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{
"metadata": {
"name": "",
"signature": "sha256:50b271f93e8a365fc7786fed67e8c1b9727566cca589daf5682afce2ce0beb0c"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"CHAPTER10:COMPRESSIBLE FLOW THROUGH NOZZLES DIFFUSERS AND WIND TUNNELS"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example E01 : Pg 345"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# All the quantities are expressed in Si units\n",
"area_ratio = 10.25; # exit to throat area ratio\n",
"p0 = 5; # reservoir pressure in atm\n",
"T0 = 333.3; # reservoir temperature\n",
"\n",
"# from appendix A, for an area ratio of 10.25\n",
"Me = 3.95; # exit mach number\n",
"pe = 0.007*p0; # exit pressure\n",
"Te = 0.2427*T0; # exit temperature\n",
"\n",
"print\"Me =\",Me\n",
"print\"pe =\",pe,\"atm\"\n",
"print\"Te =\",Te,\"K\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Me = 3.95\n",
"pe = 0.035 atm\n",
"Te = 80.89191 K\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example E02 : Pg 346"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# All the quantities are expressed in Si units\n",
"\n",
"area_ratio = 2.; # exit to throat area ratio\n",
"p0 = 1.; # reservoir pressure in atm\n",
"T0 = 288.; # reservoir temperature\n",
"\n",
"# (a)\n",
"# since M = 1 at the throat\n",
"Mt = 1.;\n",
"pt = 0.528*p0; # pressure at throat\n",
"Tt = 0.833*T0; # temperature at throat\n",
"\n",
"# from appendix A for supersonic flow, for an area ratio of 2\n",
"Me = 2.2; # exit mach number\n",
"pe = 1./10.69*p0; # exit pressure\n",
"Te = 1./1.968*T0; # exit temperature\n",
"\n",
"print\"At throat: Mt =\",Mt\n",
"print\"\\nAt throat: pt =\",pt,\"atm\"\n",
"print\"\\nAt throat: Tt = \",Tt,\"K\"\n",
"print\"\\nAt throat: For supersonic exit:\",Me\n",
"print\"\\nAt throat: pe =\",pe,\"atm\"\n",
"print\"\\nAt throat: Te = \",Te,\"K\"\n",
"\n",
"# (b)\n",
"# from appendix A for subonic flow, for an area ratio of 2\n",
"Me = 0.3; # exit mach number\n",
"pe = 1/1.064*p0; # exit pressure\n",
"Te = 1/1.018*T0; # exit temperature\n",
"\n",
"print\"\\nFor subrsonic exit:\",Me\n",
"print\"\\npe=\",pe,\"atm\"\n",
"print\"\\nTe=\",Te,\"K\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"At throat: Mt = 1.0\n",
"\n",
"At throat: pt = 0.528 atm\n",
"\n",
"At throat: Tt = 239.904 K\n",
"\n",
"At throat: For supersonic exit: 2.2\n",
"\n",
"At throat: pe = 0.0935453695042 atm\n",
"\n",
"At throat: Te = 146.341463415 K\n",
"\n",
"For subrsonic exit: 0.3\n",
"\n",
"pe= 0.93984962406 atm\n",
"\n",
"Te= 282.907662083 K\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example E03 : Pg 346"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# All the quantities are expressed in Si units\n",
"\n",
"area_ratio = 2.; # exit to throat area ratio\n",
"p0 = 1.; # reservoir pressure in atm\n",
"T0 = 288.; # reservoir temperature\n",
"pe = 0.973; # exit pressure in atm\n",
"\n",
"p_ratio = p0/pe; # ratio of reservoir to exit pressure\n",
"\n",
"# from appendix A for subsonic flow, for an pressure ratio of 1.028\n",
"Me = 0.2; # exit mach number\n",
"area_ratio_exit_to_star = 2.964; # A_exit/A_star\n",
"\n",
"# thus\n",
"area_ratio_throat_to_star = area_ratio_exit_to_star/area_ratio; # A_exit/A_star\n",
"\n",
"# from appendix A for subsonic flow, for an area ratio of 1.482\n",
"Mt = 0.44; # throat mach number\n",
"\n",
"print\"Me =\",Me\n",
"print\"Mt =\",Mt"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Me = 0.2\n",
"Mt = 0.44\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example E04 : Pg 352"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# All the quantities are expressed in SI units\n",
"import math \n",
"p0 = 30.*101000.; # reservoir pressure\n",
"T0 = 3500.; # reservoir temperature\n",
"R = 520.; # specific gas constant\n",
"gam = 1.22; # ratio of specific heats\n",
"A_star = 0.4; # rocket nozzle throat area\n",
"pe = 5529.; # rocket nozzle exit pressure equal to ambient pressure at 20 km altitude\n",
"\n",
"# (a)\n",
"# the density of air in the reservoir can be calculated as\n",
"rho0 = p0/R/T0;\n",
"\n",
"# from eq.(8.46)\n",
"rho_star = rho0*(2/(gam+1))**(1/(gam-1));\n",
"\n",
"# from eq.(8.44)\n",
"T_star = T0*2/(gam+1);\n",
"a_star = math.sqrt(gam*R*T_star);\n",
"u_star = a_star;\n",
"m_dot = rho_star*u_star*A_star;\n",
"\n",
"# rearranging eq.(8.42)\n",
"Me = math.sqrt(2/(gam-1)*(((p0/pe)**((gam-1)/gam)) - 1));\n",
"Te = T0/(1+(gam-1)/2*Me*Me);\n",
"ae = math.sqrt(gam*R*Te);\n",
"ue = Me*ae;\n",
"\n",
"# thus the thrust can be calculated as\n",
"T = m_dot*ue;\n",
"T_lb = T*0.2247;\n",
"\n",
"# (b)\n",
"# rearranging eq.(10.32)\n",
"Ae = A_star/Me*((2/(gam+1)*(1+(gam-1)/2*Me*Me))**((gam+1)/(gam-1)/2));\n",
"\n",
"print\"(a)The thrust of the rocket is:T =\" ,T/1e6, \"N\"\n",
"print\"\\n(b)The nozzle exit area is\",T_lb \n",
"\n",
"print\"\\nAe =\",Ae, \"m2\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a)The thrust of the rocket is:T = 2.1702872295 N\n",
"\n",
"(b)The nozzle exit area is 487663.540469\n",
"\n",
"Ae = 16.7097500627 m2\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example E05 : Pg 353"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# All the quantities are expressed in SI units\n",
"import math \n",
"p0 = 30.*101000.; # reservoir pressure\n",
"T0 = 3500.; # reservoir temperature\n",
"R = 520.; # specific gas constant\n",
"gam = 1.22; # ratio of specific heats\n",
"A_star = 0.4; # rocket nozzle throat area\n",
"\n",
"# the mass flow rate using the closed form analytical expression\n",
"# from problem 10.5 can be given as\n",
"m_dot = p0*A_star*math.sqrt(gam/R/T0*((2/(gam+1))**((gam+1)/(gam-1))));\n",
"\n",
"print\"The mass flow rate is: m_dot =\",m_dot, \"kg/s\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The mass flow rate is: m_dot = 586.100122081 kg/s\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example E06 : Pg 356"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# All the quantities are expressed in SI units\n",
"M = 2.; # Mach number\n",
"# for this value M, for a normal shock, from Appendix B\n",
"p0_ratio = 0.7209;\n",
"# thus\n",
"area_ratio = 1./p0_ratio;\n",
"print\"The diffuser throat to nozzle throat area ratio is: =\",area_ratio"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The diffuser throat to nozzle throat area ratio is: = 1.38715494521\n"
]
}
],
"prompt_number": 6
}
],
"metadata": {}
}
]
}
|
