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lecture_14

# lecture_14 - 2.3 Q 4 B T log 10 2.25 Tt r 2 S h...

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h o h DRAWDOWN h o h Q 4 B T W ( u ) W ( u ) 0.5772 lnu % u u 2 2 x2 ! % u 3 3 x3 ! u 4 4 x4 ! % ..... u r 2 S 4 Tt h 0 h ( Q 4 B T )( 0.5772 ln r 2 S 4 Tt ) h 0 h ( Q 4 B T )( ln(1.78) ln( r 2 S 4 Tt )) h 0 h Q 4 B T ln( 2.25 Tt r 2 S ) h 0 h 2.3 Q 4 B T log 10 2.25 Tt r 2 S ( h 0 h ) 1 Q 4 B T ln(2.25 Tt 1 r 2 S ) LECTURE 14 - TIME AND DISTANCE DRAWDOWN 14.1 THEIS EQUATION 14.2 SIMPLIFICATION OF THEIS EQUATION CONSIDER CASE WHEN “u” IS SMALL (“t” MUST BE LARGE OR “r” MUST BE SMALL) 14.3 TIME DRAWDOWN METHOD ASSUME AT TIME t . .... 1

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( h 0 h ) 2 Q 4 B T ln(2.25 Tt 2 r 2 S ) ) h ( h 0 h ) 2 ( h 0 h ) 1 Q 4 B T ln t 2 t 1 T Q 4 B) h ln t 2 t 1 T 2.3 Q 4 B) h log t 2 t 1 T 2.3 Q 4 B) h T 264 Q ) h T 35 Q ) h h 0 h ( Q 4 B T )( 0.5772 ln r 2 S 4 Tt ) ASSUME AT TIME t . .... 2 SOLVE FOR ) h BETWEEN t AND t . ....... 2 1 SOLVE FOR T (OVER ONE LOG CYCLE) Q = GPM T = GPD FT -1 h = FT Q = GPM T = FT DAY 2 -1 h = FT 14.4 SOLUTION FOR S REMEMBER. ....
0 Q 4 B T ( 0.5772 ln r 2 S 4 Tt 0 ) 0 Q 4 B T ( ln 1.78 ln r 2 S 4 Tt 0 ) 0 ( ln 1.78 ln r 2 S 4 Tt 0 ) 0 ln 2.25 Tt 0 r 2 S r 2 S 2.25 Tt 0 S 2.25 Tt 0 r 2 S 0.3 Tt 0 r 2 S Tt 0 640 r 2 WHEN t = t , h - h = 0 0 0 SOLVE FOR S or T = GPD/FT t = DAYS 0 r = FT or T = FT DAY 2 -1 t = MINUTES 0 r = FT

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h 0 h Q 4 B T ln( 2.25 Tt r 2 S ) h 0 h
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Unformatted text preview: 2.3 Q 4 B T log 10 2.25 Tt r 2 S ) h ( h &amp;amp; h ) 1 &amp;amp; ( h &amp;amp; h ) 2 Q 2 B T ln r 2 r 1 ) h ( h &amp;amp; h ) 1 &amp;amp; ( h &amp;amp; h ) 2 2.3 Q 2 B T log r 2 r 1 T 2.3 Q 2 B) h T 528 Q ) h T 70 Q ) h S 2.25 Tt r 2 14.5 DISTANCE DRAWDOWN METHOD CONSIDER CASE WHEN u IS SMALL (t LARGE OR r SMALL) AND SOLVE FOR ) h BETWEEN r AND r 2 1 SOLVE FOR T (OVER ONE LOG CYCLE) or Q = GPM T = GPD FT-1 h = FT or Q = GPM T = FT DAY 2 -1 h = FT SOLVE FOR S S 0.3 Tt r 2 S Tt 640 r 2 or T = GPD FT-1 t = DAYS r = FT or T = FT DAY 2-1 t = MINUTES r = FT...
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lecture_14 - 2.3 Q 4 B T log 10 2.25 Tt r 2 S h...

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