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clc
clear
disp('example 14.12')
pa=500 //power of unit a
pb=2000 //power of unit b
ra=2.5 //speed regulation of a
rb=2 //speed regulation of b
dl=0.01 //change in load
df=0.01 // change in frequency
pt=20 //change in tie line power
ptl=0 //let other power station has zero
pbas=2000 //assume base as 2000MW
f=50 //assume frequency
da=(dl*pa)/(df*f) //change in power w.r.t frequency
dapu=da/(pbas) // change in power w.r.t frequency in per unit
db=(dl*pb)/(df*f) //change in power in unit b
dbpu=db/pbas //change in power w.r.t frequency in per unit
raa=ra*pbas/pa //speed regulation with pbase
rbb=rb*pbas/pb //speed regulation with pbase
ba=dapu+(1/raa) //area frequency response a
bb=dbpu+(1/rbb) //area frequency response b
ma=pt/pbas //assume change in power in unit a alone due to tie power
mb=ptl/pbas //change in power in unit b
df=-(ma+mb)/(ba+bb) //change in frequency
dpp=(ba*mb-bb*ma)/(ba+bb) //change in power
disp('(a)')
printf("change in frequency is %.3fHz \n change in power between ab %.5fp.u.MW \n \t\t%.2fMW",df,dpp,dpp*pbas)
ma2=ptl/pbas //assume change in power in unit a alone due to tie power
mb2=pt/pbas //change in power in unit b
df2=-(ma2+mb2)/(ba+bb) //change in frequency
dpp2=(ba*mb2-bb*ma2)/(ba+bb) //change in power
disp('(b)')
dpba=dpp2*pbas
printf("change in frequency is %.3fHz \n change in power between ab %.5fp.u.MW \n",df2,dpp2)
printf(" change in power %fMW",dpba)
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