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 --- .../Chapter12.ipynb | 262 --------------------- 1 file changed, 262 deletions(-) delete mode 100755 Thermodynamics_An_Engineering_Approach/Chapter12.ipynb (limited to 'Thermodynamics_An_Engineering_Approach/Chapter12.ipynb') diff --git a/Thermodynamics_An_Engineering_Approach/Chapter12.ipynb b/Thermodynamics_An_Engineering_Approach/Chapter12.ipynb deleted file mode 100755 index c181d0f6..00000000 --- a/Thermodynamics_An_Engineering_Approach/Chapter12.ipynb +++ /dev/null @@ -1,262 +0,0 @@ -{ - "metadata": { - "name": "" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 12: Thermodynamic Property Relations" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 12-1 ,Page No.652" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#from Table A-17\n", - "#cp value approximately by replacing the differentials\n", - "h1=305.22;#in kJ/kg\n", - "T1=305;#in K\n", - "h2=295.17;#in kJ/kg\n", - "T2=295;#in K\n", - "\n", - "#calculations\n", - "#from the given equation we can calculate\n", - "cp=(h1-h2)/(T1-T2);\n", - "print'the cp of air at 300 K %f kJ/ kg - K'%round(cp,3)" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "the cp of air at 300 K 1.005000 kJ/ kg - K\n" - ] - } - ], - "prompt_number": 2 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 12-2 ,Page No.654" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#given data\n", - "dT=302-300;#differnce in final and intial temperatures of air in K\n", - "dv=0.87-0.86;#differnce in final and intial volumnes of air in m^3/kg\n", - "T=(302+300)/2;#average temp in K\n", - "v=(0.87+0.86)/2;#average volumne in m^3/kg\n", - "\n", - "#constants used\n", - "R=0.287;#in kJ/kg-K\n", - "\n", - "#calculations\n", - "#using eq 12-3 by diffrentiating P= R*T/v\n", - "dP= R*dT/v - R*T*dv/v**2;\n", - "print'the change in the pressure of air %f kPa'%round(dP,3)" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "the change in the pressure of air -0.491000 kPa\n" - ] - } - ], - "prompt_number": 4 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 12-5 ,Page No.659" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#given data\n", - "T=20+273.15;#temperature of refrigerant in K\n", - "\n", - "#from Table A\u201311\n", - "vf=0.0008161;#in m^3/kg\n", - "vg=0.035969;#in m^3/kg\n", - "\n", - "#calculations\n", - "#using Eq 12-22\n", - "# hfg= T*vfg*(dP/dT)sat\n", - "#(dP/dT)sat b/w 24 C - 16 C \n", - "dPT=(646.18-504.58)/(24-16);#dP/dT ; values from Table A\u201311\n", - "vfg=vg-vf;\n", - "hfg=T*vfg*dPT;\n", - "print'the value of the enthalpy of vaporization of refrigerant-134a %f kJ/kg'%round(hfg,2)" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "the value of the enthalpy of vaporization of refrigerant-134a 182.400000 kJ/kg\n" - ] - } - ], - "prompt_number": 5 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 12-6 ,Page No.660" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "from math import exp\n", - "\n", - "#given data\n", - "T2=-50+460.0;#temperature of refrigerant-134a on R \n", - "\n", - "#constants\n", - "R=0.01946;#in Btu/lbm\n", - "\n", - "#from Table A-11E\n", - "T1=-40+460.0;#converted into R from F\n", - "P1=7.432;\n", - "hfg=97.100;\n", - "\n", - "#calcualation\\\n", - "#using Equation 12\u201324\n", - "#ln(P2/P1)= hfg/R *(1/T1 - 1/T2)\n", - "P2=P1*exp(hfg/R *(1/T1 - 1/T2));\n", - "print'the saturation pressure of refrigerant-134a %f psia'%round(P2,2)" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "the saturation pressure of refrigerant-134a 5.560000 psia\n" - ] - } - ], - "prompt_number": 7 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 12-11 ,Page No.373" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "from math import log\n", - "\n", - "#given data\n", - "T1=220;#intial temperature in K\n", - "P1=5;#intial pressure in MPa\n", - "T2=300;#final temperature in K\n", - "P2=10;#final pressure in MPa\n", - "\n", - "#constants used\n", - "Ru=8.314;#on kJ/kmol- K\n", - "\n", - "#from Table A\u20131\n", - "Tcr=154.8;\n", - "Pcr=5.08;\n", - "\n", - "#calculations\n", - "\n", - "#part - a\n", - "print('part - a');\n", - "#by assuming ideal-gas behavior\n", - "#from Table A\u201319\n", - "h1=6404;\n", - "h2=8736;\n", - "s2=205.213;\n", - "s1=196.171;\n", - "h21i=h2-h1;#h2 - h1 ideal\n", - "s21i=(s2-s1)-Ru*log(P2/P1);#s2 - s1 ideal\n", - "print'the enthalpy change %i kJ/kmol'%round(h21i);\n", - "print'the entropy change %f kJ/kmol-K'%round(s21i,2);\n", - "\n", - "#part - b\n", - "print('part - b');\n", - "#by accounting for the deviation from ideal-gas behavior\n", - "TR1=T1/Tcr;\n", - "Pr1=P1/Pcr;\n", - "#from the generalized charts at each state\n", - "Zh1=0.53;\n", - "Zs1=0.25;\n", - "TR2=T2/Tcr;\n", - "Pr2=P2/Pcr;\n", - "#from the generalized charts at each state\n", - "Zh2=0.48;\n", - "Zs2=0.20;\n", - "h21=h21i-Ru*Tcr*(Zh2-Zh1);\n", - "s21=s21i-Ru*(Zs2-Zs1);\n", - "print'the enthalpy change %i in kJ/kmol'%round(h21);\n", - "print'the entropy change %f kJ/kmol-K'%round(s21,2);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "part - a\n", - "the enthalpy change 2332 kJ/kmol\n", - "the entropy change 3.280000 kJ/kmol-K\n", - "part - b\n", - "the enthalpy change 2396 in kJ/kmol\n", - "the entropy change 3.690000 kJ/kmol-K\n" - ] - } - ], - "prompt_number": 1 - } - ], - "metadata": {} - } - ] -} \ No newline at end of file -- cgit