{ "cells": [ { "cell_type": "markdown", "metadata": {}, "source": [ "# Chapter 11: The Ideal gas and mixture relationships" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 11.10: Entropy_calculations.sce" ] }, { "cell_type": "code", "execution_count": null, "metadata": { "collapsed": true }, "outputs": [], "source": [ "clc\n", "clear\n", "//Initialization of variables\n", "T2=546.6 //R\n", "T1=520 //R\n", "T3=560 //R\n", "v2=1389.2\n", "v1=186.2\n", "R0=1.986\n", "c1=5.02\n", "c2=4.97\n", "n1=1\n", "n2=2\n", "v3=1203\n", "//calculations\n", "ds1=n1*c1*log(T2/T1) + n1*R0*log(v2/v1)\n", "ds2=n2*c2*log(T2/T3)+n2*R0*log(v2/v3)\n", "ds=ds1+ds2\n", "ds3=n1*c1*log(T2/T1)+n2*c2*log(T2/T3)\n", "ds4=n2*R0*log(v2/v3)+ n1*R0*log(v2/v1)\n", "dss=ds3+ds4\n", "//results\n", "printf('Change in entropy for gas 1 = %.3f Btu/R',ds1)\n", "printf('\n Change in entropy for gas 1 = %.3f Btu/R',ds2)\n", "printf('\n Net change in entropy = %.3f Btu/R',ds)\n", "printf('\n In case 2, change in entropy = %.3f Btu/R',dss)\n", "disp('The answer is a bit different due to rounding off error in the textbook')" ] } , { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 11.11: Entropy_calculations.sce" ] }, { "cell_type": "code", "execution_count": null, "metadata": { "collapsed": true }, "outputs": [], "source": [ "clc\n", "clear\n", "//Initialization of variables\n", "m1=1 //lbm\n", "m2=0.94 //lbm\n", "M1=29\n", "M2=18\n", "p1=50 //psia\n", "p2=100 //psia\n", "t1=250 +460 //R\n", "R0=1.986\n", "cpa=6.96\n", "cpb=8.01\n", "//calculations\n", "xa = (m1/M1)/((m1/M1)+ m2/M2)\n", "xb=1-xa\n", "t2=t1*(p2/p1)^(R0/(xa*cpa+xb*cpb))\n", "d=R0/(xa*cpa+xb*cpb)\n", "k=1/(1-d)\n", "dsa=cpa*log(t2/t1) -R0*log(p2/p1)\n", "dSa=(m1/M1)*dsa\n", "dSw=-dSa\n", "dsw=dSw*M2/m2\n", "//results\n", "printf('Final remperature = %d R',t2)\n", "printf('\n Change in entropy of air = %.3f btu/mole R and %.5f Btu/R',dsa,dSa)\n", "printf('\n Change in entropy of water = %.4f btu/mole R and %.5f Btu/R',dsw,dSw)\n", "disp('The answers are a bit different due to rounding off error in textbook')" ] } , { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 11.12: Volume_and_mass_calculations.sce" ] }, { "cell_type": "code", "execution_count": null, "metadata": { "collapsed": true }, "outputs": [], "source": [ "clc\n", "clear\n", "//Initialization of variables\n", "T=250 + 460 //R\n", "p=29.825 //psia\n", "pt=50 //psia\n", "vg=13.821 //ft^3/lbm\n", "M=29\n", "R=10.73\n", "//calculations\n", "pa=pt-p\n", "V=1/M *R*T/pa\n", "ma=V/vg\n", "xa=p/pt\n", "mb=xa/M *18/(1-xa)\n", "//results\n", "printf('In case 1, volume occupied = %.2f ft^3',V)\n", "printf('\n In case 1, mass of steam = %.2f lbm steam',ma)\n", "printf('\n In case 2, mass of steam = %.3f lbm steam',mb)\n", "" ] } , { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 11.13: Percentage_calculation.sce" ] }, { "cell_type": "code", "execution_count": null, "metadata": { "collapsed": true }, "outputs": [], "source": [ "clc\n", "clear\n", "//Initialization of variables\n", "ps=0.64 //psia\n", "p=14.7 //psia\n", "M=29\n", "M2=46\n", "//calculations\n", "xa=ps/p\n", "mb=xa*9/M *M2/(1-xa)\n", "//results\n", "printf('percentage = %.1f percent',mb*100)\n", "" ] } , { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 11.14: Partial_Pressure_calculations.sce" ] }, { "cell_type": "code", "execution_count": null, "metadata": { "collapsed": true }, "outputs": [], "source": [ "clc\n", "clear\n", "//Initialization of variables\n", "ps=0.5069 //psia\n", "p=20 //psia\n", "m1=0.01\n", "m2=1\n", "M1=18\n", "M2=29\n", "//calculations\n", "xw= (m1/M1)/(m1/M1+m2/M2)\n", "pw=xw*p\n", "//results\n", "printf('partial pressure of water vapor = %.3f psia',pw)" ] } , { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 11.1: Work_calculatio.sce" ] }, { "cell_type": "code", "execution_count": null, "metadata": { "collapsed": true }, "outputs": [], "source": [ "clc\n", "clear\n", "//Initialization of variables\n", "n=1.3\n", "T1=460+60 //R\n", "P1=14.7 //psia\n", "P2=125 //psia\n", "R=1545\n", "M=29\n", "//calculations\n", "T2=T1*(P2/P1)^((n-1)/n)\n", "wrev=R/M *(T2-T1)/(1-n)\n", "//results\n", "printf('Work done = %d ft-lbf/lbm',wrev)\n", "disp('The answer is a bit different due to rounding off error in textbook')" ] } , { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 11.2: Kinetic_energy_calculation.sce" ] }, { "cell_type": "code", "execution_count": null, "metadata": { "collapsed": true }, "outputs": [], "source": [ "clc\n", "clear\n", "//Initialization of variables\n", "P2=10 //psia\n", "P1=100 //psia\n", "T1=900 //R\n", "w=50 //Btu/lbm\n", "k=1.39\n", "cp=0.2418\n", "//calculations\n", "T2=T1*(P2/P1)^((k-1)/k)\n", "T2=477\n", "KE=-w-cp*(T2-T1)\n", "//results\n", "printf('Change in kinetic energy = %.1f Btu/lbm',KE)" ] } , { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 11.3: Temperature_calculatio.sce" ] }, { "cell_type": "code", "execution_count": null, "metadata": { "collapsed": true }, "outputs": [], "source": [ "clc\n", "clear\n", "//Initialization of variables\n", "T1=900 //R\n", "P1=100 //psia\n", "P2=10 //psia\n", "//calculations\n", "disp('From table B-9')\n", "pr1=8.411\n", "pr2=pr1*P2/P1\n", "T2=468 //R\n", "//results\n", "printf('Final temperature = %d R ',T2)" ] } , { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 11.4: Temperature_work_and_Enthalpy_calculations.sce" ] }, { "cell_type": "code", "execution_count": null, "metadata": { "collapsed": true }, "outputs": [], "source": [ "clc\n", "clear\n", "//Initialization of variables\n", "cr=6\n", "p1=14.7 //psia\n", "t1=60.3 //F\n", "M=29\n", "R=1.986\n", "//calculations\n", "disp('from table b-9')\n", "vr1=158.58 \n", "u1=88.62 //Btu/lbm\n", "pr1=1.2147\n", "vr2=vr1/cr\n", "T2=1050 //R\n", "u2=181.47 //Btu/lbm\n", "pr2=14.686\n", "p2=p1*(pr2/pr1)\n", "dw=u1-u2\n", "h2=u2+T2*R/M\n", "//results\n", "printf('final temperature = %d R',T2)\n", "printf('\n final pressure = %.1f psia',p2)\n", "printf('\n work done = %.2f Btu/lbm',dw)\n", "printf('\n final enthalpy = %.1f Btu/lbm',h2)" ] } , { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 11.5: Weight_calculations.sce" ] }, { "cell_type": "code", "execution_count": null, "metadata": { "collapsed": true }, "outputs": [], "source": [ "clc\n", "clear\n", "//Initialization of variables\n", "m1=10 //lbm\n", "m2=15 //lnm\n", "p=50 //psia\n", "t=60+460 //R\n", "M1=32\n", "M2=28.02\n", "R0=10.73 \n", "//calculations\n", "n1=m1/M1\n", "n2=m2/M2\n", "x1=n1/(n1+n2)\n", "x2=n2/(n1+n2)\n", "M=x1*M1+x2*M2\n", "R=R0/M\n", "V=(n1+n2)*R0*t/p\n", "rho=p/(R0*t)\n", "rho2=M*rho\n", "p1=x1*p\n", "p2=x2*p\n", "v1=x1*V\n", "v2=x2*V\n", "//results\n", "disp('part a')\n", "printf('Mole fractions of oxygen and nitrogen are %.3f and %.3f respectively',x1,x2)\n", "disp('part b')\n", "printf('Average molecular weight = %.1f ',M)\n", "disp('part c')\n", "printf('specific gas constant = %.4f psia ft^3/lbm R',R)\n", "disp('part d')\n", "printf('volume of mixture = %.1f ft^3',V)\n", "printf('density of mixture is %.5f mole/ft^3 and %.2f lbm/ft^3',rho,rho2)\n", "disp('part e')\n", "printf('partial pressures of oxygen and nitrogen are %.2f psia and %.2f psia respectively' ,p1,p2)\n", "clc\n", "clear\n", "//Initialization of variables\n", "m1=10 //lbm\n", "m2=15 //lnm\n", "p=50 //psia\n", "t=60+460 //R\n", "M1=32\n", "M2=28.02\n", "R0=10.73 \n", "//calculations\n", "n1=m1/M1\n", "n2=m2/M2\n", "x1=n1/(n1+n2)\n", "x2=n2/(n1+n2)\n", "M=x1*M1+x2*M2\n", "R=1545/M\n", "V=(n1+n2)*R0*t/p\n", "rho=p/(R0*t)\n", "rho2=M*rho\n", "p1=x1*p\n", "p2=x2*p\n", "v1=x1*V\n", "v2=x2*V\n", "pt=p1+p2\n", "vt=v1+v2\n", "//results\n", "disp('part a')\n", "printf('Mole fractions of oxygen and nitrogen are %.3f and %.3f respectively',x1,x2)\n", "disp('part b')\n", "printf('Average molecular weight = %.1f ',M)\n", "disp('part c')\n", "printf('specific gas constant = %.4f lbf ft/lbm R',R)\n", "disp('part d')\n", "printf('volume of mixture = %.1f ft^3',V)\n", "printf('\n density of mixture is %.5f mole/ft^3 and %.3f lbm/ft^3',rho,rho2)\n", "disp('part e')\n", "printf('partial pressures of oxygen and nitrogen are %.2f psia and %.2f psia respectively' ,p1,p2)\n", "printf('\n partial volumes of oxygen and nitrogen are %.2f ft^3 and %.2f ft^3 respectively',v1,v2)\n", "printf('\n Net partial pressure in case of oxygen = %.2f psia',pt)\n", "printf('\n Net partial volume =%.2f ft^3',vt)" ] } , { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 11.6: Analysis_calculations.sce" ] }, { "cell_type": "code", "execution_count": null, "metadata": { "collapsed": true }, "outputs": [], "source": [ "\n", "clc\n", "clear\n", "//Initialization of variables\n", "m1=5.28\n", "m2=1.28\n", "m3=23.52\n", "//calculations\n", "m=m1+m2+m3\n", "x1=m1/m\n", "x2=m2/m\n", "x3=m3/m\n", "C=12/44 *m1/ m\n", "O=(32/44 *m1 + m2)/m\n", "N=m3/m\n", "sum1=(x1+x2+x3)*100\n", "sum2=(C+N+O)*100\n", "//results\n", "printf('From gravimetric analysis, co2 = %.1f percent , o2 = %.1f percent and n2 = %.1f percent',x1*100,x2*100,x3*100)\n", "printf('\n From ultimate analysis, co2 = %.2f percent , o2 = %.2f percent and n2 = %.2f percent',C*100,O*100,N*100)\n", "printf('\n Sum in case 1 = %.1f percent',sum1)\n", "printf('\n Sum in case 2 = %.1f percent',sum2)" ] } , { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 11.7: Entropy_calculations.sce" ] }, { "cell_type": "code", "execution_count": null, "metadata": { "collapsed": true }, "outputs": [], "source": [ "clc\n", "clear\n", "//Initialization of variables\n", "x1=1/3\n", "n1=1\n", "n2=2\n", "x2=2/3\n", "p=12.7 //psia\n", "cp1=7.01 //Btu/mole R\n", "cp2=6.94 //Btu/mole R\n", "R0=1.986\n", "T2=460+86.6 //R\n", "T1=460 //R\n", "p0=14.7 //psia\n", "//calculations\n", "p1=x1*p\n", "p2=x2*p\n", "ds1= cp1*log(T2/T1) - R0*log(p1/p0)\n", "ds2= cp2*log(T2/T1) - R0*log(p2/p0)\n", "S=n1*ds1+n2*ds2\n", "//results\n", "printf('Entropy of mixture = %.2f Btu/R',S)\n", "printf('\n the answer given in textbook is wrong. please check using a calculator')" ] } , { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 11.8: Internal_energy_and_entropy_calculations.sce" ] }, { "cell_type": "code", "execution_count": null, "metadata": { "collapsed": true }, "outputs": [], "source": [ "clc\n", "clear\n", "//Initialization of variables\n", "c1=4.97 //Btu/mol R\n", "c2=5.02 //Btu/mol R\n", "n1=2\n", "n2=1\n", "T1=86.6+460 //R\n", "T2=50+460 //R\n", "//calculations\n", "du=(n1*c1+n2*c2)*(T2-T1)\n", "ds=(n1*c1+n2*c2)*log(T2/T1)\n", "//results\n", "printf('Change in internal energy = %d Btu',du)\n", "printf('\n Change in entropy = %.3f Btu/R',ds)" ] } , { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 11.9: Pressure_and_temperature_calculations.sce" ] }, { "cell_type": "code", "execution_count": null, "metadata": { "collapsed": true }, "outputs": [], "source": [ "clc\n", "clear\n", "//Initialization of variables\n", "n1=1\n", "n2=2\n", "c1=5.02\n", "c2=4.97\n", "t1=60 //F\n", "t2=100 //F\n", "R0=10.73\n", "p1=30 //psia\n", "p2=10 //psia\n", "//calcualtions\n", "t=(n1*c1*t1+n2*c2*t2)/(n1*c1+n2*c2)\n", "V1= n1*R0*(t1+460)/p1\n", "V2=n2*R0*(t2+460)/p2\n", "V=V1+V2\n", "pm=(n1+n2)*R0*(t+460)/V\n", "//results\n", "printf('Pressure of mixture = %.1f psia',pm)\n", "printf('\n Mixing temperature = %.1f F',t)" ] } ], "metadata": { "kernelspec": { "display_name": "Scilab", "language": "scilab", "name": "scilab" }, "language_info": { "file_extension": ".sce", "help_links": [ { "text": "MetaKernel Magics", "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" } ], "mimetype": "text/x-octave", "name": "scilab", "version": "0.7.1" } }, "nbformat": 4, "nbformat_minor": 0 }