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{
"metadata": {
"name": ""
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 21 - Gas Compression"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1 - Pg 387"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the work done and the net heat transferred\n",
"#initialization of varaibles\n",
"import math\n",
"R=53.34\n",
"T1=540 #R\n",
"n=1.4\n",
"g=n\n",
"n2=1.3\n",
"P2=90. #psia\n",
"P1=15. #psia\n",
"cv=0.171\n",
"cp=0.24\n",
"#calculations\n",
"pv=R*T1\n",
"Wk=n*R*T1*(math.pow((P2/P1),((g-1)/g)) -1) /(n-1)\n",
"Wn=n2*R*T1*(math.pow((P2/P1),((n2-1)/n2)) -1) /(n2-1)\n",
"Wt=R*T1*math.log(P2/P1)\n",
"Q=cv*(n-n2)*778*T1*(math.pow((P2/P1),((n2-1)/n2)) -1) /(1-n2)*0.001305\n",
"#0.001305 is the conversion factor\n",
"#results\n",
"print '%s %d %s' %(\"\\n Work in case 1 =\",Wk,\"ft lb/lb\")\n",
"print '%s %d %s' %(\"\\n Work in case 2 =\",Wn,\"ft lb/lb\")\n",
"print '%s %d %s' %(\"\\n Work in case 3 =\",Wt,\"ft lb/lb\")\n",
"print '%s %.1f %s' %(\"\\n Heat transferred =\",Q,\"B/lb\")\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
" Work in case 1 = 67394 ft lb/lb\n",
"\n",
" Work in case 2 = 63914 ft lb/lb\n",
"\n",
" Work in case 3 = 51609 ft lb/lb\n",
"\n",
" Heat transferred = -16.0 B/lb\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2 - Pg 389"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Calculate the adiabatic and isothermal efficiencies\n",
"#initialization of varaibles\n",
"import math\n",
"R=53.34\n",
"T1=540 #R\n",
"n=1.4 #gamma\n",
"g=n\n",
"n2=1.3 #gamma\n",
"P2=90. #psia\n",
"P1=15. #psia\n",
"cv=0.171\n",
"#calculations\n",
"pv=R*T1\n",
"Wk=n*R*T1*(math.pow((P2/P1),((g-1)/g)) -1) /(n-1)\n",
"Wn=n2*R*T1*(math.pow((P2/P1),((n2-1)/n2)) -1) /(n2-1)\n",
"Wt=R*T1*math.log(P2/P1)\n",
"eta1=Wt/Wn\n",
"eta2=Wk/Wn\n",
"#results\n",
"print '%s %.2f' %(\"Adiabatic efficiency = \",eta2)\n",
"print '%s %.2f' %(\"\\n Isothermal efficiency = \",eta1)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Adiabatic efficiency = 1.05\n",
"\n",
" Isothermal efficiency = 0.81\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3 - Pg 389"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the heat transferred\n",
"#initialization of varaibles\n",
"import math\n",
"R=53.34\n",
"T1=540 #R\n",
"n=1.4 #gamma\n",
"g=n\n",
"n2=1.3 #gamma\n",
"P2=90 #psia\n",
"P1=15 #psia\n",
"cv=0.171\n",
"eta=0.95\n",
"cp=0.24\n",
"#calculations\n",
"pv=R*T1\n",
"Wk=n*R*T1*(math.pow((P2/P1),((g-1)/g)) -1) /(n-1)\n",
"Wn=n2*R*T1*(math.pow((P2/P1),((n2-1)/n2)) -1) /(n2-1)\n",
"Wt=R*T1*math.log(P2/P1)\n",
"Wx=-Wk/eta\n",
"dh=cp*T1*(1.52 - 1)\n",
"Q=dh+Wx/778.\n",
"#results\n",
"print '%s %.1f %s' %(\"Heat transferred =\",Q,\"B/lb\")\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Heat transferred = -23.8 B/lb\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4 - Pg 395"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the volumetric efficiency\n",
"#initialization of varaibles\n",
"import math\n",
"n=1.3\n",
"P1=15. #psia\n",
"P2=75. #psia\n",
"eta=0.5\n",
"eta2=0\n",
"#calculations\n",
"Pr=math.pow((P2/P1),(1/n))\n",
"Cl=(1-eta)/(Pr-1)\n",
"Cl2=(1-eta2)/(Pr-1)\n",
"#results\n",
"print '%s %.3f' %(\"For volumetric efficiency to be 0.5, Clearance = \",Cl)\n",
"print '%s %.3f' %(\"\\n For volumetric efficiency to be 0, Clearance = \",Cl2)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"For volumetric efficiency to be 0.5, Clearance = 0.204\n",
"\n",
" For volumetric efficiency to be 0, Clearance = 0.408\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5 - Pg 398"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Calculate the single stage and two stage efficiencies\n",
"#initialization of varaibles\n",
"import math\n",
"P1=5 #psia\n",
"P2=83.5 #psia\n",
"n=1.25\n",
"per=0.03\n",
"#calculations\n",
"nv1=1- per*(math.pow((P2/P1),(1/n)) -1)\n",
"nv2=1-per*(math.pow((math.sqrt(P2/P1)),(1/n)) -1)\n",
"#results\n",
"print '%s %.3f' %(\"For single stage machine = \",nv1)\n",
"print '%s %.3f' %(\"\\n For Two stage machine = \",nv2)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"For single stage machine = 0.745\n",
"\n",
" For Two stage machine = 0.937\n"
]
}
],
"prompt_number": 6
}
],
"metadata": {}
}
]
}
|
