Answer to Question #201199 in Physics for mazzu

Question

Water is flowing through a pipeline shown on the right. The radii of the pipe at points 1 and 2 are 𝟎.𝟓𝟎𝟎 [m] and 𝟏.𝟓𝟎 [m], respectively. The volume flow rate through the pipeline is 𝟐.𝟓𝟎 [L/s]. If the pressure at point 1 in the pipe is 𝟐𝟎𝟎 [kPa], what is the pressure at point 2, located 𝟓.𝟎𝟎 [m] above point 1?

Answer 1

$Q=2.5\ (l/s)=0.0025\ (m^3/s)$

$A_1=\pi r_1^2=3.14\cdot0.5^2=0.785\ (m^2)$

$A_2=\pi r_2^2=3.14\cdot1.5^2=7.068\ (m^2)$

$Q=A_1v_1\to v_1=Q/A_1=0.0025/0.785=0.00318\ (m/s)$

$Q=A_2v_1\to v_2=Q/A_2=0.0025/7.068=3.54\cdot10^{-4}\ (m/s)$

According to Bernoulli's equation

$\rho v_1^2/2+p_1=\rho v_2^2/2+mgh+p_2\to$

$p_2=\rho v_1^2/2+p_1-\rho v_2^2/2-\rho gh=\rho (v_1^2- v_2^2)/2+p_1-\rho gh=$

$=1000\cdot (0.00318^2- (3.54\cdot 10^{-4})^2)/2+200000-1000\cdot9.8\cdot5=$

$=151\ 000\ (Pa)=151\ (kPa)$. Answer