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"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 21 : Energy Terminology Concepts and Units"
]
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 21.1 Page no : 616"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"from scipy.integrate import quad\n",
"\n",
"# Variables\n",
"V1 = 0.1 ;\t\t\t# Volume of gas initially -[cubic metres]\n",
"V2 = 0.2 ;\t\t\t# Volume of gas finally -[cubic metres]\n",
"T1 = 300 ;\t\t\t# Temperature of gas initially -[K]\n",
"P1 = 200 ;\t\t\t# Pressure of gas finally -[kPa]\n",
"R = 8.314 ;\t\t\t# Universal gas constant \n",
"n = (P1*V1)/(T1*R) ;\t\t\t# Moles of gas taken-[kg mol]\n",
"#You are asked to calculate work by eqn. 21.1 , but you do not know the F(force) exerted by gas , so write F = P.A, multiply divide A and eqn 21.1 reduces to W= integate(P.dv)\n",
"\n",
"# Calculations and Results\n",
"# Isobaric process see fig E21.1b to see the path followed\n",
"def f(V):\n",
" return -(P1)\n",
"W= quad(f,V1,V2)[0] ;\t\t\t# Work done by gas on piston -[kJ]\n",
"print ' (a)Work done by gas on piston for isobaric process is %.0f kJ . ',W\n",
"\n",
"def f1(V):\n",
" return -(T1*R*n/V)\n",
"W= quad(f1,V1,V2)[0] ;\t\t\t# Work done by gas on piston -[kJ]\n",
"print '(b)Work done by gas on piston for isothermal process is %.2f kJ . ',W\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" (a)Work done by gas on piston for isobaric process is %.0f kJ . -20.0\n",
"(b)Work done by gas on piston for isothermal process is %.2f kJ . -13.8629436112\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 21.2 page no. 624\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"# Variables\n",
"id_ = 3. ;\t\t\t# Internal diameter of tube-[cm]\n",
"Vf = 0.001 ;\t\t\t# Volume flow rate of water in tube-[cubic meter/s]\n",
"rho = 1000. ;\t\t\t# Assumed density of water-[kg/cubic meter] \n",
"\n",
"# Calculations\n",
"rad = id_/2. ;\t\t\t# Radius of tube -[ cm]\n",
"a = 3.14*rad**2 ;\t\t\t# Area of flow of tube -[squqre centimeter]\n",
"v = Vf*(100)**2/a ;\t\t\t# Velocity of water in tube - [m/s]\n",
"KE = v**2/2. ;\t\t\t# Specific(mass=1kg) kinetic energy of water in tube -[J/kg]\n",
"\n",
"# Results\n",
"print 'Specific kinetic energy of water in tube is %.2f J/kg . '%KE\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Specific kinetic energy of water in tube is 1.00 J/kg . \n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
" Example 21.3 page no. 626\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"# Variables\n",
"# Let water level in first reservoir be the reference plane\n",
"h = 40. ;\t\t\t# Difference of water-[ft]\n",
"g = 32.2 ;\t\t\t# acceleration due to gravity-[ft/square second]\n",
"\n",
"# Calculations\n",
"PE=g*h/(32.2*778.2) ;\t\t\t#\t\t\t# Specific(mass=1kg) potential energy of water -[Btu/lbm]\n",
"\n",
"# Results\n",
"print 'Specific potential energy of water is %.4f Btu/lbm . '%PE\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Specific potential energy of water is 0.0514 Btu/lbm . \n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 21.4 page no : 629"
]
},
{
"cell_type": "code",
"collapsed": true,
"input": [
"\n",
"\n",
"# Variables\n",
"#Constant volume process \n",
"mol_air = 10. ;\t\t\t# Moles of air-[kg mol]\n",
"T1 = 60.+273 ;\t\t\t# Initial temperature of air-[K]\n",
"T2 = 30.+273 ;\t\t\t# final temperature of air-[K]\n",
"# Additional data needed\n",
"Cv = 2.1*10.**4 ; \t\t\t# Specific heat capacity of air at constant volume-[J/(kg mol*C)]\n",
"\n",
"# Calculations\n",
"def f(T):\n",
" return mol_air*Cv\n",
"del_U = quad(f,T1,T2)[0] ;\t\t\t#Change in internal energy-[J]\n",
"\n",
"# Results\n",
"print 'Change in internal energy is %.1e J . '%del_U\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Change in internal energy is -6.3e+06 J . \n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 21.7 page no : 633"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Variables\n",
"#Constant pressure process \n",
"mol_air = 10. ;\t\t\t# Moles of air-[kg mol]\n",
"T1 = 60+273 ;\t\t\t# Initial temperature of air-[K]\n",
"T2 = 30+273 ;\t\t\t# final temperature of air-[K]\n",
"\n",
"# Calculations\n",
"# Additional data needed\n",
"Cp = 2.9*10**4 ;\t\t\t# Specific heat capacity of air at constant pressure-[J/(kg mol*C)]\n",
"# Use eqn. 21.11 for del_H\n",
"def f(T):\n",
" return mol_air*Cp\n",
"\n",
"del_H = quad(f,T1,T2)[0] ;\t\t\t#Change in enthalpy-[J]\n",
"print 'Change in enthalpy is %.1e J . '%del_H\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Change in enthalpy is -8.7e+06 J . \n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "code",
"collapsed": false,
"input": [],
"language": "python",
"metadata": {},
"outputs": []
}
],
"metadata": {}
}
]
}