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{
"metadata": {
"name": "",
"signature": "sha256:15d0388b2c544d629a9ec990a9a36b6e26ea291bc7812312e5319e0c4bde1ab8"
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"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 7 - The second law"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2 - Pg 112"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the entropy in parts ab,ac and efficiency\n",
"#Initialization of variables\n",
"import math\n",
"cv=0.175 #Btu/lbm R\n",
"R0=1.986\n",
"M=29\n",
"T2=1040 #R\n",
"T1=520 #R\n",
"#calculations\n",
"cp=cv+R0/M\n",
"sab=cv*math.log(T2/T1)\n",
"sac=cp*math.log(T2/T1)\n",
"dqab=cv*(T2-T1)\n",
"dqca=cp*(T1-T2)\n",
"dqrev=T2*(sac-sab)\n",
"eta=(dqab+dqrev+dqca)/(dqab+dqrev)*100\n",
"#results\n",
"print '%s %.4f %s' %(\"Entropy in ab part =\",sab,\"Btu/lbm R\")\n",
"print '%s %.4f %s' %(\"\\n Entropy in ac part =\",sac,\" Btu/lbm R\")\n",
"print '%s %.2f %s' %(\"\\n Efficiency =\",eta,\" percent\")\n",
"print '%s' %(\"The answers are a bit different due to rounding off error in textbook\")\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Entropy in ab part = 0.1213 Btu/lbm R\n",
"\n",
" Entropy in ac part = 0.1688 Btu/lbm R\n",
"\n",
" Efficiency = 9.80 percent\n",
"The answers are a bit different due to rounding off error in textbook\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3 - Pg 115"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the change in entropy of the process\n",
"#Initialization of variables\n",
"import math\n",
"tc=32. #F\n",
"th=80. #F\n",
"mw=5 #lbm\n",
"mi=1 #lbm\n",
"P=14.7 #psia\n",
"cp=1\n",
"#calculations\n",
"t= (-144*mi+tc*mi+th*mw)/(mw+mi)\n",
"ds1=144/(tc+460)\n",
"ds2=cp*math.log((460+t)/(460+tc))\n",
"dsice=ds1+ds2\n",
"dswater=mw*cp*math.log((t+460)/(460+th))\n",
"ds=dsice+dswater\n",
"#results\n",
"print '%s %.4f %s' %(\"Change in entropy of the process =\",ds,\"Btu/R\")\n",
"print '%s' %(\"The answer is a bit different due to rounding off error in textbook\")\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Change in entropy of the process = 0.0192 Btu/R\n",
"The answer is a bit different due to rounding off error in textbook\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4 - Pg 119"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the change in available energy\n",
"#Initialization of variables\n",
"import math\n",
"cp=1\n",
"T2=60. #F\n",
"T1=100. #F\n",
"ta=32. #F\n",
"#calculations\n",
"dq=cp*(T2-T1)\n",
"ds=cp*math.log((460+T2)/(460+T1))\n",
"dE=dq-ds*(ta+460)\n",
"#results\n",
"print '%s %.1f %s' %(\"Change in available energy =\",dE,\"Btu/lbm\")\n",
"print '%s' %(\"The answer is a bit different due to rounding off error in textbook\")\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Change in available energy = -3.5 Btu/lbm\n",
"The answer is a bit different due to rounding off error in textbook\n"
]
}
],
"prompt_number": 3
}
],
"metadata": {}
}
]
}
|
