 74" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 4.12\n", + " Dryness fraction of steam after throttling is = 0.787\n" + ] + } + ], + "source": [ + "#pg 74\n", + "#calculate the dryness fraction\n", + "print('Example 4.12');\n", + "\n", + "# aim : To determine \n", + "# the dryness fraction of steam after throttling\n", + "\n", + "# given values\n", + "P1 = 1.4;# pressure before throttling, [MN/m^2]\n", + "x1 = .7;# dryness fraction before throttling\n", + "P2 = .11;# pressure after throttling, [MN/m^2]\n", + "\n", + "# solution\n", + "# from steam table\n", + "hf1 = 830.1;# [kJ/kg]\n", + "hfg1 = 1957.7;# [kJ/kg]\n", + "h1 = hf1 + x1*hfg1; # [kJ/kg]\n", + "\n", + "hf2 = 428.8;# [kJ/kg]\n", + "hfg2 = 2250.8;# [kJ/kg]\n", + "\n", + "# now for throttling,\n", + "# hf1+x1*hfg1=hf2+x2*hfg2; where x2 is dryness fraction after throttling\n", + "\n", + "x2=(h1-hf2)/hfg2; # final dryness fraction\n", + "#results\n", + "print ' Dryness fraction of steam after throttling is = ',round(x2,3)\n", + "\n", + "# End\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13: pg 75" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 4.13\n", + " (a) The condition of the resulting mixture is dry with dryness fraction = 0.965\n", + " (b) The internal diameter of the pipe (mm) = 145.96\n" + ] + } + ], + "source": [ + "#pg 75\n", + "#calculate the dryness fraction and internal diameter\n", + "print('Example 4.13');\n", + "import math\n", + "# aim : To determine \n", + "# the dryness fraction of steam \n", + "# and the internal diameter of the pipe\n", + "\n", + "# Given values\n", + "\n", + "# steam1\n", + "P1 = 2.;# pressure before throttling, [MN/m^2]\n", + "t = 300.;# temperature,[C]\n", + "ms1_dot = 2.;# steam flow rate, [kg/s]\n", + "P2 = 800.;# pressure after throttling, [kN/m^2]\n", + "\n", + "# steam2\n", + "P = 800.;# pressure, [N/m^2]\n", + "x2 = .9;# dryness fraction\n", + "ms2_dot = 5; # [kg/s]\n", + "\n", + "# solution\n", + "# (a)\n", + "# from steam table specific enthalpy of steam1 before throttling is\n", + "hf1 = 3025;# [kJ/kg]\n", + "# for throttling process specific enthalpy will same so final specific enthalpy of steam1 is\n", + "hf2 = hf1;\n", + "# hence\n", + "h1 = ms1_dot*hf2;# [kJ/s]\n", + "\n", + "# calculation of specific enthalpy of steam2\n", + "hf2 = 720.9;# [kJ/kg]\n", + "hfg2 = 2046.5;# [kJ/kg]\n", + "# hence\n", + "h2 = hf2+x2*hfg2;# specific enthalpy, [kJ/kg]\n", + "h2 = ms2_dot*h2;# total enthalpy, [kJ/s]\n", + "\n", + "# after mixing\n", + "m_dot = ms1_dot+ms2_dot;# total mass of mixture,[kg/s]\n", + "h = h1+h2;# Total enthalpy of the mixture,[kJ/s]\n", + "h = h/7;# [kJ/kg]\n", + "\n", + "# At pressure 800 N/m^2 \n", + "hf = 720.9;# [kJ/kg]\n", + "hfg = 2046.5;# [kJ/kg]\n", + "# so total enthalpy is,hf+x*hfg, where x is dryness fraction of mixture and which is equal to h\n", + "# hence\n", + "x = (h-hf)/hfg;# dryness fraction after mixing\n", + "\n", + "# (b)\n", + "# Given\n", + "C = 15;# velocity, [m/s]\n", + "# from steam table\n", + "v = .1255;# [m^/kg]\n", + "A = ms1_dot*v/C;# area, [m^2]\n", + "# using ms1_dot = A*C/v, where A is cross section area in m^2 and\n", + "# A = %pi*d^2/4, where d is diameter of the pipe \n", + "\n", + "# calculation of d\n", + "d = math.sqrt(4*A/math.pi); # diameter, [m]\n", + "#results\n", + "print ' (a) The condition of the resulting mixture is dry with dryness fraction = ',round(x,3)\n", + "print ' (b) The internal diameter of the pipe (mm) = ',round(d*1000,2)\n", + "\n", + "# End\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14: pg 78" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 4.14\n", + " The dryness fraction of the steam entering seprating calorimeter is = 0.9\n" + ] + } + ], + "source": [ + "#pg 78\n", + "#calculate the dryness fraction\n", + "print('Example 4.14');\n", + "\n", + "# aim : To estimate \n", + "# the dryness fraction\n", + "\n", + "# Given values\n", + "M = 1.8;# mass of condensate, [kg]\n", + "m = .2;# water collected, [kg]\n", + "\n", + "# solution\n", + "x = M/(M+m);# formula for calculation of dryness fraction using seprating calorimeter\n", + "#results\n", + "print ' The dryness fraction of the steam entering seprating calorimeter is = ',x\n", + "\n", + "# End\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 15: pg 80" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 4.15\n", + " The dryness fraction of steam is = 0.928\n", + "There is a calculation mistake in book so answer is not matching\n" + ] + } + ], + "source": [ + "#pg 80\n", + "#calculate the dryness fraction\n", + "print('Example 4.15');\n", + "import numpy\n", + "# aim : To determine\n", + "# the dryness fraction of the steam at 2.2 MN/m^2\n", + "\n", + "# Given values\n", + "P1 = 2.2;# [MN/m^2]\n", + "P2 = .13;# [MN/m^2]\n", + "t2 = 112;# [C]\n", + "tf2 = 150;# temperature, [C]\n", + "\n", + "# solution\n", + "# from steam table, at 2.2 MN/m^2\n", + "# saturated steam at 2 MN/m^2 Pressure\n", + "hf1 = 931;# [kJ/kg]\n", + "hfg1 = 1870;# [kJ/kg]\n", + "hg1 = 2801;# [kJ/kg]\n", + "\n", + "# for superheated steam\n", + "# at .1 MN/m^2\n", + "hg2 = 2675;# [kJ/kg]\n", + "hg2_150 = 2777;# specific enthalpy at 150 C, [kJ/kg]\n", + "tf2 = 99.6;# saturation temperature, [C]\n", + "\n", + "# at .5 MN/m^2\n", + "hg3 = 2693;# [kJ/kg]\n", + "hg3_150 = 2773;# specific enthalpy at 150 C, [kJ/kg]\n", + "tf3 = 111.4;# saturation temperature, [C]\n", + "\n", + "Table_P_h1_x = [.1,.5];# where, P in MN/m^2 and h in [kJ/kg]\n", + "Table_P_h1_y=[hg2,hg3]\n", + "hg = numpy.interp(.13,Table_P_h1_x,Table_P_h1_y);# specific entahlpy at .13 MN/m^2, [kJ/kg]\n", + "\n", + "Table_P_h2_x = [.1,.5];# where, P in MN/m^2 and h in [kJ/kg]\n", + "Table_P_h2_y =[hg2_150,hg3_150];\n", + "hg_150 = numpy.interp(.13,Table_P_h2_x,Table_P_h2_y);# specific entahlpy at .13 MN/m^2 and 150 C, [kJ/kg]\n", + "\n", + "Table_P_tf_x = [.1,.5];# where, P in MN/m^2 and h in [kJ/kg]\n", + "Table_P_tf_y = [tf2,tf3]\n", + "tf = numpy.interp(.13,Table_P_tf_x,Table_P_tf_y);# saturation temperature, [C]\n", + "\n", + "# hence\n", + "h2 = hg+(hg_150-hg)/(t2-tf)/(tf2-tf);# specific enthalpy at .13 MN/m^2 and 112 C, [kJ/kg]\n", + "\n", + "# now since process is throttling so h2=h1\n", + "# and h1 = hf1+x1*hfg1, so\n", + "x1 = (h2-hf1)/hfg1;# dryness fraction\n", + "#results\n", + "print ' The dryness fraction of steam is = ',round(x1,3)\n", + "\n", + "print 'There is a calculation mistake in book so answer is not matching'\n", + "\n", + "# End\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 16: pg 82" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 4.16\n", + " The minimum dryness fraction of steam is x = 0.939\n" + ] + } + ], + "source": [ + "#pg 82\n", + "#calculate the minimum dryness fraction\n", + "print('Example 4.16');\n", + "\n", + "# aim : To determine \n", + "# the minimum dryness fraction of steam\n", + "\n", + "# Given values\n", + "P1 = 1.8;# testing pressure,[MN/m^2]\n", + "P2 = .11;# pressure after throttling,[MN/m^2]\n", + "\n", + "# solution\n", + "# from steam table\n", + "# at .11 MN/m^2 steam is completely dry and specific enthalpy is\n", + "hg = 2680;# [kJ/kg]\n", + "\n", + "# before throttling steam is wet, so specific enthalpy is=hf+x*hfg, where x is dryness fraction\n", + "# from steam table\n", + "hf = 885.;# [kJ/kg]\n", + "hfg = 1912.;# [kJ/kg]\n", + "\n", + "# now for throttling process,specific enthalpy will same before and after\n", + "# hence\n", + "x = (hg-hf)/hfg;\n", + "#results\n", + "print ' The minimum dryness fraction of steam is x = ',round(x,3)\n", + "\n", + "# End" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 17: pg 83" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 4.17\n", + " (a) The mass of steam in the vessel (kg) = 1.569\n", + " (b) The final dryness fraction of the steam = 0.576\n", + " (c) The amount of heat transferred during cooling process (kJ) = -1377.1\n" + ] + } + ], + "source": [ + "#pg 83\n", + "#calculate the mass of steam, final dryness and amount of heat\n", + "print('Example 4.17');\n", + "\n", + "# aim : To determine the\n", + "# (a) mass of steam in the vessel\n", + "# (b) final dryness of the steam\n", + "# (c) amount of heat transferrred during the cooling process\n", + "\n", + "# Given values\n", + "V1 = .8;# [m^3]\n", + "P1 = 360.;# [kN/m^2]\n", + "P2 = 200.;# [kN/m^2]\n", + "\n", + "# solution\n", + "\n", + "# (a)\n", + "# at 360 kN/m^2\n", + "vg1 = .510;# [m^3]\n", + "m = V1/vg1;# mass of steam,[kg]\n", + "\n", + "# (b)\n", + "# at 200 kN/m^2\n", + "vg2 = .885;# [m^3/kg]\n", + "# the volume remains constant so\n", + "x = vg1/vg2;# final dryness fraction\n", + "\n", + "# (c)\n", + "# at 360 kN/m^2\n", + "h1 = 2732.9;# [kJ/kg]\n", + "# hence\n", + "u1 = h1-P1*vg1;# [kJ/kg]\n", + "\n", + "# at 200 kN/m^2\n", + "hf = 504.7;# [kJ/kg]\n", + "hfg=2201.6;#[kJ/kg]\n", + "# hence\n", + "h2 = hf+x*hfg;# [kJ/kg]\n", + "# now\n", + "u2 = h2-P2*vg1;# [kJ/kg]\n", + "# so\n", + "del_u = u2-u1;# [kJ/kg]\n", + "# from the first law of thermodynamics del_U+W=Q, \n", + "W = 0;# because volume is constant\n", + "del_U = m*del_u;# [kJ]\n", + "# hence\n", + "Q = del_U;# [kJ]\n", + "#results\n", + "print ' (a) The mass of steam in the vessel (kg) = ',round(m,3)\n", + "print ' (b) The final dryness fraction of the steam = ',round(x,3)\n", + "print ' (c) The amount of heat transferred during cooling process (kJ) = ',round(Q,1)\n", + "\n", + "# End\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 18: pg 84" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 4.18\n", + " The heat received by the steam (kJ/kg) = 380.3\n" + ] + } + ], + "source": [ + "#pg 84\n", + "#calculate the heat received by steam\n", + "print('Example 4.18');\n", + "\n", + "# aim : To determine\n", + "# the heat received by the steam per kilogram\n", + "\n", + "# Given values\n", + "# initial\n", + "P1 = 4;# pressure, [MN/m^2]\n", + "x1 = .95; # dryness fraction\n", + "\n", + "# final\n", + "t2 = 350;# temperature,[C]\n", + "\n", + "# solution\n", + "\n", + "# from steam table, at 4 MN/m^2 and x1=.95\n", + "hf = 1087.4;# [kJ/kg]\n", + "hfg = 1712.9;# [kJ/kg]\n", + "# hence\n", + "h1 = hf+x1*hfg;# [kJ/kg]\n", + "\n", + "# since pressure is kept constant ant temperature is raised so at this condition\n", + "h2 = 3095;# [kJ/kg]\n", + "\n", + "# so by energy balance\n", + "Q = h2-h1;# Heat received,[kJ/kg]\n", + "#results\n", + "print ' The heat received by the steam (kJ/kg) = ',round(Q,1)\n", + "\n", + "# End\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19: pg 85" + ] + }, + { + "cell_type": "code", + "execution_count": 19, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 4.19\n", + " (a) The Quantity of steam present (kg) = 3.0\n", + " Dryness fraction is = 0.5\n", + " The enthalpy (kJ) = 5451.9\n", + " The heat loss (kJ) = 2917.8\n", + " (b) The dryness fraction is = 0.989\n", + " The enthalpy (kJ) = 8346.3\n" + ] + } + ], + "source": [ + "#pg 85\n", + "#calculate the condition after the given cases\n", + "print('Example 4.19');\n", + "\n", + "# aim : To determine the condition of the steam after \n", + "# (a) isothermal compression to half its initial volume,heat rejected\n", + "# (b) hyperbolic compression to half its initial volume\n", + "\n", + "# Given values\n", + "V1 = .3951;# initial volume,[m^3]\n", + "P1 = 1.5;# initial pressure,[MN/m^2]\n", + "\n", + "# solution\n", + "\n", + "# (a)\n", + "# from steam table, at 1.5 MN/m^2 \n", + "hf1 = 844.7;# [kJ/kg]\n", + "hfg1 = 1945.2;# [kJ/kg]\n", + "hg1 = 2789.9;# [kJ/kg]\n", + "vg1 = .1317;# [m^3/kg]\n", + "\n", + "# calculation\n", + "m = V1/vg1;# mass of steam,[kg]\n", + "vg2b = vg1/2;# given,[m^3/kg](vg2b is actual specific volume before compression)\n", + "x1 = vg2b/vg1;# dryness fraction\n", + "h1 = m*(hf1+x1*hfg1);# [kJ]\n", + "Q = m*x1*hfg1;# heat loss,[kJ]\n", + "print ' (a) The Quantity of steam present (kg) = ',m\n", + "print ' Dryness fraction is = ',x1\n", + "print ' The enthalpy (kJ) = ',h1\n", + "print ' The heat loss (kJ) = ',Q\n", + "\n", + "# (b)\n", + "V2 = V1/2;\n", + "# Given compression is according to the law PV=Constant,so\n", + "P2 = P1*V1/V2;# [MN/m^2]\n", + "# from steam table at P2\n", + "hf2 = 1008.4;# [kJ/kg]\n", + "hfg2 = 1793.9;# [kJ/kg]\n", + "hg2 = 2802.3;# [kJ/kg]\n", + "vg2 = .0666;# [m^3/kg]\n", + "\n", + "# calculation\n", + "x2 = vg2b/vg2;# dryness fraction\n", + "h2 = m*(hf2+x2*hfg2);# [kJ]\n", + "\n", + "print ' (b) The dryness fraction is = ',round(x2,3)\n", + "print ' The enthalpy (kJ) = ',round(h2,1)\n", + "\n", + "# End\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 20: pg 88" + ] + }, + { + "cell_type": "code", + "execution_count": 20, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 4.20\n", + " (a) The mass of steam present (kg) = 5.0\n", + " (b) The work transfer (kJ) = 708.0\n", + " (c) The change in internal energy (kJ) = -1611.0\n", + "since del_U<0,so this is loss of internal energy\n", + " (d) The heat exchange between the steam and surrounding (kJ) = -903.0\n", + "since Q<0,so this is loss of heat energy to surrounding\n", + "there are minor vairations in the values reported in the book due to rounding off error\n" + ] + } + ], + "source": [ + "#pg 88\n", + "#calculate the mass of steam, work transfer, change of internal energy, heat exchange\n", + "print('Example 4.20');\n", + "\n", + "# aim : To determine the\n", + "# (a) mass of steam \n", + "# (b) work transfer\n", + "# (c) change of internal energy\n", + "# (d) heat exchange b/w the steam and surroundings\n", + "\n", + "# Given values\n", + "P1 = 2.1;# initial pressure,[MN/m**2]\n", + "x1 = .9;# dryness fraction\n", + "V1 = .427;# initial volume,[m**3]\n", + "P2 = .7;# final pressure,[MN/m**2]\n", + "# Given process is polytropic with\n", + "n = 1.25; # polytropic index\n", + "\n", + "# solution\n", + "# from steam table\n", + "\n", + "# at 2.1 MN/m**2\n", + "hf1 = 920.0;# [kJ/kg]\n", + "hfg1=1878.2;# [kJ/kg]\n", + "hg1=2798.2;# [kJ/kg]\n", + "vg1 = .0949;# [m**3/kg]\n", + "\n", + "# and at .7 MN/m**2\n", + "hf2 = 697.1;# [kJ/kg]\n", + "hfg2 = 2064.9;# [kJ/kg]\n", + "hg2 = 2762.0;# [kJ/kg]\n", + "vg2 = .273;# [m**3/kg]\n", + "\n", + "#calculations and results\n", + "# (a)\n", + "v1 = x1*vg1;# [m**3/kg]\n", + "m = V1/v1;# [kg]\n", + "print ' (a) The mass of steam present (kg) = ',round(m)\n", + "\n", + "# (b)\n", + "# for polytropic process\n", + "v2 = v1*(P1/P2)**(1/n);# [m**3/kg]\n", + "\n", + "x2 = v2/vg2;# final dryness fraction\n", + "# work transfer\n", + "W = m*(P1*v1-P2*v2)*10**3/(n-1);# [kJ]\n", + "print ' (b) The work transfer (kJ) = ',round(W)\n", + "\n", + "# (c)\n", + "# initial\n", + "h1 = hf1+x1*hfg1;# [kJ/kg]\n", + "u1 = h1-P1*v1*10**3;# [kJ/kg]\n", + "\n", + "# final\n", + "h2 = hf2+x2*hfg2;# [kJ/kg]\n", + "u2 = h2-P2*v2*10**3;# [kJ/kg]\n", + "\n", + "del_U = m*(u2-u1);# [kJ]\n", + "print ' (c) The change in internal energy (kJ) = ',round(del_U)\n", + "if(del_U<0):\n", + " print('since del_U<0,so this is loss of internal energy')\n", + "else:\n", + " print('since del_U>0,so this is gain in internal energy')\n", + "\n", + "\n", + "# (d)\n", + "Q = del_U+W;# [kJ]\n", + "print ' (d) The heat exchange between the steam and surrounding (kJ) = ',round(Q,1)\n", + "if(Q<0):\n", + " print('since Q<0,so this is loss of heat energy to surrounding')\n", + "else:\n", + " print('since Q>0,so this is gain in heat energy to the steam')\n", + "\n", + "print 'there are minor vairations in the values reported in the book due to rounding off error'\n", + "\n", + "# End\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 21: pg 91" + ] + }, + { + "cell_type": "code", + "execution_count": 21, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 4.21\n", + " (a) The volume occupied by 1 kg of steam (m^3/kg) = 0.22\n", + " (b)(1) The final dryness fraction of the steam = 0.887\n", + " (2) The change in internal energy of the steam during expansion is (kJ/kg) (This is a loss of internal energy) = -243.0\n", + " There are minor variation in the answer due to rounding off error\n" + ] + } + ], + "source": [ + "#pg 91\n", + "#calculate the volume, dryness fraction and change in internal energy\n", + "print('Example 4.21');\n", + "\n", + "# aim : To determine the \n", + "# (a) volume occupied by steam\n", + "# (b)(1) final dryness fraction of steam\n", + "# (2) Change of internal energy during expansion\n", + "\n", + "# (a)\n", + "# Given values\n", + "P1 = .85;# [mN/m**2]\n", + "x1 = .97;\n", + "\n", + "# solution\n", + "# from steam table, at .85 MN/m**2,\n", + "vg1 = .2268;# [m**3/kg]\n", + "# hence\n", + "v1 = x1*vg1;# [m**3/kg]\n", + "print ' (a) The volume occupied by 1 kg of steam (m^3/kg) = ',round(v1,2)\n", + "\n", + "# (b)(1)\n", + "P2 = .17;# [MN/m**2]\n", + "# since process is polytropic process with\n", + "n = 1.13; # polytropic index\n", + "# hence\n", + "v2 = v1*(P1/P2)**(1/n);# [m**3/kg]\n", + "\n", + "# from steam table at .17 MN/m**2\n", + "vg2 = 1.031;# [m**3/kg]\n", + "# steam is wet so\n", + "x2 = v2/vg2;# final dryness fraction\n", + "print ' (b)(1) The final dryness fraction of the steam = ',round(x2,3)\n", + "\n", + "# (2)\n", + "W = (P1*v1-P2*v2)*10**3/(n-1);# [kJ/kg]\n", + "# since process is adiabatic, so\n", + "del_u = -W;# [kJ/kg]\n", + "print ' (2) The change in internal energy of the steam during expansion is (kJ/kg) (This is a loss of internal energy) = ',round(del_u)\n", + "print' There are minor variation in the answer due to rounding off error'\n", + "\n", + "# End\n" + ] + } + ], + "metadata": { + "kernelspec": { + 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