From f270f72badd9c61d48f290c3396004802841b9df Mon Sep 17 00:00:00 2001 From: kinitrupti Date: Fri, 12 May 2017 18:53:46 +0530 Subject: Removed duplicates --- .../chapter13_3.ipynb | 375 +++++++++++++++++++++ 1 file changed, 375 insertions(+) create mode 100755 Engineering_Thermodynamics_by_O._Singh/chapter13_3.ipynb (limited to 'Engineering_Thermodynamics_by_O._Singh/chapter13_3.ipynb') diff --git a/Engineering_Thermodynamics_by_O._Singh/chapter13_3.ipynb b/Engineering_Thermodynamics_by_O._Singh/chapter13_3.ipynb new file mode 100755 index 00000000..836097b9 --- /dev/null +++ b/Engineering_Thermodynamics_by_O._Singh/chapter13_3.ipynb @@ -0,0 +1,375 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 13:One Dimensional Compressible Fluid Flow" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 13.1;pg no: 525" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 13.1, Page:525 \n", + " \n", + "\n", + "Engineering Thermodynamics by Onkar Singh Chapter 13 Example 1\n", + "mass flow rate(m)=rho*A*C\n", + "so rho*C=4*m/(%pi*d^2)\n", + "so rho=165.79/C\n", + "now using perfect gas equation,p=rho*R*T\n", + "T=P/(rho*R)=P/((165.79/C)*R)\n", + "C/T=165.79*R/P\n", + "so C=1.19*T\n", + "we know,C^2=((2*y*R)/(y-1))*(To-T)\n", + "C^2=(2*1.4*287)*(300-T)/(1.4-1)\n", + "C^2=602.7*10^3-2009*T\n", + "C^2+1688.23*C-602.7*10^3=0\n", + "solving we get,C=302.72 m/s and T=254.39 K\n", + "using stagnation property relation,\n", + "To/T=1+(y-1)*M^2/2\n", + "so M= 0.947\n", + "stagnation pressure,Po in bar= 0.472\n", + "so mach number=0.947,stagnation pressure=0.472 bar,velocity=302.72 m/s\n" + ] + } + ], + "source": [ + "#cal of mach number,stagnation pressure,velocity\n", + "#intiation of all variables\n", + "# Chapter 13\n", + "import math\n", + "print\"Example 13.1, Page:525 \\n \\n\"\n", + "print(\"Engineering Thermodynamics by Onkar Singh Chapter 13 Example 1\")\n", + "To=(27+273);#stagnation temperature in K\n", + "P=0.4*10**5;#static pressure in pa\n", + "m=3000/3600;#air flowing rate in kg/s\n", + "d=80*10**-3;#diameter of duct in m\n", + "R=287;#gas constant in J/kg K\n", + "y=1.4;#expansion constant\n", + "print(\"mass flow rate(m)=rho*A*C\")\n", + "print(\"so rho*C=4*m/(%pi*d^2)\")\n", + "4*m/(math.pi*d**2)\n", + "print(\"so rho=165.79/C\")\n", + "print(\"now using perfect gas equation,p=rho*R*T\")\n", + "print(\"T=P/(rho*R)=P/((165.79/C)*R)\")\n", + "print(\"C/T=165.79*R/P\")\n", + "165.79*R/P\n", + "print(\"so C=1.19*T\")\n", + "print(\"we know,C^2=((2*y*R)/(y-1))*(To-T)\")\n", + "print(\"C^2=(2*1.4*287)*(300-T)/(1.4-1)\")\n", + "print(\"C^2=602.7*10^3-2009*T\")\n", + "print(\"C^2+1688.23*C-602.7*10^3=0\")\n", + "print(\"solving we get,C=302.72 m/s and T=254.39 K\")\n", + "C=302.72;\n", + "T=254.39;\n", + "print(\"using stagnation property relation,\")\n", + "print(\"To/T=1+(y-1)*M^2/2\")\n", + "M=math.sqrt(((To/T)-1)/((y-1)/2))\n", + "print(\"so M=\"),round(M,3)\n", + "M=0.947;#approx.\n", + "Po=P*(1+(y-1)*M**2/2)/10**5\n", + "print(\"stagnation pressure,Po in bar=\"),round(Po,3)\n", + "print(\"so mach number=0.947,stagnation pressure=0.472 bar,velocity=302.72 m/s\")\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 13.2;pg no: 525" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 13.2, Page:525 \n", + " \n", + "\n", + "Engineering Thermodynamics by Onkar Singh Chapter 13 Example 2\n", + "mach number,M_a=(1/sin(a))=sqrt(2)\n", + "here,P/Po=0.25/1.01=0.2475.Corresponding to this P/Po ratio the mach number and T/To can be seen from air table as M=1.564 and T/To=0.6717\n", + "T=To*0.6717 in K\n", + "and C_max=M*sqrt(y*R*T) in m/s\n", + "corresponding to mach number(M_a=1.414)as obtained from experimental observation,the T/To can be seen from air table and it comes out as (T/To)=0.7145\n", + "so T=0.7145*To in K\n", + "and C_av=M_a*sqrt(y*R*T) in m/s\n", + "ratio of kinetic energy= 0.869\n", + "so ratio of kinetic energy=0.869\n" + ] + } + ], + "source": [ + "#cal of ratio of kinetic energy\n", + "#intiation of all variables\n", + "# Chapter 13\n", + "import math\n", + "print\"Example 13.2, Page:525 \\n \\n\"\n", + "print(\"Engineering Thermodynamics by Onkar Singh Chapter 13 Example 2\")\n", + "To=(273.+1100.);#stagnation temperature in K\n", + "a=45.;#mach angle over exit cross-section in degree\n", + "Po=1.01;#pressure at upstream side of nozzle in bar\n", + "P=0.25;#ststic pressure in bar\n", + "y=1.4;#expansion constant \n", + "R=287.;#gas constant in J/kg K\n", + "print(\"mach number,M_a=(1/sin(a))=sqrt(2)\")\n", + "M_a=math.sqrt(2)\n", + "M_a=1.414;#approx.\n", + "print(\"here,P/Po=0.25/1.01=0.2475.Corresponding to this P/Po ratio the mach number and T/To can be seen from air table as M=1.564 and T/To=0.6717\")\n", + "M=1.564;\n", + "print(\"T=To*0.6717 in K\")\n", + "T=To*0.6717\n", + "print(\"and C_max=M*sqrt(y*R*T) in m/s\")\n", + "C_max=M*math.sqrt(y*R*T)\n", + "print(\"corresponding to mach number(M_a=1.414)as obtained from experimental observation,the T/To can be seen from air table and it comes out as (T/To)=0.7145\")\n", + "print(\"so T=0.7145*To in K\")\n", + "T=0.7145*To\n", + "print(\"and C_av=M_a*sqrt(y*R*T) in m/s\")\n", + "C_av=M_a*math.sqrt(y*R*T)\n", + "print(\"ratio of kinetic energy=\"),round(((1./2.)*C_av**2)/((1./2.)*C_max**2),3)\n", + "print(\"so ratio of kinetic energy=0.869\")" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 13.3;pg no: 526" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 13.3, Page:526 \n", + " \n", + "\n", + "Engineering Thermodynamics by Onkar Singh Chapter 13 Example 3\n", + "From bernoulli equation,Po-P=(1/2)*rho*C^2\n", + "so Po=P+(1/2)*rho*C^2 in N/m^2\n", + "speed indicator reading shall be given by mach no.s\n", + "mach no.,M=C/a=C/sqrt(y*R*T)\n", + "using perfect gas equation,P=rho*R*T\n", + "so T=P/(rho*R)in K\n", + "so mach no.,M 0.95\n", + "considering compressibility effect,Po/P=(1+(y-1)*M^2/2)^(y/(y-1))\n", + "so stagnation pressure,Po=P*((1+(y-1)*M^2/2)^(y/(y-1)))in N/m^2\n", + "also Po-P=(1+k)*(1/2)*rho*C^2\n", + "substitution yields,k= 0.2437\n", + "so compressibility correction factor,k=0.2437\n" + ] + } + ], + "source": [ + "#cal of mach no,compressibility correction factor\n", + "#intiation of all variables\n", + "# Chapter 13\n", + "import math\n", + "print\"Example 13.3, Page:526 \\n \\n\"\n", + "print(\"Engineering Thermodynamics by Onkar Singh Chapter 13 Example 3\")\n", + "C=300.;#aircraft flying speed in m/s\n", + "P=0.472*10**5;#altitude pressure in Pa\n", + "rho=0.659;#density in kg/m^3\n", + "y=1.4;#expansion constant\n", + "R=287.;#gas constant in J/kg K\n", + "print(\"From bernoulli equation,Po-P=(1/2)*rho*C^2\")\n", + "print(\"so Po=P+(1/2)*rho*C^2 in N/m^2\")\n", + "Po=P+(1/2)*rho*C**2\n", + "print(\"speed indicator reading shall be given by mach no.s\")\n", + "print(\"mach no.,M=C/a=C/sqrt(y*R*T)\")\n", + "print(\"using perfect gas equation,P=rho*R*T\")\n", + "print(\"so T=P/(rho*R)in K\")\n", + "T=P/(rho*R)\n", + "M=C/math.sqrt(y*R*T)\n", + "print(\"so mach no.,M\"),round(M,2)\n", + "M=0.947;#approx.\n", + "print(\"considering compressibility effect,Po/P=(1+(y-1)*M^2/2)^(y/(y-1))\")\n", + "print(\"so stagnation pressure,Po=P*((1+(y-1)*M^2/2)^(y/(y-1)))in N/m^2\")\n", + "Po=P*((1+(y-1)*M**2/2)**(y/(y-1)))\n", + "print(\"also Po-P=(1+k)*(1/2)*rho*C^2\")\n", + "k=((Po-P)/((1./2.)*rho*C**2))-1.\n", + "print(\"substitution yields,k=\"),round(k,4)\n", + "print(\"so compressibility correction factor,k=0.2437\")\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 13.4;pg no: 527" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 13.4, Page:527 \n", + " \n", + "\n", + "Engineering Thermodynamics by Onkar Singh Chapter 13 Example 4\n", + "we know that,Po/P=(1+(y-1)*M^2/2)^((y)/(y-1))\n", + "so M= 1.897\n", + "so mach number,M=1.89\n" + ] + } + ], + "source": [ + "#cal of mach number\n", + "#intiation of all variables\n", + "# Chapter 13\n", + "import math\n", + "print\"Example 13.4, Page:527 \\n \\n\"\n", + "print(\"Engineering Thermodynamics by Onkar Singh Chapter 13 Example 4\")\n", + "Po=2;#total pressure in bar\n", + "P=0.3;#static pressure in bar\n", + "y=1.4;#expansion constant\n", + "print(\"we know that,Po/P=(1+(y-1)*M^2/2)^((y)/(y-1))\")\n", + "M=math.sqrt((math.exp(math.log(Po/P)/(y/(y-1)))-1)/((y-1)/2))\n", + "print(\"so M=\"),round(M,3)\n", + "print(\"so mach number,M=1.89\")\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 13.5;pg no: 527" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 13.5, Page:527 \n", + " \n", + "\n", + "Engineering Thermodynamics by Onkar Singh Chapter 13 Example 5\n", + "actual static pressure(P)=1+0.3 in bar\n", + "It is also given that,Po-P=0.6,\n", + "so Po=P+0.6 in bar\n", + "air velocity,ao=sqrt(y*R*To)in m/s\n", + "density of air,rho_o=Po/(R*To)in \n", + "considering air to be in-compressible,\n", + "Po=P+rho_o*C^2/2\n", + "so C in m/s= 235.13\n", + "for compressible fluid,Po/P=(1+(y-1)*M^2/2)^(y/(y-1))\n", + "so M=sqrt((exp(log(Po/P)/(y/(y-1)))-1)/((y-1)/2))\n", + "compressibility correction factor,k\n", + "k=(M^2/4)+((2-y)/24)*M^4\n", + "stagnation temperature,To/T=1+((y-1)/2)*M^2\n", + "so T=To/(1+((y-1)/2)*M^2) in K\n", + "density,rho=P/(R*T) in kg/m^3\n", + "substituting Po-P=(1/2)*rho*C^2(1+k)\n", + "C in m/s= 250.94\n", + "so C=250.95 m/s\n" + ] + } + ], + "source": [ + "#cal of air stream velocity\n", + "#intiation of all variables\n", + "# Chapter 13\n", + "import math\n", + "print\"Example 13.5, Page:527 \\n \\n\"\n", + "print(\"Engineering Thermodynamics by Onkar Singh Chapter 13 Example 5\")\n", + "To=305.;#stagnation temperature of air stream in K\n", + "y=1.4;#expansion constant\n", + "R=287.;#gas constant in J/kg K\n", + "print(\"actual static pressure(P)=1+0.3 in bar\")\n", + "P=1.+0.3\n", + "print(\"It is also given that,Po-P=0.6,\")\n", + "print(\"so Po=P+0.6 in bar\")\n", + "Po=P+0.6\n", + "print(\"air velocity,ao=sqrt(y*R*To)in m/s\")\n", + "ao=math.sqrt(y*R*To)\n", + "print(\"density of air,rho_o=Po/(R*To)in \")\n", + "rho_o=Po*10.**5/(R*To)\n", + "print(\"considering air to be in-compressible,\")\n", + "print(\"Po=P+rho_o*C^2/2\")\n", + "C=math.sqrt((Po-P)*10.**5*2./rho_o)\n", + "print(\"so C in m/s=\"),round(C,2)\n", + "print(\"for compressible fluid,Po/P=(1+(y-1)*M^2/2)^(y/(y-1))\")\n", + "print(\"so M=sqrt((exp(log(Po/P)/(y/(y-1)))-1)/((y-1)/2))\")\n", + "M=math.sqrt((math.exp(math.log(Po/P)/(y/(y-1)))-1)/((y-1)/2))\n", + "M=0.7567;#approx.\n", + "print(\"compressibility correction factor,k\")\n", + "print(\"k=(M^2/4)+((2-y)/24)*M^4\")\n", + "k=(M**2/4.)+((2.-y)/24.)*M**4\n", + "print(\"stagnation temperature,To/T=1+((y-1)/2)*M^2\")\n", + "print(\"so T=To/(1+((y-1)/2)*M^2) in K\")\n", + "T=To/(1+((y-1)/2)*M**2)\n", + "print(\"density,rho=P/(R*T) in kg/m^3\")\n", + "rho=P*10**5/(R*T)\n", + "print(\"substituting Po-P=(1/2)*rho*C^2(1+k)\")\n", + "C=math.sqrt((Po-P)*10.**5/((1./2.)*rho*(1.+k)))\n", + "print(\"C in m/s=\"),round(C,2)\n", + "print(\"so C=250.95 m/s\")" + ] + } + ], + "metadata": { + 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