diff options
Diffstat (limited to '416/CH2')
34 files changed, 488 insertions, 0 deletions
diff --git a/416/CH2/EX2.1/exp2_1cpp.sce b/416/CH2/EX2.1/exp2_1cpp.sce new file mode 100755 index 000000000..b9ae7ebc3 --- /dev/null +++ b/416/CH2/EX2.1/exp2_1cpp.sce @@ -0,0 +1,33 @@ +clc
+disp("example =2.1")
+printf("\n")
+disp("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
+printf("connected load = 8X100W+2X60W+2X100W=%dW\n",cl);
+disp("solution for (b)")
+disp("total wattage at different times is")
+t1=5;t2=2;t3=2;t4=9;t5=6;
+fr=[0 1 0] //12 to 5am period of duration 5h
+s=[0 2 1] //5am to 7am period of duration 2h
+t=[0 0 0] //7am to 9am period of duration 2h
+fo=[0 2 0] //9am to 6pm period of duration 9h
+fi=[4 2 0] //6pm to 12pm period of duration 6h
+w=[fr;s;t;fo;fi]
+wt=[100*w(:,1),60*w(:,2),100*w(:,3)]
+wtt=[sum(wt(1,:));sum(wt(2,:));sum(wt(3,:));sum(wt(4,:));sum(wt(5,:))]
+printf("\t%dW\n\t%dW\n\t%dW\n\t%dW\n\t%dW",wtt(1),wtt(2),wtt(3),wtt(4),wtt(5))
+printf("\nthe maximum demand is %dW\n",max(wtt))
+m=max(wtt)
+disp("solution for (c)")
+printf("\ndemand factor =%3f\n",m/cl)
+disp("solution for (d)")//energy consumed is power multiply by corresponding time
+energy=[wtt(1,1)*t1;wtt(2,1)*t2;wtt(3,1)*t3;wtt(4,1)*t4;wtt(5,1)*t5]
+printf("\t%dWh\n\t%dWh\n\t%dWh\n\t%dWh\n\t%dWh",energy(1),energy(2),energy(3),energy(4),energy(5))
+e=sum(energy)
+printf("\ntotal energy consumed during 24 hours =%dWh+%dWh+%dWh+%dWh+%dWh=%dWh\n",energy(1),energy(2),energy(3),energy(4),energy(5),e)
+disp("solution for (e)");
+ec=cl*24;
+printf("\nif all devices are used throughout the day the energy consumed in Wh is %dWh \n\t\t%.2fkWh",ec,ec/1000)
+//for 24 hours of max. load
\ No newline at end of file diff --git a/416/CH2/EX2.1/exp2_1cpp.txt b/416/CH2/EX2.1/exp2_1cpp.txt new file mode 100755 index 000000000..f0eb15ff3 --- /dev/null +++ b/416/CH2/EX2.1/exp2_1cpp.txt @@ -0,0 +1,33 @@ +
+ example =2.1
+
+
+ solution for (a)
+connected load = 8X100W+2X60W+2X100W=1120W
+
+ solution for (b)
+
+ total wattage at different times is
+ 60W
+ 220W
+ 0W
+ 120W
+ 520W
+the maximum demand is 520W
+
+ solution for (c)
+
+demand factor =0.464286
+
+ solution for (d)
+ 300Wh
+ 440Wh
+ 0Wh
+ 1080Wh
+ 3120Wh
+total energy consumed during 24 hours =300Wh+440Wh+0Wh+1080Wh+3120Wh=4940Wh
+
+ solution for (e)
+
+if all devices are used throughout the day the energy consumed in Wh is 26880Wh
+ 26.88kWh
\ No newline at end of file diff --git a/416/CH2/EX2.10/exp2_10cpp.sce b/416/CH2/EX2.10/exp2_10cpp.sce new file mode 100755 index 000000000..cd62416cb --- /dev/null +++ b/416/CH2/EX2.10/exp2_10cpp.sce @@ -0,0 +1,7 @@ +clc
+disp("example 2.10")
+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
+printf("annual load factor =energy generated during 1 year/(max. load)x8760=%.1f \n maximum load =%dkW",plp,pml)
+printf("\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)
diff --git a/416/CH2/EX2.10/exp2_10cpp.txt b/416/CH2/EX2.10/exp2_10cpp.txt new file mode 100755 index 000000000..dbb2b7b11 --- /dev/null +++ b/416/CH2/EX2.10/exp2_10cpp.txt @@ -0,0 +1,7 @@ +
+ example 2.10
+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
\ No newline at end of file diff --git a/416/CH2/EX2.11/exp2_11c.sce b/416/CH2/EX2.11/exp2_11c.sce new file mode 100755 index 000000000..719962812 --- /dev/null +++ b/416/CH2/EX2.11/exp2_11c.sce @@ -0,0 +1,12 @@ +clc
+disp("example 2.11")
+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
+printf(" sum of maximum=%dMW",p)
+printf("\n maximum demand = sum of max.demands/diversity factor =%d/%f = %fMW",p,df,md)
+printf("\n annual energy =%fMWh \n increase in maximum demand =%fMW \n installed capacity =%fMW",ae,imd,ic)
diff --git a/416/CH2/EX2.11/exp2_11c.txt b/416/CH2/EX2.11/exp2_11c.txt new file mode 100755 index 000000000..67d731a9a --- /dev/null +++ b/416/CH2/EX2.11/exp2_11c.txt @@ -0,0 +1,7 @@ +
+ example 2.11
+ 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
\ No newline at end of file diff --git a/416/CH2/EX2.12/exp2_12.png b/416/CH2/EX2.12/exp2_12.png Binary files differnew file mode 100755 index 000000000..3be5ad7e8 --- /dev/null +++ b/416/CH2/EX2.12/exp2_12.png diff --git a/416/CH2/EX2.12/exp2_12c.sce b/416/CH2/EX2.12/exp2_12c.sce new file mode 100755 index 000000000..4a59dbca5 --- /dev/null +++ b/416/CH2/EX2.12/exp2_12c.sce @@ -0,0 +1,43 @@ +clc
+disp("example 2.12")
+disp("from the above data,the durations of different loads during one week are")
+aw=[0 5 8 12 13 17 21 24] //given week timings and corresponding loads
+lw=[100 150 250 100 250 350 150]
+aen=[0 5 17 21 24] //given weakends timing and corresponding
+len=[100 150 200 150]
+saw=size(aw);saen=size(aen)
+sae=saw(1,2)-1;saen=saen(1,2)-1
+for x=1:sae //getting duration of load
+ tdw(1,x)=aw(1,x+1)-aw(1,x)
+end
+for x=1:saen
+ tden(1,x)=aen(1,x+1)-aen(1,x)
+end
+taw=5*tdw //duration of entair week
+taen=2*tden
+alw=[taw taen;lw len]
+lwen=[lw len] //arranging load in accending order
+[m n]=gsort(lwen)
+kn=size(lwen)
+kld=kn(1,2)
+
+for x=2:kld
+
+ ldcq(:,x)=alw(:,n(x))
+ if x>1 then
+ ldcq(1,x)=ldcq(1,x)+ldcq(1,x-1)
+ end
+end
+
+plot2d2(ldcq(1,:),ldcq(2,:))
+printf(" load duration \n 350MW 4x5 =20hours \n 250MW 20+8x5=60hours \n 200MW 60+4x2 =68hours \n 150MW 68+6x5+15x2 =128hours \n 100MW 128+6x5+5x2 =168hours")
+disp("the load duration curve is plotted in fig")
+disp("the total area under the load duration curve is 31600MWh which represents the energy conumption in one week.")
+eclw=ldcq(2,1)*ldcq(1,1)
+for x=2:1:kld
+ eclw=eclw+(ldcq(2,x)*(ldcq(1,x)-ldcq(1,x-1)))
+end
+lf=eclw/(max(lwen)*24*7)
+printf("total energy consumed is %dWh",eclw)
+printf("\ntotal maximum energy could consume %dWh",eclw/lf)
+printf("\nload factor =%f",lf)
diff --git a/416/CH2/EX2.12/exp2_12c.txt b/416/CH2/EX2.12/exp2_12c.txt new file mode 100755 index 000000000..ccd67ffc8 --- /dev/null +++ b/416/CH2/EX2.12/exp2_12c.txt @@ -0,0 +1,19 @@ +
+ example 2.12
+
+ from the above data,the durations of different loads d
+ uring one week are
+ load duration
+ 350MW 4x5 =20hours
+ 250MW 20+8x5=60hours
+ 200MW 60+4x2 =68hours
+ 150MW 68+6x5+15x2 =128hours
+ 100MW 128+6x5+5x2 =168hours
+ the load duration curve is plotted in fig
+
+ the total area under the load duration curve is 31600M
+ Wh which represents the energy conumption in one
+ week.
+total energy consumed is 24600
+total maximum energy could consume 58800
+load factor =0.418367
\ No newline at end of file diff --git a/416/CH2/EX2.13/exp2_13.sce b/416/CH2/EX2.13/exp2_13.sce new file mode 100755 index 000000000..9caa7d3b8 --- /dev/null +++ b/416/CH2/EX2.13/exp2_13.sce @@ -0,0 +1,6 @@ +clc
+disp("example 2.13")
+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
+printf("annual load factor =%fx%fx%f=%f.",dlf,lptmlp,mlptalp,dlf*lptmlp*mlptalp)
\ No newline at end of file diff --git a/416/CH2/EX2.13/exp2_13.txt b/416/CH2/EX2.13/exp2_13.txt new file mode 100755 index 000000000..8a67e1b35 --- /dev/null +++ b/416/CH2/EX2.13/exp2_13.txt @@ -0,0 +1,3 @@ +
+ example 2.13
+annual load factor =0.825000x0.870000x0.780000=0.559845.
\ No newline at end of file diff --git a/416/CH2/EX2.14/exp2_14.sce b/416/CH2/EX2.14/exp2_14.sce new file mode 100755 index 000000000..148e83249 --- /dev/null +++ b/416/CH2/EX2.14/exp2_14.sce @@ -0,0 +1,12 @@ +clc
+disp("example 2.14")
+disp("(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)
+printf("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)
+disp("(b)")
+peakloadonfeeder=(peakloadoftransformer1+peakloadonxmer+peakloadonxmer3)/diversitybtwxmer
+printf("peak load on feeder =(100+80+100)/1.4 =%dkW",peakloadonfeeder)
diff --git a/416/CH2/EX2.14/exp2_14.txt b/416/CH2/EX2.14/exp2_14.txt new file mode 100755 index 000000000..b596d14ff --- /dev/null +++ b/416/CH2/EX2.14/exp2_14.txt @@ -0,0 +1,9 @@ +
+ example 2.14
+
+ (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
\ No newline at end of file diff --git a/416/CH2/EX2.2/exp2_2.png b/416/CH2/EX2.2/exp2_2.png Binary files differnew file mode 100755 index 000000000..56a8f1607 --- /dev/null +++ b/416/CH2/EX2.2/exp2_2.png diff --git a/416/CH2/EX2.2/exp2_2pp.sce b/416/CH2/EX2.2/exp2_2pp.sce new file mode 100755 index 000000000..16483d5a6 --- /dev/null +++ b/416/CH2/EX2.2/exp2_2pp.sce @@ -0,0 +1,38 @@ +clc
+disp("example 2.2")
+disp("(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
+printf("maximum demand of consumer A =%1fkW \n \ndemand factor of consumer A =%2f \n \nmaximum demand of consumer B =%dkW\n \ndemand factor of consumer B = %2f",mca,mca/cla,mcb,mcb/clb)
+disp("(b)")
+printf("The variation in demand versus time curves are plotted and shown in Fig This is known as chonological load curve.")
+A=[100*ones(1,5),1100*ones(1,1),200*ones(1,2),0*ones(1,9),500*ones(1,7)]
+B=[0*ones(1,7),300*ones(1,1),1000*ones(1,2),200*ones(1,8),600*ones(1,5),0*ones(1,1)]; //time line of different periods by a and b consumers
+t=1:1:24 ;//for 24 hours ploting
+ma=max(A);mb=max(B);
+subplot(121); //matrix plotting
+plot2d2(t,A,1);
+plot2d2(t,B,2);
+xtitle("load curves of A and B/ fig 1","time","load watts")
+C=A+B;
+subplot(122);
+plot2d2(t,C,1);
+xtitle("chronological load of group / fig 2","time","load watts")
+mg=max(C); //maximum demand of group
+disp("(c)")
+printf("maximum demand of the group is %dW",mg);
+gd=(ma+mb)/mg;
+printf("group diversity factor = %3f",gd) ; //group diversity factor is sum of individual maximum consumaer load to the group max load
+disp("(d)")
+sa=sum(A)
+printf("energy consumed by A during 24 hours is =%dWh",sa)
+printf("\nit is seen that energy consumed by A is equal to the area under the chronological load curve of A \n energy consumed by B during 24 hours is")
+sb=sum(B);
+printf("300x1+100x2+200x8+600x5=%dWh",sb);
+disp("(e)");
+printf("maximum energy which A could consume in 24hours = %.2fkWh \nmaximum energy which B consume in 24 hours is =%.2fkWh",mca*24,mcb*24 );
+disp("(f)");
+printf("actual energy/maximum energy");
+mca=mca*10^3;mcb=mcb*10^3
+aemea=sa/(mca*24)
+aemeb=sb/(mcb*24)
+printf("\nfor A = %d/%d =%f \nfor b =%d/%d =%f",sa,mca*24,sa/(mca*24),sb,mcb*24,aemeb);
\ No newline at end of file diff --git a/416/CH2/EX2.2/exp2_2pp.txt b/416/CH2/EX2.2/exp2_2pp.txt new file mode 100755 index 000000000..4488cbb9b --- /dev/null +++ b/416/CH2/EX2.2/exp2_2pp.txt @@ -0,0 +1,26 @@ +
+ example 2.2
+
+ (a)
+maximum demand of consumer A =1.100000kW
+
+demand factor of consumer A =0.440000
+
+maximum demand of consumer B =1kW
+
+demand factor of consumer B = 0.333333
+ (b)
+The variation in demand versus time curves are plotted and shown in Fig This is known as chonological load curve.
+ (c)
+maximum demand of the group is 1100Wgroup diversity factor = 1.909091
+ (d)
+energy consumed by A during 24 hours is =5500Wh
+it is seen that energy consumed by A is equal to the area under the chronological load curve of A
+ energy consumed by B during 24 hours is300x1+100x2+200x8+600x5=6900Wh
+ (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 = 5500/26400 =0.208333
+for b =6900/24000 =0.287500
\ No newline at end of file diff --git a/416/CH2/EX2.3/exp2_3pp.sce b/416/CH2/EX2.3/exp2_3pp.sce new file mode 100755 index 000000000..990d2f3eb --- /dev/null +++ b/416/CH2/EX2.3/exp2_3pp.sce @@ -0,0 +1,23 @@ +clc
+disp("example 2.3")
+printf("\n")
+cola=5;na=600;ns=20;
+cls=2;clfm=10;clsm=5;cll=20;clci=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
+printf("maximum demand of each appartment =%.2fkWh \n",mdea)
+mda=(na*mdea)/gdfa
+printf("maximum demand of 600 apatments =%.2fkW \n",mda);
+datsp=mda/pdfa
+printf("demand of 600 apartments at time of the system peak =%dkW \n",datsp);
+mdtcc=((cls*ns*dfs)+(clfm*dffl)+(clsm*dfsm)+(cll*dfl)+(clci*dfci))/gdfc
+printf("maximum demand of total commertial complex=%dkW \n",mdtcc)
+dcsp=mdtcc/pdfc
+printf("demand of the commertial load at the time of the peak = %dkW\n",dcsp);
+dsltsp=nsl*clsl
+printf("demand of the street lighting at the time of the system peak =%dkW",dsltsp);
+ispd=datsp+dcsp+dsltsp
+printf("\nincrease in system peak deamand =%dkW ",ispd)
+
diff --git a/416/CH2/EX2.3/exp2_3pp.txt b/416/CH2/EX2.3/exp2_3pp.txt new file mode 100755 index 000000000..b8b0f486c --- /dev/null +++ b/416/CH2/EX2.3/exp2_3pp.txt @@ -0,0 +1,10 @@ +
+ example 2.3
+
+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
\ No newline at end of file diff --git a/416/CH2/EX2.4/exp2_4.png b/416/CH2/EX2.4/exp2_4.png Binary files differnew file mode 100755 index 000000000..1fd7fde1b --- /dev/null +++ b/416/CH2/EX2.4/exp2_4.png diff --git a/416/CH2/EX2.4/exp2_4pp.sce b/416/CH2/EX2.4/exp2_4pp.sce new file mode 100755 index 000000000..e3b921314 --- /dev/null +++ b/416/CH2/EX2.4/exp2_4pp.sce @@ -0,0 +1,22 @@ +clc
+disp("example 2.4")
+printf("\n")
+printf("the chronological load curve is plotted in fig 1 the durition of loads is as under :")
+lc=[20*ones(1,5),40*ones(1,4),80*ones(1,9),100*ones(1,4),20*ones(1,2)]
+ldc=gsort(lc);
+[mm,nn]=size(ldc)
+printf("\n")
+for i=1:nn
+printf("\t%dW",ldc(i));//arranging accending order
+end
+e=sum(ldc)
+printf("\nthe load duration curve is ploted in 2 the energy produced by plant in 24 hours \n =100x4+80x(13-4)+40(17-13)+20(24-17)=%dMWh \n",e);
+lff=e/(24*max(ldc));
+printf("load factor =1420/2400=%f=%f in persent",lff,lff*100)
+t=1:1:24
+subplot(121);
+plot2d2(t,lc);
+xtitle("chronological curve","time","load MW");
+subplot(122);
+plot2d2(t,ldc);
+xtitle("load duration curve","time","load MW");
diff --git a/416/CH2/EX2.4/exp2_4pp.txt b/416/CH2/EX2.4/exp2_4pp.txt new file mode 100755 index 000000000..d2d0e2280 --- /dev/null +++ b/416/CH2/EX2.4/exp2_4pp.txt @@ -0,0 +1,8 @@ +
+ example 2.4
+
+the chronological load curve is plotted in fig 1 the durition of loads is as under :
+ 100W 100W 100W 100W 80W 80W 80W 80W 80W 80W 80W 80W 80W 40W 40W 40W 40W 20W 20W 20W 20W 20W 20W 20W
+the load duration curve is ploted in 2 the energy produced by plant in 24 hours
+ =100x4+80x(13-4)+40(17-13)+20(24-17)=1420MWh
+load factor =1420/2400=0.591667=59.166667 in persent
\ No newline at end of file diff --git a/416/CH2/EX2.5/exp2_5.txt b/416/CH2/EX2.5/exp2_5.txt new file mode 100755 index 000000000..5815b700d --- /dev/null +++ b/416/CH2/EX2.5/exp2_5.txt @@ -0,0 +1,4 @@ +
+ example 2.5
+capacity factor =0.473360
+utilisation factor =169.004563
\ No newline at end of file diff --git a/416/CH2/EX2.5/exp2_5q.sce b/416/CH2/EX2.5/exp2_5q.sce new file mode 100755 index 000000000..cf834f488 --- /dev/null +++ b/416/CH2/EX2.5/exp2_5q.sce @@ -0,0 +1,6 @@ +clc
+disp("example 2.5")
+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/lf
+printf("capacity factor =%f",cf)
+printf("\nutilisation factor =%f",uf)
diff --git a/416/CH2/EX2.6/exp2_6.png b/416/CH2/EX2.6/exp2_6.png Binary files differnew file mode 100755 index 000000000..a876ed423 --- /dev/null +++ b/416/CH2/EX2.6/exp2_6.png diff --git a/416/CH2/EX2.6/exp2_6pp.sce b/416/CH2/EX2.6/exp2_6pp.sce new file mode 100755 index 000000000..d5415c978 --- /dev/null +++ b/416/CH2/EX2.6/exp2_6pp.sce @@ -0,0 +1,52 @@ +clc
+disp("Example 2.6")
+time=[5 9 18 22 24]
+loadt=[20 40 80 100 20] //given time and load
+k=size(time)
+k=k(1,2)
+timed(1,1)=time(1,1)
+for x=2:k //finding time duration of each load
+ timed(1,x)=time(1,x)-time(1,x-1)
+end
+[m n]=gsort(loadt) //sorting decresing order
+for x=1:k //sorting the load and timeduration correspondingly
+ timed1(1,x)=timed(1,n(x))
+end
+tim(1,1)=timed1(1,1)
+for x=2:k
+ tim(1,x)=timed1(1,x)+tim(1,x-1)
+end
+lo(1,1)=24*min(m)
+m(k+1)=[]
+printf("the energy at different load levels is as under :")
+printf("\nload=%dMW,energy=%dMWh",m(k),lo(1,1))
+y=2
+for x=k-2:-1:1
+ lo(1,y)=lo(1,y-1)+(tim(1,x))*(m(x)-m(x+1))
+ t=m(x);l=lo(1,y)
+ printf("\nload=%dMW,energy=%dMWh",t,l)
+ y=y+1
+end
+for x=1:k
+ for y=x+1:k
+ if m(1,x)==m(1,y) then
+ m(1,y)=[]
+ end
+ end
+end
+pop=gsort(m,'g','i')
+subplot(121)
+plot(lo,pop)
+xtitle("energy load curve","energy","load")
+//time=[5 9 18 22 24]
+//loadt=[20 40 80 100 20]
+printf("\nthe energy load curve is plotted in fig 1 \nthe energy supplied up to different times of the day is as under :")
+et(1,1)=time(1,1)*loadt(1,1)
+for x=2:k
+ printf("\nenergy supplied upto %d is %dMWh",time(1,x-1),et(1,x-1))
+ et(1,x)=et(1,x-1)+loadt(1,x)*(time(1,x)-time(1,x-1))
+
+end
+subplot(122)
+plot(time,et)
+xtitle("masscurve","time in hours","load in MW")
diff --git a/416/CH2/EX2.6/exp2_6pp.txt b/416/CH2/EX2.6/exp2_6pp.txt new file mode 100755 index 000000000..9aaf6b791 --- /dev/null +++ b/416/CH2/EX2.6/exp2_6pp.txt @@ -0,0 +1,13 @@ +
+ Example 2.6
+the energy at different load levels is as under :
+load=20MW,energy=480MWh
+load=40MW,energy=820MWh
+load=80MW,energy=1340MWh
+load=100MW,energy=1420MWh
+the energy load curve is plotted in fig 1
+the energy supplied up to different times of the day is as under :
+energy supplied upto 5 is 100MWh
+energy supplied upto 9 is 260MWh
+energy supplied upto 18 is 980MWh
+energy supplied upto 22 is 1380MWh
\ No newline at end of file diff --git a/416/CH2/EX2.7/exp2_7c.sce b/416/CH2/EX2.7/exp2_7c.sce new file mode 100755 index 000000000..a3926f876 --- /dev/null +++ b/416/CH2/EX2.7/exp2_7c.sce @@ -0,0 +1,15 @@ +clc
+disp("example 2.7")
+md=40;cf=0.5;uf=0.8;//maximum demand in MW;capacity factor;utility factor
+disp("(a)")
+lf=cf/uf; //load factor is ratio of capacity factor to the utility factor
+printf("load factor = capacity factor/utilisation factor =%f",lf)
+disp("(b)")
+pc=md/uf; //plant capacity is ratio of maximum demand to utility factor
+printf("plant capacity = maximum demand/utilisation factor =%dMW",pc)
+disp("(c)")
+rc=pc-md; //reserve capacity is plant capacity minus maximum demand
+printf("reserve capacity =%dMW",rc)
+disp("d")
+printf("annual energy production =%dMWh",md*lf*8760)
+
diff --git a/416/CH2/EX2.7/exp2_7c.txt b/416/CH2/EX2.7/exp2_7c.txt new file mode 100755 index 000000000..418a99042 --- /dev/null +++ b/416/CH2/EX2.7/exp2_7c.txt @@ -0,0 +1,11 @@ +
+ example 2.7
+
+ (a)
+load factor = capacity factor/utilisation factor =0.625000
+ (b)
+plant capacity = maximum demand/utilisation factor =50MW
+ (c)
+reserve capacity =10MW
+ d
+annual energy production =219000MWh
\ No newline at end of file diff --git a/416/CH2/EX2.8/exp2_8c.png b/416/CH2/EX2.8/exp2_8c.png Binary files differnew file mode 100755 index 000000000..418795901 --- /dev/null +++ b/416/CH2/EX2.8/exp2_8c.png diff --git a/416/CH2/EX2.8/exp2_8cpp.sce b/416/CH2/EX2.8/exp2_8cpp.sce new file mode 100755 index 000000000..febcc81ba --- /dev/null +++ b/416/CH2/EX2.8/exp2_8cpp.sce @@ -0,0 +1,14 @@ +clc
+disp("example 2.8")
+disp("the chronological load curve is plotted in fig 1")
+a=[0 5 9 18 20 22 24] //time in matrix format
+b=[50 50 100 100 150 80 50]//load in matrix format
+for x=1:6
+ z(1,x)=((b(1,x)+b(1,x+1))/2)*(a(1,(x+1))-a(1,x))
+end
+e=sum(z);
+printf("energy required required by the system in 24 hrs \n =50x5MWh+((100+50)/2)x4MWh +(100x9)MWh+(100+150)MWh+(150+80)MWh+(80+50)MWh \n =%dMWh",sum(z))
+dlf=e/(max(b)*24)
+printf("\ndaily load factor =2060/(150x24) =%f",dlf)
+plot(a,b)
+xtitle("load curve","time","MW")
diff --git a/416/CH2/EX2.8/exp2_8cpp.txt b/416/CH2/EX2.8/exp2_8cpp.txt new file mode 100755 index 000000000..1c7ffd485 --- /dev/null +++ b/416/CH2/EX2.8/exp2_8cpp.txt @@ -0,0 +1,8 @@ +
+ example 2.8
+
+ the chronological load curve is plotted in fig 1
+energy required required by the system in 24 hrs
+ =50x5MWh+((100+50)/2)x4MWh +(100x9)MWh+(100+150)MWh+(150+80)MWh+(80+50)MWh
+ =2060MWh
+daily load factor =2060/(150x24) =0.572222
\ No newline at end of file diff --git a/416/CH2/EX2.9/exp2_9.png b/416/CH2/EX2.9/exp2_9.png Binary files differnew file mode 100755 index 000000000..21ef42bfa --- /dev/null +++ b/416/CH2/EX2.9/exp2_9.png diff --git a/416/CH2/EX2.9/exp2_9c.sce b/416/CH2/EX2.9/exp2_9c.sce new file mode 100755 index 000000000..d9e2546e4 --- /dev/null +++ b/416/CH2/EX2.9/exp2_9c.sce @@ -0,0 +1,32 @@ +clc
+clear
+disp("example 2.9")
+disp("load duration curve in fig1")
+disp("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
+for x=1:6
+ z(1,x)=((b(1,x)+b(1,x+1))/2)*(a(1,(x+1))-a(1,x))
+end
+et=0
+for x=1:6
+ et=et+z(1,x);
+ A=a(1,(x+1))
+ ett(1,x)=et;
+ q(1,x)=a(1,x+1)
+ printf("\nfrom mid night upto %d,energy=%dMWh",A,et)
+end
+q(1,x+1)=[]
+[m n]=gsort(b)
+m(1,7)=[];m(1,6)=[]; //rearranging for mass curve
+disp("energy curve in fig 2")
+t=[0 3.88 15.88 19.88 23]
+for j=1:6
+ k(1,j)=a(1,(j+1))
+end
+subplot(121);
+plot(t,m);
+xtitle("load duration","hours","MW")
+subplot(122);
+plot(q,ett,-9);
+xtitle("energy curve","time","MWh")
diff --git a/416/CH2/EX2.9/exp2_9c.txt b/416/CH2/EX2.9/exp2_9c.txt new file mode 100755 index 000000000..561a14769 --- /dev/null +++ b/416/CH2/EX2.9/exp2_9c.txt @@ -0,0 +1,15 @@ +
+ example 2.9
+
+ load duration curve in fig1
+
+ the energy consumed upto different times is as
+
+from mid night upto 5,energy=250MWh
+from mid night upto 9,energy=550MWh
+from mid night upto 18,energy=1450MWh
+from mid night upto 20,energy=1700MWh
+from mid night upto 22,energy=1930MWh
+from mid night upto 24,energy=2060MWh
+ energy curve in fig 2
+
\ No newline at end of file |
