Water is flowing steadily through a horizontal pipe of non-uniform cross-section. If the pressure of water is `4 xx 10^4 N//m^2` at a point where cross-section is `0.02 m^2` and velocity of flow is `2 m//s`. What is the pressure at a point where cross-section reduces to `0.01 m^2` ?
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Explanation

`P_1 = 4 xx 10^4 N//m^2, a_1 = 0.02 m^2, v_1 = 2m//s`
`P_2 = ?, a_2 = 0.01 m^2, v_2 = ?`
By continuity equation
`a_1 v_1 = a_2 v_2`
`0.02 xx 2 = 0.01 v_2`
`v_2 = 4 m//s`
Applying Bernoulli's theorem
`P_1 + (1)/(2) rho v_1^2 = P_2 + (1)/(2) rho v_2^2`
`P_2 = P_1 + (1)/(2) rho (v_1^2 - v_2^2)`
=`4 xx 10^4 + (1)/(2) xx 10^3 (2^2 - 4^2)`
=`4 xx 10^4 + (1)/(2) xx 10^3 (6)(-2)`
=`3.4 xx 10^4 N//m^2`.