From 41f1f72e9502f5c3de6ca16b303803dfcf1df594 Mon Sep 17 00:00:00 2001 From: Thomas Stephen Lee Date: Fri, 4 Sep 2015 22:04:10 +0530 Subject: add/remove/update books --- Fundamentals_Of_Thermodynamics/Chapter9.ipynb | 519 -------------------------- 1 file changed, 519 deletions(-) delete mode 100755 Fundamentals_Of_Thermodynamics/Chapter9.ipynb (limited to 'Fundamentals_Of_Thermodynamics/Chapter9.ipynb') diff --git a/Fundamentals_Of_Thermodynamics/Chapter9.ipynb b/Fundamentals_Of_Thermodynamics/Chapter9.ipynb deleted file mode 100755 index c1f858fc..00000000 --- a/Fundamentals_Of_Thermodynamics/Chapter9.ipynb +++ /dev/null @@ -1,519 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:7638f00c0629ff24b9e3c96e6eb43a103a0583de57d4e01375fbb8aa638bef10" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter9:SECOND-LAW ANALYSIS FOR A CONTROL VOLUME" - ] - }, - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Ex9.1:pg-336" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#example 1\n", - "#work done by steam\n", - "\n", - "hi=3051.2 #initial specific heat of enthalpy of steam in kJ/kg\n", - "si=7.1228 #initial specific entropy of steam in kJ/kg-K\n", - "Pe=0.15 #final pressure in MPa\n", - "se=si #specific entropy in final state in kJ/kg-K\n", - "sf=1.4335 #in kJ/kg-K\n", - "sfg=5.7897 #in kJ/kg-K\n", - "vi=50.0 #velocity with which steam enters turbine in m/s\n", - "ve=200.0 #velocity with which steam leaves the turbine in m/s\n", - "xe=(se-sf)/sfg #quality of steam in final state\n", - "hf=467.1 #in kJ/kg\n", - "hfg=2226.5 #in kJ/kg\n", - "he=hf+xe*hfg #final specific heat of enthalpy of steam in kJ/kg\n", - "w=hi+vi**2/(2*1000)-he-ve**2/(2*1000) #work of steam for isentropic process in kJ/kg\n", - "print\"\\n hence, work per kilogram of steam for this isentropic process is\",round(w,1),\"KJ/kg\" " - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "\n", - " hence, work per kilogram of steam for this isentropic process is 377.5 KJ/kg\n" - ] - } - ], - "prompt_number": 5 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Ex9.2:pg-337" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#example 2\n", - "#exit velocity of steam from nozzle\n", - "\n", - "hi=3051.2 #initial specific heat of enthalpy in kJ/kg\n", - "si=7.1228 #initial specific entropy in kJ/kg-K\n", - "se=si #final specific entropy \n", - "Pe=0.3 #final pressure in MPa\n", - "print\"from steam table,various properties at final state are\"\n", - "he=2780.2 #final specific heat of enthalpy in kJ/kg-K\n", - "Te=159.1 #final temperature in celsius\n", - "vi=30.0 #velocity with which steam enters the nozzle in m/s\n", - "ve=((2*(hi-he)+(vi**2/1000))*1000)**0.5 #final velocity of steam with which it exits in m/s\n", - "print\"\\n hence,exit velocity of the steam from the nozzle is\",round(ve),\"m/s\" " - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "from steam table,various properties at final state are\n", - "\n", - " hence,exit velocity of the steam from the nozzle is 737.0 m/s\n" - ] - } - ], - "prompt_number": 9 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Ex9.3:pg-340" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#example 3\n", - "#violation of second law\n", - "\n", - "print\"from R-134a tables\"\n", - "se=1.7148 #specific entropy in final state in kJ/kg-K\n", - "si=1.7395 #initial specific entropy in kJ/kg-K \n", - "print\"therefore,se=si.The process described involves a violation of the second law and thus would be impossible.\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "from R-134a tables\n", - "therefore,se=si.The process described involves a violation of the second law and thus would be impossible.\n" - ] - } - ], - "prompt_number": 10 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Ex9.4:pg-340" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#example 4\n", - "#calculating required specific work\n", - "\n", - "Cp=1.004 #specific heat of air at constant pressure in kJ/kg-K\n", - "Ti=290 #initial temperature in kelvins\n", - "Pi=100 #initial pressure in kPa\n", - "Pe=1000 #final pressure in kPa\n", - "k=1.4 \n", - "Te=Ti*(Pe/Pi)**((k-1)/k) #final temperature in kelvins\n", - "we=Cp*(Ti-Te) #required specific work in kJ/kg\n", - "print\"\\n hence,specific work required is\",round(we),\"kJ/kg\" " - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "\n", - " hence,specific work required is -271.0 kJ/kg\n" - ] - } - ], - "prompt_number": 12 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Ex9.5:pg-342" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#example 5\n", - "#entropy generation\n", - "\n", - "h1=2865.54 #specific heat of enthalpy at state 1 in kJ/kg\n", - "h2=83.94 #specific heat of enthalpy at state 2 in kJ/kg\n", - "h3=2725.3 #specific heat of enthalpy at state 3 in kJ?kg\n", - "s1=7.3115 #specific entropy at state 1 in kJ/kg-K\n", - "s2=0.2966 #specific entropy at state 2 in kJ/kg-K\n", - "s3=6.9918 #specific entropy at state 3in kJ/kg-K\n", - "m1=2 #mass flow rate at state 1 in kg/s\n", - "m2=m1*(h1-h3)/(h3-h2) #mass flow rate at state 2 in kg/s\n", - "m3=m1+m2 #mass flow rate at state 3 in kg/s\n", - "Sgen=m3*s3-m1*s1-m2*s2 #entropy generation in the process\n", - "print\"\\n hence,entropy generated in this process is \",round(Sgen,3),\"kW/K\" " - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "\n", - " hence,entropy generated in this process is 0.072 kW/K\n" - ] - } - ], - "prompt_number": 15 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Ex9.6:pg-344" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#example 6\n", - "#work required to fill the tank\n", - "import math\n", - "T1=17+273 #initial temperature of tank in Kelvins\n", - "sT1=6.83521 #specific entropy in kJ/kg-K\n", - "R=0.287 #gas constant in kJ/kg-K\n", - "P1=100 #initial pressure in kPa\n", - "P2=1000 #final pressure in kPa\n", - "sT2=sT1+R*math.log(P2/P1) #specific entropy at temperature T2 in kJ/kg-K\n", - "T2=555.7 #from interplotation \n", - "V1=0.04 #volume of tank in m^3\n", - "V2=V1 #final volume is equal to initial volume\n", - "m1=P1*V1/(R*T1) #initial mass of air in tank in kg\n", - "m2=P2*V2/(R*T2) #final mass of air in tank in kg\n", - "Min=m2-m1 #in kg\n", - "u1=207.19 #initial specific heat of enthalpy in kJ/kg\n", - "u2=401.49 #final specific heat of enthalpy in kJ/kg\n", - "hin=290.43 #in kJ/kg\n", - "W12=Min*hin+m1*u1-m2*u2 #work required to fill the tank in kJ\n", - "print\"\\n hence,the total amount of work required to fill the tank is\",round(W12,1),\"kJ\" " - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "\n", - " hence,the total amount of work required to fill the tank is -31.9 kJ\n" - ] - } - ], - "prompt_number": 18 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Ex9.7:pg-347" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#example 7\n", - "#work required to pump water isentropically \n", - "\n", - "P1=100 #initial pressure in kPa\n", - "P2=5000 #final pressure in kPa\n", - "v=0.001004 #specific volume in m^3/kg\n", - "w=v*(P2-P1) #work required to pump water isentropically\n", - "print\"\\n hence,work required to pump water isentropically is \",round(w,2),\"kJ/kg\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "\n", - " hence,work required to pump water isentropically is 4.92 kJ/kg\n" - ] - } - ], - "prompt_number": 21 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Ex9.8:pg-348" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#example 8\n", - "#Velocity in exit flow\n", - "\n", - "print\"From Steam Tables, for liquid water at 20 C\"\n", - "vf=0.001002 #in m^3/kg\n", - "v=vf\n", - "Pi=300 #Line pressure in kPa\n", - "Po=100 #in kPa\n", - "Ve=(2*v*(Pi-Po)*1000)**0.5 #velocity in the exit flow\n", - "print\" \\n Hence, an ideal nozzle can generate upto \",round(Ve),\"m/s\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "From Steam Tables, for liquid water at 20 C\n", - " \n", - " Hence, an ideal nozzle can generate upto 20.0 m/s\n" - ] - } - ], - "prompt_number": 24 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Ex9.9:pg-351" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#example 9\n", - "#Rate of Entropy Generation\n", - "\n", - "print\"From R-410a tables,we get\"\n", - "hi=280.6 #in kJ/kg\n", - "he=307.8 #in kJ/kg\n", - "si=1.0272 #in kJ/kg\n", - "se=1.0140 #in kJ/kg\n", - "m=0.08 #flow rate of refrigerant in kg/s\n", - "P=3 #electrical power input in kW\n", - "Qcv=m*(he-hi)-P #in kW\n", - "To=30 #in Celsius\n", - "Sgen=m*(se-si)-Qcv/(To+273.2) #rate of entropy generation \n", - "print\"\\n Hence,the rate of entropy generation for this process is\",round(Sgen,5),\"kW/K\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "From R-410a tables,we get\n", - "\n", - " Hence,the rate of entropy generation for this process is 0.00166 kW/K\n" - ] - } - ], - "prompt_number": 29 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Ex9.10:pg-353" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#example 10\n", - "#turbine efficiency\n", - "\n", - "hi=3051.2 #initial specific heat of enthalpy in kJ/kg\n", - "si=7.1228 #initial specific entropy in kJ/kg-K\n", - "sf=0.7548 #in kJ/kg-K\n", - "sfg=7.2536 #in kJ/kg-K\n", - "ses=si #final specific entropy is same as the initial\n", - "xes=(si-sf)/sfg #quality of steam when it leaves the turbine\n", - "hf=225.9 #in kJ/kg\n", - "hfg=2373.1 #in kJ/kg\n", - "hes=hf+xes*hfg #final specific heat of enthalpy in kJ/kg\n", - "ws=hi-hes #work output of turbine calculated ideally in kJ/kg\n", - "wa=600 #actual work output of turbine in kJ/kg\n", - "nturbine=wa/ws #efiiciency of turbine \n", - "print\"\\n hence,efficiency of the turbine is\",round(nturbine*100,1),\"%\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "\n", - " hence,efficiency of the turbine is 80.9 %\n" - ] - } - ], - "prompt_number": 32 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Ex9.11:pg-355" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#example 11\n", - "#turbine inlet pressure\n", - "import math\n", - "hi=1757.3 #initial specific heat of enthalpy of air in kJ/kg\n", - "si=8.6905 #initial specifc entropy of airin kJ/kg-K\n", - "he=855.3 #final specific heat of enthalpy of air in kJ/kg\n", - "w=hi-he #actual work done by turbine in kJ/kg\n", - "n=0.85 #efficiency of turbine \n", - "ws=w/n #ideal work done by turbine in kJ/kg\n", - "hes=hi-ws #from first law of isentropic process\n", - "Tes=683.7 #final temperature in kelvins from air tables\n", - "ses=7.7148 #in kJ/kg-K\n", - "R=0.287 #gas constant in kJ/kg-K\n", - "Pi=100/math.e**((si-ses)/-R) #turbine inlet pressure in kPa\n", - "print\"\\n hence,turbine inlet pressure is\",round(Pi),\"kPa\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "\n", - " hence,turbine inlet pressure is 2995.0 kPa\n" - ] - } - ], - "prompt_number": 36 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Ex9.12:pg-357" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#example 12\n", - "#required work input\n", - "\n", - "Pe=150.0 #final pressure of air in kPa\n", - "Pi=100.0 #initial presure of air in kPa\n", - "k=1.4\n", - "Ti=300.0 #initial temperature of air in kelvis\n", - "Tes=Ti*(Pe/Pi)**((k-1)/k) #from second law\n", - "ws=1.004*(Ti-Tes) #from first law of isentropic process\n", - "n=0.7 #efficiency of automotive supercharger \n", - "w=ws/n #real work input in kJ/kg\n", - "Te=Ti-w/1.004 #temperature at supercharger exit in K\n", - "print\"\\n hence,required work input is \",round(w),\"kJ/kg\"\n", - "print\"\\n and exit temperature is \",round(Te,1),\"K\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "\n", - " hence,required work input is -53.0 kJ/kg\n", - "\n", - " and exit temperature is 352.6 K\n" - ] - } - ], - "prompt_number": 44 - }, - { - "cell_type": "code", - "collapsed": false, - "input": [], - "language": "python", - "metadata": {}, - "outputs": [] - } - ], - "metadata": {} - } - ] -} \ No newline at end of file -- cgit