{
"metadata": {
"name": "",
"signature": "sha256:eafe81375ae6bd6abb8d4dff5b78c1d88df56b904362e7f55dc57c603062c8b9"
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"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 4 : First Law of Thermodynamics"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.1 Page No : 78"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Variables\n",
"V1 = 0.3; \t\t\t# Initial volume in m3\n",
"V2 = 0.15; \t\t\t# Final volume in m3\n",
"P = 0.105e06; \t\t\t# Pressure in Pa\n",
"Q = -37.6e03; \t\t\t# Heat tranferred in J\n",
"\n",
"# Calculation\n",
"W = P*(V2-V1); \t\t\t# Work done\n",
"U = Q-W; \t\t\t# Internal energy change\n",
"\n",
"# Results\n",
"print \"Change in the internal energy of the system is %.2f kJ\"%(U/1000)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Change in the internal energy of the system is -21.85 kJ\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.2 Page No : 78"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Variables\n",
"Qacb = 84e03;\n",
"Wacb = 32e03;\n",
"Uba = Qacb-Wacb; \t\t\t# Ub-Ua\n",
"\n",
"# Calculation and Results\n",
"# Part (a)\n",
"Wadb = 10.5e03; \n",
"Qadb = Uba+Wadb; \n",
"print \"The heat flow into the system along the path adb\",(Qadb/1000),\"kJ\"\n",
"\n",
"# Part (b)\n",
"Wb_a = -21e03;\n",
"Uab = - Uba;\n",
"Qb_a = Uab+Wb_a;\n",
"print \"The heat liberated along the path b-a is\",round(Qb_a/1000),\"kJ\"\n",
"\n",
"# Part (c)\n",
"Wdb = 0.; \t\t\t# Constant volume\n",
"Wad = 10.4e03; \n",
"Wadb = Wdb-Wad; \n",
"Ud = 42e03;\n",
"Ua = 0.;\n",
"Qad = Ud-Ua+Wad;\n",
"Qdb = Qadb-Qad; \n",
"print \"The heat absorbed in the path ad and db are\",round(Qdb/1000),\"kJ\",\"and\",round(Qad/1000),\"kJ\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The heat flow into the system along the path adb 62.5 kJ\n",
"The heat liberated along the path b-a is -73.0 kJ\n",
"The heat absorbed in the path ad and db are 10.0 kJ and 52.0 kJ\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.3 Page No : 79"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Variables\n",
"# Process a-b\n",
"Qab = 0;\n",
"Wab = 2170; \t\t\t# in KJ/min\n",
"Eab = Qab-Wab; \n",
"\n",
"# Process b-c\n",
"Qbc = 21000;\n",
"Wbc = 0; \n",
"Ebc = Qbc-Wbc;\n",
"\n",
"# Process c-d\n",
"Qcd = -2100;\n",
"Ecd = -36600; \n",
"Wcd = Qcd-Ecd;\n",
"\n",
"# Calculation\n",
"# Process d-a\n",
"Q = -17000; \t\t\t# Total heat transfer\n",
"Qda = Q-Qab-Qbc-Qcd; \n",
"Eda = -Eab-Ebc-Ecd;\n",
"Wda = Qda-Eda;\n",
"M = [[Qab, Wab, Eab],[Qbc, Wbc ,Ebc],[Qcd, Wcd, Ecd],[Qda, Wda, Eda]];\n",
"process = [\"a-b\",\"b-c\",\"c-d\",\"d-a\"]\n",
"# Results\n",
"print \"The completed table is\"\n",
"print \" process Q W deltaE\"\n",
"for i in range(4):\n",
" print \"%10s\"%process[i],\n",
" for j in range(3):\n",
" print \"%10d\"%M[i][j],\n",
"\n",
" print \"\"\n",
"print \"\\nRate of work output : %.f kJ/min\"%(sum([M[0][0],M[1][0],M[2][0],M[3][0]]))\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The completed table is\n",
" process Q W deltaE\n",
" a-b 0 2170 -2170 \n",
" b-c 21000 0 21000 \n",
" c-d -2100 34500 -36600 \n",
" d-a -35900 -53670 17770 \n",
"\n",
"Rate of work output : -17000 kJ/min\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.4 Page No : 80"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Part (a)\n",
"import math \n",
"\n",
"# Variables\n",
"m = 3.;\n",
"V1 = 0.22; #volume m^3\n",
"P1 = 500.e03; #initial pressure Pa\n",
"P2 = 100.e03; #final pressure Pa\n",
"\n",
"# Calculation\n",
"V2 = V1*(P1/P2)**(1./1.2);\n",
"dU = 3.56*(P2*V2-P1*V1);\n",
"gama = 1.2;\n",
"W = (P2*V2-P1*V1)/(1-gama);\n",
"Q = dU+W;\n",
"\n",
"# Results\n",
"print \"Q,W and dU of the quasi static process are\",int(dU/1000),round(W/1000),round(Q/1000),\"kJ respectively\"\n",
"\n",
"# Part (b)\n",
"Qb = 30e03;\n",
"Wb = Qb-dU; \n",
"print \"Work transfer for the process is\",round(Wb/1000),\"kJ\"\n",
"\n",
"# Part (c)\n",
"print \"Wb is not equal to integral(p*dv) .since the process is not quasi static\"\n",
"\n",
"# rounding off error. please check"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Q,W and dU of the quasi static process are -92 129.0 37.0 kJ respectively\n",
"Work transfer for the process is 122.0 kJ\n",
"Wb is not equal to integral(p*dv) .since the process is not quasi static\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.5 Page No : 81"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from numpy.linalg import inv\n",
"from numpy import *\n",
"from scipy.integrate import *\n",
"\n",
"# Variables\n",
"V1 = 0.03; #initial volume m^3\n",
"P1 = 170e03; #initial pressure Pa\n",
"P2 = 400e03; #final pressure Pa\n",
"V2 = 0.06; #final volume m^3\n",
"U = 3.15*(P2*V2-P1*V1);\n",
"B = array([P1, P2]);\n",
"B= B.transpose()\n",
"\n",
"A = [[1, V1],[1, V2]];\n",
"A = array(A)\n",
"x = inv(A)*B;\n",
"\n",
"a = -60000\n",
"b = 7666666.7;\n",
"\n",
"# Calculation\n",
"def pressure(V):\n",
" return a+b*V;\n",
"\n",
"W = quad(pressure,V1,V2)[0]\n",
"Q = U+W;\n",
"\n",
"# Results\n",
"print \"The work done by the system is\",round(W/1000,2),\"kJ\"\n",
"print \"The internal energy change of the system is\",round(U/1000,1),\"J\"\n",
"print \"The heat flow into the system is\",round(Q/1000,2),\"kJ\"\n",
"\n",
"# rounding off error. please check."
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The work done by the system is 8.55 kJ\n",
"The internal energy change of the system is 59.5 J\n",
"The heat flow into the system is 68.09 kJ\n"
]
}
],
"prompt_number": 36
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.6 Page No : 82"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Process 1-2\n",
"import math \n",
"\n",
"# Variables\n",
"Q12 = 235; \t\t\t# in KJ/Kg\n",
"W12 = 0 ;\n",
"\n",
"# Calculation\n",
"U12 = Q12-W12;\n",
"# Process 2-3\n",
"Q23 = 0; \n",
"U23 = -70 ;\n",
"W23 = Q23-U23;\n",
"\n",
"# Process 3-1\n",
"Q31 = - 200; \n",
"U31 = -U12-U23;\n",
"W31 = Q31-U31;\n",
"W = W12 + W23 + W31;\n",
"Q = Q12 + Q23 + Q31;\n",
"\n",
"# Results\n",
"print \"Heat trasfer in the cycle is\",Q,\"KJ/Kg\"\n",
"print \"Work done during the the cycle is\",W,\"KJ/Kg\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Heat trasfer in the cycle is 35 KJ/Kg\n",
"Work done during the the cycle is 35 KJ/Kg\n"
]
}
],
"prompt_number": 39
}
],
"metadata": {}
}
]
}