# Ch-15, New Energy Sources¶

## example 15.1 Page 345¶

In [3]:
a=0.1  #plate area
b=3    #flux density
d=0.5  #distence between plates
v=1000  #average gas velosity
c=10    #condectivity
e=b*v*d
ir=d/(c*a)  #internal resistence
mapo=e**2/(4*ir) #maximum power output
print "E=%dV \ninternal resistence %.1fohm \nmaximum power output %dW =%.3fMW"%(e,ir,mapo,mapo/10**6)

E=1500V
internal resistence 0.5ohm
maximum power output 1125000W =1.125MW


## example 15.2 Page 345¶

In [4]:
b=4.2 #flux density
v=600  #gas velocity
d=0.6  #dimension of plate
k=0.65  #constent
e=b*v*d #open circuit voltage

 voltage E=1512V


## example 15.3 Page 346¶

In [5]:
b=4.2 #flux density
v=600  #gas velocity
d=0.6  #dimension of plate
k=0.65  #constent
sl=0.6  #length given
sb=0.35  #breath given
sh=1.7   #height given
c=60     #given condectivity
e=b*v*d #open circuit voltage
rg=d/(c*sb*sh)
vd=e-v #voltage drop in duct
i=vd/rg  #current due to voltage drop in duct
j=i/(sb*sh) #current density
si=e/(rg)  #short circuit current
sj=si/(sb*sh)  #short circuit current density
pd=j*vg     #power density
p=pd*sl*sh*sb #power
pp=e*i  #also power
pde=v*i  #power delevered is V*i
los=p-pde  #loss
eff=pde/p  #efficiency
maxp=e**2/(4*rg)
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)

 resistence of duct 0.016807ohms
voltage drop in duct 529.2V
current 31487.4A
current density 52920.000000A/m**2
short circuit current 89964.0A
short current density 151200.000000A/m**2
power 47.608949MW
loss in duct 16.663132W
efficiency is 0.650000
maximum power delivered to load 34MW


## example 15.4 Page 347¶

In [6]:
c=50 #conduntance
a=0.2 #area
d=0.24 #distence between electrodes
v=1800 #gas velosity
b=1 #flux density
k=0.7
ov=k*b*v*d
tp=c*d*a*b**2*v**2*(1-k)
eff=k
op=eff*tp
e=b*v*d
rg=d/(c*a)
si=e/rg
maxp=e**2/(4*rg)
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)

 output voltage 302.4V
total power 2.3328MW
efficiency 0.7
output power 1.632960MW
open circuit voltage 432V
internal resistence 0.024ohm
short circuit current 18000A
maximum power output is 1.944MW


## example 15.5 Page 363¶

In [7]:
from math import cos, pi
a=100 #area
spd=0.7 #sun light power density
m=1000  #weight of water collector
tp=30  #temperature of water
th2=60  #angle of incidence
cp=4186 #specific heat of water
sp=spd*cos(th2*pi/180)*a #solar power collected by collector
ei=sp*3600*10**3  #energy input in 1 hour
temp=ei/(cp*10**3)
tw=tp+temp
print " solar power collected by collector %dkW \n energy input in one hour %e J \n rise in temperature is %.1fC \n temperature of water %.1fc"%(sp,ei,temp,tw)

 solar power collected by collector 35kW
energy input in one hour 1.260000e+08 J
rise in temperature is 30.1C
temperature of water 60.1c


## example 15.6 Page 364¶

In [8]:
from math import sqrt, ceil
vo=100 #motor rated voltage
efm=0.4 #efficiency of motor pump
efi=0.85 #efficiency of inverter
v=25 #volume of water per day
ov=18 #pv pannel output module
pr=40 #power rating
ao=2000 #annual output of array
dw=1000 #density of water
en=v*dw*h*9.81 #energy needed to pump water every day
enkw=en/(3.6*10**6)  #energy in kilo watt hour
oe=efm*efi  #overall efficiency
epv=round(enkw/oe)  #energy out of pv system
de=ao/365  #daily energy output
pw=epv*10**3/de  #peak wattage of pv array
rv=vo*(pi)/sqrt(2)  #rms voltage
nm=rv/ov  #number of modules in series
nm=ceil(nm)
rpp=nm*pr #rated peak power output
np=pw/rpp #number of strings in parallel
np=round(np)
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)
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)

 energy needed o pump water every day 3.406250kWh/day
overall efficiency 0.34
energy output of pv system 10kWh/day

annual energy out of array 10Wh/Wp
daily energy output of array 5.000Wh/Wp
peak wattage of pv array 2000.00Wp
rms output voltage 222.14V
number of modules in series 13
rated peak power output of each string 520.00W
number of strings in parallel 4


## example 15.7 Page 373¶

In [9]:
from __future__ import division
from math import cos, pi
ws=20 #wind speed
rd=10 #rotor diameter
ros=30 #rotor speed
mc=0.593 #maximum value of power coefficient
p=p1/10**3
pd=p/((pi)*(rd/2)**2)  #power density
pm=p*(mc)  #maximum power
mt=(pm*10**3)/((pi)*rd*(ros/60))
print " power %.fkW \n power density %.3fkW/m**3 \n maximum power %fkW \n maximum torque %.1fN-m"%(p,pd,pm,mt)

 power 406kW
power density 5.172kW/m**3
maximum power 240.881303kW
maximum torque 15335.0N-m


## example 15.8 Page 373¶

In [10]:
cp=0.593
d=1.293
s=15
a=2/3
dp=2*d*(s**2)*a*(1-a)
dlp=760*dp/(101.3*10**3) #760 mmhg=101.3*10**3pascal then pressure in mm of hg
dpa=dlp/760 #pressure in atmosphere
print "pressure in pascal %.1fpascal \npressure in height of mercury %.2fmm-hg \npressure in atmosphere %.5fatm"%(dp,dlp,dpa)

pressure in pascal 129.3pascal
pressure in height of mercury 0.97mm-hg
pressure in atmosphere 0.00128atm


## example 15.9 Page 385¶

In [11]:
from math import floor
ng=50 #number of generator
r=30  #rated power
tg=12  #duration of generation
efg=0.9  #efficiency of generated
g=9.81   #gravity
le=5   #lenght of embankment
ro=1025 #density
ti=r/(0.9)**2
q=ti*10**(6)/(ro*g*mah) #maximum input
q=floor(q*10**2)/10**2
qw=q*ng  #total quantity of water
tcr=qw*tg*3600/2  #total capacity of resevoir
sa=tcr/mah   #surface area
wbe=sa/(le*10**6)  #wash behind embankment
avg=r/2
te=avg*tg*365*ng  #total energy output
print "quantity of water for maximum output %fm**3-sec "%(q)
print "\nsurface area of reservoir %fkm**3 "%(sa/10**6)
print "\nwash behind embankment %fkm \ntotal energy output %eMWh"%(wbe,te)

print "area of reservoir %fkm**3 "%(sa/10**6)
print "\nwash behind embankment %fkm \ntotal energy output %eMWh"%(wbe,te)

quantity of water for maximum output 368.330000m**3-sec

surface area of reservoir 39.779640km**3

wash behind embankment 7.955928km
total energy output 3.285000e+06MWh
area of reservoir 39.779640km**3

wash behind embankment 7.955928km
total energy output 3.285000e+06MWh


## example 15.10 Page 385¶

In [12]:
tc=2100  #total capacity of plant
n=60     #number of generaed
p=35     #power of generated by each generator
d=12     #duration of generation
cee=2.1  #cost of electrical energy per kWh
efft=0.85 #efficiency of turbine
effg=0.9  #efficiency of generator
g=9.81   #gravity
ro=1025   #density
acc=0.7   #assuming coal conumotion
pi=p/(efft*effg) #power input
q=pi*10**6/(h*g*ro) #quantity of water
tqr=q*n*d*3600/2  #total quantity of water in reservoir
avp=tc/2 #average output during 12h
toe=avp*d  #total energy in 12 hours
eg=toe*365 #energy generated for totel year
coe=eg*cee*10**3 #cost of electrical energy generated
sc=eg*10**3*acc #saving cost
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)

total quantity of water in reservoir 5.896832e+08m**3
energy generated per year 4.599000e+06MW
cost of electrical energy Rs9.657900e+09
saving in cost Rs.3.219300e+09