The cross-sectional area of water pipe entering the basement is `4 cm^(-2)` . The pressure at the point is `3.5 xx 10^(5) Nm^(-2)` and the speed of water is `2.2 ms^(-1)` . This pipe tapers to a cross-sectional area of `2 cm^(2)` when it reaches the second floor 10 m above . Calculate the speed and pressure at the second floor. use `g = 10 m//s^(2)`.

Explanation

Here, `a_(1) = 4 xx 10^(-4) m^(2)`
`P_(1) = 3.5 xx 10^(5) Nm^(-2) , upsilon_(1) = 2.2 ms^(-1)`
`a_(2) = 2 xx 10^(-4) m^(2) , rho = 10^(3) kg m^(-3)`,
Ac cording to equation continuity
`a_(1)upsilon_(1) = a_(2) upsilon_(2)`
or, `upsilon_(2) = (a_(1)upsilon_(2))/(a^2) = ((4 xx 10^(-4))xx2.2)/(2 xx 10^(-4)) = 4.4 ms^(-1)`.
Ac cording to Bernoulli's Theorem
`P_(1) + 1/2 rho upsilon_(1)^(2)+ rho gh_(1) = P_(2) +1/2 rho upsilon_(2)^(2)+ rho gh_(2) `
or `P_(1) - P_(2) =rho g (h_(2)-h_(1)) +1/2 rho (upsilon_(2)^(2) - upsilon_(1)^(2) )`
`= 3.5 xx 10^(5) - 10^(3) xx 10 xx 10 `
` -1/2 xx 10^(3) [(4.4)^(2) - (2.2)^(2)]`
`= 3.5 xx 10^(5) - 1 xx 10^(5) - 0.07 xx 10^(5)`
`= (3.5 xx 1.07) xx 10^(5) = 2.43 xx 10^(5) Nm^(-2)`.