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{
"metadata": {
"name": ""
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 7 : Thermodynamic Relations"
]
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 7.17 Page no : 370"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"'''\n",
"Determine the following :\n",
"(i) Work done on the copper during the process,\n",
"(ii) Change in entropy,\n",
"(iii) The heat transfer,\n",
"(iv) Change in internal energy, and\n",
"(v) (c p \u2013 c v ) for this change of state.\n",
"'''\n",
"\n",
"# Variables\n",
"B = 5.*10**(-5); \t\t\t# /K\n",
"K = 8.6*10**(-12); \t\t\t# m**2/N\n",
"v = 0.114*10**(-3); \t\t\t#m**3/kg\n",
"p2 = 800.*10**5; \t\t\t#Pa\n",
"p1 = 20.*10**5; \t\t\t#Pa\n",
"T = 288.; \t\t\t#K\n",
"\n",
"# Calculations and Results\n",
"W = -v*K/2*(p2**2-p1**2);\n",
"print (\"(i) Work done on the copper = %.3f\")%(W),(\"J/kg\")\n",
"\n",
"ds = -v*B*(p2-p1);\n",
"print (\"(ii) Change in entropy = %.3f\")% (ds), (\"J/kg K\")\n",
"\n",
"Q = T*ds;\n",
"print (\"(iii) The heat transfer = %.3f\")%(Q), (\"J/kg\")\n",
"\n",
"du = Q-W;\n",
"print (\"(iv) Change in internal energy = %.3f\")%(du),(\"J/kg\")\n",
"\n",
"R = B**2*T*v/K;\n",
"print (\"(v) cp \u2013 cv = %.3f\")%(R),(\"J/kg K\")\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(i) Work done on the copper = -3.135 J/kg\n",
"(ii) Change in entropy = -0.445 J/kg K\n",
"(iii) The heat transfer = -128.045 J/kg\n",
"(iv) Change in internal energy = -124.909 J/kg\n",
"(v) cp \u2013 cv = 9.544 J/kg K\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 7.18 Page no : 371"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"'''\n",
"Using Clausius-Claperyon\u2019s equation, find enthalpy of vapourisation.\n",
"'''\n",
"\n",
"# Variables\n",
"vg = 0.1274; \t\t\t#m**3/kg\n",
"vf = 0.001157; \t\t\t#m**3/kg\n",
"# dp/dT = 32; \t\t\t#kPa/K\n",
"T3 = 473; \t\t\t#K\n",
"\n",
"# Calculations\n",
"h_fg = 32*10**3*T3*(vg-vf)/10**3;\n",
"\n",
"# Results\n",
"print (\" enthalpy of vapourisation = %.3f\")%(h_fg),(\"kJ/kg\")\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" enthalpy of vapourisation = 1910.814 kJ/kg\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 7.19 Page no : 372"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"'''\n",
"Determine the pressure an ice skate blade must exert to allow smooth ice skate at \u2013 10\u00b0C.\n",
"'''\n",
"\n",
"import math\n",
"\n",
"# Variables\n",
"h_fg = 334.; \t\t\t#kJ/kg\n",
"v_liq = 1.; \t\t\t#m**3/kg\n",
"v_ice = 1.01; \t\t\t#m**3/kg\n",
"T1 = 273.; \t\t\t#K\n",
"T2 = 263.; \t\t\t#K\n",
"p1 = 1.013*10**5; \t\t\t#Pa\n",
"\n",
"# Calculations\n",
"p2 = (p1+h_fg*10**3/(v_ice-v_liq)*math.log(T1/T2))/10**5;\n",
"\n",
"# Results\n",
"print (\"pressure = %.3f\")%(p2),(\"bar\")\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"pressure = 13.477 bar\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 7.20 Page no : 372"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"'''\n",
"Calculate the specific volume v g of saturation mercury vapour at 0.1 bar\n",
"'''\n",
"\n",
"# Variables\n",
"h_fg = 294.54; \t\t\t#kJ/kg\n",
"p = 0.1; \t\t\t#bar\n",
"T = 523; \t\t\t#K\n",
"\n",
"# Calculations\n",
"vg = h_fg*10**3/T/(2.302*3276.6*p*10**5/T**2 - 0.652*p*10**5/T);\n",
"\n",
"# Results\n",
"print (\"specific volume = %.3f\")%(vg),(\"m**3/kg\")\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"specific volume = 2.139 m**3/kg\n"
]
}
],
"prompt_number": 3
}
],
"metadata": {}
}
]
}
|
