Lecture 05 Problems

VARIABLE LOADS 1. A Di Dies esel el power power plant plant has a ma maxi ximum mum deman demand d of 12 !" wi with th a lo load ad fa#to fa#torr of .$ .$ an and d #apa#it% of .&. Estimate the plant #apa#it%. A. 1 !" B. 160 MW  '. 1$ !" D. 2 !" Solution(  Average load Loadfactor = Peak load 0.80

=

 Average   load 120

 Average load= 96 kw

Usefactor =

0.60

=

 Average load Plant capacit  96

Plant capacit 

= 160 kw Plant capacity  2. 'al#ul 'al#ulate ate the !" pow power er #apa#it #apa#it% % of a )eo )eother thermal mal plant plant with a loa load d fa# fa#tor tor of .$2 and 12 !" pea* load. +he operation is limited to $, hours a %ear with a use fa#tor of .-. A. 1-, B. 1 '. 2// D. 145 Solution(  Average load Loadfactor = Peak load 0.82

=

 Average   load 120

 Average load= 98.4 kw

Usefactor =

0.70

=

 Average load Plant capacit 

98.4

Plant capacit 

= 144.9 kw Plant capacity  /. A 0ien euipment euipment #onsumes #onsumes , *w3hr4month *w3hr4month at 256 rated plant plant #apa#it% #apa#it%.. It operates at 25 hours7 / da%s4month. "hat is the rated #apa#it%8 A. 1-., *w 1

B. 2/. *w C. 28.90 kw D. $2. 2. 5 *w Solution(

   

 Average load=  5000

kw− hr   1 month   1 day             month   30 days  24 hrs 

 Average load= 6.94 kw Plant factor =

0.24

=

 Average load Ratingof equipmen 6.94

Ratingof equipme

= 28.93 kw Ratingof equipment  5. A #entral station station is suppl%in0 suppl%in0 ener0% ener0% to a #ommun #ommunit% it% throu0h throu0h two su9stations. su9stations. One su9statio su9station n feeds four distri9ution #ir#uits: the other7 six. +he maximum dail% re#orded demandsare( ;ower station 127 *w Su9station A &7 *w