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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 /Non-conventional_Energy_Sources_by_G._D._Rai | |
| 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 'Non-conventional_Energy_Sources_by_G._D._Rai')
13 files changed, 0 insertions, 1797 deletions
diff --git a/Non-conventional_Energy_Sources_by_G._D._Rai/Chapter10.ipynb b/Non-conventional_Energy_Sources_by_G._D._Rai/Chapter10.ipynb deleted file mode 100755 index edc9f942..00000000 --- a/Non-conventional_Energy_Sources_by_G._D._Rai/Chapter10.ipynb +++ /dev/null @@ -1,285 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:7dac7ff4d7c0b0efd562d141ece5753006bca27ad006e31a868bd655b584d707"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter10-Chemical Energy Sources"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 10.2.8.1-pg585"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "##Ex10.2.8.1;Find Reversible voltage for hydrogen oxygen fuel cell\n",
- "del_G=-237.3*10**3;##Joules/gm-mole of H2\n",
- "##Reversible voltafe E of a cell is given by =del_Wrev/nF=-del_G/nF\n",
- "##since 2 electrons are transferred per molecule of H2.thus\n",
- "n=2.;\n",
- "F=96500.;##Faraday's constant\n",
- "E=-del_G/(n*F);\n",
- "print'%s %.2f %s'%(\"Reversible voltage=\",E,\" volts\");\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Reversible voltage= 1.23 volts\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 10.2.8.2-pg585"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "##Ex10.2.8.2;calculate voltage output of cell,efficiency,electric work output,heat transfer to the surroundings\n",
- "import math\n",
- "##1] voltage output of cell\n",
- "del_G=-237.3*10**3;##Joules/gm-mole of H2\n",
- "n=2.;\n",
- "F=96500.;##Faraday's constant\n",
- "E=-del_G/(n*F);\n",
- "print'%s %.2f %s'%(\" E=\",E,\" volts\");\n",
- "##2] Efficiency\n",
- "##nmax=del_Wmax/-(del_H)25 degree celcuis = -(del_G)T/(-del_H)25\n",
- "del_G_at298k=-56690.;##unit=kcal/kg mole\n",
- "del_H_at298k=-68317.;##unit=kcal/kg mole\n",
- "nmax=del_G_at298k/del_H_at298k\n",
- "print'%s %.2f %s'%(\"\\n nmax=\",nmax,\"\")\n",
- "##3]Electric work output per mole\n",
- "F=(96500/4.184);\n",
- "del_Wrever=(n*F*E);\n",
- "print'%s %.2f %s'%(\"\\n Electric work output per mole=\",del_Wrever,\" kcal/kg mole\");\n",
- "##4] Heat transfer to the surroundings\n",
- "##the heat transfer is Q=T*del-s=del_H_at298k-del_G_at298k\n",
- "Q=del_H_at298k-del_G_at298k;\n",
- "print'%s %.2f %s'%(\"\\n The heat transfer is Q=\",Q,\" kcal/kg mole\");\n",
- "##The negative sign indicates that the heat is removed from the cell and transferred to the surrounding\n",
- "\n",
- "##value of \"Electric work output per mole\" is approximate in the text book to the real calculated value"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " E= 1.23 volts\n",
- "\n",
- " nmax= 0.83 \n",
- "\n",
- " Electric work output per mole= 56716.06 kcal/kg mole\n",
- "\n",
- " The heat transfer is Q= -11627.00 kcal/kg mole\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 10.2.8.3-pg587"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "##Ex10.2.8.3;The heat transferred to the surrounding\n",
- "import math\n",
- "del_G_at298k=-237191.;##unit=kJ/kg mole\n",
- "del_H_at298k=-285838.;##unit=kJ/kg mole\n",
- "ne=2.;\n",
- "F=96500.;##Faraday's constant\n",
- "E=-del_G_at298k/(ne*F);\n",
- "print'%s %.2f %s'%(\" E=\",E,\" volts\");\n",
- "nmax=del_G_at298k/del_H_at298k\n",
- "print'%s %.2f %s'%(\"\\n nmax=\",nmax,\"\");\n",
- "nmax=nmax*100;\n",
- "print'%s %.2f %s'%(\"=0\",nmax,\" persent\");\n",
- "##Electric work output per mole of the fule is We=-del_G kJ/kg mole\n",
- "We=del_G_at298k##kJ/kg mole\n",
- "print'%s %.2f %s'%(\"\\n Electric work output per mole of the fule is We=\",We,\" kJ/kg mole\")\n",
- "##since there is 1 mol os H2O for each mole of fule,there is also a work output of 237191 kJ/kg mole\n",
- "##Heat transferred is Q=T*del-s=del_H_at298k-del_G_at298k\n",
- "Q=del_H_at298k-del_G_at298k;\n",
- "print'%s %.2f %s'%(\"\\n The heat transfer is Q=\",Q,\" kJ/kg mole\");\n",
- "##The negative sign indicates that the heat is removed from the cell and transferred to the surrounding\n",
- "\n",
- "##value of \"Electric work output per mole\" is misprinted in the text book.\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " E= 1.23 volts\n",
- "\n",
- " nmax= 0.83 \n",
- "=0 82.98 persent\n",
- "\n",
- " Electric work output per mole of the fule is We= -237191.00 kJ/kg mole\n",
- "\n",
- " The heat transfer is Q= -48647.00 kJ/kg mole\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 10.2.8.4-pg587"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "##Ex10.2.8.4;calculate del_G,del_S,del_H;\n",
- "import math\n",
- "##We have the relation del_G=-n*F*E\n",
- "##where,del_G=gibbs free energy of the system at 1 atm and temperature(T)\n",
- "n=1.;##numbers of electons transferred per molecule of reactant\n",
- "E=0.0455;##volts ;e.m.f. of the cell\n",
- "F=96500.;##Faraday's constant\n",
- "##let X=dE/dT\n",
- "X=0.000338;\n",
- "del_G=-n*F*E;\n",
- "print'%s %.2f %s'%(\" del_G=\",del_G,\" joules\");\n",
- "##del_S = Entropy change of the system at temperature T and press p=1 atm in the case\n",
- "del_S=n*F*(X);##del_S=n*F*(dE/dT)\n",
- "print'%s %.2f %s'%(\"\\n del_S=\",del_S,\" joules/deg.\");\n",
- "##And entropy change is given by the relation del_H=nF[T(dE/dT)-E]\n",
- "T=298;\n",
- "del_H=n*F*((T*X)-E);\n",
- "print'%s %.2f %s'%(\"\\n del_H=\",del_H,\" joule\");\n",
- "\n",
- "\n",
- "##value are taken approximate in the text book to the real calculated value\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " del_G= -4390.75 joules\n",
- "\n",
- " del_S= 32.62 joules/deg.\n",
- "\n",
- " del_H= 5329.12 joule\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 10.2.8.5-pg588"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "##Ex10.2.8.5;heat transfer rate would be involved under these circumstances\n",
- "import math\n",
- "del_G_at25degree_celcius=-195500.;##unit=cal/gm mole\n",
- "del_H_at25degree_celcius=-212800.;##unit=cal/gm mole\n",
- "F=(96500/4.184);##since F=96500 coulombs/gm-mole\n",
- "n=8.\n",
- "E_at25degree_celcius=-del_G_at25degree_celcius/(n*F);##Joules/coulomb\n",
- "print'%s %.2f %s'%(\" E_at25degree_celcius=\",E_at25degree_celcius,\" volts=1.060 volts\");\n",
- "##Max. efficiency nmax=del_Wmax/-(del_H)at25 degree celcuis = -(del_G)T/(-del_H)25\n",
- "nmax=del_G_at25degree_celcius/del_H_at25degree_celcius;\n",
- "print'%s %.2f %s'%(\"\\n nmax=\",nmax,\"\");\n",
- "##voltage efficiency nv=on load voltage/open circuit voltage=Operating voltage/Theoretical voltage\n",
- "Theoretical_voltage=1.060/0.92;\n",
- "print'%s %.2f %s'%(\"\\n Theoretical_voltage=\",Theoretical_voltage,\" volts\");\n",
- "##power developed=100 kW=100*10^3 W\n",
- "power_developed=(100*10**3)*0.86;##unit=kcal/hr; since 1 watt=1 joule/sec=0.86 kcal/hr\n",
- "print'%s %.2f %s'%(\"\\n power_developed=\",power_developed,\" kcal/hr\");\n",
- "del_G=-195500.;\n",
- "##Required flow rate of Methane\n",
- "R_F_R_O_M=(power_developed*16.)/del_G;##kg/hr;\n",
- "##(methane moles)=16\n",
- "print'%s %.2f %s'%(\"\\n flow rate of Methane=\",R_F_R_O_M,\" kg/hr\");\n",
- "##Heat transfer Q=T8del_s=del_H+del_w=del_H-del_G\n",
- "Q=del_H_at25degree_celcius-del_G_at25degree_celcius;\n",
- "print'%s %.2f %s'%(\"\\n The heat transfer is Q=\",Q,\" kcal/kg mole\");\n",
- "\n",
- "##The value are approximate in the text book to the real calculated value\n",
- "##value of \"Required flow rate of methane\" is wrong in the text book.\n",
- "##value of \"Heat transfer\" is wrong in the text book.\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " E_at25degree_celcius= 1.06 volts=1.060 volts\n",
- "\n",
- " nmax= 0.92 \n",
- "\n",
- " Theoretical_voltage= 1.15 volts\n",
- "\n",
- " power_developed= 86000.00 kcal/hr\n",
- "\n",
- " flow rate of Methane= -7.04 kg/hr\n",
- "\n",
- " The heat transfer is Q= -17300.00 kcal/kg mole\n"
- ]
- }
- ],
- "prompt_number": 2
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Non-conventional_Energy_Sources_by_G._D._Rai/Chapter12.ipynb b/Non-conventional_Energy_Sources_by_G._D._Rai/Chapter12.ipynb deleted file mode 100755 index 7c2c549a..00000000 --- a/Non-conventional_Energy_Sources_by_G._D._Rai/Chapter12.ipynb +++ /dev/null @@ -1,67 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:49516c3f23d8642eab7059abb4c60afe8871c4f8eaa6c5db6c64f1493784e28a"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter12-Magneto Hydro Dynamic(MHD) Power Generation"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12.6.1-pg682"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "##Ex12.6.1.;calculate open circuit voltage and maximum power output\n",
- "B=2;##flux density;unit=Wb/m^2\n",
- "u=10**3;##average gas velocity;unit=m/second\n",
- "d=0.50;##distance between plates;unit=m\n",
- "E0=B*u*d;##Open ccircuit voltage\n",
- "print'%s %.2f %s'%(\" Open ccircuit voltage E0=\",E0,\" Volts\");\n",
- "##Generator resistance; Rg=d/sigma*A\n",
- "sigma=10;##Gaseous conductivity;unit=Mho/m\n",
- "A=0.25;##Plate Area;unit=m^2 \n",
- "Rg=d/(sigma*A);\n",
- "print'%s %.2f %s'%(\"\\n Generator resistance Rg=\",Rg,\" Ohm\");\n",
- "##Maximum power\n",
- "Maximum_power=(E0**2)/(4.*Rg);\n",
- "print'%s %.2f %s'%(\"\\n Maximum_power=\",Maximum_power,\" watts\");\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " Open ccircuit voltage E0= 1000.00 Volts\n",
- "\n",
- " Generator resistance Rg= 0.20 Ohm\n",
- "\n",
- " Maximum_power= 1250000.00 watts\n"
- ]
- }
- ],
- "prompt_number": 1
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Non-conventional_Energy_Sources_by_G._D._Rai/Chapter13.ipynb b/Non-conventional_Energy_Sources_by_G._D._Rai/Chapter13.ipynb deleted file mode 100755 index 3f16e664..00000000 --- a/Non-conventional_Energy_Sources_by_G._D._Rai/Chapter13.ipynb +++ /dev/null @@ -1,388 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:446aceff980b12efb55c191333cfc1afc3a816c084c4cb1a989aad4ec192647e"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter13-Thermo Electric Power"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.2.1-pg707"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "##Ex13.2.1.;Peltier heats absorbed and rejected\n",
- "##peltier coefficients at these junctions are aplha_p_1-2=alpha_s_1-2*T\n",
- "##Let A=alpha_s_1-2 at 373 k=55*10^-6 v/degree_k and B=alpha_s_1-2 at 273 k=50*10^-6 v/degree_k\n",
- "A=(55*10**-6);\n",
- "B=(50*10**-6);\n",
- "T1=373.;##k\n",
- "T2=273.;##k\n",
- "I=10*10**-3;##current;unit=Ampere\n",
- "alpha_p_1_2_at_373k=A*T1;\n",
- "alpha_p_1_2_at_273k=B*T2;\n",
- "print'%s %.2f %s %.2f %s'%(\" alpha_p_1_2_at_373k=\",alpha_p_1_2_at_373k,\" W/amp\"and \" \\n alpha_p_1_2_at_273k=\",alpha_p_1_2_at_273k,\" W/amp\");\n",
- "##Peltier heats absorned and rejected to be\n",
- "q2_peltier=alpha_p_1_2_at_373k*I;\n",
- "q1_peltier=alpha_p_1_2_at_273k*I;\n",
- "print'%s %.2e %s %.2e %s '%(\"\\n q2_peltier=\",q2_peltier,\" w \" and\"\\n q1_peltier=\",q1_peltier,\" W\");\n",
- "c=q2_peltier-q1_peltier;\n",
- "print(\"\\n If no other heat transfer were involved,the difference between these vaues,\");\n",
- "print'%s %.2e %s %.2e %s %.2e %s '%(\"\\n \",q2_peltier,\" \"and \"\",q1_peltier,\" W\"and \"\\n,would be supplied as electric power\",c,\"\");\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " alpha_p_1_2_at_373k= 0.02 \n",
- " alpha_p_1_2_at_273k= 0.01 W/amp\n",
- "\n",
- " q2_peltier= 2.05e-04 \n",
- " q1_peltier= 1.36e-04 W \n",
- "\n",
- " If no other heat transfer were involved,the difference between these vaues,\n",
- "\n",
- " 2.05e-04 1.36e-04 \n",
- ",would be supplied as electric power 6.87e-05 \n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.3.2-pg708"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "##Ex.13.3.2.;Find the thomson heat transferred\n",
- "import math\n",
- "import scipy\n",
- "from scipy import integrate\n",
- "\n",
- "##Let D=dalpha_s1/dT;\n",
- "D=5.4*10**-3;##unit=micro V/degree k^2\n",
- "T1=273;##unit=k\n",
- "T2=373;##unit=k\n",
- "I=10*10**-3;##unit=A\n",
- "##Thomson coefficient sigma,varies with temp. \n",
- "##sigma_1_of_T=-T*D;unit=V/degree k\n",
- "##The thomson heat is given by equation\n",
- "##qth=I*Integration of sigma_1_of_T w.r.t. T\n",
- "def fun(T):\n",
- " y=T\n",
- " return y\n",
- "Integration = scipy.integrate.quad(fun,T1,T2);\n",
- "qth=I*D*Integration[0];\n",
- "print'%s %.2f %s'%(\"The THOMSON HEAT=\",qth,\" micro W\");\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The THOMSON HEAT= 1.74 micro W\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.4.1-pg715"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "##Ex13.4.1.;Determine the efficiency of the thermoelectric generator.what will be its carnot efficiency\n",
- "import math\n",
- "TH=600.;##degree k;##temperature of the hot reservior of source\n",
- "TC=300.;##degree k;##temperature of the sink\n",
- "Z=2*(10**-3);##1/degree k;##Figure of merit for the material\n",
- "M_optimum=(1.+((Z/2.)*(TH+TC)))**0.5;\n",
- "print'%s %.2f %s'%(\" M_optimum=\",M_optimum,\"\");\n",
- "##Efficiency of the thermoelectric generator is n=(((TH-TC)/TH)*((M_optimum-1)/(M_optimum+(TC/TH)))*100;\n",
- "a=((TH-TC)/TH);\n",
- "b=(M_optimum-1)/(M_optimum+(TC/TH));\n",
- "n=a*b*100;\n",
- "print'%s %.2f %s'%(\"\\n Efficiency of the thermoelectric generator is n=\",n,\" persent\");\n",
- "##where as efficiency of the carnot cycle (reversible) nc=((TH-TC)/TH)*100\n",
- "nc=a*100;\n",
- "print'%s %.2f %s'%(\"\\n Efficiency of the carnot cycle (reversible) nc=\",nc,\" persent\");\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " M_optimum= 1.38 \n",
- "\n",
- " Efficiency of the thermoelectric generator is n= 10.07 persent\n",
- "\n",
- " Efficiency of the carnot cycle (reversible) nc= 50.00 persent\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.4.2-pg716"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "##Ex13.4.12.;Calculare maximum generator efficiency and the efficiency for maximum power,power output\n",
- "import math\n",
- "##seedbeck coefficient(alpha_s);unit=volts/degree celcius\n",
- "alpha_s1=-190.*10**-6;##n-type\n",
- "alpha_s2=190.*10**-6;##p-type\n",
- "##Specific resistivity(p);unit=Ohm-cm\n",
- "p1=1.45*10**-3;##n-type\n",
- "p2=1.8*10**-3;##p-type\n",
- "##Figure of merit(Z);unit=degree k**-1\n",
- "Z1=2.*10**-3;##n-type\n",
- "Z2=1.7*10**-3;##p-type\n",
- "\n",
- "\n",
- "##conductivity (n-type), \n",
- "k1=(alpha_s1**2.)/(p1*Z1);\n",
- "##similarly\n",
- "k2=(alpha_s2**2.)/(p2*Z2);\n",
- "print'%s %.2f %s %.2f %s'%(\" Conductivity k1=\",k1,\" W/cm degree celcius\" and \" \\n Conductivity k2=\",k2,\" W/cm degree celcius\");\n",
- "##Z_opt=((alpha_s1-alpha_s2)**2)/[(p1*k1)**2+(p2*k2)**2];\n",
- "##let\n",
- "a=(alpha_s1-alpha_s2)\n",
- "b=(p1*k1)\n",
- "c=(p2*k2)\n",
- "A=math.sqrt(b)\n",
- "B=math.sqrt(c)\n",
- "C=(A+B);\n",
- "##/therefore\n",
- "Z_opt=(a/C)**2.;\n",
- "print'%s %.2f %s'%(\"\\n Z_opt=\",Z_opt,\" degree k\");\n",
- "##Thermal conductance\n",
- "A1=2.3;##cm**2\n",
- "A2=1.303;##cm**2\n",
- "l1=1.5;##cm\n",
- "l2=0.653;##cm\n",
- "K=((k1*A1)/l1)+((k2*A2)/l2)\n",
- "print'%s %.2f %s'%(\"\\n Thermal conductance K=\",K,\" W/degree celcius\");\n",
- "##R=Resistance of the generator=R1+R2\n",
- "R=((p1*l1)/A1)+((p2*l2)/A2);\n",
- "print'%s %.2f %s'%(\"\\n Resistance of the generator R=\",R,\" ohm\");\n",
- "TH=923.;##unit=k\n",
- "TC=323.;##unit=k\n",
- "M_opt=(1.+((Z_opt/2.)*(TH+TC)))**0.5;\n",
- "print'%s %.2f %s'%(\"\\n M_opt=\",M_opt,\" ohm\");\n",
- "RL=M_opt*R;\n",
- "print'%s %.2f %s'%(\"\\n RL=\",RL,\" ohms\");\n",
- "##Optimum efficiency n_opt=(((TH-TC)/TH)*((M_opt-1)/(M_opt+(TC/TH)))*100;\n",
- "aa=((TH-TC)/TH);\n",
- "##taking M_opt=1.43\n",
- "b=(1.43-1.)/(1.43+(TC/TH));\n",
- "n_opt=aa*b*100.;\n",
- "print'%s %.2f %s'%(\"\\n Optimum efficiency n_opt=\",n_opt,\" persent\");\n",
- "##efficiency for max. power output n= (TH-TC)/TH)*m/[((1+m)**2/TH)*(KR/alpha_s_12**2)+(1+m)-(TH-TC)/2TH)]\n",
- "##Efficiency power output\n",
- "##RL=R i.e. m=1\n",
- "## let ab=(1+m)**2/TH;ac=(KR/alpha_s_12**2);ad=(TH-TC)/2TH\n",
- "m=1.;\n",
- "ab=4./TH;\n",
- "ac=1./Z_opt;\n",
- "ad=aa/2.;\n",
- "n_max=(aa/(ab*ac+2.-ad))*100.;\n",
- "print'%s %.2f %s'%(\"\\n max. power output n_max \",n_max,\" persent\")\n",
- "##Power output P_opt=I**2*RL=alpha_s12**2(TH-TC)*RL/(R+RL)**2=alpha_s12**2(TH-TC)/(1+M_opt)**2*RL\n",
- "##let at=alpha_s12**2(TH-TC);mi=(1+M_opt)**2*RL\n",
- "at=a*a*(TH-TC)*(TH-TC);\n",
- "ml=(1.+1.43)*(1.+1.43)*2.63*10**-3\n",
- "P_opt=at/ml;\n",
- "print'%s %.2f %s'%(\"\\n Power output P_opt=\",P_opt,\" watts\");\n",
- "##for max. power P_max (RL=R)\n",
- "##P_max=alpha_s12**2(TH-TC)*RL/(r+RL)**2=alpha_s12**2(TH-TC)RL*4RL\n",
- "P_max=at/(4.*1.84*10**-3);\n",
- "print'%s %.2f %s'%(\"\\n max. power P_max=\",P_max,\" watts\");\n",
- "\n",
- "\n",
- "##Many calcuating mistak are there in a following example,which is corrected in program.\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " Conductivity k1= 0.01 \n",
- " Conductivity k2= 0.01 W/cm degree celcius\n",
- "\n",
- " Z_opt= 0.00 degree k\n",
- "\n",
- " Thermal conductance K= 0.04 W/degree celcius\n",
- "\n",
- " Resistance of the generator R= 0.00 ohm\n",
- "\n",
- " M_opt= 1.47 ohm\n",
- "\n",
- " RL= 0.00 ohms\n",
- "\n",
- " Optimum efficiency n_opt= 15.70 persent\n",
- "\n",
- " max. power output n_max 16.13 persent\n",
- "\n",
- " Power output P_opt= 3.35 watts\n",
- "\n",
- " max. power P_max= 7.06 watts\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13.4.3-pg718"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "##Ex.13.4.3;maximum efficiency,no. of thermocouple in series,open ckt voltage,heat i/p and reject at full load.\n",
- "import math\n",
- "kA=0.02;##unit=watt/cm degree kelvin\n",
- "kB=0.03;##unit=watt/cm degree kelvin\n",
- "pA=0.01;##unit=ohm cm\n",
- "pB=0.012;##unit=ohm cm\n",
- "TH=1500.;##unit=degree kelvin\n",
- "TC=1000;##unit=degree kelvin\n",
- "AA=43.5;##unit=cm**2\n",
- "AB=48.6;##unit=cm**2\n",
- "LA=0.49;##unit=cm\n",
- "LB=0.49;##unit=cm\n",
- "I=20.*48.6;##Current density in the element limited to,I=20 amp/cm**2\n",
- "output=100.;##unit=kW\n",
- "##alpha_SAB at 1250 degree kelvin=0.0012 volt/degree kelvin=alpha_SA-alpha_SB\n",
- "alpha_SAB=0.0012;##unit=volt/degree kelvin\n",
- "##let\n",
- "b=(pA*kA);\n",
- "c=(pB*kB);\n",
- "A=math.sqrt(b);\n",
- "B=math.sqrt(c);\n",
- "C=(A+B);\n",
- "##figure of merit\n",
- "Z=(alpha_SAB/C)**2.;\n",
- "print'%s %.2f %s'%(\" Z=\",Z,\" degree k^-1\");\n",
- "M=(1+((Z/2.)*(TH+TC)))**0.5;\n",
- "print'%s %.2f %s'%(\"\\n M=\",M,\"\");\n",
- "##let\n",
- "aa=((TH-TC)/TH);\n",
- "bb=(M-1)/(M+(TC/TH));\n",
- "##1] MAx. efficiency of a thermoelectric converter is given by n_max=((TH-TC)/TH)*[(M-1)/(M+(TC/TH))]*100;\n",
- "n_max=aa*bb*100.;\n",
- "print'%s %.2f %s'%(\"\\n Maximum efficiency n_max=\",n_max,\" persent\");\n",
- "##2] No. of thermocouple in series\n",
- "V=alpha_SAB*(TH-TC);\n",
- "print'%s %.2f %s'%(\"\\n V=\",V,\" volt\");\n",
- "R=((pA*LA)/AA)+((pB*LB)/AB);##since R=RA+RB=((pA*LA)/AA)+((pB*LB)/AB);\n",
- "print'%s %.2f %s'%(\"\\n R=\",R,\" ohm\");\n",
- "VL=V-(R*I);\n",
- "print'%s %.2f %s'%(\"\\n VL=\",VL,\" volt\");\n",
- "##NTCS=total voltage required/voltage required by one couple\n",
- "NTCS=115./VL;\n",
- "print'%s %.2f %s'%(\"\\n No. of thermocouple in series=\",NTCS,\"\");\n",
- "##3] Open circuit voltage\n",
- "OCV=V*309.;\n",
- "print'%s %.2f %s'%(\"\\n Open circuit voltage=\",OCV,\" volt\")\n",
- "##4] Heat input and reject at full load.\n",
- "##Heat input at full load.=output/efficency=100/0.091\n",
- "HIFL=output/n_max;\n",
- "print'%s %.2f %s'%(\"\\n Heat input at full load=\",HIFL,\" kW\")\n",
- "## Heat reject at full load. =Heat input-Work output\n",
- "HRFL=HIFL-output;\n",
- "print'%s %.2f %s'%(\"\\n Heat reject at full load=\",HRFL,\"kW\")\n",
- "\n",
- "\n",
- "\n",
- "##The value of \"pB\" is misprinted\n",
- "##The values are taken in the text book is approximately equal to calculated values\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " Z= 0.00 degree k^-1\n",
- "\n",
- " M= 1.63 \n",
- "\n",
- " Maximum efficiency n_max= 9.09 persent\n",
- "\n",
- " V= 0.60 volt\n",
- "\n",
- " R= 0.00 ohm\n",
- "\n",
- " VL= 0.37 volt\n",
- "\n",
- " No. of thermocouple in series= 308.39 \n",
- "\n",
- " Open circuit voltage= 185.40 volt\n",
- "\n",
- " Heat input at full load= 11.00 kW\n",
- "\n",
- " Heat reject at full load= -89.00 kW\n"
- ]
- }
- ],
- "prompt_number": 9
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Non-conventional_Energy_Sources_by_G._D._Rai/Chapter14.ipynb b/Non-conventional_Energy_Sources_by_G._D._Rai/Chapter14.ipynb deleted file mode 100755 index eecdaf28..00000000 --- a/Non-conventional_Energy_Sources_by_G._D._Rai/Chapter14.ipynb +++ /dev/null @@ -1,116 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:f0bae5619ed9c3242b82c3fadf724693800c27dd8d5dc8811bfb1cb1c626c969"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter14-Thermionic Generation"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 14.4.1-pg738"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "##Ex.14.4.1.;Calculate the efficiency of the generator and also compare with the carnot efficiency\n",
- "import math\n",
- "##cathode work funtion \n",
- "flux_c=2.5;##unit=volts\n",
- "##anode work funtion \n",
- "flux_a=2.;##unit=volts\n",
- "##Temp. of cathode\n",
- "Tc=2000.;##unit=degree k\n",
- "##Temp. of surrounding\n",
- "Ts=1000.;##unit=degree k\n",
- "##plasma potentail drop\n",
- "flux_p=0.1;##unit=volts\n",
- "##Net output voltage\n",
- "V=flux_c-flux_a-flux_p\n",
- "print'%s %.2f %s'%(\" V=\",V,\" volt\");\n",
- "##charge of an electron\n",
- "e=1.6*10**-19;##unit=coulomb\n",
- "##boltzmann constant\n",
- "k=1.38*10**-23;##unit=joule/degree kelvin\n",
- "A=1.20*10**6;\n",
- "##one electron volt=1.6*10**-19 joule\n",
- "##The net current in the generator J=J_cathode-J_anode\n",
- "##let EC=e**(-flux_c/k*Tc)\n",
- "EC=math.e**(-(1.6*10**-19*flux_c)/(k*Tc));\n",
- "J_cathode=A*(Tc*Tc)*EC##J_cathode=A*Tc**2*e**(-flux_c/k*Tc)\n",
- "print'%s %.2f %s'%(\"\\n J_cathode=\",J_cathode,\" amp/m^2\");\n",
- "##let EA=e**(-flux_c/k*Ts)\n",
- "EA=math.e**(-(1.6*10**-19*flux_a)/(k*Ts));\n",
- "J_anode=A*(Ts**2)*EA;##J_cathode=A*Ts**2*e**(-flux_c/k*Ts)\n",
- "print'%s %.2f %s'%(\"\\n J_anode=\",J_anode,\" amp/m^2\");\n",
- "##The net current can be taken =Jc,as Ja can be neglected in comparison with Jc\n",
- "J=J_cathode;\n",
- "print'%s %.2f %s'%(\"\\n J=\",J,\" amp/m^2\");\n",
- "##The heat supplied to the cathode Qc/Ac=J(flux_c+((2*k*Tc)/e))+samestion of sigma*(Tc**4-Ts**4)\n",
- "##let QA=Qc/Ac; and\n",
- "a=2.5+((2*1.38*10**-23*2000.)/(1.6*10**-19));\n",
- "b=J*a;\n",
- "c=(0.2*5.67*(10**-12)*(10**-4)*((2000**4)-(1000**4)));\n",
- "##therefore\n",
- "QA=b+c; ##since: QA=(J*(2.5+((2*(1.38*10**-23)*2000*)/(1.6*10**-19))))+(0.2*5.67*(10**-12)*(10**-4)*((2000**4)-(1000**4)))\n",
- "print'%s %.2f %s'%(\"\\n The heat supplied to the cathode Qc/Ac=\",QA,\" watt/m^2\");\n",
- "##efficiency of the generator\n",
- "ng=((J*V)/(7.026*10**6))*100.;\n",
- "print'%s %.2f %s'%(\"\\n ng=\",ng,\" persent\");\n",
- "##carnot efficiency this device\n",
- "T1=2000;\n",
- "T2=1000;\n",
- "T=2000;\n",
- "nc=((T1-T2)/T)*100;\n",
- "print'%s %.2f %s'%(\"\\n nc=\",nc,\" persent\");\n",
- "\n",
- "\n",
- "##Value of \"The heat supplied to the cathode Qc/Ac\" is given wrong\n",
- "##value of charge e is taken wrong;corrected by giving value 1.6*10**-19\n",
- "##value of J anode is differ from calculated value. \n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " V= 0.40 volt\n",
- "\n",
- " J_cathode= 2438475.02 amp/m^2\n",
- "\n",
- " J_anode= 102.00 amp/m^2\n",
- "\n",
- " J= 2438475.02 amp/m^2\n",
- "\n",
- " The heat supplied to the cathode Qc/Ac= 6937461.44 watt/m^2\n",
- "\n",
- " ng= 13.88 persent\n",
- "\n",
- " nc= 0.00 persent\n"
- ]
- }
- ],
- "prompt_number": 2
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Non-conventional_Energy_Sources_by_G._D._Rai/Chapter2.ipynb b/Non-conventional_Energy_Sources_by_G._D._Rai/Chapter2.ipynb deleted file mode 100755 index 8b4db74e..00000000 --- a/Non-conventional_Energy_Sources_by_G._D._Rai/Chapter2.ipynb +++ /dev/null @@ -1,202 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:54c30c216e131996bdae1a02fbe32e2e6e143be1feb0a02491a5808c29f1c930"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter2-Solar Radiation and its Measurement"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.4.1-pg 59"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "##Ex2.4.1.;Detremine local solar time and declination\n",
- "import math\n",
- "##The local solar time=IST-4(standard time longitude-longitude of location)+Equation of time correstion\n",
- "##IST=12h 30min;for the purpose of calculation we are writing it as a=12h,b=29 min 60sec;\n",
- "a=12.;\n",
- "b=29.60;\n",
- "##(standard time longitude-longitude of location)=82 degree 30min - 77 degree 30min;\n",
- "##for the purpose of calculation we are writing it as\n",
- "STL3=82.5-72.5;\n",
- "##Equation of time correstion: 1 min 01 sec\n",
- "##for the purpose of calculation we are writing it as\n",
- "c=1.01;\n",
- "##The local solar time=IST-4(standard time longitude-longitude of location)+Equation of time correstion\n",
- "LST=b-STL3-c;\n",
- "print'%s %.2f %s %.2f %s'%(\" The local solar time=\",a,\"\"and \"\",LST,\" in hr.min.sec\");\n",
- "##Declination delta can be obtain by cooper's eqn : delta=23.45*sin((360/365)*(284+n))\n",
- "n=170.;##(on June 19)\n",
- "##let\n",
- "a=(360./365.)*(284.+n)\n",
- "aa=(a*math.pi)/180.\n",
- "##therefore\n",
- "delta=23.45*math.sin(aa);\n",
- "print'%s %.2f %s'%(\"\\n delta=\",delta,\" degree\");\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " The local solar time= 12.00 18.59 in hr.min.sec\n",
- "\n",
- " delta= 23.43 degree\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.4.2-pg 59"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "##Ex2.4.2.;Calculate anglr made by beam radiation with the normal to a flat collector.\n",
- "gama=0.;##since collector is pointing due south.\n",
- "##For this case we have equation : cos_(theta_t)=cos(fie-s)*cos(delta)*cos(w)+sin(fie-s)*sin(delta)\n",
- "##with the help of cooper eqn on december 1,\n",
- "n=335.;\n",
- "##let\n",
- "a=(360./365.)*(284.+n);\n",
- "aa=(a*math.pi)/180;\n",
- "##therefore\n",
- "delta=23.45*math.sin(aa);\n",
- "print'%s %.2f %s'%(\" delta=\",delta,\" degree\");\n",
- "##Hour angle w corresponding to 9.00 hour=45 Degreew\n",
- "w=45.;##degree\n",
- "##let\n",
- "a=math.cos(((28.58*math.pi)/180.)-((38.58*math.pi)/180.))*math.cos(delta*math.pi*180**-1)*math.cos(w*math.pi*180**-1);\n",
- "b=math.sin(delta*math.pi*180**-1)*math.sin(((28.58*math.pi)/180.)-((38.58*math.pi)/180.));\n",
- "##therefore\n",
- "cos_of_theta_t=a+b;\n",
- "theta_t=math.acos(cos_of_theta_t)*57.3;\n",
- "print'%s %.2f %s'%(\"\\n theta_t=\",theta_t,\" Degree\");\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " delta= -22.11 degree\n",
- "\n",
- " theta_t= 44.73 Degree\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2.7.1-pg 68"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "##Ex2.7.1.;Determine the average values of radiation on a horizontal surface\n",
- "import math\n",
- "##Declination delta for June 22=23.5 degree, sunrice hour angle ws\n",
- "delta=(23.5*math.pi)/180;##unit=radians\n",
- "fie=(10*math.pi)/180;##unit=radians\n",
- "##Sunrice hour angle ws=acosd(-tan(fie)*tan(delta))\n",
- "ws=math.acos(-math.tan(fie)*math.tan(delta));\n",
- "print'%s %.2f %s'%(\" Sunrice hour angle ws=\",ws,\" Degree\");\n",
- "n=172.;##=days of the year (for June 22)\n",
- "##We have the relation for Average insolation at the top of the atmosphere\n",
- "##Ho=(24/%pi)*Isc*[{1+0.033*(360*n/365)}*((cos (fie)*cos(delta)*sin(ws))+(2*%pi*ws/360)*sin(fie)*sin(delta))]\n",
- "Isc=1353.;##SI unit=W/m^2\n",
- "ISC=1165.;##MKS unit=kcal/hr m^2\n",
- "##let\n",
- "a=24./math.pi;\n",
- "aa=(360.*172.)/365.;\n",
- "aaa=(aa*math.pi)/180.;\n",
- "b=math.cos(aaa);\n",
- "bb=0.033*b;\n",
- "bbb=1+bb;\n",
- "c=(10*math.pi)/180.;\n",
- "c1=math.cos(c);\n",
- "cc=(23.5*math.pi)/180;\n",
- "cc1=math.cos(cc);\n",
- "ccc=(94.39*math.pi)/180;\n",
- "ccc1=math.sin(ccc);\n",
- "c=c1*cc1*ccc1;\n",
- "d=(2*math.pi*ws)/360.;\n",
- "e=(10*math.pi)/180;\n",
- "e1=math.sin(e);\n",
- "ee=(23.5*math.pi)/180;\n",
- "ee1=math.sin(ee);\n",
- "e=e1*ee1;\n",
- "##therefoe Ho in SI unit\n",
- "Ho=a*Isc*(bbb*(c+(d*e)));\n",
- "print'%s %.2f %s'%(\"\\n SI UNIT->Ho=:\",Ho,\" W/m^2\");\n",
- "Hac=Ho*(0.3+(0.51*0.55))\n",
- "print'%s %.2f %s'%(\"\\n SI UNIT->Hac=\",Hac,\" W/m^2 day\");\n",
- "ho=a*ISC*(bbb*(c+(d*e)));\n",
- "print'%s %.2f %s'%(\"\\n MKS UNIT->Ho=\",ho,\" kcal/m^2\");\n",
- "hac=ho*0.58;\n",
- "print'%s %.2f %s'%(\"\\n MKS UNIT->Hac=\",hac,\" kcal/m^2 day\");\n",
- "\n",
- "##The values are approximately same as in textbook\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " Sunrice hour angle ws= 1.65 Degree\n",
- "\n",
- " SI UNIT->Ho=: 9025.25 W/m^2\n",
- "\n",
- " SI UNIT->Hac= 5239.16 W/m^2 day\n",
- "\n",
- " MKS UNIT->Ho= 7771.19 kcal/m^2\n",
- "\n",
- " MKS UNIT->Hac= 4507.29 kcal/m^2 day\n"
- ]
- }
- ],
- "prompt_number": 1
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Non-conventional_Energy_Sources_by_G._D._Rai/Chapter2_1.ipynb b/Non-conventional_Energy_Sources_by_G._D._Rai/Chapter2_1.ipynb deleted file mode 100755 index e69de29b..00000000 --- a/Non-conventional_Energy_Sources_by_G._D._Rai/Chapter2_1.ipynb +++ /dev/null diff --git a/Non-conventional_Energy_Sources_by_G._D._Rai/Chapter2_2.ipynb b/Non-conventional_Energy_Sources_by_G._D._Rai/Chapter2_2.ipynb deleted file mode 100755 index e69de29b..00000000 --- a/Non-conventional_Energy_Sources_by_G._D._Rai/Chapter2_2.ipynb +++ /dev/null diff --git a/Non-conventional_Energy_Sources_by_G._D._Rai/Chapter2_3.ipynb b/Non-conventional_Energy_Sources_by_G._D._Rai/Chapter2_3.ipynb deleted file mode 100755 index e69de29b..00000000 --- a/Non-conventional_Energy_Sources_by_G._D._Rai/Chapter2_3.ipynb +++ /dev/null diff --git a/Non-conventional_Energy_Sources_by_G._D._Rai/Chapter3.ipynb b/Non-conventional_Energy_Sources_by_G._D._Rai/Chapter3.ipynb deleted file mode 100755 index 5038a8d0..00000000 --- a/Non-conventional_Energy_Sources_by_G._D._Rai/Chapter3.ipynb +++ /dev/null @@ -1,220 +0,0 @@ -{
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- {
- "cell_type": "heading",
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- "metadata": {},
- "source": [
- "Chapter3-Solar Energy Collectors"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.6.1-pg100"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "##Ex3.6.1.;calculate: solar altitude anglr,Incident angle,Collector efficiency\n",
- "import math\n",
- "##Solar declination :delta\n",
- "n=1\n",
- "delta=23.45*math.sin((360./365.)*(284.+n));\n",
- "print'%s %.2f %s'%(\" Solar declination delta=\",delta,\" degree\");\n",
- "fie=22.;##degree\n",
- "##solar hour angle ws=0,(at mean of 11:30 and 12:30)\n",
- "ws=0.;\n",
- "##Solar altitude anglr alpha is given by\n",
- "\n",
- "##alpha=asind(((cos(fie)*cos(delta)*cos(ws))+(sin(fie)*sin(delta)))\n",
- "##let\n",
- "a=math.cos((22*math.pi)/180.)*math.cos((-23*math.pi)/180.)*math.cos(0);\n",
- "b=math.sin((22*math.pi)/180.)*math.sin((-23*math.pi)/180.);\n",
- "##therefore\n",
- "sin_alpha=a+b;\n",
- "print'%s %.2f %s'%(\"\\n sin_aplha=\",sin_alpha,\"\");\n",
- "alpha=math.asin(sin_alpha);\n",
- "print'%s %.2f %s'%(\"\\n aplha=\",alpha,\"degree\");\n",
- "##Incident angle\n",
- "theta=(180./2.)-alpha;\n",
- "print'%s %.2f %s'%(\"\\n Incident angle=\",theta,\"degree\");\n",
- "##Rb is given by\n",
- "Rb=((math.cos(((22*math.pi)/180.)-(37*math.pi)/180.)*math.cos((-23*math.pi)/180.)*math.cos(0))+(math.sin(((22*math.pi)/180.)-(37*math.pi)/180)* math.sin((-23*math.pi)/180)))/sin_alpha;\n",
- "print'%s %.2f %s'%(\"\\n Rb=\",Rb,\"\");\n",
- "##Effective absorptance product is <t.alpha>=t.alpha/ 1-(1-alpha)*pd\n",
- "pd=0.24;##Diffuse reflectance for two glass covers\n",
- "##let TA=<t.alpha>\n",
- "TA=(0.88*0.90)/(1-(1-0.90)*pd);\n",
- "print'%s %.2f %s'%(\"\\n Effective absorptance product is <t.alpha>=\",TA,\"\");\n",
- "##Solar radiation intensity(consider beam radiation only)\n",
- "##Hb=0.5 ly/mm = 0.5 cal/cm^2 * min\n",
- "Hb=((0.5*10**4)/10**3)*60;##unit=kcal/m^2 hr\n",
- "print'%s %.2f %s'%(\"\\n Hb=\",Hb,\" kcal/m^2 hr\");\n",
- "Hb=Hb*1.163;##unit=W/m^2 hr; [since 1 kcal = 1.163 watt]\n",
- "print'%s %.2f %s'%(\"\\n Hb=\",Hb,\" W/m^2 hr\");\n",
- "##S=Hb*Rb*<t.alpha>\n",
- "S=Hb*Rb*TA;\n",
- "print'%s %.2f %s'%(\"\\n S=\",S,\" W/m^2 hr\");\n",
- "s=S/1.163;\n",
- "print'%s %.2f %s'%(\"\\n S=\",s,\" kcal/m^2 hr\");\n",
- "##Useful gain\n",
- "##qu=FR(S-UL*(Tfi-Ta))\n",
- "qu=0.810*(s-(6.80*(60-15)))\n",
- "print'%s %.2f %s'%(\"\\n qu=\",qu,\" kcal/m^2 hr\");\n",
- "##Qu=FR(S-UL*(Tfi-Ta))\n",
- "Qu=0.810*(S-(7.88*(60-15)))\n",
- "print'%s %.2f %s'%(\"\\n qu=\",Qu,\" W/m^2 hr\");\n",
- "##Collection Efficiency : nc=(qu/(Hb*Rb))*100;\n",
- "nc=(28.07/(300*Rb))*100.;\n",
- "print'%s %.2f %s'%(\"\\n Collection Efficiency=\",nc,\" persent\");\n",
- "\n",
- "\n",
- "##values of \"sine alpha\" in the textbook is taken approximate to the real values\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " Solar declination delta= -23.38 degree\n",
- "\n",
- " sin_aplha= 0.71 \n",
- "\n",
- " aplha= 0.79 degree\n",
- "\n",
- " Incident angle= 89.21 degree\n",
- "\n",
- " Rb= 1.40 \n",
- "\n",
- " Effective absorptance product is <t.alpha>= 0.81 \n",
- "\n",
- " Hb= 300.00 kcal/m^2 hr\n",
- "\n",
- " Hb= 348.90 W/m^2 hr\n",
- "\n",
- " S= 396.50 W/m^2 hr\n",
- "\n",
- " S= 340.93 kcal/m^2 hr\n",
- "\n",
- " qu= 28.29 kcal/m^2 hr\n",
- "\n",
- " qu= 33.94 W/m^2 hr\n",
- "\n",
- " Collection Efficiency= 6.68 persent\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3.9.1-pg 119"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "##calculate the useful gain,exit fluid temperature and collection efficiency\n",
- "##Optical properties are estimated as\n",
- "p=0.85;\n",
- "import math\n",
- "##(T. alpha)=0.77;let A=(T. alpha)\n",
- "A=0.77\n",
- "gama=0.94;\n",
- "Do=0.06;\n",
- "L=8;##unit=meter,##L=length of concentrator\n",
- "W=2;##W=width of concentrator in meter\n",
- "dco=0.09;##dco=diameter of transpaarent cover\n",
- "Ar= math.pi*Do*L;##Ar=area of the receiver pipe\n",
- "A_alpha=(W-dco)*L;##aperture area of the concentration\n",
- "Cp=0.30;##unit=kcal/kg degree calcius\n",
- "m=400;##unit=kg/hr,m=flow rate\n",
- "HbRb=600;##unit=kcal/hr m^2\n",
- "Tfi=150;##degree calcius\n",
- "T_alpha=25;##degree calcius\n",
- "##Heat transfer coefficient from fluid inside to surroundings,\n",
- "Uo=5.2;##unit=kcal/hr-m^2\n",
- "##Heat transfer coefficient from absorber cover surface to surroundings,\n",
- "UL=6;##unit=kcal/hr-m^2\n",
- "F=(Uo/UL);\n",
- "##Heat removed factor FR is\n",
- "##FR=((m*Cp)/(Ar*UL))*(1-(%e^-((Ar*UL*F)/(m*Cp))))\n",
- "##let X=(m*Cp)/(Ar*UL);Y=(%e^-((Ar*UL*F)/(m*Cp)))\n",
- "X=(m*Cp)/(1.51*UL*0.86);\n",
- "Y=math.e**(-1/X);\n",
- "FR=X*0.86*(1-Y);\n",
- "##Absorbed solar energy is\n",
- "S=HbRb*p*gama*A;\n",
- "print'%s %.2f %s %.2f %s'%(\" Area of the receiver pipe Ar= \",Ar,\"=1.51 m^2\"and\" \\n A_aplha= \",A_alpha,\" m^2=collection efficiency factor \");\n",
- "print'%s %.2f %s'%(\"\\n value of F= \",F,\"\");\n",
- "print'%s %.2f %s %.2f %s '%(\"\\n Heat removed factor FR=\",FR,\"\"and\" \\n Absorbed solar energy is \\n S=\",S,\" kcal/Hr m^2 .....(MKS) \");\n",
- "##for unit in S.I. , 1 kcal/Hr m^2 = 1.16298 W/m^2\n",
- "s= S*1.16298; ##in W/m^2\n",
- "print'%s %.2f %s'%(\"\\n S=\",s,\" W/m^2.....(SI)\");\n",
- "##the values of F,FR will be same in any unit,since they are factors(dimensionless)\n",
- "##Useful Gain=Qu=A_alpha*FR*(S-((Ar*UL)/A_alpha)*(Tfi-T_alpha))\n",
- "##In MKS unit\n",
- "Qu=A_alpha*FR*(S-((1.51*UL)/A_alpha)*(Tfi-T_alpha))\n",
- "print'%s %.2f %s'%(\"\\n useful gain in (MKS) Qu=\",Qu,\" kcal/hr\");\n",
- "##IN SI unit\n",
- "qu=A_alpha*FR*(s-((1.51*6.98)/A_alpha)*(Tfi-T_alpha))##UL=6.98 W/m^2 degree celcius\n",
- "print'%s %.2f %s'%(\"\\n useful gain in (SI) Qu=\",qu,\" Watt\");\n",
- "##the exit fluid temperature can be obtained from\n",
- "tci=150;##degree celcius\n",
- "tco=tci+(Qu/(m*Cp));##from Qu=mCp(tco-tc); where, tco=collector fluid temp. at outlet,tci=Fluid inlet temp.\n",
- "n=(Qu/(16*HbRb))*100;##ncollector=Qu/(A_alpha*HbRb)*100;\n",
- "print'%s %.2f %s %.2f %s'%(\"\\n collector fluid temp. at outlet tco=\",tco,\" degree celcius\"and \" \\n ncollector = \",n,\" percent \");\n",
- "\n",
- "##The values/results/answers is approximate in the text book to the real calculated value\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " Area of the receiver pipe Ar= 1.51 \n",
- " A_aplha= 15.28 m^2=collection efficiency factor \n",
- "\n",
- " value of F= 0.87 \n",
- "\n",
- " Heat removed factor FR= 0.83 369.14 kcal/Hr m^2 .....(MKS) \n",
- "\n",
- " S= 429.30 W/m^2.....(SI)\n",
- "\n",
- " useful gain in (MKS) Qu= 3753.64 kcal/hr\n",
- "\n",
- " useful gain in (SI) Qu= 4365.07 Watt\n",
- "\n",
- " collector fluid temp. at outlet tco= 181.28 \n",
- " ncollector = 39.10 percent \n"
- ]
- }
- ],
- "prompt_number": 2
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Non-conventional_Energy_Sources_by_G._D._Rai/Chapter6.ipynb b/Non-conventional_Energy_Sources_by_G._D._Rai/Chapter6.ipynb deleted file mode 100755 index 698a5158..00000000 --- a/Non-conventional_Energy_Sources_by_G._D._Rai/Chapter6.ipynb +++ /dev/null @@ -1,95 +0,0 @@ -{
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- "Chapter6-Wind Energy"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6.2.1-pg244"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "##Ex.6.2.1. \n",
- "import math\n",
- "##For air,the value of gas constant\n",
- "R=0.287 ##unit=kj/kg K\n",
- "##T=15. in degreecalcius\n",
- "T=15.+273.;##in kalvin\n",
- "RT=0.287*10**3*288.;\n",
- "P=1.01325*10**5;##unit=Pa;at 1 atm\n",
- "Vi=15.;##unit=m/s\n",
- "gc=1.;\n",
- "D=120.;##turbine diameter;unit=m\n",
- "N=40./60.;\n",
- "##Air density\n",
- "p=(P/RT);\n",
- "print'%s %.2f %s'%(\"Air density p=\",p,\" kg/M^3\");\n",
- "##1] Total_power= Ptotal=p*A*Vi^3/2*gc\n",
- "##power density =Ptotal/A=p*Vi^3/2*gc\n",
- "power_density=(1./(2.*gc))*(p*Vi**3.);\n",
- "##2] Maximum_power_density=Pmax/A=8*p*Vi^3/27*gc\n",
- "Maximum_power_density=(8./(27.*gc))*(p*Vi**3);\n",
- "print'%s %.2f %s %.2f %s '%(\"\\n power density =Ptotal/A= \",power_density,\" W/m^2\"and \" \\n Maximum power density=Pmax/A= \",Maximum_power_density,\" W/m^2\");\n",
- "##3]Assuming n=35%\n",
- "n=0.35;\n",
- "##let P/A=x\n",
- "x=n*(power_density);\n",
- "print'%s %.2f %s'%(\"\\n P/A=\",x,\" W/m^2\");\n",
- "##4]Total power P= power density * Area\n",
- "Total_power_P=724.*(math.pi/4.)*(D**2) ##Total power P= power density*(%pi/4)*D^2\n",
- "print'%s %.2f %s %.2f %s'%(\"\\n Total_power_P=\",Total_power_P,\" watt\"and \"=\",Total_power_P*10**-3,\" kW\")\n",
- "##5]Torgue at maximum efficiency\n",
- "Tmax=(2./(27.*gc))*((1.226*D*Vi*Vi*Vi)/N);##Tmax=(2/(27*gc))*((p*D*Vi*Vi*Vi)/N);\n",
- "print'%s %.2f %s'%(\"\\n Torgue at maximum efficiency=\",Tmax,\" Newton\")\n",
- "##and maximum axial thurst\n",
- "Fxmax=(3.14/(9.*gc))*1.226*D**2.*Vi**2.;##Fxmax=(%pi/(9*gc))*p*D^2*Vi^2;\n",
- "print'%s %.2f %s'%(\"\\n maximum axial thurst=\",Fxmax,\" Newton\");\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Air density p= 1.23 kg/M^3\n",
- "\n",
- " power density =Ptotal/A= 2068.65 \n",
- " Maximum power density=Pmax/A= 1225.86 W/m^2 \n",
- "\n",
- " P/A= 724.03 W/m^2\n",
- "\n",
- " Total_power_P= 8188247.09 = 8188.25 kW\n",
- "\n",
- " Torgue at maximum efficiency= 55170.00 Newton\n",
- "\n",
- " maximum axial thurst= 1385870.40 Newton\n"
- ]
- }
- ],
- "prompt_number": 5
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Non-conventional_Energy_Sources_by_G._D._Rai/Chapter7.ipynb b/Non-conventional_Energy_Sources_by_G._D._Rai/Chapter7.ipynb deleted file mode 100755 index 4db07699..00000000 --- a/Non-conventional_Energy_Sources_by_G._D._Rai/Chapter7.ipynb +++ /dev/null @@ -1,83 +0,0 @@ -{
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- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter7-Energy from Biomass"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7.15.1-pg370"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "##Ex7.15.1; calculate volume of biogas digester and power available from the digester\n",
- "##Mass of the dry input\n",
- "M0=2*5;##M0=2.5 kg/day * 5\n",
- "pm=50.;##unit=kg/m^3\n",
- "tr=20.;##retention time in days\n",
- "C=0.24;##unit=m^3 per kg;Biogas yeild.\n",
- "n=0.6;##efficiency of burner\n",
- "Hm=28.;##unit=MJ/m^3##combustion of methane\n",
- "Fm=0.8;##methane proportional\n",
- "##Fluid volume Vf is =M0/pm\n",
- "Vf=M0/pm;\n",
- "print'%s %.2f %s %.2f %s '%(\" Mass of the dry input M0=\",M0,\" kg/day\"and \" \\n Fluid volume Vf=\",Vf,\" m^3 /day\");\n",
- "##for expression Vd=Vftr,the digester volume is\n",
- "Vd=Vf*tr;\n",
- "print'%s %.2f %s'%(\"\\n Vd=\",Vd,\" m^3\");\n",
- "##volume of biogas is Vb=C*M0= biogas yield input * mass of dry input\n",
- "Vb=C*M0;\n",
- "print'%s %.2f %s'%(\"\\n volume of biogas is Vb=\",Vb,\" m^3 /day\");\n",
- "##The Power available from the digester is\n",
- "E=n*Hm*Fm*Vb;\n",
- "print'%s %.2f %s'%(\"\\n The Power available from the digester=\",E,\" Mj/day\");\n",
- "E=E*0.2728;##unit=kWh/day\n",
- "print'%s %.2f %s'%(\"=\",E,\" kWh/day\");\n",
- "E=E*41.8##unit=W(continuous thermal)\n",
- "print'%s %.2f %s'%(\"=\",E,\" W(continuous thermal)\");\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " Mass of the dry input M0= 10.00 \n",
- " Fluid volume Vf= 0.20 m^3 /day \n",
- "\n",
- " Vd= 4.00 m^3\n",
- "\n",
- " volume of biogas is Vb= 2.40 m^3 /day\n",
- "\n",
- " The Power available from the digester= 32.26 Mj/day\n",
- "= 8.80 kWh/day\n",
- "= 367.82 W(continuous thermal)\n"
- ]
- }
- ],
- "prompt_number": 2
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Non-conventional_Energy_Sources_by_G._D._Rai/Chapter8.ipynb b/Non-conventional_Energy_Sources_by_G._D._Rai/Chapter8.ipynb deleted file mode 100755 index 6e5345c6..00000000 --- a/Non-conventional_Energy_Sources_by_G._D._Rai/Chapter8.ipynb +++ /dev/null @@ -1,206 +0,0 @@ -{
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- "source": [
- "Chapter8-Geothermal Energy"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 8.5.1-pg 450"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "##Ex8.5.1.;calculate: steam flow rate,cooling water flow,plant efficiency,Heat rate\n",
- "\n",
- "##Enthalpy at point 1 at (31 kg/cm^2)=669.6 kcal/kg\n",
- "##H1=H2=H3,enthalpy remain constant during throttling\n",
- "H1=669.7;##unit= kcal/kg\n",
- "H2=669.7;##unit= kcal/kg\n",
- "H3=669.7;##unit= kcal/kg\n",
- "##At point 3,\n",
- "P3=9.55;##unit= kg/cm^2\n",
- "##specific volume\n",
- "vs3=0.22;##unit=m^3/kg\n",
- "##Entropy\n",
- "S3=1.580\n",
- "T3=190.;##unit=degree C,(degree of superheat=13 degree C)\n",
- "##S4_s at 0.34 kg/cm^2=S3\n",
- "##x4_s=0.838\n",
- "##and H4_s=hs+xL\n",
- "H4_s=72.+(0.838*556)\n",
- "print'%s %.2f %s'%(\" H4_s=\",H4_s,\" kcal/kg\")\n",
- "##Isentropic turbine work=H3-H4_s\n",
- "ITW=H3-H4_s;\n",
- "print'%s %.2f %s'%(\"\\n Isentropic turbine work=\",ITW,\" kcal/kg\");\n",
- "##Actual turbine work\n",
- "ATW=0.80*ITW;\n",
- "print'%s %.2f %s'%(\"\\n Actual turbine work=\",ATW,\" kcal/kg\");\n",
- "H4=669.7-ATW;\n",
- "print'%s %.2f %s'%(\"\\n H4=\",H4,\" kcal/kg\")\n",
- "h5_6=72;##unit= kcal/kg; (Ignoring pump work)\n",
- "##sensible heat h7=h5=25 kcal/kg\n",
- "h5=25;##unit=kcal/kg\n",
- "h7=25;##unit=kcal/kg\n",
- "##Turbine steam flow\n",
- "TSF=(250*0.860*10**6)/(ATW*0.9);\n",
- "print'%s %.2f %s'%(\"\\n Turbine steam flow=\",TSF,\" kg/hr\");\n",
- "##let\n",
- "m4=TSF;\n",
- "##Turbine volume flow\n",
- "TVF=(TSF/60)*vs3;\n",
- "print'%s %.2f %s'%(\"\\n Turbine volume flow=\",TVF,\" m^3/min\");\n",
- "##cooling water flow m7:m7(h5_6-h7)=m4(H4-h5_6)\n",
- "m7=((H4-h5_6)/(h5_6-h7))*m4;\n",
- "print'%s %.2f %s'%(\"\\n cooling water flow m7=\",m7,\" kg/hr\");\n",
- "Heat_added=H1-h5_6;\n",
- "print'%s %.2f %s'%(\"\\n Heat_added=\",Heat_added,\" kcal/kg\");\n",
- "##plant efficiency=(Actual Turbine work*nmg)/Heat added\n",
- "##nmg=combined mechanical and electrical efficiency of turbine-generator\n",
- "nmg=0.90;\n",
- "Plant_efficiency=(ATW*nmg)/Heat_added;\n",
- "plant_efficiency=Plant_efficiency*100.;\n",
- "print'%s %.2f %s'%(\"\\n Plant Efficiency nplant=\",plant_efficiency,\" persent\");\n",
- "##Plant heat rate=(860*Heat_added)/net_work\n",
- "##net_work=105.36*0.90\n",
- "Plant_heat_rate=(860./Plant_efficiency);\n",
- "print'%s %.2f %s'%(\"\\n Plant heat rate=\",Plant_heat_rate,\" kcal/kWH\");\n",
- "\n",
- "\n",
- "##The value of \"turbine steam flow\" is wrong due to calculating mistak in textbook,due to which the further value related with it is given wrong\n",
- "##The values are corrected in this program\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " H4_s= 537.93 kcal/kg\n",
- "\n",
- " Isentropic turbine work= 131.77 kcal/kg\n",
- "\n",
- " Actual turbine work= 105.42 kcal/kg\n",
- "\n",
- " H4= 564.28 kcal/kg\n",
- "\n",
- " Turbine steam flow= 2266119.59 kg/hr\n",
- "\n",
- " Turbine volume flow= 8309.11 m^3/min\n",
- "\n",
- " cooling water flow m7= 23735548.77 kg/hr\n",
- "\n",
- " Heat_added= 597.70 kcal/kg\n",
- "\n",
- " Plant Efficiency nplant= 15.87 persent\n",
- "\n",
- " Plant heat rate= 5417.84 kcal/kWH\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 8.5.2-pg460"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "##Ex8.5.2.;calculate: hot water flow,condenser cooling water flow,cycle efficiency,plant heat rate.\n",
- "import math\n",
- "H1=669.6;##unit=kcal/kg\n",
- "H2=669.6;##unit=kcal/kg\n",
- "##pressure at point 2,is 10.5 kg/cm^2;thus,\n",
- "T2=195.;##unit=degree celcius; (14 degree celcius of superheat)\n",
- "s2=1.567;\n",
- "vsup=0.27;\n",
- "x3s=0.832;\n",
- "H3s=535.;##unit=kcal/kg\n",
- "##Isentropic turbine work\n",
- "ITW=H2-H3s;\n",
- "print'%s %.2f %s'%(\" Isentropic turbine work=\",ITW,\" kcal/kg\");\n",
- "##Actual turbine work\n",
- "ATW=0.65*ITW;\n",
- "print'%s %.2f %s'%(\"\\n Actual turbine work=\",ATW,\" kcal/kg\");\n",
- "H3=669.6-ATW;\n",
- "print'%s %.2f %s'%(\"\\n H3=\",H3,\" kcal/kg\")\n",
- "##h_4-5(ignore bpump work)\n",
- "h4=72.4;##unit=kcal/kg\n",
- "##h6 at 27 degree c\n",
- "h6=27;##unit=kcal/kg\n",
- "##Turbine steam flow or hot water flow=power output/actual turbine work\n",
- "TSF=(10*10**6*0.86)/ATW;\n",
- "print'%s %.2f %s'%(\"\\n Turbine steam flow or hot water flow=\",TSF,\" kg/hr\");\n",
- "##consider cooling water flow m4:m3*(H3-h4)=m4(h4-h6)\n",
- "##or\n",
- "m4=((582.11-72.4)*0.983*10**5)/(72.4-27);\n",
- "print'%s %.2f %s'%(\"\\n cooling water flow=\",m4,\" kg/hr\");\n",
- "Heat_added=H1-h4\n",
- "print'%s %.2f %s'%(\"\\n Heat_added=\",Heat_added,\" kcal/kg\");\n",
- "##plant efficiency=Turbine work/Heat added\n",
- "Plant_efficiency=(ATW/Heat_added);\n",
- "plant_efficiency=Plant_efficiency*100;\n",
- "print'%s %.2f %s'%(\"\\n Plant Efficiency=\",plant_efficiency,\" persent\");\n",
- "##Plant heat rate=860/Plant Efficiency\n",
- "Plant_heat_rate=860./Plant_efficiency;\n",
- "print'%s %.2f %s'%(\"\\n Plant heat rate=\",Plant_heat_rate,\" kcal/kWh\");\n",
- "\n",
- "\n",
- "##The value of m3=14.03*10^5 is given wrong in the text book;the actual value is m3=11.03*10^5\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " Isentropic turbine work= 134.60 kcal/kg\n",
- "\n",
- " Actual turbine work= 87.49 kcal/kg\n",
- "\n",
- " H3= 582.11 kcal/kg\n",
- "\n",
- " Turbine steam flow or hot water flow= 98296.95 kg/hr\n",
- "\n",
- " cooling water flow= 1103623.19 kg/hr\n",
- "\n",
- " Heat_added= 597.20 kcal/kg\n",
- "\n",
- " Plant Efficiency= 14.65 persent\n",
- "\n",
- " Plant heat rate= 5870.29 kcal/kWh\n"
- ]
- }
- ],
- "prompt_number": 2
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Non-conventional_Energy_Sources_by_G._D._Rai/Chapter9.ipynb b/Non-conventional_Energy_Sources_by_G._D._Rai/Chapter9.ipynb deleted file mode 100755 index 19bad3a0..00000000 --- a/Non-conventional_Energy_Sources_by_G._D._Rai/Chapter9.ipynb +++ /dev/null @@ -1,135 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:7f40e58996047c1cb062bf18947910b9a976381238ed50505c43ba92251b46eb"
- },
- "nbformat": 3,
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- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter9-Energy from the Oceans"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.3.5.1-pg 527"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "##Ex9.3.5.1.;Calculate Energy generated\n",
- "import math\n",
- "R=12.;##unit=m; R is the range\n",
- "r=3.;##unit=m; the head below turbine stops operating\n",
- "time=(44700./2.);\n",
- "A=30*10**6;\n",
- "g=9.80;\n",
- "p=1025.;\n",
- "##The total theoretical work W=integrate('1','w',R,r);\n",
- "W=(g*p*A*((R**2)-(r**2)))/2.;\n",
- "print'%s %.2f %s'%(\" W=\",W,\" \");\n",
- "##The average power generated\n",
- "Pav=W/time;##unit=watts\n",
- "print'%s %.2f %s'%(\"\\n The average power generated=\",Pav,\" watts\");\n",
- "pav=(Pav/1000.)*3600.;##unit=kWh\n",
- "print'%s %.2f %s'%(\"\\n The average power generated=\",pav,\" kWh\")\n",
- "##the energy generated\n",
- "Energy_generated=pav*0.73\n",
- "print'%s %.2f %s'%(\"\\n Energy generated=\",Energy_generated,\" kWh\");\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " W= 20341125000000.00 \n",
- "\n",
- " The average power generated= 910117449.66 watts\n",
- "\n",
- " The average power generated= 3276422818.79 kWh\n",
- "\n",
- " Energy generated= 2391788657.72 kWh\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9.3.6.1-pg529"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "##Ex9.3.6.1;calculate power in h.p. at any instant and the yearly power output\n",
- "import math\n",
- "A=0.5*10**6;##unit=m\n",
- "h0=8.5;##unit=m\n",
- "t=3*3600.##unit=s; since t=3 hr\n",
- "p=1025.;##unit=kg/m^3\n",
- "h=8.;##unit=m\n",
- "n0=0.70;##efficiency of the generator;70%\n",
- "##volume of the basin=Ah0\n",
- "volume_of_the_basin=A*h0;\n",
- "##Average discharge Q=volume/time period\n",
- "Q=(A*h0)/t;\n",
- "print'%s %.2f %s %.2f %s '%(\" volume of the basin=\",volume_of_the_basin,\" m^3\"and \" \\n Average discharge Q=\",Q,\" m^3 /s\")\n",
- "##power at any instant\n",
- "P=((Q*p*h)/75)*n0;\n",
- "print'%s %.2f %s'%(\"\\n power at any instant P=\",P,\" h.p.\");\n",
- "##The total energy in kWh/tidal cycle\n",
- "E=P*0.736*3;\n",
- "print'%s %.2f %s'%(\"\\n The total energy in kWh/tidal cycle E=\",E,\"\");\n",
- "##Total number of tidal cycle in a year=705\n",
- "print(\"\\n Total number of tidal cycle in a year=705\");\n",
- "##Therefore Total output per annum\n",
- "Total_output_per_annum=E*705;\n",
- "print'%s %.2f %s'%(\"\\n Total output per annum= \",Total_output_per_annum,\"kWh/year\");\n",
- "\n",
- "##The value of \"power of instant\" in a text book is misprinted.\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " volume of the basin= 4250000.00 \n",
- " Average discharge Q= 393.52 m^3 /s \n",
- "\n",
- " power at any instant P= 30117.28 h.p.\n",
- "\n",
- " The total energy in kWh/tidal cycle E= 66498.96 \n",
- "\n",
- " Total number of tidal cycle in a year=705\n",
- "\n",
- " Total output per annum= 46881768.89 kWh/year\n"
- ]
- }
- ],
- "prompt_number": 3
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file |