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{
"metadata": {
"name": "",
"signature": "sha256:6ced269b2d83adfe9e452136249f918e5748b410340744d57a784dbf4aa451de"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 8 : Entropy - Available and Unavailable Energy"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 8.2 Page No : 211"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\t\t\t\n",
"# Variables\n",
"Q = 10. \t\t\t#kJ \t\t\t#heat transfered from reservoir\n",
"T = 100.+273 \t\t\t#K \t\t\t#isothermal expansion temperature\n",
"T_res = 300.+273 \t\t\t#K \t\t\t#reservoir temperature\n",
"\t\t\t\n",
"# Calculations and Results\n",
"delta_S_sys = (Q/T) \t\t\t#kJ/K \t\t\t#delta S for the system\n",
"print \"Change in entropyDelta S) for the system = %.2e kJ/K\"%(delta_S_sys);\n",
"\n",
"delta_S_res = -1*(Q/T_res) \t\t\t#kJ/K \t\t\t#delta S for the reservoir\n",
"print \"Change in entropyDelta S) for the reservoir = %.4e kJ/K\"%(delta_S_res);\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Change in entropyDelta S) for the system = 2.68e-02 kJ/K\n",
"Change in entropyDelta S) for the reservoir = -1.7452e-02 kJ/K\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 8.3 Page No : 212"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\t\t\t\n",
"# Variables\n",
"Q = 10. \t\t\t#kJ \t\t\t#heat transfered from reservoir\n",
"T = 100.+273 \t\t\t#K \t\t\t#isothermal expansion temperature\n",
"T_res = 100.+273 \t\t\t#K \t\t\t#reservoir temperature\n",
"\t\t\t\n",
"# Calculations and Results\n",
"delta_S_sys = (Q/T) \t\t\t#kJ/K \t\t\t#delta S for the system\n",
"print \"Change in entropyDelta S) for the system = %.2e kJ/K\"%(delta_S_sys)\n",
"\n",
"delta_S_res = -1*(Q/T_res) \t\t\t#kJ/K \t\t\t#delta S for the reservoir\n",
"print \"Change in entropyDelta S) for the reservoir = %.2e kJ/K\"%(delta_S_res);\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Change in entropyDelta S) for the system = 2.68e-02 kJ/K\n",
"Change in entropyDelta S) for the reservoir = -2.68e-02 kJ/K\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 8.4 Page No : 212"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\t\t\t\n",
"# Variables\n",
"Q = 1.; \t\t\t#kJ \t\t\t#heat transfered from reservoir\n",
"T = 100.+273; \t\t\t#K \t\t\t#isothermal expansion temperature\n",
"T_res = 100.+273; \t\t\t#K \t\t\t#reservoir temperature\n",
"\t\t\t\n",
"# Calculations and Results\n",
"delta_S_res = -1*(Q/T_res); \t\t\t#kJ/K \t\t\t#delta S for the reservoir\n",
"print \"Change in entropyDelta S) for the reservoir = %.2e kJ/K\"%(delta_S_res);\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Change in entropyDelta S) for the reservoir = -2.68e-03 kJ/K\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 8.12 Page No : 225"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"# Variables\n",
"pA = 120. \t\t\t#kPa \t\t\t#Pressure at location A\n",
"TA = 50.+273 \t\t\t#K \t\t\t#Temperature at location A\n",
"VA = 150. \t\t\t#m/s \t\t\t#Velocity at location A\n",
"\n",
"pB = 100. \t\t\t#kPa \t\t\t#Pressure at location B\n",
"TB = 30.+273 \t\t\t#K \t\t\t#Temperature at location B\n",
"VB = 250. \t\t\t#m/s \t\t\t#Velocity at location B\n",
"\n",
"Cp = 1.005 \t\t\t#kJ/kg\n",
"R = 0.287 \t\t\t#kJ/kgK\n",
"\t\t\t\n",
"# Calculations and Results\n",
"delta_S_sys = (Cp*math.log(TB/TA))-(R*math.log(pB/pA)) \t\t\t#kJ/kgK \t\t\t#Entropy of system\n",
"if delta_S_sys < 0 :\n",
" print \"Flow is from B to A.\";\n",
"else:\n",
" print \"Flow is from A to B.\" \n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Flow is from B to A.\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 8.13 Page No : 226"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"# Variables\n",
"mi = 5. \t\t\t#kg \t\t\t#mass of ice\n",
"Ti = 273. - 10 \t\t\t#K \t\t\t#Temperature of ice\n",
"ci = 2.1 \t\t\t#kJ/kgK \t\t\t#specific heat of ice\n",
"L = 330. \t\t\t#kJ/kg \t\t\t#Latent heat\n",
"mw = 20. \t\t\t#kg \t\t\t#mass of water\n",
"Tw = 273.+80 \t\t\t#K \t\t\t#Temperatur of water\n",
"cw = 4.2 \t\t\t#kJ/kgK \t\t\t#specific heat of water\n",
"\n",
"# calculatins and results\n",
"\n",
"#Part(a)\n",
"print \"Part a\";\n",
"Tmix = ((mi*ci*(Ti-273))-(L*mi)+(mw*cw*Tw)+(mi*cw*273))/(mw*cw+mi*cw)\n",
"print \"Temperature of the mixture when equilibrium is established between ice and water = %.f K\"%(Tmix)\n",
"#Part (b)\n",
"print \"Part b\";\n",
"delta_S_ice = mi*(ci*math.log(273/Ti)+L/273+cw*math.log(Tmix/273))\t\t\t#kJ/K \t\t\t#Entropy of ice\n",
"print \"Entropy of ice = %.2f kJ/K\"%(delta_S_ice)\n",
"#Part (c)\n",
"print \"Part c\";\n",
"delta_S_water = mw*(cw*math.log(Tmix/Tw))\t\t\t#kJ/K \t\t\t#Entropy of water\n",
"print \"Entropy of water = %.2f kJ/K\"%(delta_S_water)\n",
"#Part (d)\n",
"print \"Part d\";\n",
"delta_S_uni = delta_S_water+delta_S_ice\t\t\t#kJ/K \t\t\t#Entropy of universe\n",
"print \"Entropy of universe = %.2f kJ/K\"%(delta_S_uni)\n",
"\n",
"# note : rounding off error"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Part a\n",
"Temperature of the mixture when equilibrium is established between ice and water = 320 K\n",
"Part b\n",
"Entropy of ice = 9.79 kJ/K\n",
"Part c\n",
"Entropy of water = -8.17 kJ/K\n",
"Part d\n",
"Entropy of universe = 1.62 kJ/K\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 8.14 Page No : 230"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\t\t\t\n",
"# Variables\n",
"Q1 = 100. \t\t\t#kJ \t\t\t#Heat input\n",
"T0 = 300. \t\t\t#K \t\t\t#Surrounding temperature\n",
"\n",
"\t\t\t#Part(a)\n",
"print \"Part a\";\n",
"T1 = 1000. \t\t\t#K \t\t\t#reservoir temperature\n",
"print \"Avalable enery of 100 kJ of heat from a reservoir at 1000K = %.f kJ\"%(Q1*1-T0/T1)\n",
"print \"Unvalable enery of 100 kJ of heat from a reservoir at 1000K = %.1f kJ\"%(Q1*(1-(T0/T1)))\n",
"\t\t\t#Part(b)\n",
"print \"Part b\";\n",
"T1 = 600 \t\t\t#K \t\t\t#reservoir temperature\n",
"print \"Avalable enery of 100 kJ of heat from a reservoir at 1000K = %.f kJ\"%(Q1*1-T0/T1)\n",
"print \"Unvalable enery of 100 kJ of heat from a reservoir at 1000K = %.1f kJ\"%(Q1*(1-(T0/T1)))\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Part a\n",
"Avalable enery of 100 kJ of heat from a reservoir at 1000K = 100 kJ\n",
"Unvalable enery of 100 kJ of heat from a reservoir at 1000K = 70.0 kJ\n",
"Part b\n",
"Avalable enery of 100 kJ of heat from a reservoir at 1000K = 100 kJ\n",
"Unvalable enery of 100 kJ of heat from a reservoir at 1000K = 50.0 kJ\n"
]
}
],
"prompt_number": 14
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 8.15 Page No : 231"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\t\t\t\n",
"# Variables\n",
"T0 = 300. \t\t\t#K \t\t\t#Surrounding temperature\n",
"T1 = 1000. \t\t\t#K \t\t\t#Temperature of final reservoir\n",
"T2 = 600. \t\t\t#K \t\t\t#Temperature of intermediate reservoir\n",
"Q1 = 100. \t\t\t#kJ \t\t\t#Heat input\n",
"\t\t\t\n",
"# Calculations and Results\n",
"print \"Increase in unavaliable energy due to irreversible heat transfer = %.1f kJ\"%(Q1*(1-T0/T1)-Q1*(1-T0/T2))\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Increase in unavaliable energy due to irreversible heat transfer = 20.0 kJ\n"
]
}
],
"prompt_number": 16
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 8.16 Page No : 234"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\t\t\t\n",
"# Variables\n",
"T1 = 500. \t\t\t#K\n",
"T0 = 300. \t\t\t#K\n",
"T2 = 350. \t\t\t#K\n",
"W = 250. \t\t\t#kJ\n",
"Q1 = 1000. \t\t\t#kJ\n",
"\n",
"# Results\n",
"print \"Available energy = %.1f kJ\"%(((1-T0/T1))*Q1);\n",
"print \"Unavailable energy = %.1f kJ\"%(Q1 - (((1-T0/T1))*Q1));\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Available energy = 400.0 kJ\n",
"Unavailable energy = 600.0 kJ\n"
]
}
],
"prompt_number": 20
}
],
"metadata": {}
}
]
}
|
