Ch-2, Loads & Load Curves

example - 2.1, page - 17

In [4]:
from __future__ import division
print "solution for (a)"
nb=8; nf=2; nl=2 #given number of equipments is 8 bulbs 2 fans 2plugs
lb=100 ;lf=60 ;ll=100 #corresponding wattages
cl=nb*lb+nf*lf+nl*ll  #total connected load
print "connected load = %dW\n"%(cl) 
print "solution for (b)"
t1=5 ;t2=2 ;t3=2; t4=9; t5=6; # time periods in hours
# let = [bulb, fan, light]
w1=1*lf #12 to 5am period of duration 5h
w2=2*lf+1*ll #5am to 7am period of duration 2h
w3=0 #7am to 9am period of duration 2h
w4=2*lf #9am to 6pm period of duration 9h
w5=4*lb+2*lf #6pm to 12pm period of duration 6h
w=[w1 ,w2 ,w3, w4, w5]
print "\nthe maximum demand is %d W\n"%(max(w))
m=max(w)
print "solution for (c)"
df=m/cl
print "\ndemand factor =%0.3f\n"%(df)
print "solution for (d)"
#energy consumed is power multiply by corresponding time
energy=[w1*t1 ,w2*t2 ,w3*t3 ,w4*t4, w5*t5]
e=sum(energy)
print "\ntotal energy consumed during 24 hours = %d Wh = %0.2f kWh\n"%(e,e/1e3)
print "solution for (e)" 
ec=cl*24 
print "\nif all devices are used throughout the day the energy consumed = %d Wh = %.2fkWh"%(ec,ec/1000)     
#for 24 hours of max. load

solution for (a)
connected load = 1120W

solution for (b)

the maximum demand is 520 W

solution for (c)

demand factor =0.464

solution for (d)

total energy consumed during 24 hours = 4940 Wh = 4.94 kWh

solution for (e)

if all devices are used throughout the day the energy consumed = 26880 Wh = 26.88kWh

example - 2.2, page - 19

In [5]:
%matplotlib inline
from matplotlib.pyplot import plot, subplot, title, title, show, hist
print "(a)" 
mca=1.1 ;cla=2.5; mcb=1; clb=3;       #mca=maximum demand of consumera cla=connected load of a mcb=maximum load of consumer b clb=connected load of consumer b
print "maximum demand of consumer A = %0.1fkW \n \ndemand factor of consumer A = %0.2f \n \nmaximum demand of consumer B = %dkW\n \ndemand factor of consumer B = %0.3f"%(mca,mca/cla,mcb,mcb/clb)
print "(b)"
print "The variation in demand versus time curves are plotted and shown in Fig This is known as chronological load curve."
a=([100]*5,[1100]*1,[200]*2,[0]*9,[500]*7)
i=0
A=range(1,25)
for x in a:

    for y in x:
        
        A[i]=y
        i+=1

        
b=([0]*(7),[300]*(1),[1000]*(2),[200]*(8),[600]*(5),[0]*(1))  #time line of different periods by a and b consumers
i=0
B=range(1,25)
for x in b:

    for y in x:
        
        B[i]=y
        i+=1
from numpy import arange, nditer
t=arange(1,25,1)         #for 24 hours ploting
ma=max(A) ;mb=max(B) 
subplot(211)        #matrix plotting
plot(t,A) 
plot(t,B) 
title("load curves of A and B: fig(1) \n")
show()
C=range(1,25)
i=0
for x,y in nditer([A,B]):
   C[i]=x+y 
   i+=1 
subplot(222) 
plot(t,C)  
title("chronological load of group : fig(2) \n")
show()
mg=max(C)  #maximum demand of group
print "(c)"
print "maximum demand of the group is %dW"%(mg) 
gd=(ma+mb)/mg 
print "group diversity factor = %.3f"%(gd)   #group diversity factor is sum of individual maximum consumaer load to the group max load 
print "(d)"
sa=sum(A)
print "energy consumed by A during 24 hours is = %d Wh = %0.2f kWh"%(sa,sa/1e3)
sb=sum(B) 
print "(e)" 
print "maximum energy which A could consume in 24hours = %.2fkWh \nmaximum energy which B consume in 24 hours is =%.2fkWh"%(mca*24,mcb*24 ) 
print "(f)" 
print "actual energy/maximum energy" 
mca=mca*10**3; mcb=mcb*10**3
aemea=sa/(mca*24)
aemeb=sb/(mcb*24)
print "\nfor A = %0.4f \nfor b = %.4f"%(sa/(mca*24),aemeb) 

(a)
maximum demand of consumer A = 1.1kW 
 
demand factor of consumer A = 0.44 
 
maximum demand of consumer B = 1kW
 
demand factor of consumer B = 0.333
(b)
The variation in demand versus time curves are plotted and shown in Fig This is known as chronological load curve.

(c)
maximum demand of the group is 1100W
group diversity factor = 1.909
(d)
energy consumed by A during 24 hours is = 5500 Wh = 5.50 kWh
(e)
maximum energy which A could consume in 24hours = 26.40kWh 
maximum energy which B consume in 24 hours is =24.00kWh
(f)
actual energy/maximum energy

for A = 0.2083 
for b = 0.2875

example - 2.3 : page - 22

In [6]:
cola=5 ;na=600; ns=20 
cls=2 ;clfm=10 ;clsm=5; cll=20; i=80 
dffl=0.7; dfsm=0.8; dfl=0.65; dfci=0.5 
nsl=200 ;clsl=0.04 ;dfa=0.5; gdfa=3.0 
pdfa=1.25; gdfc=2 ;pdfc=1.6; dfs=0.8  #given col||cl=connected load,n=number,df=demand factor,gdf=group diversity factor,pdf=peak diversity factor,a=appartement,c=commertials,s=shop,sl=streetlight,fm=flourmill,sm=saw mill,l=laundry,ci=cinema complex.
mdea=cola*dfa
print "maximum demand of each appartment =%.2fkWh \n"%(mdea)
mda=(na*mdea)/gdfa
print "maximum demand of 600 apatments =%.2fkW \n"%(mda) 
datsp=mda/pdfa
print "demand of 600 apartments at time of the system peak =%dkW \n"%(datsp)  
mdtcc=((cls*ns*dfs)+(clfm*dffl)+(clsm*dfsm)+(cll*dfl)+(i*dfci))/gdfc
print "maximum demand of total commertial complex=%dkW \n"%(mdtcc)
dcsp=mdtcc/pdfc
print "demand of the commertial load at the time of the peak = %dkW\n"%(dcsp) 
dsltsp=nsl*clsl
print "demand of the street lighting at the time of the system peak =%dkW"%(dsltsp) 
ispd=datsp+dcsp+dsltsp
print "\nincrease in system peak deamand =%dkW "%(ispd)

maximum demand of each appartment =2.50kWh 

maximum demand of 600 apatments =500.00kW 

demand of 600 apartments at time of the system peak =400kW 

maximum demand of total commertial complex=48kW 

demand of the commertial load at the time of the peak = 30kW

demand of the street lighting at the time of the system peak =8kW

increase in system peak deamand =438kW

example - 2.4 : page -28

In [7]:
from __future__ import division
%matplotlib inline
from matplotlib.pyplot import plot, subplot, title, title, show, hist

print "the chronological load curve is plotted in fig 1 the durition of loads is as under :"
LC=([20]*(5),[40]*4,[80]*9,[100]*4,[20]*2)

i=0
lc=range(1,25)
for x in LC:
    for y in x:
        lc[i] = y
        i+=1
lc.sort()

nn=len(lc)

e=sum(lc)
print lc
print "\nthe load duration curve is ploted in 2 the energy produced by plant in 24 hours = %d MWh \n"%(e) 
lff=e/(24*max(lc)) 
print "load factor = %0.4f= %.2f %%"%(lff,lff*100)
t=range(1,25)
subplot(211) 
plot(t,lc) 
show()
title("chronological curve")
subplot(212) 
plot(t,lc) 
title("load duration curve")
show()

the chronological load curve is plotted in fig 1 the durition of loads is as under :
[20, 20, 20, 20, 20, 20, 20, 40, 40, 40, 40, 80, 80, 80, 80, 80, 80, 80, 80, 80, 100, 100, 100, 100]

the load duration curve is ploted in 2 the energy produced by plant in 24 hours = 1420 MWh 

load factor = 0.5917= 59.17 %

example - 2.5 : page - 29

In [8]:
from __future__ import division
lf=0.5917; ml=100; ic=125  #lf=load factor,ic=installed capacity,ml=maximum load,cf=capacity factor,uf=utillization factor
cf=(ml*lf)/ic; uf=ml/ic
print "capacity factor =%0.3f"%(cf)
print "\nutilisation factor =%.2f"%(uf)

capacity factor =0.473

utilisation factor =0.80

example - 2.6 : page - 29

In [9]:
from numpy import nditer
%matplotlib inline
from matplotlib.pyplot import plot, subplot, title, xlabel, ylabel, show

t=[24, 24-7, 24-11, 4] # Periods for running load 20,40,80,100
l = [20,40,80,100] # different loads
E= range(0,len(l))
E[0] = t[0]*l[0] # MWh
i=1
for x in E[1:]:
    E[i] = E[i-1]+(l[i]-l[i-1])*t[i]
    i+=1

print "Energy at different load levels:\nLoad(MW)\t\tEnergy(MWh)\n"

for x, y in nditer([l, E]):
    print '  ',x,'\t\t\t ',y

subplot(211)
plot(E,l,'bo')
plot(E,l)
title('Energy Load Curve')
xlabel('Energy MW Hrs')
ylabel('Load MW')
show()
######### Mass Curve #######


t = [5,4,9,4,2]
l = [20,40,80,100,20] # loads on different time of day
Es= range(0,len(l))
Es[0] = t[0]*l[0] # MWh
i=1
for x in Es[1:]:
    Es[i] = Es[i-1]+(l[i])*t[i]
    i+=1


print "Energy supplied uto different times of day:\nTime\t\tEnergy(MWh)\n"
times = ['5am', '9am', '6pm','10pm','12pm']
for x, y in nditer([times, Es]):
    print x,'\t\t\t ',y

    
td= range(0,len(t))
td[0] = t[0] # MWh
i=1
for x in td[1:]:
    td[i] = td[i-1]+t[i]
    i+=1
    
subplot(212)
plot(td,Es,'bo')
plot(td,Es)
title('Mass Curve')
xlabel('Time')
ylabel('Load MW')
show()

Energy at different load levels:
Load(MW)		Energy(MWh)

   20 			  480
   40 			  820
   80 			  1340
   100 			  1420

Energy supplied uto different times of day:
Time		Energy(MWh)

5am 			  100
9am 			  260
6pm 			  980
10pm 			  1380
12pm 			  1420

example - 2.7 : page 30

In [10]:
md=40 ;cf=0.5; uf=0.8 #maximum demand in MW capacity factor utility factor
print "(a)"
lf=cf/uf   #load factor is ratio of capacity factor to the utility  factor
print "load factor = capacity factor/utilisation factor =%0.3f"%(lf)
print "(b)"
pc=md/uf   #plant capacity is ratio of maximum demand to utility factor
print "plant capacity = maximum demand/utilisation factor =%dMW"%(pc)
print "(c)"
rc=pc-md   #reserve capacity is plant capacity minus maximum demand 
print "reserve capacity =%dMW"%(rc)
print "(d)"
print "annual energy production =%dMWh"%(md*lf*8760)

(a)
load factor = capacity factor/utilisation factor =0.625
(b)
plant capacity = maximum demand/utilisation factor =50MW
(c)
reserve capacity =10MW
(d)
annual energy production =219000MWh

example - 2.8 : pge 30

In [11]:
%matplotlib inline
from matplotlib.pyplot import plot, subplot, title, xlabel, ylabel, show
a=[0, 5, 9 ,18, 20, 22, 24] #time in matrix format
b=[50 ,50 ,100 ,100 ,150 ,80 ,50]#load in matrix format
z= range (1,7)
for x in range (1,6):
    
   z[x]=((b[x]+b[x+1])/2)*(a[(x+1)])-a[x]

e = b[0]*a[1]+(b[2]+b[1])*(a[2]-a[1])/2+(b[3]*a[2])+(b[3]+b[4])+(b[4]+b[5])+(b[5]+b[6]) # MWh

print "energy required required by the system in 24 hrs = %d MWh"%(e)
#50x5MWh+((100+50)/2)x4MWh +(100x9)MWh+(100+150)MWh+(150+80)MWh+(80+50)MWh \n =
dlf=e/(max(b)*24)
print "\ndaily load factor = %f or %0.2f %%"%(dlf,dlf*100)

#########PLOT##############
plot(a,b)
title('Load Curve')
xlabel("TIME -->")
ylabel("MW -->")
show()
#xtitle("load curve"%("time"%("MW"

energy required required by the system in 24 hrs = 2060 MWh

daily load factor = 0.572222 or 57.22 %

example - 2.9: page 31

In [12]:
%matplotlib inline
from matplotlib.pyplot import plot, subplot, title, xlabel, ylabel, show
print "load duration curve in fig1"
print "the energy consumed upto different times is as "
a=[0, 5 ,9 ,18, 20, 22, 24]  #time in matrix format
b=[50, 50 ,100 ,100 ,150 ,80 ,50] #load in matrix format

z = range(0,6)
for x in range (0,6):
   z[x]=((b[x]+b[x+1])/2)*(a[x+1]-a[x])

et=0
q = range(0,7)
m = range(0,7)
ett = range(0,6)
for x in range(0,6):
    et=et+z[x] 
    A=a[(x)]
    ett[x]=et 
    q[x]=a[x+1]
    print "\nfrom mid night upto %d,energy=%dMWh"%(A,et)
n = sorted(range(len(b)), key=lambda k: b[k], reverse=True)
m = sorted(b, reverse=True)
print "energy curve in fig 2"
t=[0, 3.88, 15.88 ,19.88, 23]
k = range(0,6)
for j in range(0,6):
    k[j]=a[(j+1)]
M =range(0,5)
#rearranging for mass  curve
for i in range(0,5):
    M[i] = m[i]
Q = range(0,6)
for i in range(0,6):
    Q[i] = q[i]
    
    
subplot(121) 
plot(t,M) 
title("load duration")
xlabel("hours")
ylabel("MW")
subplot(122) 
plot(Q,ett) 
title("energy curve")
xlabel("time")
ylabel("MWh")
show()

load duration curve in fig1
the energy consumed upto different times is as 

from mid night upto 0,energy=250MWh

from mid night upto 5,energy=550MWh

from mid night upto 9,energy=1450MWh

from mid night upto 18,energy=1700MWh

from mid night upto 20,energy=1930MWh

from mid night upto 22,energy=2060MWh
energy curve in fig 2

example - 2.10 : page 31

In [13]:
egd1=438*10**4 ;plp=0.2; pcf=0.15 #annual load duration  annual  load factor plant capacity factor
pml=egd1/(plp*8760)
pc=(pml*plp)/pcf
print "annual load factor =energy generated during 1 year/(max. load)x8760=%.1f \n maximum load =%dkW"%(plp,pml)
print "\ncapacity factor =(max.load/plant capacity)x(load factor)\n plant capacity =max.load/0.75 =%fMW \n reserve capacity =3.333-2.5=%fMW"%(pc,pc-pml)

annual load factor =energy generated during 1 year/(max. load)x8760=0.2 
 maximum load =2500kW

capacity factor =(max.load/plant capacity)x(load factor)
 plant capacity =max.load/0.75 =3333.333333MW 
 reserve capacity =3.333-2.5=833.333333MW

example - 2.11: page 32

In [14]:
p1=10; p2=6; p3=8; p4=7 #peak demands of 4 areas
df=1.5 ;lf=0.65 ;imdp=0.6 #diversity factor  annual load factor  ratio of maximum demand
p=p1+p2+p3+p4
md=p/df
ae=md*lf*8760
imd=imdp*md
ic=md+imd
print " sum of maximum=%dMW"%(p)
print "\n maximum demand = sum of max.demands/diversity factor =%d/%f = %fMW"%(p,df,md)
print "\n annual energy =%fMWh \n increase in maximum demand =%fMW \n installed capacity =%fMW"%(ae,imd,ic)

 sum of maximum=31MW

 maximum demand = sum of max.demands/diversity factor =31/1.500000 = 20.666667MW

 annual energy =117676.000000MWh 
 increase in maximum demand =12.400000MW 
 installed capacity =33.066667MW

example - 2.12 : page 32

In [15]:
from __future__ import division
%matplotlib inline
from matplotlib.pyplot import plot, title, xlabel, ylabel, show

#Arranging data for Load Duration Curve
#week days 5-9pm load
L1=350 #MW
t1=4*5 #hours
#week days 8-12am & 1-5pm load
L2=250 #MW
t2=t1+8*5 #hours
#saturday & sunday 5-9pm load
L3=200 #MW
t3=t2+4*2 #hours
#All days 150MW load
L4=150 #MW
t4=t3+6*5+15*2 #hours
#All days 100MW load
L5=100 #MW
t5=t4+6*5+5*2 #hours
A=31600 #Total Load Curve Area
LF=A/L1/24/7*100 #%#Weekly load factor
print "Weekly Load factor = %0.2f %%"%LF
print "Load Duration Curve is shown in figure." 
#Load Duration Curve
L=[L1 ,L2, L3, L4, L5] #MW
T=[t1 ,t2 ,t3 ,t4 ,t5] #hours
plot(T,L) 
title('Load Duration Curve')
xlabel('Time(Hours)')
ylabel('Load(MW)') 
show()

Weekly Load factor = 53.74 %
Load Duration Curve is shown in figure.

example - 2.13 : page 33

In [16]:
dlf=0.825    #daily load factor
lptmlp=0.87   #average daily peak load to monthly load peak
mlptalp=0.78   #average monthly peak load to annual load peak
print "annual load factor = %fx%fx%f = %0.2f"%(dlf,lptmlp,mlptalp,dlf*lptmlp*mlptalp)

annual load factor = 0.825000x0.870000x0.780000 = 0.56

example - 2.14 : page 35

In [17]:
print "(a)"
#given
transformer1_motorload=300 
transformer1_demandfactorm=0.6 
tarnsformer1_commercialload=100 
transformer1_demandfactorc=0.5 
transformer1_diversityfactor=2.3
transformer2_residentalload=500 
transformer2_demandfactor=0.4 
transformer2_diversitryfactor=2.5 
transformer3_residentalload=400 
transformer3_demandfactor=0.5 
transformer3_diversityfactor=2.0 
diversitybtwxmer=1.4
peakloadoftransformer1=((transformer1_motorload*transformer1_demandfactorm)+(tarnsformer1_commercialload*transformer1_demandfactorc))/transformer1_diversityfactor
peakloadonxmer=(transformer2_residentalload*transformer2_demandfactor)/transformer2_diversitryfactor
peakloadonxmer3=(transformer3_residentalload*transformer3_demandfactor)/(transformer3_diversityfactor)
print "peak load on transformer 1 =(300x0.6+100x0.5)/2.3 =%dkW \npeak load on transformer 2 =%dkW \n peak load on transformer 3 =%dkW"%(peakloadoftransformer1,peakloadonxmer,peakloadonxmer3)
print "(b)"
peakloadonfeeder=(peakloadoftransformer1+peakloadonxmer+peakloadonxmer3)/diversitybtwxmer
print "peak load on feeder =(100+80+100)/1.4 =%dkW"%(peakloadonfeeder)

(a)
peak load on transformer 1 =(300x0.6+100x0.5)/2.3 =100kW 
peak load on transformer 2 =80kW 
 peak load on transformer 3 =100kW
(b)
peak load on feeder =(100+80+100)/1.4 =200kW