Water flows through a horizontal pipe of which the cross - section is

1.	Water flows through a horizontal pipe of which the cross - section is not constant. Thepressure is 1cm of mercury where the velocity is 0.35m/s. Find the pressure at a point wherethe velocity is 0.65m/s.
Answer» $v_1=0.35\ m\ s^{-1}\quad\quad P_1=1\ cm\ Hg=1333\ Pa\\\\v_2=0.65\ m\ s^{-1}\quad\quad P_2=?$ By BERNOULLI's Theorem, $P_1+\dfrac {1}{2}\rho(v_1)^2=P_2+\dfrac {1}{2}\rho(v_2)^2$ Potential ENERGY is not CONSIDERED since water flows through a horizontal pipe. $1333+\dfrac {1}{2}\times 1000(0.35)^2=P_2+\dfrac {1}{2}\times1000(0.65)^2\\\\\\\underline {\underline {P_2=1183\ Pa}}$ Or, $\underline {\underline {P_2=0.887\ cm\ Hg}}$

Water flows through a horizontal pipe of which the cross - section is not constant. Thepressure is 1cm of mercury where the velocity is 0.35m/s. Find the pressure at a point wherethe velocity is 0.65m/s.

Answer»

$v_1=0.35\ m\ s^{-1}\quad\quad P_1=1\ cm\ Hg=1333\ Pa\\\\v_2=0.65\ m\ s^{-1}\quad\quad P_2=?$

By BERNOULLI's Theorem,

$P_1+\dfrac {1}{2}\rho(v_1)^2=P_2+\dfrac {1}{2}\rho(v_2)^2$

Potential ENERGY is not CONSIDERED since water flows through a horizontal pipe.

$1333+\dfrac {1}{2}\times 1000(0.35)^2=P_2+\dfrac {1}{2}\times1000(0.65)^2\\\\\\\underline {\underline {P_2=1183\ Pa}}$

Or,

$\underline {\underline {P_2=0.887\ cm\ Hg}}$

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Water flows through a horizontal pipe of which the cross - section is not constant. Thepressure is 1cm of mercury where the velocity is 0.35m/s. Find the pressure at a point wherethe velocity is 0.65m/s.​

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Water flows through a horizontal pipe of which the cross - section is not constant. Thepressure is 1cm of mercury where the velocity is 0.35m/s. Find the pressure at a point wherethe velocity is 0.65m/s.