The cross-sectional area of water pipe entering the basement is 4 x 10^(-4) m^(2). The pressure at this point is 3 x 10^(5) Nm^(-2) and the speed of water is 2 ms^(-1). This pipe tapers to a cross-sectional area of 2 x 10^(-4)m^(2) when it reaches the second floor 8 m above. Calculate the speed and pressure at the 2nd floor

The cross-sectional area of water pipe entering the basement is 4 x 10

1.	The cross-sectional area of water pipe entering the basement is 4 x 10^(-4) m^(2). The pressure at this point is 3 x 10^(5) Nm^(-2) and the speed of water is 2 ms^(-1). This pipe tapers to a cross-sectional area of 2 x 10^(-4)m^(2) when it reaches the second floor 8 m above. Calculate the speed and pressure at the 2nd floor
Answer» SOLUTION :Since `A_(1)nu_(1)=A_(2)nu_(2)` `nu_(2)=(2xx4xx10^(-4))/(2XX10^(-4))` = 4m/s Using BERNOULLI’s Theorem `P_(2)=P_(1)+(1)/(2)RHO(upsilon_(1)^(2)-upsilon_(2)^(2))+rhog(h_(1)-h_(2))` `:. nu_(2)gtnu_(1)` `h_(2)gth_(1)` `=3xx10^(5)+(1)/(2)(1000)[(2)^(2)-(4)^(2)]-1000xx9.8xx8` `=2.16xx10^(5)N//m^(2)`