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{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Chapter 06:Second Law of Thermodynamics"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex6.1:pg-138"
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"\n",
" Example 6.1\n",
"\n",
" Least rate of heat rejection is 0 kW\n"
]
}
],
"source": [
"import math\n",
"T1 = 800 # Source temperature in degree Celsius\n",
"\n",
"T2 = 30 # Sink temperature in degree Celsius\n",
"\n",
"e_max = 1-((T2+273)/(T1+273)) # maximum possible efficiency \n",
"\n",
"Wnet = 1 # in kW\n",
"\n",
"Q1 = Wnet/e_max # Least rate of heat required in kJ/s\n",
"\n",
"Q2 = Q1-Wnet # Least rate of heat rejection kJ/s\n",
"\n",
"\n",
"\n",
"print \"\\n Example 6.1\"\n",
"\n",
"print \"\\n Least rate of heat rejection is \",Q2,\" kW\"\n",
"\n",
"#The answers vary due to round off error\n",
"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex6.2:pg-139"
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"\n",
" Example 6.2\n",
"\n",
" Least Power necessary to pump the heat out is 0.31 kW\n"
]
}
],
"source": [
"import math\n",
"T1 = -15 # Source temperature in degree Celsius\n",
"\n",
"T2 = 30 # Sink temperature in degree Celsius\n",
"\n",
"Q2 = 1.75 # in kJ/sec\n",
"\n",
"print \"\\n Example 6.2\"\n",
"\n",
"W= Q2*((T2+273)-(T1+273))/(T1+273) # Least Power necessary to pump the heat out\n",
"\n",
"print \"\\n Least Power necessary to pump the heat out is \",round(W,2),\"kW\"\n",
" \n",
" #The answers vary due to round off error\n",
" \n",
" "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex6.3:pg-140"
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"\n",
" Example 6.3\n",
"\n",
"\n",
" Part A:\n",
"\n",
" The heat transfer to refrigerant is 0.0 kJ\n",
"\n",
" The heat rejection to the 40 degree reservoir is 8200.0 kJ\n",
"\n",
"\n",
" Part B:\n",
"\n",
" The heat transfer to refrigerant is 1200.0 kJ\n",
"\n",
" The heat rejection to the 40 degree reservoir is 2344.0 kJ\n"
]
}
],
"source": [
"import math\n",
"#Given \n",
"\n",
"T1 = 600 # Source temperature of heat engine in degree Celsius\n",
"\n",
"T2 = 40 # Sink temperature of heat engine in degree Celsius \n",
"\n",
"T3 = -20 # Source temperature of refrigerator in degree Celsius\n",
"\n",
"Q1 = 2000 # Heat transfer to heat engine in kJ\n",
"\n",
"W = 360 # Net work output of plant in kJ\n",
"\n",
"# Part (a)\n",
"\n",
"e_max = 1.0-((T2+273)/(T1+273)) # maximum efficiency \n",
"\n",
"W1 = e_max*Q1 # maximum work output \n",
"\n",
"COP = (T3+273)/((T2-273)-(T3-273)) # coefficient of performance of refrigerator\n",
"\n",
"W2 = W1-W # work done to drive refrigerator \n",
"\n",
"Q4 = COP*W2 # Heat extracted by refrigerator\n",
"\n",
"Q3 = Q4+W2 # Heat rejected by refrigerator\n",
"\n",
"Q2 = Q1-W1 # Heat rejected by heat engine\n",
"\n",
"Qt = Q2+Q3 # combined heat rejection by heat engine and refrigerator \n",
"\n",
"print \"\\n Example 6.3\"\n",
"\n",
"print \"\\n\\n Part A:\"\n",
"\n",
"print \"\\n The heat transfer to refrigerant is \",round(Q2,3) ,\" kJ\"\n",
"\n",
"print \"\\n The heat rejection to the 40 degree reservoir is \",round(Qt,3) ,\" kJ\"\n",
"\n",
"\n",
"\n",
"# Part (b)\n",
"\n",
"print \"\\n\\n Part B:\"\n",
"\n",
"e_max_ = 0.4*e_max # maximum efficiency\n",
"\n",
"W1_ = e_max_*Q1 # maximum work output \n",
"\n",
"W2_ = W1_-W # work done to drive refrigerator \n",
"\n",
"COP_ = 0.4*COP # coefficient of performance of refrigerator\n",
"\n",
"Q4_ = COP_*W2_ # Heat extracted by refrigerator\n",
"\n",
"Q3_ = Q4_+W2_ # Heat rejected by refrigerator\n",
"\n",
"Q2_ = Q1-W1_ # Heat rejected by heat engine\n",
"\n",
"QT = Q2_+Q3_# combined heat rejection by heat engine and refrigerator \n",
"\n",
"print \"\\n The heat transfer to refrigerant is \",round(Q2_,3) ,\" kJ\"\n",
"\n",
"print \"\\n The heat rejection to the 40 degree reservoir is \",round(QT,3) ,\" kJ\"\n",
"\n",
"#The answers vary due to round off error\n",
"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex6.5:pg-142"
]
},
{
"cell_type": "code",
"execution_count": 4,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"\n",
" Example 6.5\n",
"\n",
" The multiplication factor is 6\n"
]
}
],
"source": [
"import math\n",
"T1 = 473 # Boiler temperature in K\n",
"\n",
"T2 = 293 # Home temperature in K\n",
"\n",
"T3 = 273 # Outside temperature in K\n",
"\n",
"print \"\\n Example 6.5\"\n",
"\n",
"MF = (T2*(T1-T3))/(T1*(T2-T3)) \n",
"\n",
"print \"\\n The multiplication factor is \",MF \n",
"\n",
"#The answers vary due to round off error\n",
"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex6.6:pg-144"
]
},
{
"cell_type": "code",
"execution_count": 5,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"\n",
" Example 6.6\n",
"\n",
" Minimum area required for the collector plate is 10.0 m**2\n"
]
}
],
"source": [
"import math\n",
"T1 = 90.0 # Operating temperature of power plant in degree Celsius \n",
"\n",
"T2 = 20.0 # Atmospheric temperature in degree Celsius\n",
"\n",
"W = 1.0 # Power production from power plant in kW\n",
"\n",
"E = 1880 # Capability of energy collection in kJ/m**2 h\n",
"\n",
"\n",
"\n",
"print \"\\n Example 6.6\"\n",
"\n",
"e_max = 1.0-((T2+273.0)/(T1+273.0)) # maximum efficiency\n",
"\n",
"Qmin = W/e_max # Minimum heat requirement per second\n",
"\n",
"Qmin_ = Qmin*3600.0 # Minimum heat requirement per hour\n",
"\n",
"Amin = Qmin_/E # Minimum area requirement\n",
"\n",
"print \"\\n Minimum area required for the collector plate is \",math. ceil(Amin) ,\" m**2\"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex6.7:pg-144"
]
},
{
"cell_type": "code",
"execution_count": 6,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"\n",
" Example 6.7\n",
"\n",
" Area of the panel 0.167221895617 m**2\n"
]
}
],
"source": [
"import math\n",
"T1 = 1000 # Temperature of hot reservoir in K\n",
"\n",
"W = 1000 # Power requirement in kW\n",
"\n",
"K = 5.67e-08 # constant \n",
"\n",
"print \"\\n Example 6.7\"\n",
"\n",
"Amin = (256*W)/(27*K*T1**4) # minimum area required\n",
"\n",
"print \"\\n Area of the panel \",Amin ,\" m**2\"\n"
]
}
],
"metadata": {
"kernelspec": {
"display_name": "Python 2",
"language": "python",
"name": "python2"
},
"language_info": {
"codemirror_mode": {
"name": "ipython",
"version": 2
},
"file_extension": ".py",
"mimetype": "text/x-python",
"name": "python",
"nbconvert_exporter": "python",
"pygments_lexer": "ipython2",
"version": "2.7.11"
}
},
"nbformat": 4,
"nbformat_minor": 0
}
|
