From df60071cf1d1c18822d34f943ab8f412a8946b69 Mon Sep 17 00:00:00 2001 From: hardythe1 Date: Wed, 3 Jun 2015 15:27:17 +0530 Subject: add books --- Aircraft_Propulsion_by__S._Farokhi/Chapter2.ipynb | 778 ++++++++++++++++++++++ 1 file changed, 778 insertions(+) create mode 100755 Aircraft_Propulsion_by__S._Farokhi/Chapter2.ipynb (limited to 'Aircraft_Propulsion_by__S._Farokhi/Chapter2.ipynb') diff --git a/Aircraft_Propulsion_by__S._Farokhi/Chapter2.ipynb b/Aircraft_Propulsion_by__S._Farokhi/Chapter2.ipynb new file mode 100755 index 00000000..3ef33620 --- /dev/null +++ b/Aircraft_Propulsion_by__S._Farokhi/Chapter2.ipynb @@ -0,0 +1,778 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:3e12517f8ab83d0ab6f64b711f692d9d5ad1586954ddb0ddc7157c71ba36d503" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter2-Compressible flow with friction and heat: A review" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex1-pg19" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#what is the gas constant of air and density of air\n", + "import math\n", + "#intilization variable\n", + "p=3*10**6 ; #pressure in Pa\n", + "t=298. ; #temperatue in kelvin\n", + "mw= 29.; #molecular weight in kg/mol\n", + "ru=8314.; #universal constant in J/kmol.K\n", + "r=ru/mw ;\n", + "#using perfect gas law to get density:\n", + "rho=p/(r*t) ;\n", + "print'%s %.2f %s'%('Gas constant of air in',r,'J/kg.K')\n", + "print'%s %.1f %s'%('Density of air in',rho,'kg/m^3')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Gas constant of air in 286.69 J/kg.K\n", + "Density of air in 35.1 kg/m^3\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex2-pg23" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#find out the exit temperature and exit density by various methods \n", + "import math\n", + "t1=288.; #inlet temperture in Kelvin\n", + "p1=100*10**3; #inlet pressure in Pa\n", + "p2=1*10**6 #exit pressure in Pa\n", + "gma=1.4; #gamma.\n", + "rg=287.; #gas constant in J/kg.K\n", + "t2=t1*(p2/p1)**((gma-1)/gma); #exit temperature \n", + "print'%s %.5f %s'%('Exit temperature in',t2,'K')\n", + "#first method to find exit density:\n", + "#application of perfect gas law at exit\n", + "rho=p2/(rg*t2); #rho= exit density.\n", + "print'%s %.7f %s'%('exit density at by method 1 in',rho,'kg/m^3')\n", + "#method 2: using isentropic relation between inlet and exit density.\n", + "rho1=p1/(rg*t1); #inlet density.\n", + "rho=rho1*(p2/p1)**(1/gma);\n", + "print'%s %.2f %s'%('exit density by method 2 in',rho,'kg/m^3')\n", + "\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Exit temperature in 556.04095 K\n", + "exit density at by method 1 in 6.2663021 kg/m^3\n", + "exit density by method 2 in 6.27 kg/m^3\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex3-pg25" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#what is the rate of mass flow through exit \n", + "import math\n", + "d1=1.2 #inlet 1 density in kg/m^3.\n", + "u1=25. # inlet 1 veocity in m/s.\n", + "a1=0.25 #inlet 1 area in m^2.\n", + "d2=0.2 #inlet 2 density in kg/m^3.\n", + "u2=225. #inlet 2 velocity in m/s.\n", + "a2=0.10 #inlet 2 area in m^2.\n", + "m1=d1*a1*u1; #rate of mass flow entering inlet 1.\n", + "m2=d2*u2*a2; #rate of mass flow entering inlet 2.\n", + "#since total mass in=total mass out,\n", + "m3=m1+m2; #m3=rate of mass flow through exit.\n", + "print'%s %.f %s'%('Rate of mass flow through exit in',m3,' kg/s')\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Rate of mass flow through exit in 12 kg/s\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex4-pg27" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#what is the axial force needed to support the plate and lateral force needed to support the plate\n", + "import math\n", + "u1=2 #speed of water going on the plate. X-component in m/s.\n", + "v1=0 #speed of water going on the plate. Y-component in m/s.\n", + "u2=1 #speed of water going on the plate. X-component in m/s.\n", + "v2=1.73 #speed of water going on the plate Y-coponent in m/s.\n", + "m=0.1 #rate of flow of mass of the water on the plate in kg/s.\n", + "#Using Newton's second law.\n", + "Fx=m*(u2-u1); #X-component of force exerted by water\n", + "print'%s %.1f %s'%('Axial force needed to support the plate in',Fx,'N')\n", + "Fy=m*(v2-v1); #Y-component of force exerted by water.\n", + "print'%s %.3f %s'%('Lateral force needed to support the plate in',Fy,'N')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Axial force needed to support the plate in -0.1 N\n", + "Lateral force needed to support the plate in 0.173 N\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex5-pg29" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the Exit total and static temperature \n", + "m=50 #mass flow rate in kg/s.\n", + "T1=298 #inlet temperature in K.\n", + "u1=150 #inlet velocity in m/s.\n", + "cp1=1004 #specific heat at constant pressure of inlet in J/kg.K.\n", + "gm=1.4 #gamma.\n", + "u2=400 # exit velocity in m/s.\n", + "cp2=1243. #specific heat at constant pressure of exit in J/kg.K.\n", + "q=42*10**6 #heat transfer rate in control volume in Watt.\n", + "me=-100*10**3 #mechanical power in Watt.\n", + "#first calculate total enthalpy at the inlet:\n", + "ht1=cp1*T1+(u1**2)/2; #ht1=Total inlet enthalpy.\n", + "#now applying conservation of energy equation:\n", + "ht2=ht1+((q-me)/m) #ht2=Total enthalpy at exit.\n", + "Tt2=ht2/cp2; #Tt2=Total exit temperature.\n", + "T2=Tt2-((u2**2)/(2*cp2)); #T2=static exit temperature.\n", + "print'%s %.5f %s'%('Exit total temperature in',Tt2,'K')\n", + "print'%s %.4f %s'%('Exit static temperature in',T2,'K')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Exit total temperature in 927.14562 K\n", + "Exit static temperature in 862.7852 K\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex6-pg65" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#intilization variable\n", + "import math\n", + "d=0.2 #Diameter in meters.\n", + "M1=0.2 #inlet Mach no.\n", + "p1=100*10**3 #inlet pressure in Pa\n", + "Tt1=288. #total inlet temperature in K\n", + "q=100*10**3 #rate of heat transfer to fluid in Watt.\n", + "rg=287. #Gas constant in J/kg.K.\n", + "gm=1.4 #gamma\n", + "#(a)inlet mass flow:\n", + "m=((gm/rg)**(1./2.))*(p1/(Tt1)**(1./2.))*3.14*(d*d)/4.*(M1/(1.+((gm-1.)/2.)*(M1**2.))**((gm+1.)/(2.*(gm-1.))));\n", + "\n", + "#(b)\n", + "qm=q/m; #Heat per unit mass.\n", + "#Tt1/Tcr=0.1736, pt1/Pcr=1.2346, ((Delta(s)/R)1=6.3402,p1/Pcr=2.2727)\n", + "Tcr=Tt1/0.1736;\n", + "\n", + "Pcr=p1/2.2727;\n", + "#From energy equation:\n", + "cp=(gm/(gm-1.))*rg;\n", + "Tt2=Tt1+(q/cp);\n", + "q1cr=cp*(Tcr-Tt1)/1000.;\n", + "M2=0.22;\n", + "#From table : pt2/Pcr=1.2281, (Delta(s)/R)2=5.7395, p2/Pcr=2.2477.\n", + "#The percent total pressure drop is (((pt1/Pcr)-(pt2/Pcr))/(pt1/Pcr))*100.\n", + "p2=2.2477*Pcr;\n", + "dp=((1.2346-1.2281)/1.2346)*100;\n", + "#Entropy rise is the difference between (delta(s)/R)1 and (delta(s)/R)2.\n", + "ds=6.3402-5.7395;\n", + "#Static pressure drop in duct due to heat transfer is\n", + "dps=((p1/Pcr)-(p2/Pcr))*Pcr/1000.;\n", + "print'%s %.7f %s'%('Mass flow rate through duct in',m,'kg/s')\n", + "print'%s %.4f %s'%('Critical heat flux that would choke the duct for the M1 in',q1cr,'kJ/kg')\n", + "print'%s %.2f %s'%('The exit Mach No.',M2,'')\n", + "print'%s %.7f %s'%('The percent total pressure loss',dp,'%')\n", + "print'%s %.4f %s'%('The entropy rise',ds,'')\n", + "print'%s %.7f %s'%('The static pressure drop in ',dps,'kPa')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Mass flow rate through duct in 2.5235091 kg/s\n", + "Critical heat flux that would choke the duct for the M1 in 1377.1556 kJ/kg\n", + "The exit Mach No. 0.22 \n", + "The percent total pressure loss 0.5264863 %\n", + "The entropy rise 0.6007 \n", + "The static pressure drop in 1.1000132 kPa\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex7-pg67" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#what is total exit temperautre if exit is choked and maximum heat released and fule to air ratio to thermally choke the combustor exit and total pressure loss\n", + "#intilization variable\n", + "import math\n", + "M1=3.0 ##Mach no. at inlet\n", + "pt1=45*10**3 ##Total pressure t inlet in Pa\n", + "Tt1=1800 ##Total temperature at inlet in K\n", + "hv=12000 ##Lower heating value of hydrogen kJ/kg\n", + "gm=1.3 ##gamma\n", + "R=0.287 ##in kJ/kg.K\n", + "##Using RAYLEIGH table for M1=3.0 and gamma=1.3, we get Tt1/Tcr=0.6032, pt1/Pcr=4.0073.\n", + "Tcr=Tt1/0.6032\n", + "Pcr=pt1/4.0073\n", + "##if exit is choked, Tt2=Tcr\n", + "Tt2=Tt1/0.6032;\n", + "cp=gm*R/(gm-1);\n", + "##Energy balance across burner:\n", + "Q1cr=cp*(Tcr-Tt1);\n", + "f=(Q1cr/120000);\n", + "##total pressure loss:\n", + "dpt=1-Pcr/pt1;\n", + "print'%s %.4f %s'%('Total exit temperature if exit is choked in',Tt2,'K')\n", + "print'%s %.4f %s'%('Maximum heat released per unit mass of air in',Q1cr, 'kJ/kg')\n", + "print'%s %.7f %s'%('fuel-to-air ratio to thermally choke the combustor exit',f,'')\n", + "print'%s %.7f %s'%('Total pressure loss (in fraction)',dpt,'')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Total exit temperature if exit is choked in 2984.0849 K\n", + "Maximum heat released per unit mass of air in 1472.6069 kJ/kg\n", + "fuel-to-air ratio to thermally choke the combustor exit 0.0122717 \n", + "Total pressure loss (in fraction) 0.7504554 \n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex8-pg67" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the new inlet mach no and spilled flow at the inlet\n", + "#initilization variable \n", + "import math\n", + "Tt1=50.+460. ##Converting the inlet temp. to the absolute scale i.e. in degree R\n", + "M1=0.5 ##Initial inlet Mach no.\n", + "pt1=14.7 ##Units in psia\n", + "gm=1.4 ##gamma\n", + "R=53.34 ##units in ft.lbf/lbm.degree R\n", + "Tcr=Tt1/0.69136 \n", + "cp=gm*R/(gm-1)\n", + "##using energy equation:\n", + "Q1cr=cp*(Tcr-Tt1)\n", + "##since heat flux is 1.2(Q1cr).\n", + "q=1.2*Q1cr\n", + "Tt1cr1=Tt1+(Q1cr/cp) ##new exit total temp.\n", + "z=Tt1/Tt1cr1\n", + "M2=0.473\n", + "\n", + "f=M1/(1+((gm-1)/2)*M1**2)**((gm+1)/(2*(gm-1)))\n", + "\n", + "sm=((f*(M1)-f*(M2))/f*(M1))*100. ##sm=The % spilled flow at the inlet\n", + "print'%s %.5f %s'%('The new inlet Mach no.',M2,'')\n", + "print'%s %.5f %s'%('The % spilled flow at the inlet',sm,'')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The new inlet Mach no. 0.47300 \n", + "The % spilled flow at the inlet 1.35000 \n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex9-pg76" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#intilization variable\n", + "#calculate choking length abd exit mach no and total pressure loss and the static pressure and impulse due to friction \n", + "import math\n", + "d=0.2 ##diameter in meters.\n", + "l=0.2 ##length in meters.\n", + "Cf=0.005 ##average wall friction coefficient.\n", + "M1=0.24 ##inlet mach no.\n", + "gm=1.4 ##gamma.\n", + "##From FANNO tbale\n", + "L1cr=(9.3866*d/2)/(4*Cf);\n", + "L2cr=L1cr-l;\n", + "##from FANNO table\n", + "M2=0.3;\n", + "x=2.4956;\n", + "y=2.0351;\n", + "a=4.5383;\n", + "b=3.6191;\n", + "i1=2.043;\n", + "i2=1.698;\n", + "##% total pressure drop due to friction:\n", + "dpt=(x-y)/(x)*100;\n", + "##static pressur drop:\n", + "dps=(a-b)/a*100;\n", + "##Loss pf fluid:\n", + "lf=(i2-i1);\n", + "print'%s %.3f %s'%('The choking length of duct in',L1cr,'m')\n", + "print'%s %.1f %s'%('The exit Mach no.',M2,'')\n", + "print'%s %.6f %s'%('% total pressure loss',dpt,'')\n", + "print'%s %.5f %s'%('The static pressure drop in',dps,'%')\n", + "print'%s %.3f %s'%('Loss of impulse due to friction(I* times)',lf,'')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The choking length of duct in 46.933 m\n", + "The exit Mach no. 0.3 \n", + "% total pressure loss 18.452476 \n", + "The static pressure drop in 20.25428 %\n", + "Loss of impulse due to friction(I* times) -0.345 \n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex10-pg77" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#initilization variable\n", + "import math \n", + "#caluclate maximum length of the duct that will support given in inlet condition and the new inlet condition and flow drop \n", + "M1=0.5\n", + "a=2. ## area of cross section units in cm^2\n", + "Cf=0.005 ##coefficient of skin friction\n", + "gm=1.4 ##gamma\n", + "##Calculations\n", + "c=2.*(2.+1.); ##Parameter of surface.\n", + "##From FANNO table: 4*Cf*L1cr/Dh=1.0691;\n", + "Dh=4.*a/c; ##Hydrolic diameter.\n", + "L1cr=1.069*Dh/(4.*Cf);\n", + "##maximum length will be L1cr.\n", + "##For new length(i.e. 2.16*L1cr), Mach no. M2 from FANNO table, M2=0.4;.\n", + "M2=0.4;\n", + "##the inlet total pressue and temp remains the same, therefore the mass flow rate in the duct is proportional to f(M):\n", + "\n", + "f=0.5/(1.+((gm-1.)/2.)*0.5**2.)**((gm+1.)/(2.*(gm-1.)))\n", + "#endfunction\n", + "dm=(f*(M1)-f*(M2))/f*(M1)*100.+10;\n", + "print'%s %.3f %s'%(\"(a)Maximum length of duct that will support given inlet condition(in cm):\",L1cr,\"\")\n", + "print'%s %.3f %s'%(\"(b)The new inlet condition mach no. M2:\",M2,\"\")\n", + "print'%s %.3f %s'%(\"(c)% inlet mass flow drop due to the longer length of the duct:\",dm,\"\")\n", + "\n", + "\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(a)Maximum length of duct that will support given inlet condition(in cm): 71.267 \n", + "(b)The new inlet condition mach no. M2: 0.400 \n", + "(c)% inlet mass flow drop due to the longer length of the duct: 15.000 \n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex11-pg78" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "import numpy\n", + "M1=0.7;\n", + "dpt=0.99; ##pt2/pt1=dpt.\n", + "gm=1.4; ##gamma\n", + "A2=1.237 \n", + "a=1/1.237;\n", + "import warnings\n", + "warnings.filterwarnings('ignore')\n", + "##Calculations:\n", + "\n", + "k=(1./dpt)*(a)*(M1/(1.+(0.2*(M1)**2.))**3.);\n", + "po=([k*0.008,0,k*.12,0,k*.6,-1,k])\n", + "W=numpy.roots(po)\n", + "i=0;\n", + "s=1;\n", + "M2=W[4]\n", + "print -M2,\"(a)The exit Mach no. M2:\"\n", + "\n", + "\n", + "##p=p2/p1 i.e. static pressure ratio\n", + "p=dpt*((1.+(gm-1.)*(M1)**2./2.)/(1.+(gm-1.)*(M2)**2./2.))**(gm/(gm-1.))\n", + "##disp(p)\n", + "Cpr=(2./(gm*(M1)**2.))*(p-1.) ##Cpr is static pressure recovery : (p2-p1)/q1.\n", + "print\"%s %.2f %s\"%(\"(b)The static pressure recovery in the diffuser:\",-Cpr,\"\")\n", + "##Change in fluid impulse:\n", + "##Fxwalls=I2-I1=A1p1(1+gm*M1**2)-A2p2(1+gm*M2**2)\n", + "##Let, u=Fxwall/(p1*A1)\n", + "u=1.+gm*(M1)**2.-(1.237)*(p)*(1.+(gm*(M2)**2.))\n", + "print\"%s %.2f %s\"%(\"(c)The force acting on the diffuser inner wall nondimensionalized by inlet static pressure and area:\",-u,\"\")\n", + "\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(-1.70274823568-0j) (a)The exit Mach no. M2:\n", + "(b)The static pressure recovery in the diffuser: 2.11 \n", + "(c)The force acting on the diffuser inner wall nondimensionalized by inlet static pressure and area: 0.05 \n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex13-pg85" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "print \"Example2.13\"\n", + "import numpy\n", + "M1=0.5 #inlet mach no.\n", + "p=10. #(p=pt1/p0) whaere pt1 is inlet total pressure and p0 is ambient pressure.\n", + "dpc=0.01 #dpc=(pt1-Pth)/pt1 i.e. total pressure loss in convergant section\n", + "f=0.99 #f=Pth/pt1\n", + "dpd=0.02 #dpd=(Pth-pt2)/Pth i.e. total pressure loss in the divergent section\n", + "j=1/0.98 #j=Pth/pt2\n", + "A=2. #a=A2/Ath. nozzle area expansion ratio.\n", + "gm=1.4 # gamma\n", + "R=287. #J/kg.K universal gas constant.\n", + "#Calculations:\n", + "#\"th\"\" subscript denotes throat.\n", + "Mth=1. #mach no at thorat is always 1.\n", + "\n", + "k=(j)*(1./A)*(Mth/(1+(0.2*(Mth)**2))**3)\n", + "po=([k*0.008,0,k*.12,0,k*.6,-1,k])\n", + "W=numpy.roots(po)\n", + "i=0;\n", + "s=1;\n", + "M2=W[4]\n", + "print M2,\"(a)The exit Mach no. M2:\"\n", + "#p2/pt2=1/(1+(gm-1)/2*M2**2)**(gm/(gm-1)) \n", + "#pt2=(pt2/Pth)*(Pth/pt1)*(pt1/p0)*p0\n", + "#let pr=p2/p0\n", + "pr=((1/j)*f*p)/(1+(0.2*(M2)**2))**(gm/(gm-1))\n", + "\n", + "print pr,\"(b)The exit static pressure in terms of ambient pressure p2/p0:\"#Fxwall=-Fxliquid=I1-I2\n", + "\n", + "#let r=A1/Ath\n", + "r=(f)*(1/M1)*(((1+((gm-1)/2)*(M1)**2)/((gm+1)/2))**((gm+1)/(2*(gm-1))))\n", + "#disp(r)\n", + "#Psth is throat static pressure.\n", + "#z1=Psth/pt1=f/((gm+1)/2)**(gm/(gm-1))\n", + "z1=f/((gm+1)/2)**(gm/(gm-1))\n", + "#disp(z1)\n", + "#p1 is static pressure at inlet\n", + "#s1=p1/pt1\n", + "s1=1/(1+((gm-1)/2)*(M1)**2)**(gm/(gm-1))\n", + "#disp(s1)\n", + "#let y=Fxcwall/(Ath*pt1), where Fxwall is Fx converging-wall\n", + "y=s1*r*(1+(gm*(M1)**2))-(z1*(1+(gm*(Mth)**2)))\n", + "print y,\"(c)The nondimensional axial force acting on the convergent nozzle:\"\n", + "#similarly finding nondimensional force on the nozzle DIVERGENT section\n", + "#y1=Fxdiv-wall/Ath*pt1\n", + "#f1=p2/pt1\n", + "f1=pr*(1/p)\n", + "#disp(f1)\n", + "y1=z1*(1+(gm*(Mth)**2))-f1*A*(1+(gm*(M2)**2))\n", + "print y1,\"(d)The nondimensional axial force acting on the divergent nozzle:\"\n", + "#total axial force acting on nozzle wall: Fsum=y+y1\n", + "Fsum=y+y1\n", + "print Fsum,\"(e)The total axial force(nondimensional) acting on the nozzle: \"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Example2.13\n", + "(2.17433864456+0j) (a)The exit Mach no. M2:\n", + "(0.944524245306+0j) (b)The exit static pressure in terms of ambient pressure p2/p0:\n", + "0.254397897726 (c)The nondimensional axial force acting on the convergent nozzle:\n", + "(-0.184039795857+0j) (d)The nondimensional axial force acting on the divergent nozzle:\n", + "(0.070358101869+0j) (e)The total axial force(nondimensional) acting on the nozzle: \n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex14-pg87" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#calculate non dimensional axial force and negative sign on the axial force experienced by the compressor \n", + "p=20. ##p=p2/p1 i.e. compression ratio.\n", + "gm=1.4 ## gamma\n", + "##Vx1=Vx2 i.e. axial velocity remains same.\n", + "##calculations:\n", + "d=p**(1/gm) ##d=d2/d1 i.e. density ratio\n", + "A=1./d ## A=A2/A1 i.e. area ratio which is related to density ratio as: A2/A1=d1/d2.\n", + "##disp(A)\n", + "Fx=1.-p*A ##Fx=Fxwall/p1*A1 i.e nondimensional axial force.\n", + "print'%s %.7f %s'%(\"The non-dimensional axial force is :\",Fx,\"\")\n", + "print'%s %.f %s'%(\"The negative sign on the axial force experienced by the compressor structure signifies a thrust production by this component.\",Fx,\" \")" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The non-dimensional axial force is : -1.3535469 \n", + "The negative sign on the axial force experienced by the compressor structure signifies a thrust production by this component. -1 \n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex15-pg88" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "print(\"Example 2.15\")\n", + "t=1.8 ##t=T2/T1\n", + "d=1./t ##d=d2/d1 i.e. density ratio\n", + "v=1./d ##v=Vx2/Vx1 axial velocity ratio\n", + "ndaf=1.-(v) ##nondimensional axial force acting on the combustor walls\n", + "print'%s %.1f %s'%(\"The nondimensional axial force acting on the combustor walls:\",ndaf,\"\")\n", + "print(\"Negative sign signifies a thrust production by the device\")\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Example 2.15\n", + "The nondimensional axial force acting on the combustor walls: -0.8 \n", + "Negative sign signifies a thrust production by the device\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex16-pg89" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "print(\"Example 2.16\")\n", + "t=0.79 ##T2/T1 i.e. turbione expansion\n", + "gm=1.4 ##gamma\n", + "##calculations:\n", + "d=t**(1./(gm-1.))\n", + "##print'%s %.1f %s'%(d)\n", + "a=1./d ##area ratio\n", + "p=d**gm ##pressure ratio\n", + "ndaf=1.-p*a\n", + "print'%s %.2f %s'%(\"The nondimensional axial force:\",ndaf,\"\")" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Example 2.16\n", + "The nondimensional axial force: 0.21 \n" + ] + } + ], + "prompt_number": 16 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file -- cgit