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| author | kinitrupti | 2017-05-12 18:40:35 +0530 |
|---|---|---|
| committer | kinitrupti | 2017-05-12 18:40:35 +0530 |
| commit | d36fc3b8f88cc3108ffff6151e376b619b9abb01 (patch) | |
| tree | 9806b0d68a708d2cfc4efc8ae3751423c56b7721 /Elements_of_Thermodynamics_and_heat_transfer | |
| parent | 1b1bb67e9ea912be5c8591523c8b328766e3680f (diff) | |
| download | Python-Textbook-Companions-d36fc3b8f88cc3108ffff6151e376b619b9abb01.tar.gz Python-Textbook-Companions-d36fc3b8f88cc3108ffff6151e376b619b9abb01.tar.bz2 Python-Textbook-Companions-d36fc3b8f88cc3108ffff6151e376b619b9abb01.zip | |
Revised list of TBCs
Diffstat (limited to 'Elements_of_Thermodynamics_and_heat_transfer')
16 files changed, 0 insertions, 5465 deletions
diff --git a/Elements_of_Thermodynamics_and_heat_transfer/Chapter_1.ipynb b/Elements_of_Thermodynamics_and_heat_transfer/Chapter_1.ipynb deleted file mode 100755 index 608b71cc..00000000 --- a/Elements_of_Thermodynamics_and_heat_transfer/Chapter_1.ipynb +++ /dev/null @@ -1,230 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:b4cfb785282e641d646d41001a3d0f6e3b05cd9780e1e3c87c417c701544b8b7"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 1 - Units and Dimensions"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1 - Pg 4"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the force required to accelerate\n",
- "#Initialization of variables\n",
- "gc=32.1739 #lbm ft/lbf s^2\n",
- "m=10 #lbm\n",
- "a=10. #ft/s^2\n",
- "#calculations\n",
- "F=m*a/gc\n",
- "#results\n",
- "print '%s %.3f %s' %(\"Force to accelerate =\",F,\"lbf\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Force to accelerate = 3.108 lbf\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2 - Pg 4"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the force required to accelerate\n",
- "#Initialization of variables\n",
- "gc=32.1739 #lbm ft/lbf s^2\n",
- "m=10. #lbm\n",
- "a=gc #ft/s^2\n",
- "#calculations\n",
- "F=m*a/gc\n",
- "#results\n",
- "print '%s %d %s' %(\"Force to accelerate =\",F,\"lbf\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Force to accelerate = 10 lbf\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3 - Pg 4"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the force required to accelerate\n",
- "#Initialization of variables\n",
- "gc=32.1739 #lbm ft/lbf s^2\n",
- "F=5.00e-9 #lbf hr/ft^2\n",
- "#calculations\n",
- "F2=F*3600*gc\n",
- "#results\n",
- "print '%s %.2e %s' %(\"Force required =\",F2,\"lbm/ft sec\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Force required = 5.79e-04 lbm/ft sec\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 4 - Pg 7"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the velocity\n",
- "#Initialization of variables\n",
- "v=88. #ft/s\n",
- "#calculations\n",
- "v2=v*3600./5280.\n",
- "#results\n",
- "print '%s %d %s' %(\"velocity =\",v2,\"mph\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "velocity = 60 mph\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5 - Pg 7"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the velocity\n",
- "#Initialization of variables\n",
- "v=88. #ft/s\n",
- "#calculations\n",
- "v2=v*1/5280*3600.\n",
- "#results\n",
- "print '%s %d %s' %(\"velocity =\",v2,\"mph\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "velocity = 60 mph\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6 - Pg 9"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the density of water and also the specific weight\n",
- "#Initialization of variables\n",
- "rho=62.305 #lbf/ft^2\n",
- "g=32.1739 #ft/s^2\n",
- "#calculations\n",
- "gam=rho/g\n",
- "#results\n",
- "print '%s %.3f %s' %(\"Density of water in this system =\",gam,\"lbf/ft^2\")\n",
- "print '%s %.3f %s' %(\"\\n Specific weight =\",rho,\"lbf/ft^2\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Density of water in this system = 1.937 lbf/ft^2\n",
- "\n",
- " Specific weight = 62.305 lbf/ft^2\n"
- ]
- }
- ],
- "prompt_number": 6
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Elements_of_Thermodynamics_and_heat_transfer/Chapter_10.ipynb b/Elements_of_Thermodynamics_and_heat_transfer/Chapter_10.ipynb deleted file mode 100755 index ff3609da..00000000 --- a/Elements_of_Thermodynamics_and_heat_transfer/Chapter_10.ipynb +++ /dev/null @@ -1,151 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:1bfddf04d127361fe3f8fbf16a545bcc1144ba2e77c1a2b0127e1628ecf77a23"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 10 - The pvT relationships"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1 - Pg 164"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the pressure in ideal gas and vanderwaals case\n",
- "#Initialization of variables\n",
- "m=1 #lbm\n",
- "T1=212+460. #R\n",
- "sv=0.193 #ft^3/lbm\n",
- "M=44\n",
- "a=924.2 #atm ft^2 /mole^2\n",
- "b=0.685 #ft^3/mol\n",
- "R=0.73 #atm ft^3/R mol\n",
- "#calculations\n",
- "v=sv*M\n",
- "p=R*T1/v\n",
- "p2=R*T1/(v-b) -a/v**2\n",
- "#results\n",
- "print '%s %.1f %s' %(\"In ideal gas case, pressure =\",p,\" atm\")\n",
- "print '%s %.1f %s' %(\"\\n In vanderwaals equation, pressure =\",p2,\" atm\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "In ideal gas case, pressure = 57.8 atm\n",
- "\n",
- " In vanderwaals equation, pressure = 50.0 atm\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2 - Pg 166"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the volume\n",
- "#Initialization of variables\n",
- "m=1 #lbm\n",
- "p=50.9 #atm\n",
- "t=212+460 #R\n",
- "R=0.73\n",
- "#calculations\n",
- "pc=72.9 #atm\n",
- "tc=87.9 +460 #R\n",
- "pr=p/pc\n",
- "Tr=t/tc\n",
- "z=0.88\n",
- "v=z*R*t/p\n",
- "#results\n",
- "print '%s %.3f %s' %(\"volume =\",v,\"ft^3/mole\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "volume = 8.481 ft^3/mole\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3 - Pg 167"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Pressure\n",
- "#Initialization of variables\n",
- "t=212+460. #R\n",
- "v=0.193 #ft^3/lbm\n",
- "M=44.\n",
- "R=0.73\n",
- "#calculations\n",
- "tc=87.9+460 #F\n",
- "zc=0.275\n",
- "vc=1.51 #ft^3/mol\n",
- "tr=t/tc\n",
- "vr=v*M/vc\n",
- "vrd=vr*zc\n",
- "z=0.88\n",
- "p=z*R*t/(M*v)\n",
- "#results\n",
- "print '%s %.1f %s' %(\"Pressure =\",p,\"atm\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Pressure = 50.8 atm\n"
- ]
- }
- ],
- "prompt_number": 3
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Elements_of_Thermodynamics_and_heat_transfer/Chapter_11.ipynb b/Elements_of_Thermodynamics_and_heat_transfer/Chapter_11.ipynb deleted file mode 100755 index 4aed469b..00000000 --- a/Elements_of_Thermodynamics_and_heat_transfer/Chapter_11.ipynb +++ /dev/null @@ -1,696 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:37c7e0de5d214bd8e1262734b18b139b442cc87b722d8f6c3ad97dd3d8903ad3"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 11 - The ideal gas and mixture relationships"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1 - Pg 184"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the work done\n",
- "#Initialization of variables\n",
- "n=1.3\n",
- "T1=460+60. #R\n",
- "P1=14.7 #psia\n",
- "P2=125. #psia\n",
- "R=1545.\n",
- "M=29.\n",
- "#calculations\n",
- "T2=T1*(P2/P1)**((n-1)/n)\n",
- "wrev=R/M *(T2-T1)/(1-n)\n",
- "#results\n",
- "print '%s %d %s' %(\"Work done =\",wrev,\"ft-lbf/lbm\")\n",
- "print '%s' %(\"The answer is a bit different due to rounding off error in textbook\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Work done = -58988 ft-lbf/lbm\n",
- "The answer is a bit different due to rounding off error in textbook\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2 - Pg 184"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Change in kinetic energy\n",
- "#Initialization of variables\n",
- "P2=10 #psia\n",
- "P1=100 #psia\n",
- "T1=900 #R\n",
- "w=50 #Btu/lbm\n",
- "k=1.39\n",
- "cp=0.2418\n",
- "#calculations\n",
- "T2=T1*(P2/P1)**((k-1)/k)\n",
- "T2=477\n",
- "KE=-w-cp*(T2-T1)\n",
- "#results\n",
- "print '%s %.1f %s' %(\"Change in kinetic energy =\",KE,\"Btu/lbm\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Change in kinetic energy = 52.3 Btu/lbm\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3 - Pg 190"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Final temperature\n",
- "#Initialization of variables\n",
- "T1=900 #R\n",
- "P1=100 #psia\n",
- "P2=10 #psia\n",
- "#calculations\n",
- "print '%s' %(\"From table B-9\")\n",
- "pr1=8.411\n",
- "pr2=pr1*P2/P1\n",
- "T2=468 #R\n",
- "#results\n",
- "print '%s %.1f %s' %(\"Final temperature =\",T2,\"R \")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "From table B-9\n",
- "Final temperature = 468.0 R \n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 4 - Pg 190"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the final temperature and pressure, work done and enthalpy\n",
- "#Initialization of variables\n",
- "cr=6\n",
- "p1=14.7 #psia\n",
- "t1=60.3 #F\n",
- "M=29\n",
- "R=1.986\n",
- "#calculations\n",
- "print '%s' %(\"from table b-9\")\n",
- "vr1=158.58 \n",
- "u1=88.62 #Btu/lbm\n",
- "pr1=1.2147\n",
- "vr2=vr1/cr\n",
- "T2=1050 #R\n",
- "u2=181.47 #Btu/lbm\n",
- "pr2=14.686\n",
- "p2=p1*(pr2/pr1)\n",
- "dw=u1-u2\n",
- "h2=u2+T2*R/M\n",
- "#results\n",
- "print '%s %d %s' %(\"final temperature =\",T2,\"R\")\n",
- "print '%s %.1f %s' %(\"\\n final pressure =\",p2,\" psia\")\n",
- "print '%s %.2f %s' %(\"\\n work done =\",dw,\" Btu/lbm\")\n",
- "print '%s %.1f %s' %(\"\\n final enthalpy =\",h2,\" Btu/lbm\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "from table b-9\n",
- "final temperature = 1050 R\n",
- "\n",
- " final pressure = 177.7 psia\n",
- "\n",
- " work done = -92.85 Btu/lbm\n",
- "\n",
- " final enthalpy = 253.4 Btu/lbm\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5 - Pg 193"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the mole fractions of oxygen and nitrogen, Average molecular weight and partial pressures, densities, volumes\n",
- "#Initialization of variables\n",
- "m1=10. #lbm\n",
- "m2=15. #lnm\n",
- "p=50. #psia\n",
- "t=60.+460 #R\n",
- "M1=32.\n",
- "M2=28.02\n",
- "R0=10.73 \n",
- "#calculations\n",
- "n1=m1/M1\n",
- "n2=m2/M2\n",
- "x1=n1/(n1+n2)\n",
- "x2=n2/(n1+n2)\n",
- "M=x1*M1+x2*M2\n",
- "R=1545/M\n",
- "V=(n1+n2)*R0*t/p\n",
- "rho=p/(R0*t)\n",
- "rho2=M*rho\n",
- "p1=x1*p\n",
- "p2=x2*p\n",
- "v1=x1*V\n",
- "v2=x2*V\n",
- "#results\n",
- "print '%s' %(\"part a\")\n",
- "print '%s %.3f %s %.3f %s' %(\"Mole fractions of oxygen and nitrogen are\",x1,\" and\",x2,\" respectively\")\n",
- "print '%s' %(\"part b\")\n",
- "print '%s %.1f' %(\"Average molecular weight = \",M)\n",
- "print '%s' %(\"part c\")\n",
- "print '%s %.2f %s' %(\"specific gas constant =\",R,\"psia ft^3/lbm R\")\n",
- "print '%s' %(\"part d\")\n",
- "print '%s %.1f %s' %(\"volume of mixture =\",V,\"ft^3\")\n",
- "print '%s %.5f %s %.3f %s' %(\"density of mixture is\",rho, \"mole/ft^3 and\",rho2, \"lbm/ft^3\")\n",
- "print '%s' %(\"part e\")\n",
- "print '%s %.2f %s %.2f %s' %(\"partial pressures of oxygen and nitrogen are\",p1,\"psia and\",p2,\"psia respectively\")\n",
- "print '%s %.2f %s %.2f %s' %(\"partial volumes of oxygen and nitrogen are\",v1,\"ft^3 and\",v2,\"ft^3 respectively\")"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "part a\n",
- "Mole fractions of oxygen and nitrogen are 0.369 and 0.631 respectively\n",
- "part b\n",
- "Average molecular weight = 29.5\n",
- "part c\n",
- "specific gas constant = 52.40 psia ft^3/lbm R\n",
- "part d\n",
- "volume of mixture = 94.6 ft^3\n",
- "density of mixture is 0.00896 mole/ft^3 and 0.264 lbm/ft^3\n",
- "part e\n",
- "partial pressures of oxygen and nitrogen are 18.43 psia and 31.57 psia respectively\n",
- "partial volumes of oxygen and nitrogen are 34.87 ft^3 and 59.74 ft^3 respectively\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6 - Pg 195"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the gravimetric and ultimate analysis\n",
- "#Initialization of variables\n",
- "m1=5.28\n",
- "m2=1.28\n",
- "m3=23.52\n",
- "#calculations\n",
- "m=m1+m2+m3\n",
- "x1=m1/m*100\n",
- "x2=m2/m*100\n",
- "x3=m3/m*100\n",
- "C=12./44 *m1/ m*100\n",
- "O=(32./44 *m1 + m2)/m*100\n",
- "N=m3/m*100\n",
- "#results\n",
- "print '%s %.1f %s %.1f %s %.1f %s' %(\"From gravimetric analysis, co2 =\",x1,\"percent , o2 =\",x2,\"percent and n2 =\",x3,\"percent\")\n",
- "print '%s %.2f %s %.2f %s %.2f %s' %(\"\\n From ultimate analysis, co2 =\",C,\"percent , o2 =\",O,\" percent and n2 =\",N,\" percent\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "From gravimetric analysis, co2 = 17.6 percent , o2 = 4.3 percent and n2 = 78.2 percent\n",
- "\n",
- " From ultimate analysis, co2 = 4.79 percent , o2 = 17.02 percent and n2 = 78.19 percent\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7 - Pg 197"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Entropy of the mixture\n",
- "#Initialization of variables\n",
- "import math\n",
- "x1=1/3.\n",
- "n1=1.\n",
- "n2=2.\n",
- "x2=2/3.\n",
- "p=12.7 #psia\n",
- "cp1=7.01 #Btu/mole R\n",
- "cp2=6.94 #Btu/mole R\n",
- "R0=1.986\n",
- "T2=460+86.6 #R\n",
- "T1=460. #R\n",
- "p0=14.7 #psia\n",
- "#calculations\n",
- "p1=x1*p\n",
- "p2=x2*p\n",
- "ds1= cp1*math.log(T2/T1) - R0*math.log(p1/p0)\n",
- "ds2= cp2*math.log(T2/T1) - R0*math.log(p2/p0)\n",
- "S=n1*ds1+n2*ds2\n",
- "#results\n",
- "print '%s %.2f %s' %(\"Entropy of mixture =\",S,\"Btu/R\")\n",
- "print '%s' %(\"\\n the answer given in textbook is wrong. please check using a calculator\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Entropy of mixture = 8.27 Btu/R\n",
- "\n",
- " the answer given in textbook is wrong. please check using a calculator\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 8 - Pg 198"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the change in internal energy and entropy\n",
- "#Initialization of variables\n",
- "import math\n",
- "c1=4.97 #Btu/mol R\n",
- "c2=5.02 #Btu/mol R\n",
- "n1=2\n",
- "n2=1\n",
- "T1=86.6+460 #R\n",
- "T2=50.+460 #R\n",
- "#calculations\n",
- "du=(n1*c1+n2*c2)*(T2-T1)\n",
- "ds=(n1*c1+n2*c2)*math.log(T2/T1)\n",
- "#results\n",
- "print '%s %d %s' %(\"Change in internal energy =\",du,\" Btu\")\n",
- "print '%s %.3f %s' %(\"\\n Change in entropy =\",ds,\"Btu/R\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Change in internal energy = -547 Btu\n",
- "\n",
- " Change in entropy = -1.037 Btu/R\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9 - Pg 198"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Pressure of the mixture\n",
- "#Initialization of variables\n",
- "n1=1\n",
- "n2=2\n",
- "c1=5.02\n",
- "c2=4.97\n",
- "t1=60. #F\n",
- "t2=100. #F\n",
- "R0=10.73\n",
- "p1=30. #psia\n",
- "p2=10. #psia\n",
- "#calcualtions\n",
- "t=(n1*c1*t1+n2*c2*t2)/(n1*c1+n2*c2)\n",
- "V1= n1*R0*(t1+460)/p1\n",
- "V2=n2*R0*(t2+460)/p2\n",
- "V=V1+V2\n",
- "pm=(n1+n2)*R0*(t+460)/V\n",
- "#results\n",
- "print '%s %.1f %s' %(\"Pressure of mixture =\",pm,\" psia\")\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Pressure of mixture = 12.7 psia\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 10 - Pg 199"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the change in entropy of gas \n",
- "#Initialization of variables\n",
- "import math\n",
- "T2=546.6 #R\n",
- "T1=520 #R\n",
- "T3=560 #R\n",
- "v2=1389.2\n",
- "v1=186.2\n",
- "R0=1.986\n",
- "c1=5.02\n",
- "c2=4.97\n",
- "n1=1\n",
- "n2=2\n",
- "v3=1203\n",
- "#calculations\n",
- "ds1=n1*c1*math.log(T2/T1) + n1*R0*math.log(v2/v1)\n",
- "ds2=n2*c2*math.log(T2/T3)+n2*R0*math.log(v2/v3)\n",
- "ds=ds1+ds2\n",
- "#results\n",
- "print '%s %.3f %s' %(\"Change in entropy for gas 1 =\",ds1,\" Btu/R\")\n",
- "print '%s %.3f %s' %(\"\\n Change in entropy for gas 1 =\",ds2,\"Btu/R\")\n",
- "print '%s %.3f %s' %(\"\\n Net change in entropy =\",ds,\"Btu/R\")\n",
- "print '%s' %(\"The answer is a bit different due to rounding off error in the textbook\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Change in entropy for gas 1 = 4.242 Btu/R\n",
- "\n",
- " Change in entropy for gas 1 = 0.331 Btu/R\n",
- "\n",
- " Net change in entropy = 4.572 Btu/R\n",
- "The answer is a bit different due to rounding off error in the textbook\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 11 - Pg 200"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Final temperature and change in entropy of air and water.\n",
- "#Initialization of variables\n",
- "import math\n",
- "m1=1. #lbm\n",
- "m2=0.94 #lbm\n",
- "M1=29.\n",
- "M2=18.\n",
- "p1=50. #psia\n",
- "p2=100. #psia\n",
- "t1=250 +460. #R\n",
- "R0=1.986\n",
- "cpa=6.96\n",
- "cpb=8.01\n",
- "#calculations\n",
- "xa = (m1/M1)/((m1/M1)+ m2/M2)\n",
- "xb=1-xa\n",
- "t2=t1*(p2/p1)**(R0/(xa*cpa+xb*cpb))\n",
- "d=R0/(xa*cpa+xb*cpb)\n",
- "k=1/(1.-d)\n",
- "dsa=cpa*math.log(t2/t1) -R0*math.log(p2/p1)\n",
- "dSa=(m1/M1)*dsa\n",
- "dSw=-dSa\n",
- "dsw=dSw*M2/m2\n",
- "#results\n",
- "print '%s %d %s' %(\"Final temperature =\",t2,\" R\")\n",
- "print '%s %.3f %s %.5f %s' %(\"\\n Change in entropy of air =\",dsa,\" btu/mole R and\",dSa, \"Btu/R\")\n",
- "print '%s %.4f %s %.5f %s' %(\"\\n Change in entropy of water =\",dsw,\"btu/mole R and\",dSw,\" Btu/R\")\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": [
- "Final temperature = 851 R\n",
- "\n",
- " Change in entropy of air = -0.115 btu/mole R and -0.00395 Btu/R\n",
- "\n",
- " Change in entropy of water = 0.0757 btu/mole R and 0.00395 Btu/R\n",
- "The answers are a bit different due to rounding off error in textbook\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12 - Pg 202"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the volume occupied and mass of steam\n",
- "#Initialization of variables\n",
- "T=250. + 460 #R\n",
- "p=29.825 #psia\n",
- "pt=50 #psia\n",
- "vg=13.821 #ft^3/lbm\n",
- "M=29.\n",
- "R=10.73\n",
- "#calculations\n",
- "pa=pt-p\n",
- "V=1/M *R*T/pa\n",
- "ma=V/vg\n",
- "xa=p/pt\n",
- "mb=xa/M *18./(1.-xa)\n",
- "#results\n",
- "print '%s %.2f %s' %(\"In case 1, volume occupied =\",V,\" ft^3\")\n",
- "print '%s %.2f %s' %(\"\\n In case 1, mass of steam =\",ma,\" lbm steam\")\n",
- "print '%s %.3f %s' %(\"\\n In case 2, mass of steam =\",mb,\" lbm steam\")\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "In case 1, volume occupied = 13.02 ft^3\n",
- "\n",
- " In case 1, mass of steam = 0.94 lbm steam\n",
- "\n",
- " In case 2, mass of steam = 0.918 lbm steam\n"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13 - Pg 203"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calcualte the percentage\n",
- "#Initialization of variables\n",
- "ps=0.64 #psia\n",
- "p=14.7 #psia\n",
- "M=29.\n",
- "M2=46.\n",
- "#calculations\n",
- "xa=ps/p\n",
- "mb=xa*9./M *M2/(1-xa)*100\n",
- "#results\n",
- "print '%s %.1f %s' %(\"percentage =\",mb,\"percent\")\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "percentage = 65.0 percent\n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14 - Pg 203"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the partial pressure of water vapor\n",
- "#Initialization of variables\n",
- "ps=0.5069 #psia\n",
- "p=20 #psia\n",
- "m1=0.01\n",
- "m2=1\n",
- "M1=18.\n",
- "M2=29.\n",
- "#calculations\n",
- "xw= (m1/M1)/(m1/M1+m2/M2)\n",
- "pw=xw*p\n",
- "#results\n",
- "print '%s %.3f %s' %(\"partial pressure of water vapor =\",pw,\"psia\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "partial pressure of water vapor = 0.317 psia\n"
- ]
- }
- ],
- "prompt_number": 14
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Elements_of_Thermodynamics_and_heat_transfer/Chapter_12.ipynb b/Elements_of_Thermodynamics_and_heat_transfer/Chapter_12.ipynb deleted file mode 100755 index e7d6658a..00000000 --- a/Elements_of_Thermodynamics_and_heat_transfer/Chapter_12.ipynb +++ /dev/null @@ -1,203 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:fb635a7892259910462c375c5cd6000ddf39095d8593a85d6b876d44f34c4760"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 12 - Non steady flow, friction and availability"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1 - Pg 210"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the work done\n",
- "#Initialization of variables\n",
- "p1=100 #psia\n",
- "p2=14.7 #psia\n",
- "k=1.4\n",
- "T1=700 #R\n",
- "R=10.73/29.\n",
- "V=50\n",
- "cv=0.171\n",
- "cp=0.24\n",
- "R2=1.986/29.\n",
- "#calculations\n",
- "T2=T1/ (p1/p2)**((k-1)/k)\n",
- "m1=p1*V/(R*T1)\n",
- "m2=p2*V/(R*T2)\n",
- "Wrev= cv*(m1*T1 - m2*T2) - (m1-m2)*(T2)*cp\n",
- "#results\n",
- "print '%s %d %s' %(\"Work done in case 1 =\",Wrev,\"Btu\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Work done in case 1 = 572 Btu\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2 - Pg 212"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Engine efficiency and Effectiveness\n",
- "#Initialization of variables\n",
- "p1=400 #psia\n",
- "t1=600 #F\n",
- "h1=1306.9 #Btu/lbm\n",
- "b1=480.9 #Btu/lbm\n",
- "p2=50 #psia\n",
- "h2=1122 #Btu/lbm\n",
- "h3=1169.5 #Btu/lbm\n",
- "b3=310.9 #Btu/lbm\n",
- "#calculations\n",
- "print '%s' %(\"All the values are obtained from Mollier chart,\")\n",
- "dw13=h1-h3\n",
- "dw12=h1-h2\n",
- "dasf=b3-b1\n",
- "etae=dw13/dw12*100\n",
- "eta=abs(dw13/dasf)*100\n",
- "dq=dw13+dasf\n",
- "#results\n",
- "print '%s %.1f %s' %(\"Engine efficiency =\",etae,\" percent\")\n",
- "print '%s %.1f %s' %(\"\\n Effectiveness =\",eta,\" percent\")\n",
- "print '%s %.1f %s' %(\"\\n Loss of available energy =\",dq,\" Btu/lbm\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "All the values are obtained from Mollier chart,\n",
- "Engine efficiency = 74.3 percent\n",
- "\n",
- " Effectiveness = 80.8 percent\n",
- "\n",
- " Loss of available energy = -32.6 Btu/lbm\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3 - Pg 214"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the friction of the process per pound of air\n",
- "#Initialization of variables\n",
- "p1=100. #psia\n",
- "p2=10. #psia\n",
- "n=1.3\n",
- "T1=800. #R\n",
- "cv=0.172\n",
- "R=1.986/29\n",
- "#calculations\n",
- "T2=T1*(p2/p1)**((n-1)/n)\n",
- "dwir=cv*(T1-T2)\n",
- "dwr=R*(T2-T1)/(1-n)\n",
- "dq=dwr-dwir\n",
- "#results\n",
- "print '%s %.1f %s' %(\"The friction of the process per pound of air =\",dq,\" Btu/lbm\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The friction of the process per pound of air = 18.6 Btu/lbm\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 4 - Pg 215"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the Friction\n",
- "#Initialization of variables\n",
- "ms=10 #lbm\n",
- "den=62.3 #lbm/ft^3\n",
- "A1=0.0218 #ft^2\n",
- "A2=0.00545 #ft^2\n",
- "p2=50. #psia\n",
- "p1=100. #psia\n",
- "gc=32.2 #ft/s^2\n",
- "dz=30. #ft\n",
- "#calculations\n",
- "V1=ms/(A1*den)\n",
- "V2=ms/(A2*den)\n",
- "df=-144./den*(p2-p1) - (V2**2 -V1**2)/(2*gc) - dz\n",
- "#results\n",
- "print '%s %.1f %s' %(\"Friction =\",df,\"ft-lbf/lbm\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Friction = 72.9 ft-lbf/lbm\n"
- ]
- }
- ],
- "prompt_number": 4
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Elements_of_Thermodynamics_and_heat_transfer/Chapter_13.ipynb b/Elements_of_Thermodynamics_and_heat_transfer/Chapter_13.ipynb deleted file mode 100755 index b1a173c6..00000000 --- a/Elements_of_Thermodynamics_and_heat_transfer/Chapter_13.ipynb +++ /dev/null @@ -1,502 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:34e2c9e74989754957ab71e3380ee9241975584f70767576f7f879e815515185"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 13 - Fluid flow"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1 - Pg 223"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the velocity and area\n",
- "#Initialization of variables\n",
- "import math\n",
- "import numpy\n",
- "h1=1329.1 #Btu/lbm\n",
- "v1=6.218 #ft^3/lbm\n",
- "J=778.\n",
- "g=32.174\n",
- "m=1.\n",
- "#calculations\n",
- "p=([80., 60., 54.6, 40., 20.])\n",
- "h=([ 1304.1, 1273.8, 1265, 1234.2, 1174.8])\n",
- "v=([ 7.384, 9.208, 9.844, 12.554, 21.279])\n",
- "Fc=1.\n",
- "b=len(p)\n",
- "V2=numpy.zeros(b)\n",
- "A=numpy.zeros(b)\n",
- "for i in range (1,b):\n",
- "\tV2[i]=round(Fc*math.sqrt(2*J*g*(h1-h[i])),2)\n",
- "\tA[i]=round(m*v[i] /V2[i],5)\n",
- "\n",
- "V2 = 0+ V2\n",
- "A =0+ A\n",
- "#results\n",
- "print '%s' %('velocity (ft/s)= ')\n",
- "print(V2)\n",
- "print '%s' %('Area (ft^2)= ')\n",
- "print(A)\n",
- "#The initial values of velocity and area are 0 and infinity respectively\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "velocity (ft/s)= \n",
- "[ 0. 1663.87 1791.37 2179.67 2779.33]\n",
- "Area (ft^2)= \n",
- "[ 0. 0.00553 0.0055 0.00576 0.00766]\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2 - Pg 228"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Area required in both cases\n",
- "#Initialization of variables\n",
- "import math\n",
- "n=1.4\n",
- "p1=50. #psia\n",
- "J=778.\n",
- "cp=0.24\n",
- "T1=520. #R\n",
- "k=n\n",
- "R=1545/29.\n",
- "m=1.\n",
- "p2=10. #psia\n",
- "#calculations\n",
- "rpt=(2/(n+1))**(n/(n-1))\n",
- "pt=p1*rpt\n",
- "Vtrev=223.77*math.sqrt(cp*T1*(1- rpt**((k-1)/k)))\n",
- "v1=R*T1/p1/144\n",
- "vt=v1*(p1/pt)**(1./k)\n",
- "At=m*vt/Vtrev\n",
- "V2rev=223.77*math.sqrt(cp*T1*(1-(p2/p1)**((k-1)/k)))\n",
- "v2=v1*(p1/p2)**(1/k)\n",
- "A2=m*v2/V2rev\n",
- "#results\n",
- "print '%s %.5f %s' %(\"Area required =\",At,\" ft^2\")\n",
- "print '%s %.5f %s' %(\"\\n Area in case 2 =\",A2,\" ft^2\")\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Area required = 0.00595 ft^2\n",
- "\n",
- " Area in case 2 = 0.00800 ft^2\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3 - Pg 231"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Throat area\n",
- "#Initialization of variables\n",
- "J=778.\n",
- "g=32.2\n",
- "pc=54.6 #psia\n",
- "h1=1329.1 #Btu/lbm\n",
- "h2=1265. #btu/lbm\n",
- "V2rev=1790. #ft/s\n",
- "cv=0.99\n",
- "m=1 #lbm\n",
- "cv2=0.96\n",
- "#calculations\n",
- "V2d=cv*V2rev\n",
- "hd=cv**2 *(h1-h2)\n",
- "h2d=h1-hd\n",
- "v2d=9.946\n",
- "A2d=m*v2d/V2d\n",
- "#results\n",
- "print '%s %.4f %s' %(\"Throat area in case 2 =\",A2d,\" ft^2\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Throat area in case 2 = 0.0056 ft^2\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 4 - Pg 234"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the mass flow rate\n",
- "#Initialization of variables\n",
- "import math\n",
- "p1=50. #psia\n",
- "pr=0.58\n",
- "#calculations\n",
- "p=p1*pr\n",
- "s1=1.6585\n",
- "h1=1174.1 #Btu/lbm\n",
- "sf=0.3680\n",
- "sfg=1.3313\n",
- "hfg=945.3\n",
- "vg=13.746\n",
- "hf=218.82\n",
- "x= (s1-sf)/sfg\n",
- "v2=vg*x\n",
- "h2=hf+x*hfg\n",
- "V2rev=223.77*math.sqrt(h1-h2)\n",
- "m=math.pi/4 *1/144. *V2rev/v2\n",
- "#results\n",
- "print '%s %.3f %s' %(\"mass flow rate =\",m,\" lbm/sec\")\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "mass flow rate = 0.572 lbm/sec\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5 - Pg 234"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Mass flow rate and Meta stable under cooling\n",
- "#Initialization of variables\n",
- "import math\n",
- "k=1.31\n",
- "p1=7200. #lbf/ft**2\n",
- "v1=8.515 #ft**3/lbm\n",
- "pr=0.6\n",
- "m1=0.574\n",
- "T1=741. #R\n",
- "#calculations\n",
- "V2rev=8.02*math.sqrt(k/(k-1) *p1*v1*(1- (pr)**((k-1)/k)))\n",
- "v2=v1*(1/pr)**(1/k)\n",
- "m=math.pi/4 *1/144 *V2rev/v2\n",
- "C=m/m1\n",
- "T2=T1*(0.887)\n",
- "t=250+460. #R\n",
- "dt=t-T2\n",
- "#results\n",
- "print '%s %.3f %s' %(\"Mass flow rate =\",m,\" lbm/sec\")\n",
- "print '%s %d %s' %(\"\\n Meta stable under cooling =\",dt,\"F\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Mass flow rate = 0.597 lbm/sec\n",
- "\n",
- " Meta stable under cooling = 52 F\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6 - Pg 240"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the average velocity and mass flow rate\n",
- "#Initialization of variables\n",
- "import math\n",
- "zm=0.216\n",
- "pm=62.3 #lbm/ft**2\n",
- "p1=0.0736 #lbm/ft**2\n",
- "g=32.2\n",
- "d=4.\n",
- "#calculations\n",
- "H=zm*(pm-p1)/12/p1\n",
- "V=math.sqrt(2*g*H)\n",
- "m=math.pi/4 *d**2 *V*p1\n",
- "#results\n",
- "print '%s %.1f %s' %(\"average velocity =\",V,\" ft/sec\")\n",
- "print '%s %.1f %s' %(\"\\n mass flow rate =\",m,\" lbm/sec\")\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "average velocity = 31.3 ft/sec\n",
- "\n",
- " mass flow rate = 29.0 lbm/sec\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7 - Pg 244"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the area of throat and area of exit\n",
- "#Initialization of variables\n",
- "import math\n",
- "p0=50. #psia\n",
- "T0=520. #R\n",
- "rho0=0.259 #lbm/ft^3\n",
- "p2=10. #psia\n",
- "mf=1. #lbm\n",
- "#calculations\n",
- "print '%s' %(\"From table B-17,\")\n",
- "pr=0.528\n",
- "Tr=0.833\n",
- "rhor=0.634\n",
- "ps=pr*p0\n",
- "Ts=Tr*T0\n",
- "rhos=rho0*rhor\n",
- "Vs=49.1*math.sqrt(Ts)\n",
- "As=mf/(Vs*rhos)\n",
- "p2r=p2/p0\n",
- "M2=1.71\n",
- "V2=1.487*Vs\n",
- "T2=0.632*Ts\n",
- "A2=As*1.35\n",
- "rho2=rhos*0.317\n",
- "#results\n",
- "print '%s %.5f %s' %(\"Area of throat =\",As,\"ft^2\")\n",
- "print '%s %.5f %s' %(\"\\n Area of exit =\",A2,\"ft^2\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "From table B-17,\n",
- "Area of throat = 0.00596 ft^2\n",
- "\n",
- " Area of exit = 0.00805 ft^2\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 8 - Pg 247"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Length of the pipe\n",
- "#Initialization of variables\n",
- "M1=0.2\n",
- "M2=0.4\n",
- "D=0.5 #ft\n",
- "f=0.015\n",
- "#calculations\n",
- "f1=14.5\n",
- "f2=2.31\n",
- "dl=(f1-f2)*D/f\n",
- "#results\n",
- "print '%s %.1f %s' %(\"Length of pipe =\",dl,\"ft\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Length of pipe = 406.3 ft\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9 - Pg 248"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the change in entropy\n",
- "#Initialization of variables\n",
- "import math\n",
- "py=20. #psia\n",
- "px=3.55 #psia\n",
- "R=1.986/29\n",
- "#calculations\n",
- "pr=py/px\n",
- "print '%s' %(\"from table B-19\")\n",
- "Mx=2\n",
- "My=0.577\n",
- "pr2=0.721\n",
- "ds=R*math.log(1./pr2)\n",
- "#results\n",
- "print '%s %.4f %s' %(\"Change in entropy =\",ds,\"Btu/lbm R\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "from table B-19\n",
- "Change in entropy = 0.0224 Btu/lbm R\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 10 - Pg 249"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Internal and net thrust\n",
- "#Initialization of variables\n",
- "M1=0.5\n",
- "M2=1.\n",
- "A1=0.5 #ft^2\n",
- "A2=1. #ft^2\n",
- "p1=14.7 #psia\n",
- "p2=14.7 #psia\n",
- "k=1.4\n",
- "#calculations\n",
- "thru=p2*144*A2*(1+k*M2**2)-p1*144*A1*(1+k*M1**2)\n",
- "net=thru-p1*144*(A2-A1)\n",
- "#results\n",
- "print '%s %d %s' %(\"Internal thrust =\",thru,\"lbf\")\n",
- "print '%s %d %s' %(\"\\n Net thrust =\",net,\" lbf\")\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": [
- "Internal thrust = 3651 lbf\n",
- "\n",
- " Net thrust = 2593 lbf\n",
- "The answers are a bit different due to rounding off error in textbook\n"
- ]
- }
- ],
- "prompt_number": 10
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Elements_of_Thermodynamics_and_heat_transfer/Chapter_14.ipynb b/Elements_of_Thermodynamics_and_heat_transfer/Chapter_14.ipynb deleted file mode 100755 index f953d99a..00000000 --- a/Elements_of_Thermodynamics_and_heat_transfer/Chapter_14.ipynb +++ /dev/null @@ -1,547 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:6c33bc146fd289fab423bb4c093c8f33bd27dc001258cc7dbd47055d0bcb3468"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 14 - Psychrometrics"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1 - Pg 257"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the partial pressures, dew temperature and density of air, water and specific humidity, Degree of saturation\n",
- "#Initialization of variables\n",
- "t1=80+460 #R\n",
- "ps=0.5069 #psia\n",
- "print '%s' %(\"from steam tables,\")\n",
- "vs=633.1 #ft^3/lbm\n",
- "phi=0.3\n",
- "R=85.6\n",
- "Ra=53.3\n",
- "p=14.696\n",
- "#calculations\n",
- "tdew=46. #F\n",
- "pw=phi*ps\n",
- "rhos=1/vs\n",
- "rhow=phi*rhos\n",
- "rhow2= pw*144/(R*t1)\n",
- "pa=p-pw\n",
- "rhoa= pa*144/(Ra*t1)\n",
- "w=rhow/rhoa\n",
- "mu=phi*(p-ps)/(p-pw)\n",
- "Ws=0.622*(ps/(p-ps))\n",
- "mu2=w/Ws\n",
- "#results\n",
- "print '%s' %(\"part a\")\n",
- "print '%s %.5f %s' %(\"partial pressure of water =\",pw,\"psia\")\n",
- "print '%s %d %s' %(\"\\n dew temperature =\",tdew,\"F\")\n",
- "print '%s' %(\"part b\")\n",
- "print '%s %.6f %s' %(\"density of water =\",rhow,\"lbm/ft^3\")\n",
- "print '%s %.6f %s' %(\"\\n in case 2, density of water =\",rhow2,\"lbm/ft^3\")\n",
- "print '%s %.6f %s' %(\"\\n density of air =\",rhoa,\"lbm/ft^3\")\n",
- "print '%s' %(\"part c\")\n",
- "print '%s %.4f %s' %(\"specific humidity =\",w,\"lbm steam/lbm air\")\n",
- "print '%s' %(\"part d\")\n",
- "print '%s %.3f' %(\"In method 1, Degree of saturation = \",mu)\n",
- "print '%s %.3f' %(\"\\n In method 2, Degree of saturation = \",mu2)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "from steam tables,\n",
- "part a\n",
- "partial pressure of water = 0.15207 psia\n",
- "\n",
- " dew temperature = 46 F\n",
- "part b\n",
- "density of water = 0.000474 lbm/ft^3\n",
- "\n",
- " in case 2, density of water = 0.000474 lbm/ft^3\n",
- "\n",
- " density of air = 0.072765 lbm/ft^3\n",
- "part c\n",
- "specific humidity = 0.0065 lbm steam/lbm air\n",
- "part d\n",
- "In method 1, Degree of saturation = 0.293\n",
- "\n",
- " In method 2, Degree of saturation = 0.293\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2 - Pg 258"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the change in moisture content and change in moisture content\n",
- "#Initialization of variables\n",
- "p=14.696 #psia\n",
- "ps=0.0808 #psia\n",
- "ps2=0.5069 #psia\n",
- "phi2=0.5\n",
- "phi=0.6\n",
- "grain=7000.\n",
- "#calculations\n",
- "pw=phi*ps\n",
- "w1=0.622*pw/(p-pw)\n",
- "pw2=phi2*ps2\n",
- "w2=0.622*pw2/(p-pw2)\n",
- "dw=w2-w1\n",
- "dwg=dw*grain\n",
- "#results\n",
- "print '%s %.6f %s' %(\"change in moisture content =\",dw,\" lbm water/lbm dry air\")\n",
- "print '%s %.2f %s' %(\"\\n in grains, change =\",dwg,\" grains water/lbm dry air\")\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": [
- "change in moisture content = 0.008857 lbm water/lbm dry air\n",
- "\n",
- " in grains, change = 62.00 grains water/lbm dry air\n",
- "The answers are a bit different due to rounding off error in textbook\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3 - Pg 264"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the humidity ratio and relative humidity\n",
- "#Initialization of variables\n",
- "t1=80. #F\n",
- "t2=60. #F\n",
- "p=14.696 #psia\n",
- "ps=0.507 #psia\n",
- "pss=0.256 #psia\n",
- "cp=0.24\n",
- "#calculations\n",
- "ws=0.622*pss/(p-pss)\n",
- "w=(cp*(t2-t1) + ws*1060)/(1060+ 0.45*(t1-t2))\n",
- "pw=w*p/(0.622+w)\n",
- "phi=pw/ps*100\n",
- "td=46. #F\n",
- "#results\n",
- "print '%s %.4f %s' %(\"\\n humidity ratio =\",w,\"lbm/lbm dry air\")\n",
- "print '%s %.1f %s' %(\"\\n relative humidity =\",phi,\" percent\")\n",
- "print '%s %d %s' %(\"\\n Dew point =\",td,\"F\")\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- " humidity ratio = 0.0064 lbm/lbm dry air\n",
- "\n",
- " relative humidity = 29.7 percent\n",
- "\n",
- " Dew point = 46 F\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 4 - Pg 264"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the enthalpy and sigma function\n",
- "#Initialization of variables\n",
- "W=0.0065 #lbm/lbm of dry air\n",
- "t=80. #F\n",
- "td=60. #F\n",
- "#calculations\n",
- "H=0.24*t+W*(1060+0.45*t)\n",
- "sig=H-W*(td-32)\n",
- "Ws=0.0111\n",
- "H2=0.24*td+Ws*(1060+0.45*td)\n",
- "sig2=H2-Ws*(td-32)\n",
- "#results\n",
- "print '%s %.2f %s' %(\"In case 1, enthalpy =\",H,\" Btu/lbm dry air\")\n",
- "print '%s %.2f %s' %(\"\\n In case 1, sigma function =\",sig,\" Btu/lbm dry air\")\n",
- "print '%s %.2f %s' %(\"\\n In case 2, enthalpy =\",H2,\" Btu/lbm dry air\")\n",
- "print '%s %.2f %s' %(\"\\n In case 2, sigma function =\",sig2,\" Btu/lbm dry air\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "In case 1, enthalpy = 26.32 Btu/lbm dry air\n",
- "\n",
- " In case 1, sigma function = 26.14 Btu/lbm dry air\n",
- "\n",
- " In case 2, enthalpy = 26.47 Btu/lbm dry air\n",
- "\n",
- " In case 2, sigma function = 26.15 Btu/lbm dry air\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5 - Pg 264"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Enthalpy and heat added\n",
- "#Initialization of variables\n",
- "t1=30. #F\n",
- "t2=60. #F\n",
- "t3=80. #F\n",
- "W1=0.00206\n",
- "W2=0.01090\n",
- "#calculations\n",
- "cm1=0.24+0.45*W1\n",
- "H1=cm1*t1+W1*1060\n",
- "cm2=0.24+0.45*W2\n",
- "H2=cm2*t3+W2*1060\n",
- "hf=t2-32\n",
- "dq=H2-H1-(W2-W1)*hf\n",
- "#results\n",
- "print '%s %.2f %s' %(\"In case 1, Enthalpy =\",H1,\" Btu/lbm dry air\")\n",
- "print '%s %.2f %s' %(\"\\n In case 2, Enthalpy =\",H2,\" Btu/lbm dry air\")\n",
- "print '%s %.2f %s' %(\"\\n Heat added =\",dq,\" Btu/lbm dry air\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "In case 1, Enthalpy = 9.41 Btu/lbm dry air\n",
- "\n",
- " In case 2, Enthalpy = 31.15 Btu/lbm dry air\n",
- "\n",
- " Heat added = 21.49 Btu/lbm dry air\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6 - Pg 265"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the partial pressure and dew temperature, density of air, water and specific humidity\n",
- "#Initialization of variables\n",
- "pw=0.15#psia\n",
- "print '%s' %(\"using psychrometric charts,\")\n",
- "tdew=46 #F\n",
- "#calculations\n",
- "va=13.74 #ft^3/lbm dry air\n",
- "rhoa=1./va\n",
- "V=13.74\n",
- "mw=46/7000.\n",
- "rhow=mw/V\n",
- "w=0.00657\n",
- "#results\n",
- "print '%s' %(\"part a\")\n",
- "print '%s %.2f %s' %(\"partial pressure of water =\",pw,\" psia\")\n",
- "print '%s %d %s' %(\"\\n dew temperature =\",tdew,\"F\")\n",
- "print '%s' %(\"part b\")\n",
- "print '%s %.6f %s' %(\"density of water =\",rhow,\"lbm/ft^3\")\n",
- "print '%s %.4f %s' %(\"\\n density of air =\",rhoa,\"lbm/ft^3\")\n",
- "print '%s' %(\"part c\")\n",
- "print '%s %.5f %s' %(\"specific humidity =\",w,\"lbm steam/lbm air\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "using psychrometric charts,\n",
- "part a\n",
- "partial pressure of water = 0.15 psia\n",
- "\n",
- " dew temperature = 46 F\n",
- "part b\n",
- "density of water = 0.000478 lbm/ft^3\n",
- "\n",
- " density of air = 0.0728 lbm/ft^3\n",
- "part c\n",
- "specific humidity = 0.00657 lbm steam/lbm air\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7 - Pg 266"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Enthalpy change\n",
- "#Initialization of variables\n",
- "W1=0.00206 #lbm/lbm dry air\n",
- "W2=0.01090 #lbm/lbm dry air\n",
- "t=60 #F\n",
- "#calculations\n",
- "dw=W1-W2\n",
- "\n",
- "hs=144.4\n",
- "hs2=66.8-32\n",
- "w1=14.4 #Btu/lbm\n",
- "ws1=20 #Btu/lbm\n",
- "w2=76.3 #Btu/lbm\n",
- "ws2=98.5 #Btu/lbm\n",
- "dwh1=-(w1-ws1)/7000. *hs\n",
- "H1=9.3+dwh1\n",
- "dwh2=(w2-ws2)/7000. *hs2\n",
- "H2=31.3+dwh2\n",
- "dwc=dw*(t-32)\n",
- "dq=H2-H1+dwc\n",
- "#results\n",
- "print '%s %.2f %s' %(\"Enthalpy change =\",dq,\" Btu/lbm dry air\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Enthalpy change = 21.53 Btu/lbm dry air\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 8 - Pg 267"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the humidity and Temperature\n",
- "#Initialization of variables\n",
- "print '%s' %(\"From psychrometric charts,\")\n",
- "va1=13 #ft^3/lbm dry air\n",
- "va2=13.88 #ft^3/lbm dry air\n",
- "flow=2000. #cfm\n",
- "#calculations\n",
- "ma1= flow/va1\n",
- "ma2=flow/va2\n",
- "t=62.5# F\n",
- "phi=0.83 #percent\n",
- "#results\n",
- "print '%s %.2f' %(\"humidity = \",phi)\n",
- "print '%s %.1f %s' %(\"\\n Temperature =\",t,\" F\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "From psychrometric charts,\n",
- "humidity = 0.83\n",
- "\n",
- " Temperature = 62.5 F\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9 - Pg 270"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the dry bulb temperature and percent humidity\n",
- "#Initialization of variables\n",
- "t=90 #F\n",
- "ts=67.2 #F\n",
- "phi=0.3\n",
- "per=0.8\n",
- "#calculations\n",
- "dep=t-ts\n",
- "dt=dep*per\n",
- "tf=t-dt\n",
- "print '%s' %(\"from psychrometric charts,\")\n",
- "phi2=0.8\n",
- "#results\n",
- "print '%s %.2f %s' %(\"Dry bulb temperature =\",tf,\" F\")\n",
- "print '%s %.2f' %(\"\\n percent humidity = \",phi2)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "from psychrometric charts,\n",
- "Dry bulb temperature = 71.76 F\n",
- "\n",
- " percent humidity = 0.80\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 10 - Pg 271"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the cooling range and Approach. Also calculate the amount of water cooled and percentage of water lost \n",
- "#Initialization of variables\n",
- "m=1. #lbm\n",
- "t1=100. #F\n",
- "t2=75. #F\n",
- "db=65. #F\n",
- "print '%s' %(\"From psychrometric charts,\")\n",
- "t11=82 #F\n",
- "phi1=0.4\n",
- "H1=30. #Btu/lbm dry air\n",
- "w1=65. #grains/lbm dry air\n",
- "w2=250. #grains/lbm dry air\n",
- "#calculations\n",
- "cr=t1-t2\n",
- "appr=t2-db\n",
- "dmf3=(w2-w1)*0.0001427\n",
- "hf3=68.\n",
- "hf4=43.\n",
- "H2=62.2\n",
- "H1=30.\n",
- "mf4= (H1-H2+ dmf3*hf3)/(hf4-hf3)\n",
- "per=dmf3/(dmf3+mf4)*100\n",
- "#results\n",
- "print '%s %d %s' %(\"cooling range =\",cr,\"F\")\n",
- "print '%s %d %s' %(\"\\n Approach =\",appr,\"F\")\n",
- "print '%s %.3f %s' %(\"\\n amount of water cooled per pound of dry air =\",mf4,\"lbm dry air/lbm dry air\")\n",
- "print '%s %.2f %s' %(\"\\n percentage of water lost by evaporation =\",per,\"percent\")\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "From psychrometric charts,\n",
- "cooling range = 25 F\n",
- "\n",
- " Approach = 10 F\n",
- "\n",
- " amount of water cooled per pound of dry air = 1.216 lbm dry air/lbm dry air\n",
- "\n",
- " percentage of water lost by evaporation = 2.12 percent\n"
- ]
- }
- ],
- "prompt_number": 10
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Elements_of_Thermodynamics_and_heat_transfer/Chapter_15.ipynb b/Elements_of_Thermodynamics_and_heat_transfer/Chapter_15.ipynb deleted file mode 100755 index 5c710163..00000000 --- a/Elements_of_Thermodynamics_and_heat_transfer/Chapter_15.ipynb +++ /dev/null @@ -1,291 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:d49c3ebbe462c89b25518f64eebe9040738bf2ef82ee8e4dec85ee0229afc06b"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 15 - Vapor cycles and processes"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1 - Pg 276"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Thermal efficiency and Furnace efficiency\n",
- "#Initialization of variables\n",
- "p1=600. #psia\n",
- "p2=0.2563 #psia\n",
- "t1=486.21 #F\n",
- "t2=60. #F\n",
- "fur=0.75\n",
- "#calculations\n",
- "print '%s' %(\"from steam tables,\")\n",
- "h1=1203.2\n",
- "hf1=471.6\n",
- "hfg1=731.6\n",
- "h2=1088\n",
- "hf2=28.06\n",
- "hfg2=1059.9\n",
- "s1=1.4454\n",
- "sf1=0.6720\n",
- "sfg1=0.7734\n",
- "s2=2.0948\n",
- "sf2=0.0555\n",
- "sfg2=2.0393\n",
- "xd=(s1-sf2)/sfg2\n",
- "hd=hf2+xd*hfg2\n",
- "xa=0.3023\n",
- "ha=hf2+xa*hfg2\n",
- "wbc=0\n",
- "wda=0\n",
- "wcd=h1-hd\n",
- "wab=ha-hf1\n",
- "W=wab+wcd+wbc+wda\n",
- "Wrev=hfg1- (t2+459.7)*sfg1\n",
- "etat=(t1-t2)/(t1+459.7)*100\n",
- "eta=fur*etat\n",
- "#results\n",
- "print '%s %d %s' %(\"Thermal efficiency =\",etat,\"percent\")\n",
- "print '%s %.1f %s' %(\"\\n Furnace efficiency =\",eta,\" percent\")\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "from steam tables,\n",
- "Thermal efficiency = 45 percent\n",
- "\n",
- " Furnace efficiency = 33.8 percent\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2 - Pg 277"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Thermal and Overall efficiency\n",
- "#Initialization of variables\n",
- "dhab=-123.1\n",
- "etac=0.5\n",
- "ha=348.5\n",
- "etaf=0.75\n",
- "eta=0.85\n",
- "hf=471.6\n",
- "hfg=731.6\n",
- "hc=1203.2\n",
- "dhcd=452.7\n",
- "#calculations\n",
- "dwabs=dhab/etac\n",
- "hbd=ha-dwabs\n",
- "dwcds=dhcd*eta\n",
- "dqa=hc-hbd\n",
- "etat=(dwcds+dwabs)/dqa*100.\n",
- "eta=etat*etaf\n",
- "#results\n",
- "print '%s %.1f %s' %(\"Thermal efficiency =\",etat,\" percent\")\n",
- "print '%s %.1f %s' %(\"\\n Overall efficiency = \",eta,\" percent\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Thermal efficiency = 22.8 percent\n",
- "\n",
- " Overall efficiency = 17.1 percent\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3 - Pg 277"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Thermal and Overall efficiency\n",
- "#Initialization of variables\n",
- "t=60 #F\n",
- "J=778.16\n",
- "p1=600 #psia\n",
- "p2=0.2563 #psia\n",
- "etaf=0.85 \n",
- "#calculations\n",
- "print '%s' %(\"From steam tables,\")\n",
- "vf=0.01604 #ft^3/lbm\n",
- "dw=-vf*(p1-p2)*144/J\n",
- "ha=28.06 #Btu/lbm\n",
- "hb=29.84 #Btu/lbm\n",
- "hd=1203.2 #Btu/lbm\n",
- "he=750.5 #Btu/lbm\n",
- "dqa=hd-hb\n",
- "dqr=ha-he\n",
- "dw=dqa+dqr\n",
- "dwturb=hd-he\n",
- "dwpump=ha-hb\n",
- "etat=dw/dqa*100.\n",
- "eta=etat*etaf\n",
- "#results\n",
- "print '%s %.1f %s' %(\"Thermal efficiency =\",etat,\" percent\")\n",
- "print '%s %.1f %s' %(\"\\n Overall efficiency =\",eta,\"percent\")\n",
- " \n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "From steam tables,\n",
- "Thermal efficiency = 38.4 percent\n",
- "\n",
- " Overall efficiency = 32.7 percent\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 4 - Pg 278"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Thermal and Overall efficiency\n",
- "#Initialization of variables\n",
- "dhab=-1.78\n",
- "etac=0.5\n",
- "ha=28.06\n",
- "eta=0.85\n",
- "hf=471.6\n",
- "hfg=731.6\n",
- "hd=1203.2\n",
- "dhcd=452.7\n",
- "#calculations\n",
- "dwabs=dhab/etac\n",
- "hbd=ha-dwabs\n",
- "dwcds=dhcd*eta\n",
- "dqa=hd-hbd\n",
- "etat=(dwcds+dwabs)/dqa*100.\n",
- "eta=etat*eta\n",
- "#results\n",
- "print '%s %.1f %s' %(\"Thermal efficiency =\",etat,\"percent\")\n",
- "print '%s %.1f %s' %(\"\\n Overall efficiency =\",eta,\"percent\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Thermal efficiency = 32.5 percent\n",
- "\n",
- " Overall efficiency = 27.7 percent\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5 - Pg 287"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the thermal efficiency\n",
- "#Initialization of variables\n",
- "p2=600 #psia\n",
- "p1=44 #psia\n",
- "te=486.21 #F\n",
- "tb=273.1 #F\n",
- "J=778.16\n",
- "p3=0.25 #psia\n",
- "#calculations\n",
- "hc=241.9\n",
- "hj=834.6\n",
- "y=1-0.805\n",
- "v1=0.0172\n",
- "v2=0.016\n",
- "ha=28.06\n",
- "hd=hc+v1*(p2-p1)*144/J\n",
- "hb=ha+v2*(p1-p3)*144/J\n",
- "hh=1374\n",
- "Qa=hh-hd\n",
- "Qr=(ha-hj)*(1-y)\n",
- "etat=(Qa+Qr)/Qa*100.\n",
- "#results\n",
- "print '%s %.1f %s' %(\"thermal efficiency =\",etat,\"percent\")\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "thermal efficiency = 42.6 percent\n"
- ]
- }
- ],
- "prompt_number": 5
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Elements_of_Thermodynamics_and_heat_transfer/Chapter_16.ipynb b/Elements_of_Thermodynamics_and_heat_transfer/Chapter_16.ipynb deleted file mode 100755 index e73c5d78..00000000 --- a/Elements_of_Thermodynamics_and_heat_transfer/Chapter_16.ipynb +++ /dev/null @@ -1,556 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:b48de029dcc8d0a67a0785422dda0e79a8ad980134c0bbc3f290af08b9a74588"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 16 - Combustion"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1 - Pg 299"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Molecule\n",
- "#Initialization of variables\n",
- "per=85.\n",
- "#calculations\n",
- "a=per/12.\n",
- "b=100-per\n",
- "ad=1.13*a\n",
- "bd=1.13*b+1\n",
- "#results\n",
- "print '%s %d %s %d' %(\"Molecule is C\",ad,\"H\",bd)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Molecule is C 8 H 17\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2 - Pg 299"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Oxygen and Nitrogen content\n",
- "#Initialization of variables\n",
- "per=0.071\n",
- "#calculations\n",
- "O2=8.74\n",
- "N2=per/2 + 3.76*O2\n",
- "#results\n",
- "print '%s %.2f %s %.2f' %(\"Oxygen =\",O2,\"and Nitrogen =\",N2)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Oxygen = 8.74 and Nitrogen = 32.90\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 4 - Pg 302"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the air fuel ratio\n",
- "#Initialization of variables\n",
- "N2=78.1\n",
- "M=29.\n",
- "co2=8.7\n",
- "co=8.9\n",
- "x4=0.3\n",
- "x5=3.7\n",
- "x6=14.7\n",
- "#calculations\n",
- "O2=N2/3.76\n",
- "Z=(co2+co+x4)/8\n",
- "AF=(O2+N2)*M/(Z*113)\n",
- "#results\n",
- "print '%s %.1f %s' %(\"Air fuel ratio =\",AF,\"lbm air/lbm fuel\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Air fuel ratio = 11.3 lbm air/lbm fuel\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5 - Pg 303"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the air fuel ratio\n",
- "#Initialization of variables\n",
- "N2=78.1\n",
- "M=29\n",
- "ba=2.12\n",
- "x4=0.3\n",
- "x5=3.7\n",
- "x6=14.7\n",
- "#calculations\n",
- "O2=N2/3.76\n",
- "O2=N2/3.76\n",
- "Z=(x4*4+x5*2+x6*2)/17\n",
- "AF=(O2+N2)*M/(Z*113)\n",
- "#results\n",
- "print '%s %.1f %s' %(\"Air fuel ratio =\",AF,\"lbm air/lbm fuel\")\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Air fuel ratio = 11.4 lbm air/lbm fuel\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6 - Pg 303"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the air fuel ratio\n",
- "#Initialization of variables\n",
- "N2=78.1\n",
- "M=29\n",
- "ba=2.12\n",
- "x4=0.3\n",
- "x5=3.7\n",
- "x6=14.7\n",
- "#calculations\n",
- "O2=N2/3.76\n",
- "c=14.7\n",
- "b= x4*4 + x5*2 + x6*2\n",
- "a=b/ba\n",
- "AF=(O2+N2)*M/(a*12. + b)\n",
- "#results\n",
- "print '%s %.1f %s' %(\"Air fuel ratio =\",AF,\"lbm air/lbm fuel\")"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Air fuel ratio = 11.3 lbm air/lbm fuel\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7 - Pg 304"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the air fuel ratio\n",
- "#Initialization of variables\n",
- "N2=78.1\n",
- "M=29\n",
- "ba=2.12\n",
- "co2=8.7\n",
- "co=8.9\n",
- "x4=0.3\n",
- "x5=3.7\n",
- "x6=14.7\n",
- "#calculations\n",
- "O2=N2/3.76\n",
- "c=14.7\n",
- "Z=2.238\n",
- "X=(Z*17-x4*4-x5*2)/2\n",
- "a=co2+co/2+x4+x6/2\n",
- "b=3.764*a\n",
- "AF=(O2+N2)*M/(Z*113)\n",
- "#results\n",
- "print '%s %.1f %s' %(\"Air fuel ratio =\",AF,\"lbm air/lbm fuel\")\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Air fuel ratio = 11.3 lbm air/lbm fuel\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 8 - Pg 305"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the air fuel ratio\n",
- "#Initialization of variables\n",
- "x1=8.7\n",
- "x2=8.9\n",
- "x3=0.3\n",
- "N=78.1\n",
- "z=113\n",
- "M=29\n",
- "#calculations\n",
- "co2=(x1+x2+x3)*100/(N+x1+x2+x3)\n",
- "a=2.325\n",
- "AF=103*M/(a*z)\n",
- "#results\n",
- "print '%s %.2f' %(\"Air fuel ratio = \",AF)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Air fuel ratio = 11.37\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9 - Pg 308"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Higher heating value\n",
- "#Initialization of variables\n",
- "dH=-2369859 #Btu\n",
- "r=1.986\n",
- "dn=5.5\n",
- "T=536.7 #R\n",
- "#calculations\n",
- "dQ=dH+dn*r*T\n",
- "#results\n",
- "print '%s %d %s' %(\"Higher heating value =\",dQ,\" Btu\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Higher heating value = -2363996 Btu\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 10 - Pg 308"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Lower heating value\n",
- "#Initialization of variables\n",
- "y=13\n",
- "x=12\n",
- "M2=18\n",
- "M=170\n",
- "p=0.4593\n",
- "vfg=694.9\n",
- "J=778.2\n",
- "m=9*18\n",
- "u1=-2363996 #Btu\n",
- "#calculations\n",
- "z=y*M2/M\n",
- "hfg=1050.4 #Btu/lbm\n",
- "ufg= hfg- p*vfg*144/J\n",
- "dU=ufg*m \n",
- "Lhv=u1+dU\n",
- "#results\n",
- "print '%s %d %s' %(\"Lower heating value =\",Lhv,\"Btu/lbm\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Lower heating value = -2203398 Btu/lbm\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 11 - Pg 309"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Heat of the reaction\n",
- "#Initialization of variables\n",
- "n1=8\n",
- "n2=9\n",
- "n3=1\n",
- "n4=12.5\n",
- "U11=3852\n",
- "U12=115\n",
- "U21=3009\n",
- "U22=101\n",
- "U31=24773\n",
- "U32=640\n",
- "U41=2539\n",
- "U42=83\n",
- "H=-2203389\n",
- "#calculations\n",
- "dU1=n1*(U11-U12)+n2*(U21-U22)\n",
- "dU2=n3*(U31-U32)+n4*(U41-U42)\n",
- "Q=H+dU1-dU2\n",
- "#results\n",
- "print '%s %d %s' %(\"Heat of reaction =\",Q,\"Btu\")\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Heat of reaction = -2202154 Btu\n"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12 - Pg 310"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the final temperature\n",
- "#Initialization of variables\n",
- "n1=8\n",
- "n2=9\n",
- "n3=47\n",
- "h1=118\n",
- "h2=104\n",
- "h3=82.5\n",
- "Q=2203279 #Btu\n",
- "#calculations\n",
- "U11=n1*h1+n2*h2+n3*h3\n",
- "U12=U11+Q\n",
- "T2=5271 #R\n",
- "#results\n",
- "print '%s %d %s' %(\"Upon interpolating, T2 =\",T2,\" R\")\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Upon interpolating, T2 = 5271 R\n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13 - Pg 313"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the percentage of dissociation\n",
- "#Initialization of variables\n",
- "import numpy\n",
- "from numpy import roots\n",
- "kp=5. \n",
- "#calculations\n",
- "p1=[24.,0,3,-2]\n",
- "vec=numpy.roots(p1)\n",
- "x=numpy.real(vec[2]*100.)\n",
- "p2=[249.,0,3,-2]\n",
- "vec2=numpy.roots(p2)\n",
- "y=numpy.real(vec2[2]*100.)\n",
- "#results\n",
- "print '%s %.1f %s' %(\"percentage of dissociation =\",x,\" percent\")\n",
- "print '%s %.d %s' %(\"\\n If pressure =10 . degree of dissociation =\",y,\"percent\")\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "percentage of dissociation = 34.3 percent\n",
- "\n",
- " If pressure =10 . degree of dissociation = 18 percent\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14 - Pg 314"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Extent of reaction\n",
- "#Initialization of variables\n",
- "import numpy\n",
- "from numpy import roots\n",
- "p=[24.,48,7,-4]\n",
- "vec=numpy.roots(p)\n",
- "x=numpy.real(vec[2] *100)\n",
- "#results\n",
- "print '%s %d %s' %(\"Extent of reaction=\",100-x,\"percent\")\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Extent of reaction= 78 percent\n"
- ]
- }
- ],
- "prompt_number": 15
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Elements_of_Thermodynamics_and_heat_transfer/Chapter_18.ipynb b/Elements_of_Thermodynamics_and_heat_transfer/Chapter_18.ipynb deleted file mode 100755 index 8c56b4c9..00000000 --- a/Elements_of_Thermodynamics_and_heat_transfer/Chapter_18.ipynb +++ /dev/null @@ -1,240 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:f6da3400e962f069ab8b21139258f5f7cf61e5a3c236ae304fff64bfb1520423"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 18 - Refrigeration"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1 - Pg 347"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the cop, hp required, work of compression and expansion\n",
- "#Initialization of variables\n",
- "Ta=500. #R\n",
- "Tr=540. #R\n",
- "#calculations\n",
- "cop=Ta/(Tr-Ta)\n",
- "hp=4.71/cop\n",
- "print '%s' %(\"From steam tables,\")\n",
- "ha=48.02\n",
- "hb=46.6\n",
- "hc=824.1\n",
- "hd=886.9\n",
- "Wc=-(hd-hc)\n",
- "We=-(hb-ha)\n",
- "#results\n",
- "print '%s %.1f' %(\"Coefficient of performance =\",cop)\n",
- "print '%s %.3f %s' %(\"\\n horsepower required per ton of refrigeration =\",hp,\" hp/ton refrigeration\")\n",
- "print '%s %.1f %s' %(\"\\n Work of compression =\",Wc,\" Btu/lbm\")\n",
- "print '%s %.2f %s' %(\"\\n Work of expansion =\",We,\"Btu/lbm\")"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "From steam tables,\n",
- "Coefficient of performance = 12.5\n",
- "\n",
- " horsepower required per ton of refrigeration = 0.377 hp/ton refrigeration\n",
- "\n",
- " Work of compression = -62.8 Btu/lbm\n",
- "\n",
- " Work of expansion = 1.42 Btu/lbm\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2 - Pg 349"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the cop\n",
- "#Initialization of variables\n",
- "x=0.8\n",
- "he=26.28 #Btu/lbm\n",
- "hb=26.28 #Btu/lbm\n",
- "pe=98.76 #psia\n",
- "pc=51.68 #psia\n",
- "hc=82.71 #Btu/lbm\n",
- "hf=86.80+0.95\n",
- "#calculations\n",
- "dwisen=-(hf-hc)\n",
- "dwact=dwisen/x\n",
- "hd=hc-dwact\n",
- "cop=(hc-hb)/(hd-hc)\n",
- "#results\n",
- "print '%s %.2f' %(\"Coefficient of performance = \",cop)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Coefficient of performance = 8.96\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3 - Pg 351"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the work done, hp required, relative efficiency and mass flow rate, Compressor capacity\n",
- "#Initialization of variables\n",
- "hc=613.3#btu/lbm\n",
- "hb=138.9#btu/lbm\n",
- "ha=138.9#btu/lbm\n",
- "hd=713.4 #btu/lbm\n",
- "ta=464.7 #R\n",
- "t0=545.7 #R\n",
- "v=8.150 #ft^3/lbm\n",
- "#calculations\n",
- "Qa=hc-hb\n",
- "Qr=ha-hd\n",
- "Wcd=Qa+Qr\n",
- "cop=abs(Qa/Wcd)\n",
- "hp=abs(4.71/cop)\n",
- "carnot=abs(ta/(t0-ta))\n",
- "rel=abs(cop/carnot)\n",
- "mass=200/Qa\n",
- "C=mass*v\n",
- "#results\n",
- "print '%s %.1f %s' %(\"Work done =\",Wcd,\"Btu/lbm\")\n",
- "print '%s %.3f %s' %(\"\\n horsepower required per ton of refrigeration =\",hp,\" hp/ton refrigeration\")\n",
- "print '%s %.2f' %(\"\\n Coefficient of performance actual = \",cop)\n",
- "print '%s %.3f' %(\"\\n Ideal cop = \",carnot)\n",
- "print '%s %.3f' %(\"\\n relative efficiency =\",rel)\n",
- "print '%s %.3f %s' %(\"\\n Mass flow rate =\",mass,\"lbm/min ton\")\n",
- "print '%s %.2f %s' %(\"\\n Compressor capacity =\",C,\"cfm/ton\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Work done = -100.1 Btu/lbm\n",
- "\n",
- " horsepower required per ton of refrigeration = 0.994 hp/ton refrigeration\n",
- "\n",
- " Coefficient of performance actual = 4.74\n",
- "\n",
- " Ideal cop = 5.737\n",
- "\n",
- " relative efficiency = 0.826\n",
- "\n",
- " Mass flow rate = 0.422 lbm/min ton\n",
- "\n",
- " Compressor capacity = 3.44 cfm/ton\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 4 - Pg 355"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Pressure ratio, Heat, Water make up required and Volume of vapor entering ejector\n",
- "#Initialization of variables\n",
- "pc=0.6982 #psia\n",
- "pe=0.1217 #psia\n",
- "m=200 #gal/min\n",
- "qual=0.98\n",
- "h1=23.07 #Btu/lbm\n",
- "h2=8.05 #Btu/lbm\n",
- "hw=1071.3\n",
- "#calculations\n",
- "rp=pc/pe\n",
- "m2=m/0.01602 *0.1388 #Conversion of units \n",
- "m2=1670\n",
- "dh=15.02\n",
- "Qa=m2*(h1-h2)\n",
- "h3=h2 + qual*hw\n",
- "m3=Qa/(h3-h1)\n",
- "v=0.016+ qual*2444\n",
- "C=m3*v\n",
- "#results\n",
- "print '%s %.2f' %(\"Pressure ratio =\",rp)\n",
- "print '%s %d %s' %(\"\\n Heat =\",Qa,\"Btu/min\")\n",
- "print '%s %.2f %s' %(\"\\n Water make up required =\",m3,\"lbm/min\")\n",
- "print '%s %d %s' %(\"\\n Volume of vapor entering ejector =\",C,\"cfm\")\n",
- "print '%s' %(\"The answers are a bit different due to rounding off error in textbook\")"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Pressure ratio = 5.74\n",
- "\n",
- " Heat = 25083 Btu/min\n",
- "\n",
- " Water make up required = 24.24 lbm/min\n",
- "\n",
- " Volume of vapor entering ejector = 58054 cfm\n",
- "The answers are a bit different due to rounding off error in textbook\n"
- ]
- }
- ],
- "prompt_number": 1
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Elements_of_Thermodynamics_and_heat_transfer/Chapter_19.ipynb b/Elements_of_Thermodynamics_and_heat_transfer/Chapter_19.ipynb deleted file mode 100755 index e14fdd84..00000000 --- a/Elements_of_Thermodynamics_and_heat_transfer/Chapter_19.ipynb +++ /dev/null @@ -1,671 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:6dda6a8235cbc517967b1f5033dd3771eb08f1241054f5c218f55be7d9c59b8c"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 19 - Fundamentals of heat transfer"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1 - Pg 381"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the experimental value of thermal conductivity and Required temperature\n",
- "#Initialization of variables\n",
- "import math\n",
- "r1=1.12 #in\n",
- "r2=3.06 #in\n",
- "t1=203 #F\n",
- "t2=184 #F\n",
- "r3=2.09 #in\n",
- "po=11.1 #watts\n",
- "#calculations\n",
- "km=po*3.413*(12/r1-12/r2)/(4*math.pi*(t1-t2))\n",
- "dt=po*3.413*(12/r1-12/r3)/(4*math.pi*km)\n",
- "t3d=t1-dt\n",
- "#results\n",
- "print '%s %.2f %s' %(\"The experimental value of thermal conductivity =\",km,\"Btu/hr ft F\")\n",
- "print '%s %.1f %s' %(\"\\n Required temperature =\",t3d,\" F\")"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The experimental value of thermal conductivity = 1.08 Btu/hr ft F\n",
- "\n",
- " Required temperature = 189.1 F\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2 - Pg 383"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the heat loss and Temperature required\n",
- "#Initialization of variables\n",
- "import math\n",
- "r1=4.035 #in\n",
- "r2=4.312 #in\n",
- "r3=5.312 #in\n",
- "r4=6.812 #in\n",
- "k12=25 #Btu/hr ft F\n",
- "k23=0.05 #Btu/hr ft F\n",
- "k34=0.04 #Btu/hr ft F\n",
- "t1=625. #F\n",
- "t4=125. #F\n",
- "l=100. #ft\n",
- "hr=1.7 #Btu/hr ft^2 F\n",
- "#calculations\n",
- "Rs=1/(2.*math.pi*l) *(math.log(r2/r1) /k12+math.log(r3/r2) /k23 +math.log(r4/r3) /k34)\n",
- "Qd=(t1-t4)/Rs\n",
- "dt=Qd*12/(hr*math.pi*2*l*6.812)\n",
- "t0=t4-dt\n",
- "#results\n",
- "print '%s %d %s' %(\"Heat loss =\",Qd,\"Btu/hr\")\n",
- "print '%s %d %s' %(\"\\n Temperature required =\",t0,\"F\")\n",
- "print '%s' %(\"The answers given in the textbook are a bit different due to rounding off error\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Heat loss = 30231 Btu/hr\n",
- "\n",
- " Temperature required = 75 F\n",
- "The answers given in the textbook are a bit different due to rounding off error\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3 - Pg 396"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Coefficient of heat transfer\n",
- "#Initialization of variables\n",
- "import math\n",
- "dout=1 #in\n",
- "d1=0.049 #in\n",
- "t1=70. #F\n",
- "t2=80. #F\n",
- "rho=62.2 #lbm/ft^3\n",
- "mum=2.22 #lbm/ft hr\n",
- "k=0.352 #Btu/hr ft F\n",
- "cp=1 #Btu/lbm F\n",
- "vel=500000. #lbm/hr\n",
- "n=100. #tubes\n",
- "#calculations\n",
- "D=dout-2*d1\n",
- "t=(t1+t2)/2.\n",
- "V=vel/n *4*144/(math.pi*D**2 *rho)\n",
- "Re=rho*V*D/(mum*12)\n",
- "Pr=cp*mum/k\n",
- "Nu=0.023*Re**0.8 *Pr**0.4\n",
- "hc=Nu*k*12/D\n",
- "#results\n",
- "print '%s %d %s' %(\"Coefficient of heat transfer =\",hc,\"Btu/hr ft^2 F\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Coefficient of heat transfer = 1040 Btu/hr ft^2 F\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 4 - Pg 397"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Coefficient of heat transfer and Percentage change\n",
- "#Initialization of variables\n",
- "import math\n",
- "d1=0.5 #ft\n",
- "t1=200. #F\n",
- "t2=80. #F\n",
- "ta=400. #F\n",
- "rho=0.0662 #lbm/ft**3\n",
- "mum=0.0483 #lbm/ft hr\n",
- "k=0.0167 #Btu/hr ft F\n",
- "cp=0.2408 #Btu/lbm F\n",
- "rho2=0.0567 #lbm/ft**3\n",
- "mum2=0.0542 #lbm/ft hr\n",
- "k2=0.0190 #Btu/hr ft F\n",
- "cp2=0.2419 #Btu/lbm F\n",
- "g=32.17\n",
- "#calculations\n",
- "ti=(t1+t2)/2.\n",
- "bet=1/(460.+ti)\n",
- "Pr1=cp*mum/k\n",
- "Gr1=d1**3 *rho**2 *3600**2 *g*bet*(t1-t2)/mum**2\n",
- "Gr1pr1=Gr1*Pr1\n",
- "hc1=k/d1 *0.53*(Gr1pr1)**0.25\n",
- "Q1=hc1*(t1-t2)\n",
- "tf=(ta+t2)/2.\n",
- "bet2=1/(460.+tf)\n",
- "Pr2=cp2*mum2/k2\n",
- "Gr2=d1**3 *rho2**2 *3600**2 *g*bet2*(ta-t2)/mum2**2\n",
- "Gr2pr2=Gr2*Pr2\n",
- "hc2=k2/d1 *0.53*(Gr2pr2)**0.25\n",
- "Q2=hc2*(ta-t2)\n",
- "per=100*(Q2-Q1)/Q1\n",
- "#results\n",
- "print '%s %.3f %s' %(\"Coefficient of heat transfer in case 1=\",hc1,\" Btu/hr ft^2 F\")\n",
- "print '%s %.3f %s' %(\"\\n Coefficient of heat transfer in case 2 =\",hc2,\"Btu/hr ft^2 F\")\n",
- "print '%s %d %s' %(\"\\n Percentage change =\",per,\"percent\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Coefficient of heat transfer in case 1= 1.076 Btu/hr ft^2 F\n",
- "\n",
- " Coefficient of heat transfer in case 2 = 1.312 Btu/hr ft^2 F\n",
- "\n",
- " Percentage change = 225 percent\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5 - Pg 398"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Temperature o wing surface and Heat transfer convective\n",
- "#Initialization of variables\n",
- "chord=40. #ft\n",
- "v=1200. #mph\n",
- "t1=80. #F\n",
- "t2=200. #F\n",
- "mu=0.0447 #lbm/ft hr\n",
- "rho=5280. #lbm/ft**3\n",
- "cp=0.2404 #Btu/lbm F\n",
- "k=0.0152 #Btu/hr ft F\n",
- "J=778.\n",
- "gc=32.17 #ft/s**2\n",
- "mu2=0.0514 #lbm/ft hr\n",
- "k2=0.0179 #Btu/hr ft F\n",
- "cp2=0.2414 #Btu/lbm F\n",
- "#calculations\n",
- "Re=rho*v*chord*0.0735/mu\n",
- "r=(mu*cp/k)**(1./3.)\n",
- "tav=t1+ r*v**2 *rho**2 /(2*gc*J*cp*3600**2)\n",
- "ts=t1+ 0.5*(t2-t1)+ 0.22*(tav-t1)\n",
- "Re2=v*rho*chord*0.0610/mu2\n",
- "Pr2=cp2*mu2/k2\n",
- "hc=cp2*v*rho*0.0610 *0.037*Re2**(-0.2) *Pr2**(-0.667)\n",
- "Q2=hc*(t2-tav)\n",
- "#results\n",
- "print '%s %.1f %s' %(\"Temperature of wing surface =\",tav,\" F\")\n",
- "print '%s %d %s' %(\"\\n Heat transfer convective =\",Q2,\"Btu/hr ft^2\")\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": [
- "Temperature of wing surface = 309.3 F\n",
- "\n",
- " Heat transfer convective = -9711 Btu/hr ft^2\n",
- "The answers are a bit different due to rounding off error in textbook\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6 - Pg 413"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the percent of radiation emitted by surface and percent absorbed\n",
- "#Initialization of variables\n",
- "import math\n",
- "r1=1. #in\n",
- "r2=5. #in\n",
- "F12=1.\n",
- "#calculations\n",
- "F21=4*math.pi*r1**2 *F12/(4*math.pi*r2**2)*100\n",
- "F22=(1-F21/100.)*100.\n",
- "#results\n",
- "print '%s %d %s' %(\"Percent of radiation emitted by surface 2 on small sphere =\",F21,\" percent\")\n",
- "print '%s %d %s' %(\"\\n Remaining\",F22, \"percent is absorbed by inner surface of larger sphere\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Percent of radiation emitted by surface 2 on small sphere = 4 percent\n",
- "\n",
- " Remaining 96 percent is absorbed by inner surface of larger sphere\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7 - Pg 413"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Net exchange of radiation\n",
- "#Initialization of variables\n",
- "short=2. #ft\n",
- "apart=3. #ft\n",
- "lon=4. #ft\n",
- "T1=2260. #R\n",
- "T2=530. #R\n",
- "sigma=0.1714\n",
- "#calculations\n",
- "A1=short*lon\n",
- "ratio=short/apart\n",
- "print '%s' %(\"from curve 3\")\n",
- "F=0.165\n",
- "Q12=A1*F*sigma*((T1/100)**4 -(T2/100)**4)\n",
- "#results\n",
- "print '%s %d %s' %(\"Net exchange of radiation =\",Q12,\"Btu/hr\")\n",
- "print '%s' %(\"The answer in the textbook is a bit different due to rounding off error in textbook.\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "from curve 3\n",
- "Net exchange of radiation = 58844 Btu/hr\n",
- "The answer in the textbook is a bit different due to rounding off error in textbook.\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 8 - Pg 416"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Net exchange of radiation\n",
- "#Initialization of variables\n",
- "F=0.51\n",
- "A1=8 #ft^2\n",
- "sigma=0.1714\n",
- "T1=2260. #R\n",
- "T2=530. #R\n",
- "#calculations\n",
- "Q12=A1*F*sigma*((T1/100)**4 -(T2/100)**4)\n",
- "#results\n",
- "print '%s %d %s' %(\"Net exchange of radiation =\",Q12,\"Btu/hr\")\n",
- "print '%s' %(\"The answer in the textbook is a bit different due to rounding off error in textbook.\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Net exchange of radiation = 181881 Btu/hr\n",
- "The answer in the textbook is a bit different due to rounding off error in textbook.\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9 - Pg 417"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Net exchange of radiation\n",
- "#Initialization of variables\n",
- "F=0.51\n",
- "A1=8. #f^2\n",
- "sigma=0.1714\n",
- "T1=2260. #R\n",
- "T2=530. #R\n",
- "#calculations\n",
- "F12=1/(1/0.51 +(1/0.9 -1) +(1/0.6 -1))\n",
- "Q12=A1*F12*sigma*((T1/100)**4 -(T2/100)**4)\n",
- "#results\n",
- "print '%s %d %s' %(\"Net exchange of radiation =\",Q12,\"Btu/hr\")\n",
- "print '%s' %(\"The answer in the textbook is a bit different due to rounding off error in textbook.\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Net exchange of radiation = 130225 Btu/hr\n",
- "The answer in the textbook is a bit different due to rounding off error in textbook.\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 10 - Pg 418"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Percentage change in total heat transfer\n",
- "#Initialization of variables\n",
- "em=0.79\n",
- "sigma=0.1714\n",
- "T1=660. #R\n",
- "T2=540. #R\n",
- "T3=860. #R\n",
- "#calculations\n",
- "Q1=em*sigma*((T1/100)**4 -(T2/100)**4)\n",
- "Q2=em*sigma*((T3/100)**4 -(T2/100)**4)\n",
- "Qh1=129+Q1\n",
- "Qh2=419+Q2\n",
- "per=100*(Qh2-Qh1)/Qh1\n",
- "#results\n",
- "print '%s %.1f %s' %(\"Percentage change in total heat transfer =\",per,\"percent\")\n",
- "print '%s' %(\"The answer in the textbook is a bit different due to rounding off error in textbook.\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Percentage change in total heat transfer = 285.7 percent\n",
- "The answer in the textbook is a bit different due to rounding off error in textbook.\n"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 11 - Pg 419"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Error in probe reading\n",
- "#Initialization of variables\n",
- "Tp=12.57\n",
- "Tw=10.73\n",
- "ep=0.8\n",
- "sig=0.1714\n",
- "hc=7\n",
- "#calculations\n",
- "dt=ep*sig*(Tp**4-Tw**4)/hc\n",
- "#results\n",
- "print '%s %d %s' %(\"Error in probe reading =\",dt,\"F\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Error in probe reading = 229 F\n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12 - Pg 420"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Heat transfer in both cases\n",
- "#Initialization of variables\n",
- "import math\n",
- "l=6 #ft\n",
- "d1=0.55 #in\n",
- "d2=0.75 #in\n",
- "h1=280. #Btu/hr ft^2 F\n",
- "h2=2000. #Btu/fr ft^2 F\n",
- "k=220. #Btu/hr ft F\n",
- "t2=212. #F\n",
- "t1=60. #F\n",
- "f=500. #Btu/hr ft^2 F\n",
- "#calculations\n",
- "A2=math.pi*d1*l/12\n",
- "A3=math.pi*d2*l/12\n",
- "Rt=1/(h1*A2) + 1/(h2*A3) +math.log(d2/d1) /(2*math.pi*k*l)\n",
- "Q=(t2-t1)/Rt\n",
- "Rt2=Rt+ 1/(f*A2)\n",
- "Q2=(t2-t1)/Rt2\n",
- "#results\n",
- "print '%s %d %s' %(\"Heat transfer =\",Q,\"Btu/hr\")\n",
- "print '%s %d %s' %(\"\\n Heat transfer in case 2=\",Q2,\" Btu/hr\")\n",
- "print '%s' %(\"The answer in the textbook is a bit different due to rounding off error in textbook.\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Heat transfer = 33074 Btu/hr\n",
- "\n",
- " Heat transfer in case 2= 21994 Btu/hr\n",
- "The answer in the textbook is a bit different due to rounding off error in textbook.\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13 - Pg 422"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Overall Heat transfer coefficient\n",
- "#Initialization of variables\n",
- "import math\n",
- "l=6 #ft\n",
- "d1=0.55 #in\n",
- "d2=0.75 #in\n",
- "h1=280. #Btu/hr ft^2 F\n",
- "h2=2000. #Btu/fr ft^2 F\n",
- "k=220. #Btu/hr ft F\n",
- "t2=212. #F\n",
- "t1=60. #F\n",
- "#calculations\n",
- "A2=math.pi*d1*l/12\n",
- "A3=math.pi*d2*l/12\n",
- "Rt=1/(h1*A2) + 1/(h2*A3) +math.log(d2/d1) /(2*math.pi*k*l)\n",
- "U3=1/(A3*Rt)\n",
- "#results\n",
- "print '%s %.1f %s' %(\"Overall Heat transfer coefficient =\",U3,\"Btu/hr ft^2 F\")\n",
- "print '%s' %(\"The answer in the textbook is a bit different due to rounding off error in textbook.\")\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Overall Heat transfer coefficient = 184.7 Btu/hr ft^2 F\n",
- "The answer in the textbook is a bit different due to rounding off error in textbook.\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14 - Pg 427"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Parameters X and Z\n",
- "#Initialization of variables\n",
- "t1=300. #F\n",
- "t2=260. #F\n",
- "t3=200. #F\n",
- "t4=160. #F\n",
- "#calculations\n",
- "X=(t2-t4)/(t1-t4)\n",
- "Z=(t1-t3)/(t2-t4)\n",
- "#results\n",
- "print '%s %.3f %s %.1f %s' %(\"Parameters X and Z are\",X, \"and\",Z,\"respectively\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Parameters X and Z are 0.714 and 1.0 respectively\n"
- ]
- }
- ],
- "prompt_number": 16
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Elements_of_Thermodynamics_and_heat_transfer/Chapter_2.ipynb b/Elements_of_Thermodynamics_and_heat_transfer/Chapter_2.ipynb deleted file mode 100755 index 511042b0..00000000 --- a/Elements_of_Thermodynamics_and_heat_transfer/Chapter_2.ipynb +++ /dev/null @@ -1,102 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:e3ffdae85ed1c10a54629e535e9e044264a512de563ad3fb48c98b93c4a65432"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 2 - Fundamental Concepts"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1 - Pg 18"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the potential energy\n",
- "#Initialization of variables\n",
- "z=100 #ft\n",
- "m=32.1739 #lbm\n",
- "#calculations\n",
- "PE=m*z\n",
- "#results\n",
- "print '%s %.2f %s' %(\"Potential energy =\",PE,\"ft-lbm\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Potential energy = 3217.39 ft-lbm\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2 - Pg 21"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the absolute energy of this mixture and the change in mass\n",
- "#Initialization of variables\n",
- "m0=18.016 #lbm\n",
- "gc=32.1739 #lbm ft/lbf sec^2\n",
- "c=186000*5280\n",
- "dU=94.4*10**6 #ft-lbf\n",
- "#calculations\n",
- "U=m0/gc *c**2\n",
- "dm= -dU*gc/c**2\n",
- "#results\n",
- "print '%s %.2e %s' %(\"Absolute energy of this mixture =\",U,\" ft-lbf\")\n",
- "print '%s' %(\"\\n In case b, there is no change in mass\")\n",
- "print '%s %.2e %s' %(\"\\n Change in mass =\",dm,\" lbm\")\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": [
- "Absolute energy of this mixture = 5.40e+17 ft-lbf\n",
- "\n",
- " In case b, there is no change in mass\n",
- "\n",
- " Change in mass = -3.15e-09 lbm\n",
- "The answers are a bit different due to rounding off error in textbook.\n"
- ]
- }
- ],
- "prompt_number": 2
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Elements_of_Thermodynamics_and_heat_transfer/Chapter_20.ipynb b/Elements_of_Thermodynamics_and_heat_transfer/Chapter_20.ipynb deleted file mode 100755 index c603307c..00000000 --- a/Elements_of_Thermodynamics_and_heat_transfer/Chapter_20.ipynb +++ /dev/null @@ -1,377 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:dd9802b686cdfc5d04064f27577856b20913db7526956cd0d6229f433fb9f632"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 20 - Advanced topics in heat transfer"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1 - Pg 437"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Surface temperature of transmission line, rate of heat generation and max temperature in the line\n",
- "#Initialization of variables\n",
- "import math\n",
- "heat=54.5 #Btu/hr ft\n",
- "d=0.811 #in\n",
- "h=2.5 #Btu/hr ft**2 F\n",
- "ts=100 #F\n",
- "km=220 #Btu/hr ft F\n",
- "#calculations\n",
- "t2=heat*12/(h*math.pi*d) +ts\n",
- "w=heat*4*144/(math.pi*d**2)\n",
- "t1=w*(d/2)**2 /(4*144*km) + t2\n",
- "#results\n",
- "print '%s %.1f %s' %(\"Surface temperature of transmission line =\",t2,\" F\")\n",
- "print '%s %d %s' %(\"\\n Rate of heat generation per unit volume of wire =\",w,\"Btu/hr ft^2\")\n",
- "print '%s %.2f %s' %(\"\\n Max. temperature in the line =\",t1,\" F\")\n",
- "print '%s' %(\"The answers in the textbook are a bit different due to rounding off errors\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Surface temperature of transmission line = 202.7 F\n",
- "\n",
- " Rate of heat generation per unit volume of wire = 15192 Btu/hr ft^2\n",
- "\n",
- " Max. temperature in the line = 202.70 F\n",
- "The answers in the textbook are a bit different due to rounding off errors\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2 - Pg 442"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the heat rate, total hourly loss and approx. temp of the tip of the fin and total heat loss\n",
- "#Initialization of variables\n",
- "import math\n",
- "d1=1. #in\n",
- "l=1. #ft\n",
- "r=0.5 #ft\n",
- "L=0.5 #in\n",
- "Ts=430. #F\n",
- "Ta=170. #F\n",
- "dela=0.0125 #ft\n",
- "h=10. #Btu/hr ft^2 F\n",
- "eta=0.77\n",
- "eta2=0.94\n",
- "n=60. #fins\n",
- "thick=0.025 #in\n",
- "k2=132. #Btu/hr ft F\n",
- "#calculations\n",
- "Q=h*math.pi*d1**2 *(Ts-Ta)/12\n",
- "rate=(r+L)/r\n",
- "k=26 #Btu/hr ft F\n",
- "Lt=L/12 *(h*12/(k*dela))**(1/2)\n",
- "dtm=eta*(Ts-Ta)\n",
- "As=2*math.pi*((2*d1)**2 -d1**2)/4\n",
- "Q1=h*n*As*dtm/144\n",
- "Q2=h*math.pi*d1*(12-60*thick)*(Ts-Ta)/144\n",
- "Qt=Q1+Q2\n",
- "al=0.8\n",
- "tl=Ta+(Ts-Ta)/math.cosh(al)\n",
- "al2=r/12 *(h*12*2/(k2*thick))\n",
- "dtm2=eta2*(Ts-Ta)\n",
- "Q12=h*n*As*dtm2/144\n",
- "Qt2=Q12+Q2\n",
- "#results\n",
- "print '%s %.1f %s' %(\"Heat rate per foot of bare tube =\",Q,\"Btu/hr\")\n",
- "print '%s %.1f %s' %(\"\\n Total hourly heat loss per foot of finned tube =\",Qt,\"Btu/hr\")\n",
- "print '%s %d %s' %(\"\\n Approx. temp for tip of the fin =\",tl,\"F\")\n",
- "print '%s %.1f %s' %(\"\\n In case of Al, Total beat loss =\",Qt2,\" Btu/hr\")\n",
- "print '%s' %(\"The answers in the textbook are a bit different due to rounding off errors\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Heat rate per foot of bare tube = 680.7 Btu/hr\n",
- "\n",
- " Total hourly heat loss per foot of finned tube = 4526.5 Btu/hr\n",
- "\n",
- " Approx. temp for tip of the fin = 364 F\n",
- "\n",
- " In case of Al, Total beat loss = 5394.4 Btu/hr\n",
- "The answers in the textbook are a bit different due to rounding off errors\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3 - Pg 444"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Length required\n",
- "#Initialization of variables\n",
- "import math\n",
- "tl=125. #F\n",
- "t0=80. #F\n",
- "t1=1000. #F\n",
- "d=1. #in\n",
- "k=25. #Btu/hr ft F\n",
- "k2=0.0208\n",
- "Nu=18.\n",
- "#calculations\n",
- "byal=(tl-t0)/(t1-t0)\n",
- "al=math.acosh(1/byal)\n",
- "b=math.pi*d/12.\n",
- "A=math.pi*d**2 /(4*144)\n",
- "tm=(tl+t1)/2. +460\n",
- "hr=0.79*0.1714*((tm/100)**4 - ((t0+460)/100)**4)/(tm-460-t0)\n",
- "hc=Nu*k2*12/d\n",
- "a=((hc+hr)*b/(k*A))**(0.5)\n",
- "L=al/a\n",
- "#results\n",
- "print '%s %.2f %s' %(\"Length required =\",L,\" ft\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Length required = 0.99 ft\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5 - Pg 452"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the time required\n",
- "#Initialization of variables\n",
- "import math\n",
- "c=0.0947 #Btu/lbm F\n",
- "rho=0.0551 #lbm/ft**3\n",
- "mu=0.0553 #lbm/hr ft\n",
- "t1=440. #F\n",
- "ts=400. #F\n",
- "t2=80. #F\n",
- "d=0.1 #in\n",
- "k=0.0194 #Btu/hr ft**2 F\n",
- "rho2=558. #lbm/ft**3\n",
- "v=10. #ft/s\n",
- "#calculations\n",
- "Re=d*3600*v*rho/(12*mu)\n",
- "Nu=0.37*Re**0.6\n",
- "hc=k*Nu*12/d\n",
- "ex=math.log((t1-ts)/(t1-t2))\n",
- "tau=-ex*d*rho2*c/(12*6*hc)\n",
- "time=tau*3600\n",
- "#results\n",
- "print '%s %d %s' %(\"Time required =\",time,\"sec\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Time required = 22 sec\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6 - Pg 456"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Cooling time and Center temperature\n",
- "#Initialization of variables\n",
- "h=2 #Btu/hr ft**2 F\n",
- "delta=1/6.\n",
- "t=125. #F\n",
- "t0=100. #F\n",
- "ti=350. #F\n",
- "k=0.167 #Btu/hr ft F\n",
- "rho=80. #lbm/ft**3\n",
- "c=0.4 #Btu/lbm F\n",
- "#calculations\n",
- "Bi=h*delta/k\n",
- "tr=(t-t0)/(ti-t0)\n",
- "tau=1.5*delta**2 *rho*c/k\n",
- "tr2=0.21\n",
- "tc=tr2*(ti-t0) + t0\n",
- "#results\n",
- "print '%s %.2f %s' %(\"Cooling time =\",tau,\" hr\")\n",
- "print '%s %d %s' %(\"\\n Center temperature =\",tc,\" F\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Cooling time = 7.98 hr\n",
- "\n",
- " Center temperature = 152 F\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7 - Pg 458"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate if the thin layero f insulation would increase the heat dissipation from wire\n",
- "#Initialization of variables\n",
- "h=2.5 #Btu/hr ft^2 F\n",
- "kc=0.1 #Btu/hr ft F\n",
- "r1=0.811/2\n",
- "#calculations\n",
- "r2c=kc/h *12\n",
- "#results\n",
- "if r2c>=r1:\n",
- " print '%s %.2f %s' %(\"Thin layer of insulation would increase the heat dissipation from wire, r2c =\",r2c,\"in\")\n",
- "else:\n",
- " print '%s %.2f %s' %(\"Thin layer of insulation would decrease the heat dissipation from wire. r2c=\",r2c,\"in\")\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Thin layer of insulation would increase the heat dissipation from wire, r2c = 0.48 in\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 8 - Pg 465"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the heat transfer from surfaces 1,2,3 and Temperature of surface R\n",
- "#Initialization of variables\n",
- "F12=0.19\n",
- "F13=F12\n",
- "FR3=F13\n",
- "F2R=0.38\n",
- "J1=1714.\n",
- "Wb2=0.1714\n",
- "#calculations\n",
- "print '%s' %(\"Upon solving the simultaneous equations\")\n",
- "Q1=1774 #Btu/hr ft\n",
- "Q2=-547 #Btu/r ft\n",
- "Q3=-1227 #Btu/hr ft\n",
- "J2=548 #Btu/hr ft^2\n",
- "Tr=909 #R\n",
- "#results\n",
- "print '%s %d %s' %(\"Heat transfer rate from surface 1 =\",Q1,\" Btu/hr ft\")\n",
- "print '%s %d %s' %(\"\\n Heat transfer rate from surface 2 =\",Q2,\" Btu/hr ft\")\n",
- "print '%s %d %s' %(\"\\n Heat transfer rate from surface 3 =\",Q3,\" Btu/hr ft\")\n",
- "print '%s %d %s' %(\"\\n Temperature of surface R =\",Tr,\"R\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Upon solving the simultaneous equations\n",
- "Heat transfer rate from surface 1 = 1774 Btu/hr ft\n",
- "\n",
- " Heat transfer rate from surface 2 = -547 Btu/hr ft\n",
- "\n",
- " Heat transfer rate from surface 3 = -1227 Btu/hr ft\n",
- "\n",
- " Temperature of surface R = 909 R\n"
- ]
- }
- ],
- "prompt_number": 7
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Elements_of_Thermodynamics_and_heat_transfer/Chapter_5.ipynb b/Elements_of_Thermodynamics_and_heat_transfer/Chapter_5.ipynb deleted file mode 100755 index 20adaee3..00000000 --- a/Elements_of_Thermodynamics_and_heat_transfer/Chapter_5.ipynb +++ /dev/null @@ -1,146 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:ce73d802673a1107e52bf47eb7b2328d569cd7caf2e704d6bf5c2e992f46aa53"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 5 - The first law and the dynamic open system"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1 - Pg 68"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the work done by the system and power\n",
- "#Initialization of variables\n",
- "rate= 5 #lbm/sec\n",
- "Q=50 #Btu/s\n",
- "h2=1020 #Btu/lbm\n",
- "h1=1000 #Btu/lbm\n",
- "V2=50 #ft/s\n",
- "V1=100 #ft/s\n",
- "J=778\n",
- "g=32.2 #ft/s^2\n",
- "gc=g\n",
- "Z2=0\n",
- "Z1=100. #ft\n",
- "#calculations\n",
- "dw=Q/rate -(h2-h1) -(V2**2- V1**2)/(2*gc*J) -g/gc *(Z2-Z1)/J\n",
- "power=dw*rate\n",
- "#results\n",
- "print '%s %.1f %s' %(\"work done by the system =\",dw,\"Btu/lbm\")\n",
- "print '%s %.1f %s' %(\"\\n Power =\",power,\"Btu/s\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "work done by the system = -9.7 Btu/lbm\n",
- "\n",
- " Power = -48.6 Btu/s\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2 - Pg 69"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the area of inlet pipe\n",
- "#Initialization of variables\n",
- "V=100 #ft/s\n",
- "v=15. #lbm/ft^3\n",
- "m=5. #lbm/s\n",
- "#calculations\n",
- "A=m*v/V\n",
- "#results\n",
- "print '%s %.2f %s' %(\"Area of inlet pipe =\",A,\"ft^2\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Area of inlet pipe = 0.75 ft^2\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3 - Pg 79"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the final temperature of the steam and change in temperature\n",
- "#Initialization of variables\n",
- "P=100. #psia\n",
- "#calculations\n",
- "print '%s' %(\"From table B-4\")\n",
- "h=1187.2 #Btu/lbm\n",
- "t1=328 #F\n",
- "t2=540 #F\n",
- "dt=t2-t1\n",
- "#results\n",
- "print '%s %d %s' %(\"Final temperature of the steam =\",t2,\"F\")\n",
- "print '%s %d %s' %(\"\\n Change in temperature =\",dt,\"F\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "From table B-4\n",
- "Final temperature of the steam = 540 F\n",
- "\n",
- " Change in temperature = 212 F\n"
- ]
- }
- ],
- "prompt_number": 3
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Elements_of_Thermodynamics_and_heat_transfer/Chapter_7.ipynb b/Elements_of_Thermodynamics_and_heat_transfer/Chapter_7.ipynb deleted file mode 100755 index 92d0279f..00000000 --- a/Elements_of_Thermodynamics_and_heat_transfer/Chapter_7.ipynb +++ /dev/null @@ -1,162 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:15d0388b2c544d629a9ec990a9a36b6e26ea291bc7812312e5319e0c4bde1ab8"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 7 - The second law"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2 - Pg 112"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the entropy in parts ab,ac and efficiency\n",
- "#Initialization of variables\n",
- "import math\n",
- "cv=0.175 #Btu/lbm R\n",
- "R0=1.986\n",
- "M=29\n",
- "T2=1040 #R\n",
- "T1=520 #R\n",
- "#calculations\n",
- "cp=cv+R0/M\n",
- "sab=cv*math.log(T2/T1)\n",
- "sac=cp*math.log(T2/T1)\n",
- "dqab=cv*(T2-T1)\n",
- "dqca=cp*(T1-T2)\n",
- "dqrev=T2*(sac-sab)\n",
- "eta=(dqab+dqrev+dqca)/(dqab+dqrev)*100\n",
- "#results\n",
- "print '%s %.4f %s' %(\"Entropy in ab part =\",sab,\"Btu/lbm R\")\n",
- "print '%s %.4f %s' %(\"\\n Entropy in ac part =\",sac,\" Btu/lbm R\")\n",
- "print '%s %.2f %s' %(\"\\n Efficiency =\",eta,\" percent\")\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": [
- "Entropy in ab part = 0.1213 Btu/lbm R\n",
- "\n",
- " Entropy in ac part = 0.1688 Btu/lbm R\n",
- "\n",
- " Efficiency = 9.80 percent\n",
- "The answers are a bit different due to rounding off error in textbook\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3 - Pg 115"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the change in entropy of the process\n",
- "#Initialization of variables\n",
- "import math\n",
- "tc=32. #F\n",
- "th=80. #F\n",
- "mw=5 #lbm\n",
- "mi=1 #lbm\n",
- "P=14.7 #psia\n",
- "cp=1\n",
- "#calculations\n",
- "t= (-144*mi+tc*mi+th*mw)/(mw+mi)\n",
- "ds1=144/(tc+460)\n",
- "ds2=cp*math.log((460+t)/(460+tc))\n",
- "dsice=ds1+ds2\n",
- "dswater=mw*cp*math.log((t+460)/(460+th))\n",
- "ds=dsice+dswater\n",
- "#results\n",
- "print '%s %.4f %s' %(\"Change in entropy of the process =\",ds,\"Btu/R\")\n",
- "print '%s' %(\"The answer is a bit different due to rounding off error in textbook\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Change in entropy of the process = 0.0192 Btu/R\n",
- "The answer is a bit different due to rounding off error in textbook\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 4 - Pg 119"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the change in available energy\n",
- "#Initialization of variables\n",
- "import math\n",
- "cp=1\n",
- "T2=60. #F\n",
- "T1=100. #F\n",
- "ta=32. #F\n",
- "#calculations\n",
- "dq=cp*(T2-T1)\n",
- "ds=cp*math.log((460+T2)/(460+T1))\n",
- "dE=dq-ds*(ta+460)\n",
- "#results\n",
- "print '%s %.1f %s' %(\"Change in available energy =\",dE,\"Btu/lbm\")\n",
- "print '%s' %(\"The answer is a bit different due to rounding off error in textbook\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Change in available energy = -3.5 Btu/lbm\n",
- "The answer is a bit different due to rounding off error in textbook\n"
- ]
- }
- ],
- "prompt_number": 3
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Elements_of_Thermodynamics_and_heat_transfer/Chapter_8.ipynb b/Elements_of_Thermodynamics_and_heat_transfer/Chapter_8.ipynb deleted file mode 100755 index d35614c2..00000000 --- a/Elements_of_Thermodynamics_and_heat_transfer/Chapter_8.ipynb +++ /dev/null @@ -1,142 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:efb3cb597d2918b83a867fe2c379b3952acba398aec14a9edf762b5c40043583"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 8 - Second and third law topics"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1 - Pg 125"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the value of dp/ds at constant volume\n",
- "#Initialization of variables\n",
- "P=500 #psia\n",
- "T=700. #F\n",
- "J=778.\n",
- "#calculations\n",
- "dpds=1490 *144/J\n",
- "#results\n",
- "print '%s %d %s' %(\"dp by ds at constant volume =\",dpds,\" F/ft^3/lbm\")\n",
- "print '%s' %(\"The answer is a bit different due to rounding off error in textbook\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "dp by ds at constant volume = 275 F/ft^3/lbm\n",
- "The answer is a bit different due to rounding off error in textbook\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2 - Pg 131"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the thermal efficiency\n",
- "#Initialization of variables\n",
- "import math\n",
- "cp=0.25 #Btu/lbm R\n",
- "T0=520. #R\n",
- "T1=3460. #R\n",
- "#calculations\n",
- "dq=cp*(T0-T1)\n",
- "ds=cp*math.log(T0/T1)\n",
- "dE=dq-T0*ds\n",
- "eta=dE/dq*100.\n",
- "#results\n",
- "print '%s %.1f %s' %(\"Thermal efficiency =\",eta,\"percent\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Thermal efficiency = 66.5 percent\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3 - Pg 134"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Loss of available energy\n",
- "#Initialization of variables\n",
- "import math\n",
- "cp=0.25 #Btu/lbm R\n",
- "T0=520. #R\n",
- "T1=3460. #R\n",
- "dG=21069. #Btu/lbm\n",
- "dH=21502. #Btu/lbm\n",
- "#calculations\n",
- "dq=cp*(T0-T1)\n",
- "ds=cp*math.log(T0/T1)\n",
- "dE=dq-T0*ds\n",
- "eta=dE/dq\n",
- "dw=eta*dH\n",
- "de=-dG+dw\n",
- "#results\n",
- "print '%s %d %s' %(\"Loss of available energy =\",de,\"Btu/lbm\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Loss of available energy = -6774 Btu/lbm\n"
- ]
- }
- ],
- "prompt_number": 3
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Elements_of_Thermodynamics_and_heat_transfer/Chapter_9.ipynb b/Elements_of_Thermodynamics_and_heat_transfer/Chapter_9.ipynb deleted file mode 100755 index 48194c0d..00000000 --- a/Elements_of_Thermodynamics_and_heat_transfer/Chapter_9.ipynb +++ /dev/null @@ -1,449 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:071044d67259dbca809a68700f00a2b22bbcf445c19f6ea5840022a417d2395b"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 9 - Properties of the pure substance"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1 - Pg 149"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Internal energy\n",
- "#Initialization of variables\n",
- "T=32 #F\n",
- "m=1 #lbm\n",
- "J=778.16\n",
- "#calculations\n",
- "print '%s' %(\"From steam tables,\")\n",
- "hf=0 \n",
- "p=0.08854 #psia\n",
- "vf=0.01602 #ft^3/lbm\n",
- "u=hf-p*144*vf/J\n",
- "#results\n",
- "print '%s %.7f %s' %(\"Internal energy =\",u,\" Btu/lbm\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "From steam tables,\n",
- "Internal energy = -0.0002625 Btu/lbm\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2 - Pg 149"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Change in entropy\n",
- "#Initialization of variables\n",
- "P=40 #psia\n",
- "#calculations\n",
- "print '%s' %(\"from steam tables,\")\n",
- "hf=200.8 #Btu/lbm\n",
- "hg=27 #Btu/lbm\n",
- "T=495. #R\n",
- "ds=(hf-hg)/T\n",
- "#results\n",
- "print '%s %.3f %s' %(\"Change in entropy =\",ds,\"Btu/lbm R\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "from steam tables,\n",
- "Change in entropy = 0.351 Btu/lbm R\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3 - Pg 149"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the specific enthalpy\n",
- "#Initialization of variables\n",
- "x=0.35\n",
- "T=18. #F\n",
- "#calculations\n",
- "print '%s' %(\"From table B-14,\")\n",
- "hf=12.12 #Btu/lbm\n",
- "hg=80.27 #Btu.lbm\n",
- "hfg=-hf+hg\n",
- "h=hf+x*hfg\n",
- "#results\n",
- "print '%s %.1f %s' %(\"specific enthalpy =\",h,\"Btu/lbm\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "From table B-14,\n",
- "specific enthalpy = 36.0 Btu/lbm\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 4 - Pg 149"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the heat required\n",
- "#Initialization of variables\n",
- "x=0.35\n",
- "T=18 #F\n",
- "T2=55.5 #F\n",
- "#calculations\n",
- "print '%s' %(\"From table B-14,\")\n",
- "hf=12.12 #Btu/lbm\n",
- "hg=80.27 #Btu.lbm\n",
- "hfg=-hf+hg\n",
- "h=hf+x*hfg\n",
- "h2=85.68 #Btu/lbm\n",
- "dh=h2-h\n",
- "#results\n",
- "print '%s %.2f %s' %(\"Heat required =\",dh,\"Btu/lbm\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "From table B-14,\n",
- "Heat required = 49.71 Btu/lbm\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5 - Pg 149"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the enthalpy and Quality\n",
- "#Initialization of variables\n",
- "P=1460. #psia\n",
- "T=135. #F\n",
- "P2=700. #psia\n",
- "#calculations\n",
- "print '%s' %(\"From mollier chart,\")\n",
- "h=120 #Btu/lbm\n",
- "x=0.83\n",
- "#results\n",
- "print '%s %d %s' %(\"enthalpy =\",h,\"Btu/lbm\")\n",
- "print '%s %.2f' %(\"\\n Quality =\",x)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "From mollier chart,\n",
- "enthalpy = 120 Btu/lbm\n",
- "\n",
- " Quality = 0.83\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6 - Pg 149"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Heat transferred\n",
- "#Initialization of variables\n",
- "m=1 #lbm\n",
- "P1=144. #psia\n",
- "P2=150. #psia\n",
- "T1=360. #F\n",
- "J=778.16\n",
- "#calculations\n",
- "print '%s' %(\"From table 3,\")\n",
- "v1=3.160 #ft^3/lbm\n",
- "h1=1196.5 #Btu/lbm\n",
- "u1=h1-P1*144.*v1/J\n",
- "h2=1211.4 #Btu/lbm\n",
- "u2=h2-P2*144.*v1/J\n",
- "dq=u2-u1\n",
- "#results\n",
- "print '%s %.1f %s' %(\"Heat transferred =\",dq,\"Btu/lbm\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "From table 3,\n",
- "Heat transferred = 11.4 Btu/lbm\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7 - Pg 150"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the work done and work required\n",
- "#Initialization of variables\n",
- "T1=100. #F\n",
- "P2=1000. #psia\n",
- "x=0.6\n",
- "J=778.16\n",
- "#calculations\n",
- "print '%s' %(\"From table 3,\")\n",
- "v=0.01613 #ft^3/lbm\n",
- "P1=0.9 #psia\n",
- "wrev=-v*(P2-P1)*144/J\n",
- "dv=0.000051 #ft^3/lbm\n",
- "wcomp=(P2+P1)/2 *dv*144/J\n",
- "wact=wrev/x\n",
- "#results\n",
- "print '%s %.1f %s' %(\"Work done =\",wrev,\"Btu/lbm\")\n",
- "print '%s %.1f %s' %(\"\\n In case 2, work required =\",wact,\" Btu/lbm\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "From table 3,\n",
- "Work done = -3.0 Btu/lbm\n",
- "\n",
- " In case 2, work required = -5.0 Btu/lbm\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 8 - Pg 150"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Heat transferred\n",
- "#Initialization of variables\n",
- "pa=1000. #atm\n",
- "ta=100. #F\n",
- "#calculations\n",
- "hf=67.97 #Btu/lbm\n",
- "w=3 #Btu/lbm\n",
- "ha=hf+w\n",
- "print '%s' %(\"from steam table 2,\")\n",
- "hc=1191.8 #Btu/lbm\n",
- "qrev=hc-ha\n",
- "#results\n",
- "print '%s %.1f %s' %(\"Heat transferred =\",qrev,\" Btu/lbm\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "from steam table 2,\n",
- "Heat transferred = 1120.8 Btu/lbm\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9 - Pg 151"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the work done and Final state pressure\n",
- "#Initialization of variables\n",
- "P1=144 #psia\n",
- "T1=400 #F\n",
- "y=0.7\n",
- "#calculations\n",
- "print '%s' %(\"From steam tables,\")\n",
- "h1=1220.4 #Btu/lbm\n",
- "s1=1.6050 #Btu/lbm R\n",
- "s2=1.6050 #Btu/lbm R\n",
- "P2=3 #psia\n",
- "sf=0.2008 #Btu/lbm R\n",
- "sfg=1.6855 #Btu/lbm R\n",
- "x=(s1-sf)/sfg\n",
- "hf=109.37 #Btu/lbm\n",
- "hfg=1013.2 #Btu/;bm\n",
- "h2=hf+x*hfg\n",
- "work=h1-h2\n",
- "dw=y*work\n",
- "h2d=h1-dw\n",
- "#results\n",
- "print '%s %d %s' %(\"Work done =\",work,\"Btu/lbm\")\n",
- "print '%s %.1f %s' %(\"\\n work done in case 2 =\",dw,\"Btu/lbm\")\n",
- "print '%s %d %s' %(\"\\n Final state pressure =\",P2,\"psia\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "From steam tables,\n",
- "Work done = 266 Btu/lbm\n",
- "\n",
- " work done in case 2 = 186.8 Btu/lbm\n",
- "\n",
- " Final state pressure = 3 psia\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 10 - Pg 154"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Quality of wet steam\n",
- "#Initialization of variables\n",
- "pb=14.696 #psia\n",
- "pa=150 #psia\n",
- "tb=300 #F\n",
- "#calculations\n",
- "print '%s' %(\"From steam tables,\")\n",
- "hb=1192.8 #Btu/lbm\n",
- "ha=hb\n",
- "hf=330.51 #Btu/lbm\n",
- "hfg=863.6 #Btu/lbm\n",
- "x=(ha-hf)/hfg*100\n",
- "#results\n",
- "print '%s %.1f %s' %(\"Quality of wet steam =\",x,\"percent\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "From steam tables,\n",
- "Quality of wet steam = 99.8 percent\n"
- ]
- }
- ],
- "prompt_number": 10
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file |