Pipeline

How much percent has changed (reduced) area of pipe cross-section, if circular shape changed to square with same perimeter?

Result

p =  21.46 %

Solution:

p=100S1S2S1=100(1S2S1)= =100(1a2πr2)=100(1π4)=21.46%p = 100 \dfrac{S_1-S_2}{S_1} = 100(1- \dfrac{ S_2}{S_1}) = \ \\ = 100(1- \dfrac{ a^2}{\pi r^2}) = 100(1- \dfrac{ \pi }{ 4}) = 21.46 \%



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