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{
"metadata": {
"name": "",
"signature": "sha256:543241e2957635d03b859da9db1a20b76f4037f0de553827d6103edb92a77b0c"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Ch-15, New Energy Sources"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"example 15.1 Page 345"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"a=0.1 #plate area\n",
"b=3 #flux density\n",
"d=0.5 #distence between plates\n",
"v=1000 #average gas velosity\n",
"c=10 #condectivity\n",
"e=b*v*d\n",
"ir=d/(c*a) #internal resistence\n",
"mapo=e**2/(4*ir) #maximum power output\n",
"print \"E=%dV \\ninternal resistence %.1fohm \\nmaximum power output %dW =%.3fMW\"%(e,ir,mapo,mapo/10**6)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"E=1500V \n",
"internal resistence 0.5ohm \n",
"maximum power output 1125000W =1.125MW\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"example 15.2 Page 345"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"b=4.2 #flux density\n",
"v=600 #gas velocity\n",
"d=0.6 #dimension of plate\n",
"k=0.65 #constent\n",
"e=b*v*d #open circuit voltage\n",
"vg=e/d #voltage gradient\n",
"v=k*e #voltage across load\n",
"vgg=v/d #voltage gradient due to load voltage\n",
"print \" voltage E=%dV \\n voltage gradient %dV/m \\n voltage across load %.1fV \\n voltage gradient due to load voltage %dv\"%(e,vg,v,vgg)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" voltage E=1512V \n",
" voltage gradient 2520V/m \n",
" voltage across load 982.8V \n",
" voltage gradient due to load voltage 1638v\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"example 15.3 Page 346"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"b=4.2 #flux density\n",
"v=600 #gas velocity\n",
"d=0.6 #dimension of plate\n",
"k=0.65 #constent\n",
"sl=0.6 #length given\n",
"sb=0.35 #breath given\n",
"sh=1.7 #height given\n",
"c=60 #given condectivity\n",
"e=b*v*d #open circuit voltage\n",
"vg=e/d #voltage gradient\n",
"v=k*e #voltage across load\n",
"vgg=v/d #voltage gradient due to load voltage\n",
"rg=d/(c*sb*sh)\n",
"vd=e-v #voltage drop in duct\n",
"i=vd/rg #current due to voltage drop in duct\n",
"j=i/(sb*sh) #current density\n",
"si=e/(rg) #short circuit current\n",
"sj=si/(sb*sh) #short circuit current density\n",
"pd=j*vg #power density\n",
"p=pd*sl*sh*sb #power \n",
"pp=e*i #also power\n",
"pde=v*i #power delevered is V*i\n",
"los=p-pde #loss\n",
"eff=pde/p #efficiency\n",
"maxp=e**2/(4*rg)\n",
"print \" resistence of duct %fohms \\n voltage drop in duct %.1fV \\n current %.1fA \\n current density %fA/m**2 \\n short circuit current %.1fA \\n short current density %fA/m**2 \\n power %fMW \\n power delivered to load %fW \\n loss in duct %fW \\n efficiency is %f \\n maximum power delivered to load %dMW\"%(rg,vd,i,j,si,sj,p/10**6,pde/10**6,los/10**6,eff,maxp/10**6) "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" resistence of duct 0.016807ohms \n",
" voltage drop in duct 529.2V \n",
" current 31487.4A \n",
" current density 52920.000000A/m**2 \n",
" short circuit current 89964.0A \n",
" short current density 151200.000000A/m**2 \n",
" power 47.608949MW \n",
" power delivered to load 30.945817W \n",
" loss in duct 16.663132W \n",
" efficiency is 0.650000 \n",
" maximum power delivered to load 34MW\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"example 15.4 Page 347"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"c=50 #conduntance\n",
"a=0.2 #area\n",
"d=0.24 #distence between electrodes\n",
"v=1800 #gas velosity\n",
"b=1 #flux density\n",
"k=0.7 \n",
"ov=k*b*v*d\n",
"tp=c*d*a*b**2*v**2*(1-k)\n",
"eff=k\n",
"op=eff*tp\n",
"e=b*v*d\n",
"rg=d/(c*a)\n",
"si=e/rg\n",
"maxp=e**2/(4*rg)\n",
"print \" output voltage %.1fV \\n total power %.4fMW \\n efficiency %.1f \\n output power %fMW \\n open circuit voltage %dV \\n internal resistence %.3fohm \\n short circuit current %dA \\n maximum power output is %.3fMW\"%(ov,tp/10**6,eff,op/10**6,e,rg,si,maxp/10**6)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" output voltage 302.4V \n",
" total power 2.3328MW \n",
" efficiency 0.7 \n",
" output power 1.632960MW \n",
" open circuit voltage 432V \n",
" internal resistence 0.024ohm \n",
" short circuit current 18000A \n",
" maximum power output is 1.944MW\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"example 15.5 Page 363"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from math import cos, pi\n",
"a=100 #area\n",
"spd=0.7 #sun light power density\n",
"m=1000 #weight of water collector\n",
"tp=30 #temperature of water\n",
"th2=60 #angle of incidence\n",
"cp=4186 #specific heat of water\n",
"sp=spd*cos(th2*pi/180)*a #solar power collected by collector\n",
"ei=sp*3600*10**3 #energy input in 1 hour\n",
"temp=ei/(cp*10**3)\n",
"tw=tp+temp\n",
"print \" solar power collected by collector %dkW \\n energy input in one hour %e J \\n rise in temperature is %.1f`C \\n temperature of water %.1f`c\"%(sp,ei,temp,tw)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" solar power collected by collector 35kW \n",
" energy input in one hour 1.260000e+08 J \n",
" rise in temperature is 30.1`C \n",
" temperature of water 60.1`c\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"example 15.6 Page 364"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from math import sqrt, ceil\n",
"vo=100 #motor rated voltage\n",
"efm=0.4 #efficiency of motor pump\n",
"efi=0.85 #efficiency of inverter\n",
"h=50 #head of water\n",
"v=25 #volume of water per day\n",
"ov=18 #pv pannel output module\n",
"pr=40 #power rating\n",
"ao=2000 #annual output of array\n",
"dw=1000 #density of water\n",
"en=v*dw*h*9.81 #energy needed to pump water every day\n",
"enkw=en/(3.6*10**6) #energy in kilo watt hour\n",
"oe=efm*efi #overall efficiency\n",
"epv=round(enkw/oe) #energy out of pv system\n",
"de=ao/365 #daily energy output\n",
"pw=epv*10**3/de #peak wattage of pv array\n",
"rv=vo*(pi)/sqrt(2) #rms voltage\n",
"nm=rv/ov #number of modules in series\n",
"nm=ceil(nm)\n",
"rpp=nm*pr #rated peak power output\n",
"np=pw/rpp #number of strings in parallel\n",
"np=round(np)\n",
"print \" energy needed o pump water every day %fkWh/day \\n overall efficiency %.2f \\n energy output of pv system %dkWh/day \"%(enkw,oe,epv)\n",
"print \"\\n annual energy out of array %dWh/Wp \\n daily energy output of array %.3fWh/Wp \\n peak wattage of pv array %.2fWp \\n rms output voltage %.2fV\\n number of modules in series %d \\n rated peak power output of each string %.2fW \\n number of strings in parallel %d\"%(epv,de,pw,rv,nm,rpp,np)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" energy needed o pump water every day 3.406250kWh/day \n",
" overall efficiency 0.34 \n",
" energy output of pv system 10kWh/day \n",
"\n",
" annual energy out of array 10Wh/Wp \n",
" daily energy output of array 5.000Wh/Wp \n",
" peak wattage of pv array 2000.00Wp \n",
" rms output voltage 222.14V\n",
" number of modules in series 13 \n",
" rated peak power output of each string 520.00W \n",
" number of strings in parallel 4\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"example 15.7 Page 373"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from __future__ import division\n",
"ws=20 #wind speed\n",
"rd=10 #rotor diameter\n",
"ros=30 #rotor speed\n",
"ad=1.293 #air density\n",
"mc=0.593 #maximum value of power coefficient\n",
"p1=0.5*ad*(pi)*(rd**2)*(ws**3)/4 #power\n",
"p=p1/10**3\n",
"pd=p/((pi)*(rd/2)**2) #power density\n",
"pm=p*(mc) #maximum power\n",
"mt=(pm*10**3)/((pi)*rd*(ros/60))\n",
"print \" power %.fkW \\n power density %.3fkW/m**3 \\n maximum power %fkW \\n maximum torque %.1fN-m\"%(p,pd,pm,mt)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" power 406kW \n",
" power density 5.172kW/m**3 \n",
" maximum power 240.881303kW \n",
" maximum torque 15335.0N-m\n"
]
}
],
"prompt_number": 7
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"example 15.8 Page 373"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"cp=0.593\n",
"d=1.293\n",
"s=15\n",
"a=2/3\n",
"dp=2*d*(s**2)*a*(1-a)\n",
"dlp=760*dp/(101.3*10**3) #760 mmhg=101.3*10**3pascal then pressure in mm of hg\n",
"dpa=dlp/760 #pressure in atmosphere\n",
"print \"pressure in pascal %.1fpascal \\npressure in height of mercury %.2fmm-hg \\npressure in atmosphere %.5fatm\"%(dp,dlp,dpa)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"pressure in pascal 129.3pascal \n",
"pressure in height of mercury 0.97mm-hg \n",
"pressure in atmosphere 0.00128atm\n"
]
}
],
"prompt_number": 8
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"example 15.9 Page 385"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from math import floor\n",
"ng=50 #number of generator\n",
"r=30 #rated power \n",
"mah=10 #maximum head\n",
"mih=1 #minimum head\n",
"tg=12 #duration of generation\n",
"efg=0.9 #efficiency of generated\n",
"g=9.81 #gravity\n",
"le=5 #lenght of embankment\n",
"ro=1025 #density\n",
"ti=r/(0.9)**2\n",
"q=ti*10**(6)/(ro*g*mah) #maximum input\n",
"q=floor(q*10**2)/10**2\n",
"qw=q*ng #total quantity of water\n",
"tcr=qw*tg*3600/2 #total capacity of resevoir\n",
"sa=tcr/mah #surface area \n",
"wbe=sa/(le*10**6) #wash behind embankment\n",
"avg=r/2\n",
"te=avg*tg*365*ng #total energy output\n",
"print \"quantity of water for maximum output %fm**3-sec \"%(q)\n",
"print \"\\nsurface area of reservoir %fkm**3 \"%(sa/10**6)\n",
"print \"\\nwash behind embankment %fkm \\ntotal energy output %eMWh\"%(wbe,te) \n",
"\n",
"print \"area of reservoir %fkm**3 \"%(sa/10**6)\n",
"print \"\\nwash behind embankment %fkm \\ntotal energy output %eMWh\"%(wbe,te) "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"quantity of water for maximum output 368.330000m**3-sec \n",
"\n",
"surface area of reservoir 39.779640km**3 \n",
"\n",
"wash behind embankment 7.955928km \n",
"total energy output 3.285000e+06MWh\n",
"area of reservoir 39.779640km**3 \n",
"\n",
"wash behind embankment 7.955928km \n",
"total energy output 3.285000e+06MWh\n"
]
}
],
"prompt_number": 9
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"example 15.10 Page 385"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"tc=2100 #total capacity of plant\n",
"n=60 #number of generaed\n",
"p=35 #power of generated by each generator\n",
"h=10 #head of water\n",
"d=12 #duration of generation\n",
"cee=2.1 #cost of electrical energy per kWh\n",
"efft=0.85 #efficiency of turbine\n",
"effg=0.9 #efficiency of generator\n",
"g=9.81 #gravity\n",
"ro=1025 #density\n",
"acc=0.7 #assuming coal conumotion\n",
"pi=p/(efft*effg) #power input\n",
"q=pi*10**6/(h*g*ro) #quantity of water\n",
"tqr=q*n*d*3600/2 #total quantity of water in reservoir\n",
"avp=tc/2 #average output during 12h\n",
"toe=avp*d #total energy in 12 hours\n",
"eg=toe*365 #energy generated for totel year\n",
"coe=eg*cee*10**3 #cost of electrical energy generated\n",
"sc=eg*10**3*acc #saving cost \n",
"print \"total quantity of water in reservoir %em**3 \\nenergy generated per year %eMW \\ncost of electrical energy Rs%e \\nsaving in cost Rs.%e \"%(tqr,eg,coe,sc)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"total quantity of water in reservoir 5.896832e+08m**3 \n",
"energy generated per year 4.599000e+06MW \n",
"cost of electrical energy Rs9.657900e+09 \n",
"saving in cost Rs.3.219300e+09 \n"
]
}
],
"prompt_number": 10
}
],
"metadata": {}
}
]
}
|