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» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :Since `A_(1)nu_(1)=A_(2)nu_(2)` <br/> `nu_(2)=(2xx4xx10^(-4))/(<a href="https://interviewquestions.tuteehub.com/tag/2xx10-1840192" style="font-weight:bold;" target="_blank" title="Click to know more about 2XX10">2XX10</a>^(-4))` = 4m/s <br/> Using <a href="https://interviewquestions.tuteehub.com/tag/bernoulli-395852" style="font-weight:bold;" target="_blank" title="Click to know more about BERNOULLI">BERNOULLI</a>’s Theorem <br/> `P_(2)=P_(1)+(1)/(2)<a href="https://interviewquestions.tuteehub.com/tag/rho-623364" style="font-weight:bold;" target="_blank" title="Click to know more about RHO">RHO</a>(upsilon_(1)^(2)-upsilon_(2)^(2))+rhog(h_(1)-h_(2))`<br/> `:. nu_(2)gtnu_(1)` <br/> `h_(2)gth_(1)` <br/> `=3xx10^(5)+(1)/(2)(1000)[(2)^(2)-(4)^(2)]-1000xx9.8xx8` <br/> `=2.16xx10^(5)N//m^(2)`</body></html>