Ch-4, Tariffs & Power Factor Improvement

example 4.1 page 59

In [3]:
day=30 #days 
pll=40 ;nll=5; tll=3 #light load 
pfl=100; nfl=3 ;tfl=5 #fan load
prl=1*1000 #refrigerator
pml=1*1000; nml=1 #misc. load 
t1=2.74 ;t11=15#tariff
t2=2.70 ;t22=25  #tariff on 25 units
tr=2.32  #reamaining units
tc=7.00 #constant charge
dis=0.05#discount for prompt payment
te=(pll*nll*tll+pfl*nfl*tfl)*day+prl*day+pml*day
tee=te/1000
mb=tc+tr*(tee-t11-t22)+t1*t11+t2*t22
nmb=mb*(1-dis)
print "total energy consumption in %d day %dunits \nthe monthly bill Rs%.2f \nnet monthly bill Rs%.2f"%(day,tee,mb,nmb)

total energy consumption in 30 day 123units 
the monthly bill Rs308.16 
net monthly bill Rs292.75

example 4.2 Page 59

In [4]:
l=100 #connected load
md=80 #maximum demand
wt=0.6  #working time
c=6000  #constant cost
t=700  #cost on per kW
re=1.8 #rate
ec=l*wt*8760#electricity consumption per year
teb=c+md*t+re*ec #total electricity bill per year
print " energy consumption %dkWh \n total electricity bill per year Rs%d"%(ec,teb)

 energy consumption 525600kWh 
 total electricity bill per year Rs1008080

example 4.3 Page 60

In [5]:
md=160 ;lff=0.7 ;dfc=1.7#maximum demand #load factor#diversity factor bt consumers
ic=200 #installed capacity
ccp=30000#capital cost of plant per kW
ctds=1800*10**6 #capital cost of transmission and distribution
idi=0.11 #interest,depreciation insurance and taxes on capital investiment
fmc=30*10**6  #fixed managerial and general maintanance cost
ol=236*10**6 #operating labour,maintanance and suppies
cm=90*10**6 #cost of metering,billing and collection
eca=0.05  #energy consumed by auxillary
el=0.15#energy loss and maintanance
p=0.25
lf=0.8#load factor
ap=0.5 #addition energy for profit
print 'a'
print " capital cost of plant Rs%e \n total capital cost Rs%e\n interest,depereiation system Rs%e "%(ccp*ic*10**3,ccp*ic*10**3+ctds,(ccp*ic*10**3+ctds)*idi)
print "\n sum of maximum demand of consumers energy prodused %dMW \n energy produced %ekWh \n energy consumed by auxilliries %ekWh\n energy output %ekWH \n energy sold to consumer %ekWh\n"%(md*dfc,md*8760*lff*10**3,md*8760*lff*eca*10**3,md*8760*lff*10**3*(1-eca),md*8760*lff*10**3*(1-eca)*(1-el))
print '(b)fixed cost'
idetc=(ccp*ic*10**3+ctds)*idi
tot=idetc+fmc 
print " interest, deprecition etc Rs%e per year\n managerial and maintence Rs%.eper year \n total \t Rs%e "%(idetc,fmc,tot)
pro=p*tot
gtot=tot+pro
print "\n profit@%d \tRs%eper year \n grand total Rs%e per year"%(p*100,pro,gtot)
print 'Operating cost'
tot2=ol+cm
pro2=tot2*p
gtot2=tot2+pro2
print " Operating labour,supplies maintenance etc Rs.%eper year \n metering,billing etc Rs%eper year\n total\t\tRs%e per year\n profit \t Rs%eper year \n grand total \t Rs%e per year"%(ol,cm,tot2,pro2,gtot2)
print 'tariff'
co=gtot/(md*dfc*1000)
es=md*8760*lff*10**3*(1-eca)*(1-el)
cs=gtot2/es
print " cost per kW \tRs%e \n cost per kWh \tRs%e"%(co,cs)
print '(b)'
ep=md*1000*8760*lf
print " energy produced %ekWh \n energy consumed by auxiliaries %ekWh/year \n energy output of plant %ekWh \n energy sold to consumer %ekWh"%(ep,ep*eca,ep*(1-eca),ep*(1-eca)*(1-el))
estc=ep*(1-eca)*(1-el)

a
 capital cost of plant Rs6.000000e+09 
 total capital cost Rs7.800000e+09
 interest,depereiation system Rs8.580000e+08 

 sum of maximum demand of consumers energy prodused 272MW 
 energy produced 9.811200e+08kWh 
 energy consumed by auxilliries 4.905600e+07kWh
 energy output 9.320640e+08kWH 
 energy sold to consumer 7.922544e+08kWh

(b)fixed cost
 interest, deprecition etc Rs8.580000e+08 per year
 managerial and maintence Rs3e+07per year 
 total 	 Rs8.880000e+08 

 profit@25 	Rs2.220000e+08per year 
 grand total Rs1.110000e+09 per year
Operating cost
 Operating labour,supplies maintenance etc Rs.2.360000e+08per year 
 metering,billing etc Rs9.000000e+07per year
 total		Rs3.260000e+08 per year
 profit 	 Rs8.150000e+07per year 
 grand total 	 Rs4.075000e+08 per year
tariff
 cost per kW 	Rs4.080882e+03 
 cost per kWh 	Rs5.143550e-01
(b)
 energy produced 1.121280e+09kWh 
 energy consumed by auxiliaries 5.606400e+07kWh/year 
 energy output of plant 1.065216e+09kWh 
 energy sold to consumer 9.054336e+08kWh

example 4.4 Page 61

In [6]:
v=230 ;ec=2020 #voltage #energy consumption
i=40; pf=1 ;t=2; c=3.5; rc=1.8 ;mon=30 #current/power factor/time/cost/reamining cost/month
ecd=v*i*pf*t*mon/1000 #energy corresponding to maximum demand
cost=ecd*c
ren=ec-ecd
rcost=ren*rc
tmb=cost+rcost
at=tmb/ec
print " energy corresponding to maximum demand %dkWh \n cost of above energy Rs%d \n remaining energy %dkWh \n cost of reamaining energy Rs%.1f \n total monthly bill Rs.%.1f\n avarage tariff Rs%.3fper kWh"%(ecd,cost,ren,rcost,tmb,at)

 energy corresponding to maximum demand 552kWh 
 cost of above energy Rs1932 
 remaining energy 1468kWh 
 cost of reamaining energy Rs2642.4 
 total monthly bill Rs.4574.4
 avarage tariff Rs2.265per kWh

example 4.5 Page 62

In [7]:
t1=3000 ;t11=0.9 #cost equation 
t2=3  #rate
x=t1/(t2-t11)
print "if energy consumption per month is more than %.1fkWh,\ntariff is more suitable"%(x)

if energy consumption per month is more than 1428.6kWh,
tariff is more suitable

example 4.6 Page 62

In [8]:
aec=201500 #annual energy consumption
lf=0.35#load factor constnt
t=4000#tariff
tmd=1200#tariff for maximum demand
t3=2.2
lfb=0.55 #load factor improved
ecd=0.25#energy consumption reduced
md=aec/(8760*lf)
yb=t+md*tmd+t3*aec
mdb=aec/(8760*lfb)
ybb=t+mdb*tmd+t3*aec
ne=aec*(1-ecd)
md3=ne/(8760*lf)
ybc=t+md3*tmd+t3*ne
aeca=yb/aec
aecb=ybb/aec
aecc=ybc/ne
print 'a'
print "maximum demand %.2fkW \n yearly bill Rs.%d per year \n(b)\n maximum demand %.2fkW \n yearly bill Rs.%dper year"%(md,yb,mdb,ybb)
print "c"
print " new energy %dkWh \n maximum demand %.2fkW \n yearly bill Rs.%dper year \n average energy cost in case a Rs%.4fper kWh \n average energy cost in case b Rs%.3fper kWh\n average energy cost in case c Rs%.3fper kWh "%(ne,md3,ybc,aeca,aecb,aecc)

a
maximum demand 65.72kW 
 yearly bill Rs.526164 per year 
(b)
 maximum demand 41.82kW 
 yearly bill Rs.497486per year
c
 new energy 151125kWh 
 maximum demand 49.29kW 
 yearly bill Rs.395623per year 
 average energy cost in case a Rs2.6112per kWh 
 average energy cost in case b Rs2.469per kWh
 average energy cost in case c Rs2.618per kWh

example 4.7 Page 66

In [9]:
pl1=20 ;pf1=0.8; t1=2000#load in MVA #power factor #duration
pl2=10 ;pf2=0.8 ;t2=1000#load in MVA #power factor #duration
pl3=2 ;pf3=0.8 ;t3=500#load in MVA #power factor #duration
pt=20 #/transformar power rating
fte=0.985 ;ste=0.99 #/full load efficiency for first and second transformer
ftl=120 ;stl=90 #core  loss inKW for first and  second transformer
cst=200000 #cost of second transformer with compared with first transformer
aid=0.15 #annual  interest and depreciation
ce=0.8 #cost of energy
tfl=pt*(1-fte)*1000#total full load
fle=tfl-ftl #full load copper loss
elc=fle*t1+(fle*t2/(pt/pl2)**2)+(fle*t3/(pt/pl3)**2) #energy loss due to copper loss
eli=ftl*(t1+t2+t3)#energy loss due to iron loss
celo=(elc+eli)*ce #cost of energy loss
print " \nfirst transformer : "
print " total full load losses %dkW \n full load copper losses %dkW \n energy loss due to copper losses %dkWh/year\n energy loss due to iron losses %dkWh/year \n cost of energy losses Rs%dper year"%(tfl,fle,elc,eli,celo)
stfl=pt*(1-ste)*1000#total full load
sle=stfl-stl#full load copper loss
selc=sle*t1+(sle*t2/(pt/pl2)**2)+(sle*t3/(pt/pl3)**2)#energy loss due to copper loss
seli=stl*(t1+t2+t3)#energy loss due to iron loss
scelo=(selc+seli)*ce#cost of energy loss
print " \nsecond transformer  :"
print " total full load losses %dkW \n full load copper losses %dkW \n energy loss due to copper losses %dkWh/year\n energy loss due to iron losses %dkWh/year \n cost of energy losses Rs%dper year"%(stfl,sle,selc,seli,scelo)
aidc=stfl*aid*1000
tybc=aidc+scelo
print " additional interest and depreciation due to higher cost of second transformer Rs%d \n total yearly charges for second transformer Rs%d per year"%(aidc,tybc)

 
first transformer : 
 total full load losses 300kW 
 full load copper losses 180kW 
 energy loss due to copper losses 405900kWh/year
 energy loss due to iron losses 420000kWh/year 
 cost of energy losses Rs660720per year
 
second transformer  :
 total full load losses 200kW 
 full load copper losses 110kW 
 energy loss due to copper losses 248050kWh/year
 energy loss due to iron losses 315000kWh/year 
 cost of energy losses Rs450440per year
 additional interest and depreciation due to higher cost of second transformer Rs30000 
 total yearly charges for second transformer Rs480440 per year

example 4.8 Page 67

In [10]:
from math import asin, acos, tan, cos
p=500 #load
pf=0.8#power factor
t=400 #tariff
md=100 #maximum demand tariff
ccb=600 #cost of capacitor bank
id=0.11#interest and deprecistion
sd=ccb*id/t#sin(ph2)
d2=asin(sd)
pf2=cos(d2)
kva=p*(tan(acos(pf))-tan(d2))
print " the most economic power factor %.3f lagging \n kvar requirement %.2fkVAR"%(pf2,kva)

 the most economic power factor 0.986 lagging 
 kvar requirement 291.35kVAR

example 4.9 Page 68

In [11]:
from math import asin, acos, tan, cos
l1=300 #load and power factor for three different loads
pf1=1 
l2=1000 
pf2=0.9 
l3=1500 
pf3=0.8
print " for %dkW unit power factor load \n power factor angle %.f\n reactive power %.fkvr"%(l1,acos(pf1),l1*(tan(acos(pf1))))
print " \nfor %dkW unit power factor load \n power factor angle %.2f\n reactive power %.2fkvr"%(l2,acos(pf2),l2*(tan(acos(pf2))))
print " \nfor %dkW unit power factor load \n power factor angle %.2f\n reactive power %.2fkvr"%(l3,acos(pf3),l3*(tan(acos(pf3))))
tl=l1+l2+l3
tt=l3*(tan(acos(pf3)))+l2*(tan(acos(pf2)))+l1*(tan(acos(pf1)))
print "\n total kW \t%dkW\n total kVAR %.1fkVAR \n total kVA %.2fkVA \n overall power factor %.3flagging"%(tl,tt,(tl**2+tt**2)**0.5,tl/(tl**2+tt**2)**0.5)
print "\n the maximum unity power factor load which yhe station can supply is equal to the kVA i.e.%.2fkVR"%((tl**2+tt**2)**0.5)

 for 300kW unit power factor load 
 power factor angle 0
 reactive power 0kvr
 
for 1000kW unit power factor load 
 power factor angle 0.45
 reactive power 484.32kvr
 
for 1500kW unit power factor load 
 power factor angle 0.64
 reactive power 1125.00kvr

 total kW 	2800kW
 total kVAR 1609.3kVAR 
 total kVA 3229.54kVA 
 overall power factor 0.867lagging

 the maximum unity power factor load which yhe station can supply is equal to the kVA i.e.3229.54kVR

example 4.10 Page 68

In [12]:
from math import asin, acos, tan, cos,sqrt, pi
v=400#voltage
i=25#/current
pf=0.8#at power factor
pf2=0.9#over all power factor
kw=v*i*pf*sqrt(3)/1000
print "kw rating of induction motor %.2fkW"%(kw)
dm=acos(pf)
rp=kw*tan(dm)
print "\n power factor angle %.2f \n reactive power %.2fkVR"%(dm,rp)
fdm=acos(pf2)
rp2=kw*tan(fdm)
print "\n final power factor %.2f \n final reactance power %.2fkVR"%(fdm,rp2)
ckvb=rp-rp2
cc=ckvb*1000/(sqrt(3)*v)
vc=v/sqrt(3)
xc=vc/cc
f=50
cec=1*10**(6)/(xc*2*pi*f)
print "\n kvar rating of capacitor bank %.4f \n current through each capacitor %.2fA\n voltage across each capacitor %.2f \n reactance of each capacitor %.2fohm \n capacitance of each capacitance %.2fuf"%(ckvb,cc,vc,xc,cec)

kw rating of induction motor 13.86kW

 power factor angle 0.64 
 reactive power 10.39kVR

 final power factor 0.45 
 final reactance power 6.71kVR

 kvar rating of capacitor bank 3.6813 
 current through each capacitor 5.31A
 voltage across each capacitor 230.94 
 reactance of each capacitor 43.46ohm 
 capacitance of each capacitance 73.24uf

example 4.11 Page 69

In [13]:
from math import asin, acos, tan, cos,sqrt, pi
v=400#line voltage
i=50 #line current
pf=0.8 #at power factor
pf2=0.95 # overall power factor
sm=25 #hp of synchronous motor 
e=0.9#efficiency
kwri=v*i*pf*sqrt(3)/1000
kvari=v*i*sqrt(3)/1000
karri=(-kwri**2+kvari**2)**0.5
kwsm=sm*735.5/(e*1000)
tkw=kwri+kwsm
print " kw rating of installation %.1fkW \n kVA rating of installation %.2fkva \n kVAR rating %.2fkvar \n kw input to synchrounous motor %.2fkw \n total kw=%.2f\n"%(kwri,kvari,karri,kwsm,tkw)
pd=acos(pf2)
tkr=tkw*tan(pd)
krsm=tkr-karri
kasm=(kwsm**2+krsm**2)**0.5
pfsm=kwsm/kasm
if krsm<0:
    ch='capacitor'
    ich='leading'
else:
    ch='inductive'
    ich='lagging'

print " overall power factor angle %.2fkw \n total kvar %.2fkvar \n kvar of synchrounous motor %.2fkvar %s \n kva of synchrounous motor %.2fkva \n power factor of synchrounous motor %.2f %s"%(pd,tkr,krsm,ch,kasm,pfsm,ich)

 kw rating of installation 27.7kW 
 kVA rating of installation 34.64kva 
 kVAR rating 20.78kvar 
 kw input to synchrounous motor 20.43kw 
 total kw=48.14

 overall power factor angle 0.32kw 
 total kvar 15.82kvar 
 kvar of synchrounous motor -4.96kvar capacitor 
 kva of synchrounous motor 21.02kva 
 power factor of synchrounous motor 0.97 leading

example 4.12 page 69

In [14]:
from math import asin, acos, tan, cos,sqrt, pi,sin, atan
psm=100 #power of synchrounous motors
pim=200 #power of inducion motor
v=400 #voltage 
pff=0.71; pp=-1#power factor
rsm=0.1 #resistance of synchrounous motor
rt=0.03 #resistance of cable
#pf(1)=1 ;p(1)=1 #power factor in a
#pf(2)=0.8 ;p(2)=1 #power factor in b
#pf(3)=0.6 ;p(3)=1 #power factor in c
pf =[1,0.8,0.6]
p = [1,1,1]

i1=pim*1000/(v*pff*sqrt(3))
i11=i1*(complex(pff,pp*sin(acos(pff))))
i2f=psm*1000/(v*sqrt(3))
ch=['a', 'b', 'c']
it = range(0,3)
opf = range(0,3)
for i in range(0,3):
    print "\n (%s)"%(ch[i])
    d=acos(pf[i])
    #it(i)=i11(1)+complex(i2f,(p(i)*i2f*tand(d)))
    it[i]=i11+complex(i2f,(p[i]*i2f*tan(d)))
    opf[i]=cos(atan((it[i]).imag/(it[i]).real))
    clsm=3*((i2f)**2)*rsm
    clt=3*(abs(it[i])**2)*rt/1000
    print "\n total current =",it[i]
    print "\n overall power factor %.3f lagging \n copper losses in synchrounous motor %.fW \n copper losses in cable %.2fKW"%(opf[i],clsm,clt)

print "(d)"
print "copper loss of synchronous motor this is evidently minimum when tand=%d cosd=%d"%(0,1)

 (a)

 total current = (433.012701892-286.317624651j)

 overall power factor 0.834 lagging 
 copper losses in synchrounous motor 6250W 
 copper losses in cable 24.25KW

 (b)

 total current = (433.012701892-178.064449178j)

 overall power factor 0.925 lagging 
 copper losses in synchrounous motor 6250W 
 copper losses in cable 19.73KW

 (c)

 total current = (433.012701892-93.8675349216j)

 overall power factor 0.977 lagging 
 copper losses in synchrounous motor 6250W 
 copper losses in cable 17.67KW
(d)
copper loss of synchronous motor this is evidently minimum when tand=0 cosd=1

example 4.13 Page 71

In [15]:
from math import atan,tan,acos,pi
p=2#constant output in MW
pf=0.9#power factor
pa=10#load
pb=5
pfb=0.8#power factor at load of 5MW
td=tan(acos(pf))
go=p*(1-td*1J)
op=0.8
tp=tan(acos(pfb))
print "power factor of indection generator is leading therefor induction generator output %d%.2fiMVA /n (a) \n"%(go.real,go.imag)
tl=pa*(1+tp*1J)
sg=tl-go
da=atan(sg.imag/sg.real)
print " total load %d+%.1fiMW \n synchronous generator load %d+%.3fiMW \n\t\t=%.2fMW at angle %.2f \n power factor of synchronous generator is %.2flagging"%(tl.real,tl.imag,sg.real,sg.imag,abs(sg),da,cos(da))
tl1=pb*(1+tp*1J)
sg1=tl1-go
da1=atan(sg1.imag/sg1.real)
print "(b)"
print " total load %d+%.1fiMW \n synchronous generator load %d+%.3fiMW \n\t\t=%.2fMW at angle %.2f \n power factor of synchronous generator is %.2flagging"%(tl1.real,(tl1).imag,sg1.real,(sg1).imag,abs(sg1),da1,cos(da1))

power factor of indection generator is leading therefor induction generator output 2-0.97iMVA /n (a) 

 total load 10+7.5iMW 
 synchronous generator load 8+8.469iMW 
		=11.65MW at angle 0.81 
 power factor of synchronous generator is 0.69lagging
(b)
 total load 5+3.7iMW 
 synchronous generator load 3+4.719iMW 
		=5.59MW at angle 1.00 
 power factor of synchronous generator is 0.54lagging

example 4.14 page 71

In [16]:
from math import atan,tan,acos,pi,sqrt,sin
c=40*10**(-6) #bank of capacitors in farads
v=400 #line voltage
i=40#/line current
pf=0.8#power factor
f=50#line frequency
xc=1/(2*pi*f*c)
ic=v/(sqrt(3)*xc)
il=i*(pf-sin(acos(pf))*1J)
til=il+1J*ic
od=atan(til.imag/til.real)
opf=cos(od)
nlol=(abs(od)/i)**2
print "(a)"
print " line current of capacitor bank %.1fA \n load current %d%diA \n total line current %d%.1fjA \n overall p.f %.3f \n new line loss to old line loss  %.3f"%(ic,il.real,il.imag,(til).real,(til).imag,opf,nlol)
pcb=(v/xc)
print "\n phase current of capacitor bank %.3fA"%(pcb)
lcb=pcb*sqrt(3)
print "\n line current of capacitor bank %.1fA"%(lcb)
tcu=il+lcb*1J
print "\n total current =",tcu
print "%.1fjA =%.2fA at an angle %.2f"%(tcu.imag,abs(tcu),atan(tcu.imag/tcu.real))
pf2=cos(atan(tcu.imag/(tcu).real))
print "\n power factor %.1f \n ratio of new line loss to original loss %.3f"%(pf2,(abs(tcu)/i)**2)

(a)
 line current of capacitor bank 2.9A 
 load current 32-23iA 
 total line current 32-21.1jA 
 overall p.f 0.835 
 new line loss to old line loss  0.000

 phase current of capacitor bank 5.027A

 line current of capacitor bank 8.7A

 total current = (32-15.2937630517j)
-15.3jA =35.47A at an angle -0.45

 power factor 0.9 
 ratio of new line loss to original loss 0.786

example 4.15 Page 72

In [17]:
from math import atan,tan,acos,pi,sqrt
p=30 #b.h.p of induction motor
f=50#line frequency
v=400#line voltage
e=0.85#effiency
pf=0.8 #power factor
i=p*746/(v*e*pf*sqrt(3))
i=i*complex(pf,-sin(acos(pf)))
ccb=(i).imag/sqrt(3)
xc=v/ccb
c=10**6/(2*f*pi*xc)
prl=((abs(i)**2-i.real**2)/abs(i)**2)*100
print " current drawn by motor is %.1fA \n "%abs(i)
print "the line loss will be minimum when i is munimum.the minimum value of i is",i,"A"
print " and occurs when the capacitor bank draws a line current of %djA \n capacitor C %.2fuf \n percentage loss reduction %d"%(i.imag,abs(c),prl)

 current drawn by motor is 47.5A 
 
the line loss will be minimum when i is munimum.the minimum value of i is (38.0032324249-28.5024243187j) A
 and occurs when the capacitor bank draws a line current of -28jA 
 capacitor C 130.95uf 
 percentage loss reduction 35

example 4.16 page 72

In [18]:
from math import atan,tan,acos,pi,sqrt
po=666.66 #power
f=50 #frequency
v=400 #voltage
pf=0.8  ;p=-1#power factor
pf2=0.95 ;p2=-1#improved power factor
vc=2200 #capacitor voltage 
rc=vc
il=po*1000/(v*pf*sqrt(3))
il1=il*(complex(pf,p*sin(acos(pf))))
i2c=il*pf
tad=tan(acos(pf2))
i2=complex(i2c,i2c*tad*p2)
print " load current i1 =",il1,"A "
print "\n load current current on improved power factor =",i2,"A"
print "(a)"
ic=abs(il1-i2)
ilc=ic*v/vc
pic=ilc/sqrt(3)
xc=vc/pic
ca=10**6/(2*pi*f*xc)
print " line current of %dV capacitor bank %.2fA\n line current of %d capacitor bank %.2fA \n phase current of capacitor bank %.2fA \n reactance %.2f \n capacitance %.2fF*10**(-6)"%(v,ic,vc,ilc,pic,xc,ca)
print "(b)"
kr=3*vc*pic/1000
print " kVA rating %.1fkVA \n kVA rating of transformer to convert %dV to %dV will be the same as the kVA rating of capacitor bank"%(kr,v,vc)
pl=100*(abs(il1)**2-abs(i2)**2)/abs(il1)**2
print "percentage reduction in losses %d percent"%(pl)
print "(d)"
pi=ic/sqrt(3)
xcc=v/pi
cc=1*10**6/(2*pi*f*xcc)
roc=ca/cc
print " phase current %.1fA \n reactance %.2fohm \n capasitance %.2f*10**-6F \n ratio of capacitance %.3f"%(pi,xcc,cc,roc)

 load current i1 = (962.240826145-721.680619609j) A 

 load current current on improved power factor = (962.240826145-316.273264908j) A
(a)
 line current of 400V capacitor bank 405.41A
 line current of 2200 capacitor bank 73.71A 
 phase current of capacitor bank 42.56A 
 reactance 51.70 
 capacitance 61.57F*10**(-6)
(b)
 kVA rating 280.9kVA 
 kVA rating of transformer to convert 400V to 2200V will be the same as the kVA rating of capacitor bank
percentage reduction in losses 29 percent
(d)
 phase current 234.1A 
 reactance 1.71ohm 
 capasitance 25.00*10**-6F 
 ratio of capacitance 2.463

example 4.17 Page 74

In [19]:
from math import acos,pi,sqrt,sin
v1=132#line voltage at primary
v2=11#line voltage at secondary
p=10 #power
pf=0.8 #power factor
mva=p*(complex(pf,sin(acos(pf))))
print " MVA rating of secondary = %dMVA =%d+%djMVA \n "%(p,abs(mva),mva.imag)
print "\n since the power factor at primary terminals is unity,rating of primary need be %dMVA only \n the tertiary will supply capacitor curren.since p.f is to be raised to 1 ,the mav compensation needed is 6MVA so rating of teritiary is %dMVA"%(abs(mva),mva.imag)

 MVA rating of secondary = 10MVA =9+5jMVA 
 

 since the power factor at primary terminals is unity,rating of primary need be 9MVA only 
 the tertiary will supply capacitor curren.since p.f is to be raised to 1 ,the mav compensation needed is 6MVA so rating of teritiary is 5MVA

example 4.18 Page 74

In [20]:
from math import sin,acos,pi,sqrt,acos,cos
v=11 #line voltage
f=50#line frequency
l=400 #load of alternator
pf=0.8 #power factor
e=0.85#efficiency
p=l/pf
lo=l+p*sin(acos(pf))*1J
print "a"
print "when pf is rased to 1 the alternator can supply %dkW for the same value of armture current hence it can supply %dKW to synchronous motor"%(p,p-l)
print "b"
print "b.h.p =%.2fHP"%(100*e/0.746)
kvam=p-lo
td=atan((kvam).imag/(kvam).real)
pff=cos(td)
print "\ncosd=%.3fleading"%(pff)

a
when pf is rased to 1 the alternator can supply 500kW for the same value of armture current hence it can supply 100KW to synchronous motor
b
b.h.p =113.94HP

cosd=0.316leading

example 4.19 Page 74

In [21]:
from math import tan,acos
kw=100  #let kw=100kw
pf=0.6 #power foctor
pf2=0.8 #power factor
kvar=kw*tan(acos(pf))
kvar2=kw*tan(acos(pf2))
ckar=((kvar-kvar2))/10
ck=round(ckar)*10
print " capacitor kVAR required for %dkW\n load for same power factor improvement %dKVAR"%(round(ckar),ck)

 capacitor kVAR required for 6kW
 load for same power factor improvement 60KVAR

example 4.20 Page 75

In [22]:
from __future__ import division
from math import tan,acos
p=160 #kva for transformer 
pf=0.6 #power factor
el=96 #effective load
eli=120 #effective load increase
rc=eli*(tan(acos(pf))-tan(acos(eli/p)))
opf=eli/p
print " required capacitor kVAR %dKVAR \n overall power factor %.2f \n it is seen that point d is on %.2f line"%(rc,opf,opf)

 required capacitor kVAR 54KVAR 
 overall power factor 0.75 
 it is seen that point d is on 0.75 line

example 4.21 page 76

In [23]:
from math import atan,tan,acos,pi,sqrt,sin
md=800 #maximum demand
pf=0.707 #power factor
c=80 #cost
p=200 #power
e=0.99#efficiency
pff=0.8 #fulload pf
ikva=md/pf
iafc=(round(ikva*100)*(c)/100)
rsm=ikva*pf
act=p*(0.7355)/e
at=-act*sin(acos(pff))
tkw=rsm+act
tkvr=rsm+at
tkva=(tkw**2+tkvr**2)**0.5
ikvad=tkva-ikva
infc=ikvad*c
print " initial kVA %.2fkVA \n initial annual fixed charges Rs%.1f \n after installation of synchronous motor reactive power of induction motor %dkVars\n active power input of synchrounous motor %.2fkW\n reactive power input to synchrounous motor %.2fKVAR \n total kW %.2fKW \n total kVars %.2fkVARS \n total kVA %.2fkVA \n increase in KVA demand %.2fkVA\n increase in annual fixed charges Rs%.1f "%(ikva,iafc,rsm,act,at,tkw,tkvr,tkva,ikvad,infc)

 initial kVA 1131.54kVA 
 initial annual fixed charges Rs90523.2 
 after installation of synchronous motor reactive power of induction motor 800kVars
 active power input of synchrounous motor 148.59kW
 reactive power input to synchrounous motor -89.15KVAR 
 total kW 948.59KW 
 total kVars 710.85kVARS 
 total kVA 1185.38kVA 
 increase in KVA demand 53.84kVA
 increase in annual fixed charges Rs4306.9

example 4.22 Page 77

In [24]:
t=16#working  time
d=300 #working days
hv=1 ;hvmd=50 #tariff on high voltage
lv=1.1 ;lvmd=60 #tariff on low voltage
al=250#avarage load
pf=0.8#power factor
md=300 #maximum demand
hvec=500#cost of hv equipment
l=0.05 #loss of hv equipment
id=0.12  #interest and deprecistion
ter=al*md*t 
mdv=md/pf
print " total energy requirement %2.2ekWH \n maximum demand %dKVA"%(ter,mdv)
print "(a)HV supply"
chv=mdv*hvec
idc=chv*id
ere=ter/(1-l)
dch=mdv*hvmd
ech=round(ere*hv/1000)*1000
tanc=ech+dch+idc
print " cost of HV equipment Rs%e\n interest and depreciation charges Rs%d \n energy received %ekWh\n demand charges Rs%d \n energy charges Rs%2e \n total annual cost Rs%d"%(chv,idc,ere,dch,ech,tanc)
print "(b) LV supply"
lvdc=mdv*lvmd
lvec=ter*lv
lvtac=lvec+lvdc
lvdac=lvtac-tanc
print " demand charges Rs%d \n energy charges Rs%2.e \n total annual cost Rs%d \n difference in annual cost Rs%d"%(lvdc,lvec,lvtac,lvdac)

 total energy requirement 1.20e+06kWH 
 maximum demand 375KVA
(a)HV supply
 cost of HV equipment Rs1.875000e+05
 interest and depreciation charges Rs22500 
 energy received 1.263158e+06kWh
 demand charges Rs18750 
 energy charges Rs1.263000e+06 
 total annual cost Rs1304250
(b) LV supply
 demand charges Rs22500 
 energy charges Rs1e+06 
 total annual cost Rs1342500 
 difference in annual cost Rs38250