{
"metadata": {
"name": ""
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 5 : Second law of thermodynamics and its applications"
]
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 5.2 Page No : 161"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"import math \n",
"\n",
"# Variables\n",
"Q = 1000.;\t\t \t #amount of energy absorbed by the heat engine in kJ/s\n",
"W = 650.;\t\t\t #work delivered by the heat engine in kW\n",
"T_source = 500. \t\t #temperature of the source in degree celsius\n",
"T_sink = 25.\t\t\t #temperature of the sink in degree celsius\n",
"\n",
"# Calculations\n",
"n_claimed = W/Q\n",
"T1 = T_source+273.15\n",
"T2 = T_sink+273.15\n",
"n_carnot = 1-(T2/T1)\n",
"\n",
"# Results\n",
"print \" The efficiency of the Carnot engine = %0.3f \"%(n_carnot);\n",
"print \" The efficiency of the engine claimed by the inventor = %0.2f \"%(n_claimed);\n",
"if n_claimed0 or S_G == 0 :\n",
" print \" As the first and second law of thermodynamics are satisfied, the device is theoretically feasible \"\n",
"else:\n",
" print \" As both the first and second law or either the first or second law of thermodynamics \\\n",
" are not satisfied, the device is not feasible \"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" The LHS of the equation applied to the flow device to check if the first law of thermodynamics is satisfied = 2745.4 kJ\n",
" The RHS of the equation applied to the flow device to check if the first law of thermodynamics is satisfied = 2745.5 kJ\n",
" The entropy generated by applying the second law of thermodynamics to the flow device = 0.3552 kJ/kgK\n",
" As the first and second law of thermodynamics are satisfied, the device is theoretically feasible \n"
]
}
],
"prompt_number": 21
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 5.18 Page No : 185"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Variables\n",
"Pi = 30.\t\t\t #pressure of superheated steam entering the turbine in bar\n",
"Ti = 300.\t\t\t #temperature of superheated steam entering the turbine in degree celsius\n",
"Pe = 0.1\t\t\t #pressure at which steam exits the turbine in bar\n",
"Xe = 0.9\t\t\t #quality of steam at the exit (no unit)(for the actual turbine)\n",
"\n",
"# Calculations\n",
"#For superheated steam at Pi and Ti\n",
"hi = 2995.1;\t\t\t #enthalpy of superheated steam at the entrance in kJ/kg\n",
"si = 6.5422;\t\t\t #entropy of superheated steam at the entrance in kJ/kgK\n",
"\n",
"#For steam at Pe\n",
"hf = 191.83;\t\t\t #enthalpy of saturated liquid in kJ/kg\n",
"hg = 2584.8;\t\t\t #enthalpy of saturated vapour in kJ/kg\n",
"sf = 0.6493;\t\t\t #entropy of saturated liquid in kJ/kgK\n",
"sg = 8.1511;\t\t\t #entropy of saturated vapour in kJ/kgK\n",
"\n",
"X2 = (si-sf)/(sg-sf)\n",
"h2 = (hf*(1-X2))+(X2*hg)\n",
"he = (hf*(1-Xe))+(Xe*hg)\n",
"n_T = (hi-he)/(hi-h2)\n",
"\n",
"# Results\n",
"print \" The isentropic efficiency of the turbine = %f \"%(n_T);\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" The isentropic efficiency of the turbine = 0.703395 \n"
]
}
],
"prompt_number": 22
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 5.19 Page No : 186"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Variables\n",
"Ti = 25.\t\t\t #temperature of air taken in by the adiabatic air compressor in degree celsius\n",
"Pi = 0.1\t\t\t #pressure of air taken in by the adiabatic air compressor in MPa\n",
"Pe = 1.\t \t\t #discharge pressure of air in MPa\n",
"n_c = 0.8\t\t\t #isentropic efficiency of the compressor (no unit)\n",
"gaamma = 1.4\t\t #ratio of molar specific heat capacities (no unit)\n",
"R = 8.314\t\t\t #universal gas constant in J/molK\n",
"\n",
"# Calculations\n",
"Ti = Ti+273.15\n",
"Te = Ti*(((Pe*10**6)/(Pi*10**6))**((gaamma-1)/gaamma))\n",
"W_s = (((R*gaamma)/(gaamma-1))*(Te-Ti))*10**-3;\t\t\t\n",
"Ws = W_s/n_c\n",
"Te_actual = ((Ws*10**3*(gaamma-1))/(R*gaamma))+Ti\n",
"\n",
"# Results\n",
"print \" The exit temperature of air = %0.2f K\"%(Te_actual);\n",
"print \" The power consumed by the compressor = %f kW/mol\"%(Ws);"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" The exit temperature of air = 645.01 K\n",
" The power consumed by the compressor = 10.093262 kW/mol\n"
]
}
],
"prompt_number": 23
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 5.20 Page No : 187"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Variables\n",
"Ti = 30.\t\t\t #temperature of saturated liquid water in degree celsius\n",
"m = 500.\t\t\t #mass flow rate of water being pumped in kg/s\n",
"P2 = 3. \t\t\t #preesure maintained in the boiler in MPa\n",
"n_p = 0.75;\t\t\t #isentropic efficiency of the pump (no unit)\n",
"\n",
"# Calculations\n",
"#For saturated liquid water at Ti\n",
"vf = 0.0010043\n",
"P1 = 4.241;\t\t\n",
"\n",
"Ws_m = (vf*((P2*10**6)-(P1*10**3)))*10**-3\n",
"Ws_act_m = Ws_m/n_p;\t\t\t \n",
"P = ((Ws_act_m*10**3)*m)*10**-6;\n",
"\n",
"# Results\n",
"print \" The power consumed by the pump = %d MW\"%(P);"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" The power consumed by the pump = 2 MW\n"
]
}
],
"prompt_number": 24
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 5.21 Page No : 188"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Variables\n",
"Pi = 3.\t\t\t #pressure of dry saturated steam entering the nozzle in bar\n",
"Xe = 0.98\t\t #quality of steam exiting the nozzle (no unit)\n",
"Pe = 2.\t\t\t #pressure of steam exiting the nozzle in bar\n",
"\n",
"# Calculations\n",
"#For steam at Pi\n",
"hi = 2724.7\n",
"he = 2652.8\n",
"V2_2_s = hi-he\n",
"\n",
"#For steam at Pe\n",
"hf = 504.70\n",
"hg = 2706.3\n",
"he_act = ((1-Xe)*hf)+(Xe*hg)\n",
"V2_2 = hi-he_act;\t\t\t\n",
"n_N = (V2_2)/(V2_2_s)\t\t\n",
"\n",
"# Results\n",
"print \" The isentropic efficiency of the nozzle = %0.3f \"%(n_N);\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" The isentropic efficiency of the nozzle = 0.868 \n"
]
}
],
"prompt_number": 25
}
],
"metadata": {}
}
]
}