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diff --git a/Engineering_Thermodynamics_by__O._Singh/chapter13.ipynb b/Engineering_Thermodynamics_by__O._Singh/chapter13.ipynb new file mode 100755 index 00000000..836097b9 --- /dev/null +++ b/Engineering_Thermodynamics_by__O._Singh/chapter13.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": {
+ "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": "2.7.9"
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+}
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