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{
"metadata": {
"name": "",
"signature": "sha256:efb3cb597d2918b83a867fe2c379b3952acba398aec14a9edf762b5c40043583"
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"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 8 - Second and third law topics"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1 - Pg 125"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the value of dp/ds at constant volume\n",
"#Initialization of variables\n",
"P=500 #psia\n",
"T=700. #F\n",
"J=778.\n",
"#calculations\n",
"dpds=1490 *144/J\n",
"#results\n",
"print '%s %d %s' %(\"dp by ds at constant volume =\",dpds,\" F/ft^3/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": [
"dp by ds at constant volume = 275 F/ft^3/lbm\n",
"The answer is a bit different due to rounding off error in textbook\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2 - Pg 131"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the thermal efficiency\n",
"#Initialization of variables\n",
"import math\n",
"cp=0.25 #Btu/lbm R\n",
"T0=520. #R\n",
"T1=3460. #R\n",
"#calculations\n",
"dq=cp*(T0-T1)\n",
"ds=cp*math.log(T0/T1)\n",
"dE=dq-T0*ds\n",
"eta=dE/dq*100.\n",
"#results\n",
"print '%s %.1f %s' %(\"Thermal efficiency =\",eta,\"percent\")\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Thermal efficiency = 66.5 percent\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3 - Pg 134"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the Loss of available energy\n",
"#Initialization of variables\n",
"import math\n",
"cp=0.25 #Btu/lbm R\n",
"T0=520. #R\n",
"T1=3460. #R\n",
"dG=21069. #Btu/lbm\n",
"dH=21502. #Btu/lbm\n",
"#calculations\n",
"dq=cp*(T0-T1)\n",
"ds=cp*math.log(T0/T1)\n",
"dE=dq-T0*ds\n",
"eta=dE/dq\n",
"dw=eta*dH\n",
"de=-dG+dw\n",
"#results\n",
"print '%s %d %s' %(\"Loss of available energy =\",de,\"Btu/lbm\")\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Loss of available energy = -6774 Btu/lbm\n"
]
}
],
"prompt_number": 3
}
],
"metadata": {}
}
]
}
|
