{ "metadata": { "name": "", "signature": "sha256:1e66a4aaf6aa5b1578af922356299d8af3b4aded7460ea4a450b6cc816355a1b" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter 06:Second Law of Thermodynamics" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex6.1:pg-138" ] }, { "cell_type": "code", "collapsed": false, "input": [ "T1 = 800 # Source temperature in degree Celsius\n", "\n", "T2 = 30 # Sink temperature in degree Celsius\n", "\n", "e_max = 1-((T2+273)/(T1+273)) # maximum possible efficiency \n", "\n", "Wnet = 1 # in kW\n", "\n", "Q1 = Wnet/e_max # Least rate of heat required in kJ/s\n", "\n", "Q2 = Q1-Wnet # Least rate of heat rejection kJ/s\n", "\n", "\n", "\n", "print \"\\n Example 6.1\"\n", "\n", "print \"\\n Least rate of heat rejection is \",Q2,\" kW\"\n", "\n", "#The answers vary due to round off error\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "\n", " Example 6.1\n", "\n", " Least rate of heat rejection is 0 kW\n" ] } ], "prompt_number": 5 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex6.2:pg-139" ] }, { "cell_type": "code", "collapsed": false, "input": [ "T1 = -15 # Source temperature in degree Celsius\n", "\n", "T2 = 30 # Sink temperature in degree Celsius\n", "\n", "Q2 = 1.75 # in kJ/sec\n", "\n", "print \"\\n Example 6.2\"\n", "\n", "W= Q2*((T2+273)-(T1+273))/(T1+273) # Least Power necessary to pump the heat out\n", "\n", "print \"\\n Least Power necessary to pump the heat out is \",round(W,2),\"kW\"\n", " \n", " #The answers vary due to round off error\n", " \n", " " ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "\n", " Example 6.2\n", "\n", " Least Power necessary to pump the heat out is 0.31 kW\n" ] } ], "prompt_number": 10 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex6.3:pg-140" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Given \n", "\n", "T1 = 600 # Source temperature of heat engine in degree Celsius\n", "\n", "T2 = 40 # Sink temperature of heat engine in degree Celsius \n", "\n", "T3 = -20 # Source temperature of refrigerator in degree Celsius\n", "\n", "Q1 = 2000 # Heat transfer to heat engine in kJ\n", "\n", "W = 360 # Net work output of plant in kJ\n", "\n", "# Part (a)\n", "\n", "e_max = 1.0-((T2+273)/(T1+273)) # maximum efficiency \n", "\n", "W1 = e_max*Q1 # maximum work output \n", "\n", "COP = (T3+273)/((T2-273)-(T3-273)) # coefficient of performance of refrigerator\n", "\n", "W2 = W1-W # work done to drive refrigerator \n", "\n", "Q4 = COP*W2 # Heat extracted by refrigerator\n", "\n", "Q3 = Q4+W2 # Heat rejected by refrigerator\n", "\n", "Q2 = Q1-W1 # Heat rejected by heat engine\n", "\n", "Qt = Q2+Q3 # combined heat rejection by heat engine and refrigerator \n", "\n", "print \"\\n Example 6.3\"\n", "\n", "print \"\\n\\n Part A:\"\n", "\n", "print \"\\n The heat transfer to refrigerant is \",round(Q2,3) ,\" kJ\"\n", "\n", "print \"\\n The heat rejection to the 40 degree reservoir is \",round(Qt,3) ,\" kJ\"\n", "\n", "\n", "\n", "# Part (b)\n", "\n", "print \"\\n\\n Part B:\"\n", "\n", "e_max_ = 0.4*e_max # maximum efficiency\n", "\n", "W1_ = e_max_*Q1 # maximum work output \n", "\n", "W2_ = W1_-W # work done to drive refrigerator \n", "\n", "COP_ = 0.4*COP # coefficient of performance of refrigerator\n", "\n", "Q4_ = COP_*W2_ # Heat extracted by refrigerator\n", "\n", "Q3_ = Q4_+W2_ # Heat rejected by refrigerator\n", "\n", "Q2_ = Q1-W1_ # Heat rejected by heat engine\n", "\n", "QT = Q2_+Q3_# combined heat rejection by heat engine and refrigerator \n", "\n", "print \"\\n The heat transfer to refrigerant is \",round(Q2_,3) ,\" kJ\"\n", "\n", "print \"\\n The heat rejection to the 40 degree reservoir is \",round(QT,3) ,\" kJ\"\n", "\n", "#The answers vary due to round off error\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "\n", " Example 6.3\n", "\n", "\n", " Part A:\n", "\n", " The heat transfer to refrigerant is 0.0 kJ\n", "\n", " The heat rejection to the 40 degree reservoir is 8200.0 kJ\n", "\n", "\n", " Part B:\n", "\n", " The heat transfer to refrigerant is 1200.0 kJ\n", "\n", " The heat rejection to the 40 degree reservoir is 2344.0 kJ\n" ] } ], "prompt_number": 14 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex6.5:pg-142" ] }, { "cell_type": "code", "collapsed": false, "input": [ "T1 = 473 # Boiler temperature in K\n", "\n", "T2 = 293 # Home temperature in K\n", "\n", "T3 = 273 # Outside temperature in K\n", "\n", "print \"\\n Example 6.5\"\n", "\n", "MF = (T2*(T1-T3))/(T1*(T2-T3)) \n", "\n", "print \"\\n The multiplication factor is \",MF \n", "\n", "#The answers vary due to round off error\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "\n", " Example 6.5\n", "\n", " The multiplication factor is 6\n" ] } ], "prompt_number": 15 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex6.6:pg-144" ] }, { "cell_type": "code", "collapsed": false, "input": [ "T1 = 90.0 # Operating temperature of power plant in degree Celsius \n", "\n", "T2 = 20.0 # Atmospheric temperature in degree Celsius\n", "\n", "W = 1.0 # Power production from power plant in kW\n", "\n", "E = 1880 # Capability of energy collection in kJ/m**2 h\n", "\n", "\n", "\n", "print \"\\n Example 6.6\"\n", "\n", "e_max = 1.0-((T2+273.0)/(T1+273.0)) # maximum efficiency\n", "\n", "Qmin = W/e_max # Minimum heat requirement per second\n", "\n", "Qmin_ = Qmin*3600.0 # Minimum heat requirement per hour\n", "\n", "Amin = Qmin_/E # Minimum area requirement\n", "\n", "print \"\\n Minimum area required for the collector plate is \",math. ceil(Amin) ,\" m**2\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "\n", " Example 6.6\n", "\n", " Minimum area required for the collector plate is 10.0 m**2\n" ] } ], "prompt_number": 21 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex6.7:pg-144" ] }, { "cell_type": "code", "collapsed": false, "input": [ "T1 = 1000 # Temperature of hot reservoir in K\n", "\n", "W = 1000 # Power requirement in kW\n", "\n", "K = 5.67e-08 # constant \n", "\n", "print \"\\n Example 6.7\"\n", "\n", "Amin = (256*W)/(27*K*T1**4) # minimum area required\n", "\n", "print \"\\n Area of the panel \",Amin ,\" m**2\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "\n", " Example 6.7\n", "\n", " Area of the panel 0.167221895617 m**2\n" ] } ], "prompt_number": 23 } ], "metadata": {} } ] }