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{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter10:IRREVERSIBILITY AND AVAILABILITY"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex10.1:pg-386"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#example 1\n",
"#Calculating reversible work\n",
"\n",
"#Form the Steam Tables,the inlet and the exit state properties are \n",
"hi=171.95 #initial specific heat of enthalpy in kJ/kg\n",
"si=0.5705 #initial specific entropy in kJ/kg-K\n",
"se=2.1341 #final specific entropy in kJ/kg-K\n",
"he=765.34 #final specific heat of enthalpy in kJ/kg-K\n",
"m=5 #mass flow rate of feedwater in kg/s\n",
"q1=900/m #heat added by one of the sources in kJ/kg\n",
"q2=he-hi-q1 #second heat transfer in kJ/kg\n",
"To=25+273.3 #Temp. of the surroundings in K\n",
"T1=100+273.2 #temp. of reservoir of one of the source in K\n",
"T2=200+273.2 #temp. of reservoir of second source in K\n",
"wrev=To*(se-si)-(he-hi)+q1*(1-To/T1)+q2*(1-To/T2) #reversible work in kJ/kg\n",
"print\"\\n Hence, the irreversibility is\",round(wrev,1),\"kJ/kg\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
" Hence, the irreversibility is 62.0 kJ/kg\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex10.2:pg-387"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#example 2\n",
"#Calculating reversible work\n",
"import math\n",
"#Form the Steam Tables,the inlet and the exit state properties are\n",
"hi=298.6 #initial specific heat of enthalpy in kJ/kg\n",
"si=6.8631 #initial specific entropy in kJ/kg-K\n",
"se=7.4664 #final specific entropy in kJ/kg-K\n",
"he=544.7 #final specific heat of enthalpy in kJ/kg-K\n",
"q=-50 #heat lost to surroundings in kJ/kg\n",
"w=hi-he+q #work in kJ/kg\n",
"To=25+273.2 #Temp. of the surroundings in K\n",
"P1=100 #Pressure of ambient air in kPa\n",
"P2=1000 #Final pressure of air after compression in kPa\n",
"R=0.287 #Universal gas constant in kJ/kg-K\n",
"wrev=To*(se-si-R*math.log(P2/P1))-(he-hi)+q*(1-To/To)#reversible work for the given change of state in kJ/kg\n",
"i=wrev-w #irreversibility in kJ/kg\n",
"print\"\\n Hence, the irreversibility is\",round(i,1),\"kJ/kg\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
" Hence, the irreversibility is 32.8 kJ/kg\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex10.3:pg-390"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#example 3\n",
"#Calculating reversible work and irreversibility\n",
"\n",
"#Form the Steam Tables at state 1\n",
"u1=1243.5 #initial specific internal energy in kJ/kg\n",
"s1=4.4819 #initial specific entropy in kJ/kg-K\n",
"v1=28.895 #initial specific volume in m^3/kg\n",
"v2=2*v1 #final specific volume in kg/m^3\n",
"u2=u1 #initial specific internal energy in kJ/kg\n",
"#These two independent properties, v2 and u2 , fix state 2.The final temp. is calculated by interplotation using the data for T2=5C and v2,x=0/3928 and u=948.5 kJ/kg. For T2=10C and v2, x=0.5433 and u=1317 kJ/kg\n",
"T2=9.1+273.2 #final temp. in K\n",
"x2=0.513 #quality in final state\n",
"s2=4.644 #final specific entropy in kJ/kg\n",
"V1=1 #volume of part of A in m^3\n",
"m=V1/v1 #mass flow rate in kg/s\n",
"To=20+273.2 #Room temperature in K\n",
"Wrev=To*m*(s2-s1) #reversible work in kJ\n",
"I=Wrev #irreversibility of the process\n",
"print\"\\n The irreversibility is \",round(I,3),\"kJ\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
" The irreversibility is 1.645 kJ\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "code",
"collapsed": false,
"input": [],
"language": "python",
"metadata": {},
"outputs": []
}
],
"metadata": {}
}
]
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