Waste Water Engineering - Civil Engineering Questions and Answers

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For having central angle Î±, the area of cross-section of sewers running partially full, is

A = $\frac { D^2} { 2 }$ [ $\frac { nα } { 180^0 }$ -- $\frac { sin α } { 2 } ]$

A = $\frac { D^2} { 4 }$ [ $\frac { nα } { 360^0 }$ - $\frac { sin α } { 2 } ]$

A = $\frac { D^2} { 4 }$ [ $\frac { nα } { 360^0 }$ - $\frac { cos α } { 2 } ]$

A = $\frac { D^2} { 2 }$ [ $\frac { nα } { 360^0 }$ - $\frac { cos α } { 2 } ]$

Answer : Option B

Explanation :

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