{ "metadata": { "name": "" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter 9 : Second law of thermodynamics" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 9.1 pageno : 308" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "# Variables\n", "t2 = 300;\t\t\t#temperature of the math.sink in K\n", "n1 = 0.4;\t\t\t#efficiency of the engine\n", "n2 = 0.6;\t\t\t#efficiency of the engine\n", "\n", "# Calculations\n", "t1 = t2/(1-n1);\t\t\t#temperature of the source in K\n", "t3 = t2/(1-n2);\t\t\t#temperature of the source in K\n", "\n", "# Result\n", "print 'the temperature of the source when 0.4 efficiency is %3.2f K \\\n", "\\nthe temperature of the source when 0.6 efficiency is %3.2f K'%(t1,t3)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "the temperature of the source when 0.4 efficiency is 500.00 K \n", "the temperature of the source when 0.6 efficiency is 750.00 K\n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 9.2 pageno : 308" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "# Variables\n", "t2 = 273.;\t\t\t#temperature of the math.sink in K\n", "t1 = 373.;\t\t\t#temperature of the source in K\n", "q1 = 840.;\t\t\t#heat supplied in joules\n", "j = 4.2;\t\t\t#joukes constant in erg/cal\n", "\n", "# Calculations\n", "w = (q1/t1)*(t1-t2);\t\t\t#work done in joules\n", "q2 = (q1/j)*(t2/t1);\t\t\t#heat rejected in calories\n", "n = 1-(t2/t1);\t\t\t#efficiency of the engine\n", "\n", "# Result\n", "print 'work done is %3.f j \\\n", "\\nheat rejected is %3.f cal \\\n", "\\nthe efficiency of the engine is %3.1f %%'%(w,q2,n*100)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "work done is 225 j \n", "heat rejected is 146 cal \n", "the efficiency of the engine is 26.8 %\n" ] } ], "prompt_number": 3 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 9.3 pageno : 309" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "# Variables\n", "t1 = 90.;\t\t\t#temperature of the oxygen boils in K\n", "t2 = 20.;\t\t\t#temperature of the liquid hydrogen in K\n", "t3 = 300.;\t\t\t#temperature of the sink in K\n", "\n", "# Calculations\n", "n = (t1-t2)/t1;\t\t\t#efficiency of the engine\n", "t4 = t3/(1-n);\t\t\t#temperature of the source in K\n", "\n", "# Result\n", "print 'the efficiency of the engine is %3.2f \\\n", "\\nthe temperature of the source is %3.2f K'%(n,t4)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "the efficiency of the engine is 0.78 \n", "the temperature of the source is 1350.00 K\n" ] } ], "prompt_number": 4 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 9.4 pageno : 309" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "# Variables\n", "t1 = 373.;\t \t \t#temperature of the source in K\n", "t2 = 273.;\t\t \t #temperature of the sink in K\n", "w = 1200*10**5*980;\t\t\t#work done in ergs\n", "j = 4.18*10**7;\t\t \t#joules constant in ergs/cal\n", "\n", "# Calculations\n", "q = (w/j)*(t1/(t1-t2));\t\t\t#heat added in cal\n", "\n", "# Result\n", "print 'the heat added is %3.2f cal'%(round(q,-1))\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "the heat added is 10490.00 cal\n" ] } ], "prompt_number": 6 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 9.5 pageno : 309" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "# Variables\n", "t1 = 273.;\t\t\t#temperature of the source in K\n", "t2 = 290.;\t\t\t#temperature of the sink in K\n", "l = 8*10.**11;\t\t\t#latent of fusion in ergs/cal\n", "\n", "# Calculations\n", "n = (t2-t1)/t1;\t\t\t#efficiency of the engine\n", "w = n*l;\t\t\t#energy to be supplied in ergs\n", "\n", "# Result\n", "print 'efficiency of the engine is %.2f %% \\\n", "\\nenergy to be supplied is %.3e ergs'%(n*100,w)\n", "print \"Note: answer in book are wrong please calculate manually.\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "efficiency of the engine is 6.23 % \n", "energy to be supplied is 4.982e+10 ergs\n", "Note: answer in book are wrong please calculate manually.\n" ] } ], "prompt_number": 12 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 9.6 pageno : 309" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "\n", "# Variables\n", "t1 = 373;\t\t\t#temperature in K\n", "t2 = 273;\t\t\t#temperature of math.sink in K\n", "q = 10**4;\t\t\t#heat taken at higher temperature in cal\n", "j = 4.2*10**7;\t\t\t#joules consmath.tant in ergs/cal\n", "\n", "# Calculations\n", "w = q*j*(t1-t2)/t1;\t\t\t#work done in ergs\n", "\n", "# Result\n", "print 'work done is %.1e ergs'%(w)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "work done is 1.1e+11 ergs\n" ] } ], "prompt_number": 14 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 9.7 page no : 310" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "import math\n", "\n", "# Variables\n", "p = 100*746/4.2;\t\t\t#power developed in cal/sec\n", "t1 = 300.;\t\t\t#temperature of the sink in K\n", "t2 = 500.\n", "\n", "# Calculations\n", "te = 1 - (t1/t2)\n", "Q1 = p * 100/40 # heat supplied\n", "Q2 = Q1 * 0.6\n", "\n", "\n", "# Result\n", "print \"Thermal efficiency = %.f %%\"%(te*100)\n", "print \"Power developed by the engine %.2f calories/sec\"%p\n", "print \"If Q1 heat supplied , Q1 = %.2e cal/sec\"%Q1\n", "print \"Q2 = %.2e cal/sec\"%Q2" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Thermal efficiency = 40 %\n", "Power developed by the engine 17761.90 calories/sec\n", "If Q1 heat supplied , Q1 = 4.44e+04 cal/sec\n", "Q2 = 2.66e+04 cal/sec\n" ] } ], "prompt_number": 23 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 9.8 page no : 310" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math \n", "\n", "# Variables\n", "l = 964.8;\t\t\t#latent heat of steam in B.Th.U per lb\n", "q = 4*15*l*778;\t\t\t#heat developed in ft lbs\n", "w = 30000*60;\t\t\t#work done is ft lbs\n", "pv = 12*1.013*10**6*10**3 \n", "T = 600 # K\n", "\n", "# Calculations\n", "n = (w/q)*100;\t\t\t#efficiency of the engine\n", "p = 100-n;\t\t\t#percentage of heat wasted\n", "T2 = 600./(6**.4)\n", "R = pv/T\n", "W = R * (T - T2) * 2.303 * math.log10(6)\n", "e = 1 - (T2/T)\n", "# Result\n", "print \"Lowest temperature T2 = %.f K\"%T2\n", "print \"Work done W = %.2e ergs\"%W\n", "print \"Efficiency = %.1f %%\"%(e*100)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Lowest temperature T2 = 293 K\n", "Work done W = 1.11e+10 ergs\n", "Efficiency = 51.2 %\n" ] } ], "prompt_number": 34 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 9.9 page no : 311" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# Variables\n", "l=964.8; #latent heat of steam in B.Th.U per lb\n", "q=4*15*l*778; #heat developed in ft lbs\n", "w=33000*60; #work done is ft lbs\n", "\n", "#CALCULATIONS\n", "n=(w/q)*100; #efficiency of the engine\n", "p=100-n; #percentage of heat wasted\n", "\n", "# Results\n", "print ('the percentage of the heat wasted is %3.2f'%p)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "the percentage of the heat wasted is 95.60\n" ] } ], "prompt_number": 49 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 9.10 page no : 311" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "# Variables\n", "ip = 16.3*500*778/33000;\t\t\t# Variables power of the engine in HP\n", "me = 0.72;\t\t\t#mechanical efficiency of the engine\n", "bhp = 31;\t\t\t#brake horse power in b.h.p\n", "ihp = bhp/me;\t\t\t#indicated horse power in HP\n", "\n", "# Calculations\n", "i = ihp/ip;\t\t\t#indicated thermal efficiency\n", "\n", "# Result\n", "print 'the indicted thermal efficiency is %3.2f %%'%(i*100)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "the indicted thermal efficiency is 22.41 %\n" ] } ], "prompt_number": 4 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 9.11 pageno : 312" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "# Variables\n", "p = 200.;\t\t\t#horse power of steam engine in lbs coal per hour\n", "j = 770.;\t\t\t#joules constant in ft lbs per B.Th.U\n", "\n", "# Calculations\n", "w = 12500*p*j;\t\t\t#equivalent work in ft.lb.per.hr\n", "hp = w/(60*33000);\t\t\t#horse power\n", "\n", "# Result\n", "print 'horse power of the engine is %3.2f'%(hp)\n", "print \"Note : answer in book is wrong. Please check manually.\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "horse power of the engine is 972.22\n", "Note : answer in book is wrong. Please check manually.\n" ] } ], "prompt_number": 5 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 9.12 pageno : 312" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "# Variables\n", "t1 = 340.;\t\t\t#temperature of the atmosphere in K\n", "t2 = 612.;\t\t\t#temperature of the compression stroke in K\n", "y = 1.39;\t\t\t#adiabatic expansion \n", "t3 = 2040.;\t\t\t#temperature after consmtant volume ignition in K\n", "\n", "# Calculations\n", "d = (t2/t1)**(1/(y-1))\t\t\t#density in gm/cc\n", "n = 1-(1/d)**(y-1);\t\t \t#efficiency of the engine\n", "p = ((d)**(y))*(t3/t2);\t\t\t#maximum temperature of the temperature in atm\n", "\n", "# Result\n", "print 'the maximum pressure of the engine is %3.f atm'%(p)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "the maximum pressure of the engine is 27 atm\n" ] } ], "prompt_number": 43 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 9.13 pageno : 313" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "# Variables\n", "t1 = 915;\t\t\t#temperature at the beggining in K\n", "t2 = 2040;\t\t\t#temperature at the end in K\n", "d = 12.6;\t\t\t#adiabatic expansion ratio\n", "y = 1.39;\t\t\t#coefficent of expansion\n", "\n", "# Calculations\n", "x = t2/t1 \t\t\t#ratio temparatures\n", "n = 1-(1/d)**(y-1)*((x**y)-1)/(y*(x-1));\t\t\t#efficiency of the engine\n", "\n", "# Result\n", "print 'the efficiency of the engine is %3.3f'%(n)\n", "print \"Note : answer slighty different because of rounding error\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "the efficiency of the engine is 0.566\n", "Note : answer slighty different because of rounding error\n" ] } ], "prompt_number": 40 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 9.14 pageno : 313" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "# Variables\n", "p1 = 15.;\t\t\t#intial pressure in lb/sq.inch\n", "dv = 15.;\t\t\t#ratio of intial to final volume\n", "t1 = 520.;\t\t\t#temperature at intial in K\n", "y = 1.4;\t\t\t#coefficient of expansion\n", "\n", "# Calculations\n", "p2 = p1*(dv)**(y);\t\t\t#final pressure in lb/sq.inch\n", "t2 = t1*(dv)**(y-1);\t\t\t#final temperatire in K\n", "\n", "# Result\n", "print 'the final pressure is %3.2f lb/sq.inch \\\n", "\\nthe final temperature is %.f K'%(p2,t2)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "the final pressure is 664.69 lb/sq.inch \n", "the final temperature is 1536 K\n" ] } ], "prompt_number": 6 } ], "metadata": {} } ] }