From 435840cef00c596d9e608f9eb2d96f522ea8505a Mon Sep 17 00:00:00 2001 From: hardythe1 Date: Tue, 5 May 2015 14:21:39 +0530 Subject: add books --- .../Chapter6.ipynb | 308 --------------------- 1 file changed, 308 deletions(-) delete mode 100755 Thermodynamics:_From_concepts_to_applications/Chapter6.ipynb (limited to 'Thermodynamics:_From_concepts_to_applications/Chapter6.ipynb') diff --git a/Thermodynamics:_From_concepts_to_applications/Chapter6.ipynb b/Thermodynamics:_From_concepts_to_applications/Chapter6.ipynb deleted file mode 100755 index 0e7c0436..00000000 --- a/Thermodynamics:_From_concepts_to_applications/Chapter6.ipynb +++ /dev/null @@ -1,308 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:b8ccd5165760dffa261415477a99582312dff66f312e3d719c763faed6ba4a02" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter6 -control volume" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example1-pg 83" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "#calculate diameter for given variable\n", - "##initialisation of variables\n", - "R= 8.314 ##J/mol K\n", - "M= 29 ##gms\n", - "T= 80 ##C\n", - "p= 104 ##/kPa\n", - "v= 30 ##m/sec\n", - "m= 8000 ##kg/h\n", - "##CALCULATIONS\n", - "V= R*(273.15+T)/(M*p)\n", - "A= m*V/(3600*v)\n", - "D=math.sqrt(4*A/math.pi)\n", - "##RESULTS\n", - "print'%s %.5f %s'%('diameter = ',D,'m^2')\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "diameter = 0.30301 m^2\n" - ] - } - ], - "prompt_number": 1 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example2-pg 88" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "#calculate intial areas\n", - "##initialisation of variables\n", - "R= 8.314 ##J/mol K\n", - "M= 29. ##gms\n", - "T1= 230. ##C\n", - "p1= 30. ##/bar\n", - "k= 1.4\n", - "T2= 180. ##C\n", - "v1= 10. ##m/s\n", - "p2= 20. ##bar\n", - "m2= 0.84 ##kg/s\n", - "##CALCULATIONS\n", - "V1= R*(273.15+T1)/(M*p1*100)\n", - "cp= k*R/((k-1)*M)\n", - "A= m2*V1*10*10*10*10/v1\n", - "v2= math.sqrt(v1*v1+2*cp*10*10*10*(T1-T2))\n", - "V2= R*(273.15+T2)/(M*p2*100)\n", - "A2= m2*V2*10*10*10*10/v2\n", - "##RESULTS\n", - "print'%s %.1f %s'%('inlet area =',A,'cm^2')\n", - "print'%s %.2f %s'%('inlet area =',A2,'cm^2')\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "inlet area = 40.4 cm^2\n", - "inlet area = 1.72 cm^2\n" - ] - } - ], - "prompt_number": 2 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example3-pg 89" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "#calculate quantity x and specific volume\n", - "##initialisation of variables\n", - "h= 2676.2 ##kJ/kg\n", - "hf= 721.11 ##kJ/kg\n", - "hg= 2679.1 ##kJ/kg\n", - "vf= 0.001115 ##m^3/kg\n", - "vg= 0.2404 ##m^3/kg\n", - "##CALCULATIONS\n", - "x= (h-hf)/(hg-hf)\n", - "v1= vf+x*(vg-vf)\n", - "##RESULTS\n", - "print'%s %.4f %s'%('quantity =',x,'')\n", - "print'%s %.4f %s'%('specific volume = ',v1,'m^3/kg')\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "quantity = 0.9985 \n", - "specific volume = 0.2400 m^3/kg\n" - ] - } - ], - "prompt_number": 4 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example4-pg 92" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "#calculate heat pump and reate heat interaction and work of the turbine and at another point rate of heat interaction and all four areas at given poin\n", - "##initialisation of variables\n", - "h4= 419.05 ##kJ/kg\n", - "h1= 434.92 ##kJ/kg\n", - "m= 2.5 ##kg/s\n", - "h2= 3272.4 ##kJ/kg\n", - "h3= 2601.7 ##kJ/kg\n", - "v1= 0.001401 ##m^3/kg\n", - "V1= 5 ##m/s\n", - "v2= 0.03817 ##m^3/kg\n", - "V2= 20. ##m/s\n", - "v3= 0.8415 ##m^3/kg\n", - "V3= 100. ##m/s\n", - "v4= 0.00104 ##m^3/kg\n", - "V4= 5 ##m/s\n", - "##CALCULATIONS\n", - "W41= m*(h4-h1)\n", - "Q12= m*(h2-h1)\n", - "W23= m*(h2-h3)\n", - "Q34= m*(h4-h3)\n", - "A1= m*v1*10*10*10*10/V1\n", - "A2= m*v2*10*1010*10/V2\n", - "A3= m*v3*10*1010*10/V3\n", - "A4= m*v4*10*1010*10/V4\n", - "##RESULTS\n", - "print'%s %.1f %s'%('rate of pump =',W41,'kW')\n", - "print'%s %.f %s'%('rate of heat ineraction =',Q12,'kW')\n", - "print'%s %.1f %s'%('rate of work of the turbine =',W23,'W')\n", - "print'%s %.f %s'%('rate of heat ineraction =',Q34,'kW')\n", - "print'%s %.2f %s'%('area =',A1,'cm^2')\n", - "print'%s %.2f %s'%('area =',A2,'cm^2')\n", - "print'%s %.2f %s'%('area =',A3,'cm^2')\n", - "print'%s %.2f %s'%('area =',A4,'cm^2')\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "rate of pump = -39.7 kW\n", - "rate of heat ineraction = 7094 kW\n", - "rate of work of the turbine = 1676.8 W\n", - "rate of heat ineraction = -5457 kW\n", - "area = 7.00 cm^2\n", - "area = 481.90 cm^2\n", - "area = 2124.79 cm^2\n", - "area = 52.52 cm^2\n" - ] - } - ], - "prompt_number": 5 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example5-pg 96" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "#calculate mass of helium and temperature of helium\n", - "##initialisation of variables\n", - "m1= 0.03 ##kg\n", - "R= 8.314 ##J/mol K\n", - "T1= 300. ##C\n", - "p1= 120. ##kPa\n", - "k= 5./3.\n", - "M=4. ##kg\n", - "p2= 600. ##kPa\n", - "##CALCULATIONS\n", - "V= m1*R*(273.15+T1)/(p1*M)\n", - "m2= m1*((p2/p1)+k-1)/k\n", - "T2= p2*V*M/(m2*R)\n", - "##RESULTS\n", - "print'%s %.3f %s'%('mass of helium =',m2,'kg')\n", - "print'%s %.1f %s'%('temperature of helium =',T2,'K')\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "mass of helium = 0.102 kg\n", - "temperature of helium = 842.9 K\n" - ] - } - ], - "prompt_number": 7 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Eaxmple 6-pg97" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "#calculate volume of container and pressure\n", - "##initialisation of variables\n", - "m1= 0.03 ##kg\n", - "v1= 2.1977 ##m^3/kg\n", - "h2= 3073.8 ##kJ/kg\n", - "h1= 3061.6 ##kJ/kg\n", - "p2= 600. ##kPa\n", - "p1= 120. ##kPa\n", - "##CALCULATIONS\n", - "V=m1*v1\n", - "r= ((h2-h1)/v1)+p2-p1\n", - "##RESULTS\n", - "print'%s %.5f %s'%('volume of container =',V,'m^3')\n", - "print'%s %.2f %s'%('pressure =',r,'kPa')\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "volume of container = 0.06593 m^3\n", - "pressure = 485.55 kPa\n" - ] - } - ], - "prompt_number": 8 - } - ], - "metadata": {} - } - ] -} \ No newline at end of file -- cgit