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diff --git a/Thermodynamics_for_Engineers/Chapter_20_2.ipynb b/Thermodynamics_for_Engineers/Chapter_20_2.ipynb deleted file mode 100755 index a4aa5154..00000000 --- a/Thermodynamics_for_Engineers/Chapter_20_2.ipynb +++ /dev/null @@ -1,323 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:48ba3152e0507fdec1fb7b61f0c70216b3a6f652a3d1e87f4355bb850157326b"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 20 - Vapor Power Cycles"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1 - Pg 420"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the thermal efficiency in both cases\n",
- "#Initalization of variables\n",
- "Qs=825.1 #Btu/lb\n",
- "ds=0.9588\n",
- "t1=101.74 #F\n",
- "th=400.95 #F\n",
- "#calculations\n",
- "Qr=ds*(t1+459.69)\n",
- "work=Qs-Qr\n",
- "eta=work/Qs*100.\n",
- "eta2=(th-t1)/(th+459.69) *100.\n",
- "#results\n",
- "print '%s %.2f %s' %(\"In case 1, Thermal efficiency =\",eta,\"percent\")\n",
- "print '%s %.2f %s' %(\"\\n In case 2, Thermal efficiency =\",eta2,\" percent\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "In case 1, Thermal efficiency = 34.76 percent\n",
- "\n",
- " In case 2, Thermal efficiency = 34.77 percent\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2 - Pg 425"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the thermal efficiency in both cases\n",
- "#Initalization of variables\n",
- "s2=1.5263\n",
- "sfg=1.8456\n",
- "sf=1.9782 \n",
- "h2=1201.1 #Btu/lb\n",
- "hf=1106 #Btu/lb\n",
- "hfg=1036.3 #Btu/lb\n",
- "v=0.01616 #m^3/kg\n",
- "p2=250 #psia\n",
- "p1=1#psia\n",
- "J=778\n",
- "#calculations\n",
- "x3=1+ (s2-sf)/sfg\n",
- "h3=hf-(1-x3)*hfg\n",
- "h4=69.7\n",
- "Wp=v*144*(p2-p1)/J\n",
- "h1=h4+Wp\n",
- "etat=((h2-h3)-Wp)/(h2-h1) *100.\n",
- "eta2=(h2-h3)/(h2-h4)*100.\n",
- "#results\n",
- "print '%s %.2f %s' %(\"\\n In case 1, Efficieny =\",etat,\"percent\")\n",
- "print '%s %.2f %s' %(\"\\n In case 2, Efficieny =\",eta2,\"percent\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- " In case 1, Efficieny = 30.79 percent\n",
- "\n",
- " In case 2, Efficieny = 30.83 percent\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3 - Pg 428"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the steam rate, enthalpy of exhaust steam, internal engine efficiency and excess heat removed\n",
- "#Initalization of variables\n",
- "p=40000 #kW\n",
- "ef=0.98\n",
- "rate=302000. #lb\n",
- "s3=1.6001\n",
- "h2=1490.1\n",
- "loss=600.\n",
- "v=400. #ft/s\n",
- "g=32.2 #ft/s^2\n",
- "J=778.\n",
- "#calculations\n",
- "out=p/(0.746*ef)\n",
- "srate=rate/out\n",
- "X=-(s3-1.9782)/1.8456\n",
- "h3=1106 - X*1036.3\n",
- "theoturb=h2-h3\n",
- "intturb=(out+loss)*2544/rate\n",
- "Ie=intturb/theoturb *100\n",
- "h3d=h2-intturb-v*v /(2*g*J)\n",
- "hexa=h3d+ v*v /(2*g*J)\n",
- "excess=rate*(hexa-h3)\n",
- "#results\n",
- "print '%s %.2f %s' %(\"Steam rate =\",srate,\"lb/shaft hp-hr\")\n",
- "print '%s %.1f %s' %(\"\\n Internal engine efficiency =\",Ie,\" percent\")\n",
- "print '%s %.1f %s' %(\"\\n Enthalpy of exhaust steam =\",h3d,\" Btu/lb\")\n",
- "print '%s %d %s' %(\"\\n Excess heat to be removed =\",excess,\"Btu/hr\")\n",
- "print '%s' %(\"The answers are a bit different due to rounding off error in textbook\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Steam rate = 5.52 lb/shaft hp-hr\n",
- "\n",
- " Internal engine efficiency = 78.1 percent\n",
- "\n",
- " Enthalpy of exhaust steam = 1021.0 Btu/lb\n",
- "\n",
- " Excess heat to be removed = 39395745 Btu/hr\n",
- "The answers are a bit different due to rounding off error in textbook\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 4 - Pg 436"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the thermal efficiency\n",
- "#Initalization of variables\n",
- "s2=1.5263\n",
- "sf=1.6993\n",
- "sfg=1.3313\n",
- "hf=1164.1 #Btu/lb\n",
- "hfg=945.3 #Btu/lb\n",
- "h2=1201.1 #Btu/lb\n",
- "h1=852.3 #Btu/lb\n",
- "#calculations\n",
- "X3=-(s2-sf)/sfg\n",
- "h3=hf-X3*hfg\n",
- "h4=218.82\n",
- "h6=h4\n",
- "h5=69.7\n",
- "x=(h4-h5)/(h3-h5)\n",
- "W= h2-h3+ (1-x)*(h3-h1)\n",
- "Qs=h2-h4\n",
- "eff=W/Qs *100\n",
- "#results\n",
- "print '%s %.2f %s' %(\"Thermal efficiency =\",eff,\"percent\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Thermal efficiency = 32.56 percent\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5 - Pg 442"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the thermal efficiency\n",
- "#Initalization of variables\n",
- "h6=157.933 #Btu/lb\n",
- "s2=0.11626\n",
- "sf=0.16594\n",
- "sfg=0.14755\n",
- "hf=139.095 #Btu/lb\n",
- "hfg=126.98 #Btu/lb\n",
- "h5=12.016 #Btu/lb\n",
- "h2=1201.1 #Btu/lb\n",
- "h1=69.7 #Btu/lb\n",
- "w=348.8 #Btu/lb\n",
- "m=0.0745 #lb\n",
- "#calculations\n",
- "x7=-(s2-sf)/sfg\n",
- "h7=hf-x7*hfg\n",
- "dh6=h6-h7\n",
- "mr=(h7-h5)/(h2-h1)\n",
- "work=w*m\n",
- "tw=work+dh6\n",
- "dh65=h6-h5\n",
- "eff=tw/dh65 *100\n",
- "#results\n",
- "print '%s %.2f %s' %(\"Thermal efficiency =\",eff,\"percent\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Thermal efficiency = 60.02 percent\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6 - Pg 443"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the available portion of heat transferred\n",
- "#Initalization of variables\n",
- "import math\n",
- "m=1 #lb\n",
- "cp=0.26\n",
- "t2=1800+460. #R\n",
- "t1=400.95+460 #R\n",
- "x=0.6\n",
- "sink=100+460. #R\n",
- "tm=2600+460. #R\n",
- "#calculations\n",
- "Q=m*cp*(t2-t1)\n",
- "ds=m*cp*math.log((t2/t1))\n",
- "tds=ds*(sink)\n",
- "avail=Q-tds\n",
- "hf=Q*x/(1-x)\n",
- "av2=hf*(tm-sink)/(tm)\n",
- "Qt=Q+hf\n",
- "av=avail+av2\n",
- "per=av/Qt *100.\n",
- "#results\n",
- "print '%s %.1f %s' %(\"Available portion of heat transferred =\",per,\"percent\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Available portion of heat transferred = 73.6 percent\n"
- ]
- }
- ],
- "prompt_number": 6
- }
- ],
- "metadata": {}
- }
- ]
-}
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