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InterviewSolution
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Water is flowing with a speed of `2m//s` in a horizontal pipe with cross-sectional area decreasing from `2xx10^(-2) m^(2)` to `0.01 m^(2)` at pressure `4xx10^(4)` pascal. What will be the pressure at smaller cross-section? |
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Answer» Here, `upsilon_(1) = 2 ms^(-1), a_(1)=2xx10^(-2)m^(2), P_(1) = 4xx10^(4) pascal , a_(2) = 0.01 m^(2) , P_(2)=?` As, `a_(1)upsilon_(1) = a_(2) upsilon_(2) or upsilon_(2) =(a_(1)upsilon_(1))/(a_(2)) = ( 2xx 10^(-2) xx 2)/(0.01) = 4ms^(-1)` Now, `P_(1) + 1/2 rho upsilon_(1)^(2) = P_(2) + 1/2 rho upsilon_(2)^(2) or P_(2)= 1P_(1)+1/2 rho (upsilon_(1)^(2)-upsilon_(2)^(2))` `:. P_(2) = 4xx10^(4) +1/2 xx 10^(3) (2^(2)-4^(2)) = 4 xx 10^(4) -6 xx 10^(3) = 3.4 xx 10^(4)` pascal. |
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