{
"cells": [
 {
		   "cell_type": "markdown",
	   "metadata": {},
	   "source": [
       "# Chapter 13: One Dimensional Compressible Fluid Flow"
	   ]
	},
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 13.1: Engineering_Thermodynamics_by_Onkar_Singh_Chapter_13_Example_1.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Display mode\n",
"mode(0);\n",
"// Display warning for floating point exception\n",
"ieee(1);\n",
"clear;\n",
"clc;\n",
"disp('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",
"disp('mass flow rate(m)=rho*A*C')\n",
"disp('so rho*C=4*m/(%pi*d^2)')\n",
"4*m/(%pi*d^2)\n",
"disp('so rho=165.79/C')\n",
"disp('now using perfect gas equation,p=rho*R*T')\n",
"disp('T=P/(rho*R)=P/((165.79/C)*R)')\n",
"disp('C/T=165.79*R/P')\n",
"165.79*R/P\n",
"disp('so C=1.19*T')\n",
"disp('we know,C^2=((2*y*R)/(y-1))*(To-T)')\n",
"disp('C^2=(2*1.4*287)*(300-T)/(1.4-1)')\n",
"disp('C^2=602.7*10^3-2009*T')\n",
"disp('C^2+1688.23*C-602.7*10^3=0')\n",
"disp('solving we get,C=302.72 m/s and T=254.39 K')\n",
"C=302.72;\n",
"T=254.39;\n",
"disp('using stagnation property relation,')\n",
"disp('To/T=1+(y-1)*M^2/2')\n",
"disp('so M=sqrt(((To/T)-1)/((y-1)/2))')\n",
"M=sqrt(((To/T)-1)/((y-1)/2))\n",
"M=0.947;//approx.\n",
"disp('stagnation pressure,Po=P*(1+(y-1)*M^2/2)in bar')\n",
"Po=P*(1+(y-1)*M^2/2)/10^5\n",
"disp('so mach number=0.947,stagnation pressure=0.472 bar,velocity=302.72 m/s')\n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 13.2: Engineering_Thermodynamics_by_Onkar_Singh_Chapter_13_Example_2.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Display mode\n",
"mode(0);\n",
"// Display warning for floating point exception\n",
"ieee(1);\n",
"clear;\n",
"clc;\n",
"disp('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",
"disp('mach number,M_a=(1/sin(a))=sqrt(2)')\n",
"M_a=sqrt(2)\n",
"M_a=1.414;//approx.\n",
"disp('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",
"disp('T=To*0.6717 in K')\n",
"T=To*0.6717\n",
"disp('and C_max=M*sqrt(y*R*T) in m/s')\n",
"C_max=M*sqrt(y*R*T)\n",
"disp('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",
"disp('so T=0.7145*To in K')\n",
"T=0.7145*To\n",
"disp('and C_av=M_a*sqrt(y*R*T) in m/s')\n",
"C_av=M_a*sqrt(y*R*T)\n",
"disp('ratio of kinetic energy=((1/2)*C_av^2)/((1/2)*C_max^2)')\n",
"((1/2)*C_av^2)/((1/2)*C_max^2)\n",
"disp('so ratio of kinetic energy=0.869')\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 13.3: Engineering_Thermodynamics_by_Onkar_Singh_Chapter_13_Example_3.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Display mode\n",
"mode(0);\n",
"// Display warning for floating point exception\n",
"ieee(1);\n",
"clear;\n",
"clc;\n",
"disp('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",
"disp('From bernoulli equation,Po-P=(1/2)*rho*C^2')\n",
"disp('so Po=P+(1/2)*rho*C^2 in N/m^2')\n",
"Po=P+(1/2)*rho*C^2\n",
"disp('speed indicator reading shall be given by mach no.s')\n",
"disp('mach no.,M=C/a=C/sqrt(y*R*T)')\n",
"disp('using perfect gas equation,P=rho*R*T')\n",
"disp('so T=P/(rho*R)in K')\n",
"T=P/(rho*R)\n",
"disp('so mach no.,M')\n",
"M=C/sqrt(y*R*T)\n",
"M=0.947;//approx.\n",
"disp('considering compressibility effect,Po/P=(1+(y-1)*M^2/2)^(y/(y-1))')\n",
"disp('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",
"disp('also Po-P=(1+k)*(1/2)*rho*C^2')\n",
"disp('substitution yields,k=')\n",
"k=((Po-P)/((1/2)*rho*C^2))-1\n",
"disp('so compressibility correction factor,k=0.2437')"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 13.4: Engineering_Thermodynamics_by_Onkar_Singh_Chapter_13_Example_4.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Display mode\n",
"mode(0);\n",
"// Display warning for floating point exception\n",
"ieee(1);\n",
"clear;\n",
"clc;\n",
"disp('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",
"disp('we know that,Po/P=(1+(y-1)*M^2/2)^((y)/(y-1))')\n",
"disp('so M=sqrt((exp(log(Po/P)/(y/(y-1)))-1)/((y-1)/2))')\n",
"M=sqrt((exp(log(Po/P)/(y/(y-1)))-1)/((y-1)/2))\n",
"disp('so mach number,M=1.89')"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 13.5: Engineering_Thermodynamics_by_Onkar_Singh_Chapter_13_Example_5.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Display mode\n",
"mode(0);\n",
"// Display warning for floating point exception\n",
"ieee(1);\n",
"clear;\n",
"clc;\n",
"disp('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",
"disp('actual static pressure(P)=1+0.3 in bar')\n",
"P=1+0.3\n",
"disp('It is also given that,Po-P=0.6,')\n",
"disp('so Po=P+0.6 in bar')\n",
"Po=P+0.6\n",
"disp('air velocity,ao=sqrt(y*R*To)in m/s')\n",
"ao=sqrt(y*R*To)\n",
"disp('density of air,rho_o=Po/(R*To)in ')\n",
"rho_o=Po*10^5/(R*To)\n",
"disp('considering air to be in-compressible,')\n",
"disp('Po=P+rho_o*C^2/2')\n",
"disp('so C=sqrt((Po-P)*2/rho_o)in m/s')\n",
"C=sqrt((Po-P)*10^5*2/rho_o)\n",
"disp('for compressible fluid,Po/P=(1+(y-1)*M^2/2)^(y/(y-1))')\n",
"disp('so M=sqrt((exp(log(Po/P)/(y/(y-1)))-1)/((y-1)/2))')\n",
"M=sqrt((exp(log(Po/P)/(y/(y-1)))-1)/((y-1)/2))\n",
"M=0.7567;//approx.\n",
"disp('compressibility correction factor,k')\n",
"disp('k=(M^2/4)+((2-y)/24)*M^4')\n",
"k=(M^2/4)+((2-y)/24)*M^4\n",
"disp('stagnation temperature,To/T=1+((y-1)/2)*M^2')\n",
"disp('so T=To/(1+((y-1)/2)*M^2) in K')\n",
"T=To/(1+((y-1)/2)*M^2)\n",
"disp('density,rho=P/(R*T) in kg/m^3')\n",
"rho=P*10^5/(R*T)\n",
"disp('substituting Po-P=(1/2)*rho*C^2(1+k)')\n",
"disp('C=sqrt((Po-P)/((1/2)*rho*(1+k)))in m/s')\n",
"C=sqrt((Po-P)*10^5/((1/2)*rho*(1+k)))\n",
"disp('so C=250.95 m/s')"
   ]
   }
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