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{
"metadata": {
"name": ""
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter No - 4 : Availability\n",
" "
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example No : 4.1 - Page No : 122"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from __future__ import division\n",
"# Given data\n",
"Q= 1000 # in kJ\n",
"T1= 1000 # in K\n",
"T2= 400 # in K\n",
"delta_Qsource= -Q/T1 # in kJ/K\n",
"delta_Qsystem= Q/T2 # in kJ/K\n",
"delta_Qnet=delta_Qsystem+delta_Qsource # in kJ/K\n",
"print \"The entropy production accompanying the heat transfer = %0.1f kJ/K \" %delta_Qnet\n",
"T0= 300 # in K\n",
"Q1= Q-T0*abs(delta_Qsource) # in kJ\n",
"Q2= Q-T0*abs(delta_Qsystem) # in kJ\n",
"LossOfEnergy= Q1-Q2 # in kJ\n",
"print \"The decrease in available energy after heat transfer = %0.0f kJ \" %LossOfEnergy"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The entropy production accompanying the heat transfer = 1.5 kJ/K \n",
"The decrease in available energy after heat transfer = 450 kJ \n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example No : 4.2 - Page No : 122"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from scipy.integrate import quad\n",
"# Given data\n",
"T1= 800+273 # in K\n",
"T2= 400+273 # in K\n",
"T3= 179+273 # in K\n",
"T0= 25+273 # in K\n",
"Q= 2018.4 # heat taken by water in kJ/kg\n",
"# Formula mCp*(T1-T2)= Q\n",
"mCp= Q/(T1-T2)\n",
"def integrand(t):\n",
" return 1/t\n",
"ans, err = quad(integrand, T1, T2)\n",
"delta_Qgas= mCp*ans # in kJ/K\n",
"delta_Qwater= Q/T3 # in kJ/K\n",
"delta_Qnet= delta_Qwater+delta_Qgas # in kJ/K\n",
"print \"Net entropy changes = %0.3f kJ/K \" %delta_Qnet\n",
"E1= T0*abs(delta_Qgas) # Original unavailable energy in kJ\n",
"E2= T0*delta_Qwater #unavailable energy after heat transfer in kJ\n",
"E= E2-E1 # in increase in unavailable energy in kJ\n",
"print \"The increase in unavailable energy = %0.2f kJ \" %E\n",
"\n",
"# Note: In the book, the calculation is not accurate"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Net entropy changes = 2.112 kJ/K \n",
"The increase in unavailable energy = 629.28 kJ \n"
]
}
],
"prompt_number": 2
}
],
"metadata": {}
}
]
}
|
