Fixed Structures and Multi-Body Interaction2Descripción completa
Descripción: Structural six minute problems for civil PE exam
Fixed Structures and Multi-Body Interaction2Full description
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Lecture 4Full description
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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 0ien euipment euipment #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(
= 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
=
600
+ 1500 + 1000 + 2900 + 2200 + 3000 9000
= 1.247 ,. +he annual pea* load on a 1,7 1,73*w 3*w power power plant is 17, 17, *". *". +wo +wo su9stations su9stations are 9ein0 supplied 9% this plant. Annual ener0% dispat#hed throu0h su9station A is 2-7,7 *"3hr with a pea* at $7 *w7 while 1&7,7 *"3hr are 9ein0 sent throu0h su9station B with a pea* at &7&, *w. =e0le#tin0 line losses7 >nd the #apa#it% fa#tor of the power plant. A. ./,/ B. .,// 2
B. 2/. *w C. 28.90 kw D. $2. 2. 5 *w Solution(
= 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
=
600
+ 1500 + 1000 + 2900 + 2200 + 3000 9000
= 1.247 ,. +he annual pea* load on a 1,7 1,73*w 3*w power power plant is 17, 17, *". *". +wo +wo su9stations su9stations are 9ein0 supplied 9% this plant. Annual ener0% dispat#hed throu0h su9station A is 2-7,7 *"3hr with a pea* at $7 *w7 while 1&7,7 *"3hr are 9ein0 sent throu0h su9station B with a pea* at &7&, *w. =e0le#tin0 line losses7 >nd the #apa#it% fa#tor of the power plant. A. ./,/ B. .,// 2
C. 0.335 D. .,/, Solution( actualenergy produced produced max imum possibl possible energy for thesame perio perio
Plant capacity factor =
Plant capacity factor =
277500 500 7000
+ 1675007000 kw− hr
(157000 000 kw) ( 8760 hr )
335 Plant capacity factor = 0.335
&. A power plant plant is said to hae hae had a use fa#tor fa#tor of 5$.,6 and a #apa#it #apa#it% % fa#tor of 52.56. 52.56. ?ow man% hours did it operate durin0 the %ear8 A. 7660 B. $-& '. $2 D. 172 Solution( actualkwhr produced produced Plant usefactor = × operatinghour kw plant plant capacity Plant capacity factor =
actualkwhr produced produced × 8760 hours kw plant plant capacity